![[Algebraic.Number.Theory].(Jürgen.Neukirch.).Pdf](http://data.docslib.org/img/7b78af1153510c18134595f96abb2455-1.webp)
Springer Berlin Heidelberg New York Barcelona Hong Kong London Milan Paris Singapore Jiirgen Neukirch
.Algebraic Number Theory
Translated from the German by Norbert Schappacher
With 16 Figures
Springer
G,° w... z Algebraische Zahlentheorie F-67o84 Strasbourg,France 7, rueRenDescartes Universit LouisPasteur U.F.R. deMathematiqueetd'Informatique Norbert Schappacher Translator: Jiirgen Neukircht
ISBN 3-540-65399-6(hc.:alk.paper) Includes bibliographicalreferencesandindex. p. cm.-(GrundlehrendermathematischenWissenschaften;322 theory /JurgenNeukirch;translatedfromtheGermanbyNorbertSchappacher. Neukirch, ]urgen,1937-.[Algebraischezahlentheorie.English]Algebraicnumber Library ofCongressCataloging-in-PublicationData ISBN 3-540-54273-6Springer-VerlagBerlinHeidelbergNewYork The originalGermaneditionwaspublishedin1992underthetitle e-mail: [email protected] way, andstorageindatabanks.Duplicationofthispublicationorpartsthereof is illustrations, recitation,broadcasting,reproductiononmicrofilmorinany other of thematerialisconcerned,specificallyrightstranslation,reprinting, reuseof This workissubjecttocopyright.Allrightsarereserved,whetherthewhole orpart ISBN 3-540-65399-6Springer-VerlagBerlinHeidelbergNewYork ISSN 0072-7830 Mathematics SubjectClassification(1991):n-XX,14-XX QA247.N51713 1999512'.74-dC2199-21030CIP 1. Algebraicnumbertheory. SPIN: 10425040 Raymond Seroul,Strasbourg Photocomposed fromthetranslator'sLATEX filesaftereditingandreformattingby Cover design:MetaDesignplusGmbH, Berlin Printed inGermany © Springer-VerlagBerlinHeidelberg1999 Law. Springer-Verlag. ViolationsareliableforprosecutionundertheGermanCopyright 1965, initscurrentversion,andpermissionforusemustalwaysbeobtained from permitted onlyundertheprovisionsofGermanCopyrightLawSeptember 9, ran
41/3143-5 43210Printedonacid-free paper 'C7 I. Title
II. Series. acv .V.. able toimplantthisenthusiasmintothemindsofhisstudents. favorite subjectinmathematics.Hewasenthusiasticaboutit,andhealso What elsecanbesaid? about hisintentions,thecontentofbookandpersonalviewsubject. contains Neukirch'sPrefacetotheGermanedition.Therehehimselfspeaks have beensomuchbetter,ifhecoulddonethishimself. translation ofJurgenNeukirch'sbookonAlgebraicNumberTheory.Itwould Regensburg crewtookover.Inthemeantimemanymembersofthisteamteach come togethertolearnasubjectwhichisnotnecessarilytheirownspeciality. the soccerfield). Arbeitsgemeinschaft" anddemonstratedtheirstrength(mathematicallyon occasions whenthisequipeshowedupinthemeetingsof"Oberwolfach theory andarithmeticalgebraicgeometry.Irememberverywellthemany eager andmotivatedtodiscusslearnthenewestdevelopmentsinnumber that thesubjectanditsbeautyjustifiedhighesteffortsotheywerealways growing tree. pear tobeputinhisgrave,sothatlaterthechildrencanpick thepearsfrom away thepearsfromhisgardentochildren.Whenhe diesheasksfora Theodor Fontane.Ittellsthestoryofanoblemanwhoalways generouslygives often heardfromherfather:HerrvonRibbeckauf imHavellandby that itwillattractmanyofthem. source foryoungstudentstostudyAlgebraicNumberTheory, andIamsure at Germanuniversities. Always attheend,whenmostdifficulttalkshadtobedelivered, time tocome. seeds initforatreetogrow from whichthe"children"canpickfruitsin He attractedthem,theygatheredaroundhiminRegensburg.toldthem It becomesclearfromhisPrefacethatNumberTheorywasNeukirch's But itisalsoverydifficultformetowritethis"Geleitwort":Thebook During themeetingsof"OberwolfachArbeitsgemeinschaft"people It isaverysadmomentformetowritethis"Geleitwort"theEnglish This is-IbelieveagoodwayofthinkingNeukirch's book:Thereare At Neukirch'sfuneralhisdaughterChristianerecitedthe poemwhichshe We findthischarismaofJurgenNeukirchinthebook.Itwill beamotivating r~' ... "7. Off. chi .'' G3. Foreword 3-' 'r7 .J;
G. Harder 7s' Cep expertise ofTEXtothefinal preparation ofthecamera-readymanuscript. happy. We sincerelyhopethatthebookpublishedherewouldhave madeitsauthor this Englishedition.KayWingberggraciouslyhelpedtocheck afewofthem. had beenaccumulatedthankstoattentivereaders,weretaken intoaccountfor proofreading thefirstfourchaptersoftranslation.Without herknowledge,
fair numberofsmallcorrectionsandmodificationstheGerman originalthat responsibility, andenergy,thisbookwouldnotbewhat it is.Inparticular,a patience totheoriginalGermanbook,andshehadassisted JurgenNeukirchin Neukirch hadtoleaveundone.Shealreadyapplied her infinitecareand Maria Strobelwhotookituponherselftofinishasbestshe couldwhatJurgen I aloneamresponsibleforalldeficienciesthatremain. to judgewhetherImanagedrenderhisideasfaithfully. thus savingtheEnglishlanguagefromvariousinfractions.Butneedlesstosay, edition). Timewilldeliveritsverdict. Arakelov divisor,etc.(aterminologythatNeukirchhadavoidedintheGerman Neukirch's blessing:Icallrepletedivisor,ideal,etc.,whatisusuallycalled beauty, butonlythepreciselyadequateexpressionofanidea.Itisforreader he warnedmethatineveryoneofthesepassages,wasnotseekingpoetic serious useoftheGermanlanguage.WhenIstartedtoworkontranslation, had pepperedhisbookwithmoremeditativeparagraphswhichmakerather sentences couldnotsimplybecopiedintoEnglish.However,JurgenNeukirch to integrateinvolvedargumentsanddisplayedformulaeintocomprehensive and wasthusstraightforwardtotranslate,eventhoughtheauthor'sdesire Zahlentheorie, backin1991,no-oneimaginedthathewouldnotlivetoseethe the firstfourchaptersofit. last chaptersintheFallof1996),andhestillhadtimetogocarefullythrough English edition.Hedidseetherawversionoftranslation(Igavehim
Hearty thanksgotoRaymond Seroul,Strasbourg,forapplyinghiswonderful After JurgenNeukirch'suntimelydeathearlyin1997, itwasMsEva- I ammuchindebtedtoFrazerJarvisforgoingthroughmyentiremanuscript, There isoneneologismthatIproposeinthistranslation,withJurgen The bulkofthetextconsistsdetailedtechnicalmathematicalprose When IfirstacceptedJurgenNeukirch'srequesttotranslatehisAlgebraische
}.. G1.
b.0 u"'
.-y Translator's Note 1-1
cad
may CAD h.0 C14 p., .C. -Urr
can viii Translator's Note
Thanks go to the Springer staff for seeing this project through until it was finally completed. Among them I want to thank especially Joachim Heinze for interfering rarely, but effectively, over the years, with the realization of this translation.
Strasbourg, March 1999 Norbert Schappacher to numbertheoryacommunityofdedicatedfollowers. satisfactory waysofreasoning-allthesefeatureshaveattimesattracted undiscovered, merelydivined;lastbutnotleast,thecharmofitsparticularly clarity ofitsstatements,thearcanetouchinlaws,betheydiscoveredor harmony. Thepossibilityofformulatingitsbasicproblemssimply,thepeculiar autonomous insettingitsgoals,andthusmanagestoprotectundisturbed no obligationtoserveneedsthatdonotoriginatewithinitself,itisessentially position, similartotheonethatmathematicsholdsamongsciences.Under progressively emphasizestheoreticalaspectsthatrelyon modemconcepts. of algebraicnumberfields,presupposingonlybasicalgebra (itstartswith provide thestudentwithanessentiallyself-containedintroductiontotheory we especiallyrecommendtothereader. us pickoutinparticularNumberTheorybyS.I.BoREVICZandI.R.SAFAREVJ( first attitude,whichisorientedtowardsspecificproblems.Amongthem,let exert ononeanother.Severalbeautifultextbooksillustratethesuccessof they growparticularlyeffectivethroughthemutualinspirationalinfluence the apparentvarietyofarithmeticphenomena.Bothattitudesarejustified,and conceptual clarity,nevertiringofsearchingforthedeep-lyingreasonsbehind for theconcreteresulttheydesire.Otherswillstriveamoreuniversal, science. Somewillpushthetheoreticaldevelopmentonlyasfarisnecessary related totheconjecturesofBirch andSwinnerton-Dyer-islargelybasedon in thecontextofWellconjectures, theMordellconjecture,ofproblems decades inconjunctionwith`arithmeticalgebraicgeometry'. Theimmense the revolutionarydevelopmentthatnumbertheoryhasundergone inthelast from aunifiedtheoreticalpointofviewseemsimperative today,asaresultof number theoryproper.Thedesiretopresent asmuchpossible used, inorderthatthereadershouldnotlosesightof concretegoalsof Still, indoingso,aspecialeffortismadetolimittheamount ofabstraction the equation2=1+1).Butunliketextbooksalluded toabove,it [14]: abookwhichisextremelyrichincontent,yeteasytoread,and the unconditionalanduniversal applicationoftheconceptualapproach. success thatthisnewgeometricperspectivehasbroughtabout -forinstance, Number Theory,amongthemathematicaldisciplines,occupiesanidealized CAD The presentbookwasconceivedwithadifferentobjectiveinmind.Itdoes But differentnumbertheoristsmaydedicatethemselvesdifferentlytotheir Preface totheGermanEdition
..+. O<-
CAD ti Sao (RD vii CAD °s' (a7 CAD ~'k
functions havegainedcentral importanceinrecentdecades,buttextbooksdo some remarks,andtoaddbeneficial exercises. that somewhatdrypresentation withquiteafewexamples,toreferahead gave metheopportunitytomodify andemendmyearliertreatment,toenrich like atorso,sufferingfromanunacceptablelackofcompleteness. Thisalso with itsimportantconsequencesforthetheoryofL-serieswould haveappeared on algebraicnumberfieldswithoutthecrowningconclusion ofclassfieldtheory end, afterlongconsideration,therewassimplynootherchoice. Asourcebook long before,anothertreatmentofthistheoryposedobvious questions.Butinthe chapters IV-VI.Sincemybook[107]onthissubjecthad beenpublishednot back toHASSE,andisclearlyhighlightedforinstanceinS.LANG's textbook[94]. for arithmeticsurfaces.Thecorrespondingideainthenumber fieldcasegoes seemed allthemorejustifiedasARAKELOVhimselfintroduced hisdivisorsonly introducing tooheavyaterminologyforthiselementarymaterial.Mydecision would haveentailedattachingthenameofArakelovtomanyotherconcepts, to employtheterm"Arakelovdivisor"althoughitisnowwidelyused.This has receivedanextensivetreatmentinatextbook.ButIfinallydecidednot probably thefirsttimethatthisapproach,withallitsintricatenormalizations, point ofview",whichhasacquiredmuchimportanceinrecentyears.Itis of theRiemann-Rochtheorem.Thepresentationisdominatedby"Arakelov's the classicalnotionsandresultswiththeoryofalgebraiccurvesidea three sectionsofchapterIII,theaimwhichistopresentperfectanalogy algebraic numberfields.Thesefoundationsarefinallysummedupinthefirst down thefoundationsofglobaltheoryandsecondlocal principles thathaveguidedmewhilewritingthebook.Thefirstchapterlays in amoreentertainingmanner.Iwouldhoweverliketoemphasizefewbasic chapter inthisforeword;simplyturningpageswillyieldthesameinformation curves. place geometricinterpretationtothefore,inspiritoftheoryalgebraic more far-reachingargumenttotheclevertrick,andmadeaparticulareffort why Ipreferred,wheneveritwasfeasible,thefunctorialpointofviewand theory -oratleastavoidinganyobstructiontosuchanintegration.Thisis integrating thetheoryofalgebraicnumberfieldsintohigherdimensional focus, borrowingemphasesandargumentsfromthehigherpointofview,thus way whichtakesthesedevelopmentsintoaccount,takingthemasthedistant the 1-dimensionalcase.ButIthoughtitnecessarytopresenttheoryina deliberately confinesitselftothetheoryofalgebraicnumberfields,i.e., book becausetheyrequirehigherdimensionaltheories,whereasthe x A lotofworkwentintothelast chapteronzetafunctionsandL-series.These °,3 It wasnotwithouthesitationthatIdecidedtoincludeClass FieldTheoryin Let meforegotheusualhabitofdescribingcontenteachindividual It istruethatthoseimpressiveresultscanhardlybetoucheduponinthis {y,
b9° w,_ `°G `G: cam' 7°' ono 0p0o ."7 L1. '°''' ".J^'' a;- ,.. L3. Preface totheGermanEdition
Cep .'C7 CD' CD- CI. ...."' ... .-i
for time,section10onhigher ramificationmaybeomittedcompletely. restricted tocompletevalued fields,andthusjoinedwithsection4.Ifpressed the timehascometogiveorders anadequatetreatment. algebraic K-theory,whichcannot beconstructedwithoutsingularschemes, important foralgebraicnumberfields,andsincethisis evenmoretrueof of singularalgebraiccurves.Ascohomologytheorybecomes increasingly diophantine problems.Butmostimportantly,theyrepresent theanalogues orders constitutetheringsofmultipliers,whichplayaneminent roleinmany sequel, Iconsideritsinsertioninthebookparticularlyimportant. Foronething, course. Althoughtheresultsofsection12onordersare irrelevantforthe yet sufficientroomfortheclassfieldtheoryofchaptersIV-VI. including infiniteGaloistheory).Thefollowingtermwill thenprovidescarce, first threechapterscaneasilybepresentedinoneacademic year(ifpossible a firsttestinthisrespect.Withbitofcarefulplanning,thebasiccontent theory. YetIthinkthatthebookhasnotlostcharacter.Infact,itpassed material whichhadtobecoveredinviewoftheintrinsicnecessities This intentionwasincreasinglyjeopardizedbytheunexpectedgrowthof on algebraicnumbertheory,butalsoasaconvenienttextbookforcourse. volume, devotedtothecohomologyofalgebraicnumberfields. with thisexclusionwillbestrongenoughtobringaboutasequelthepresent neither beassumednorreasonablydevelopedhere.Perhapsthedissatisfaction becomes trulyconvincingonlyifoneusesetalecohomology-whichcan theory isgeometry.Idobelieve,however,thatinthiscasethegeometricaspect been aparticularlybeautifulvindicationofouroft-repeatedthesisthatnumber it mirrorsimportantgeometricpropertiesofalgebraiccurves,wouldhave been undertakeninanyexistingtextbook. theory totallygermanetoalgebraicnumberfields,thesubjectofthisbook.Since of ARTIN'sL-serieswiththeirfunctionalequation-whichsurprisinglyhasnot been done,therewastheobviousopportunityofgivingathoroughpresentation but forallitsvariousadvantagescriedoutamoderntreatment.Thishaving to HECKE'sapproach,whichisnoteasyunderstandintheoriginalversion, been sufficientlyrepeatedinotherplaces.InsteadIhavepreferredtoturnback clarity ofTATE'sownpresentationcouldhardlybeimprovedupon,andithasalso suited themoreconceptualorientationofbookperfectlywell.Infact, to HeckeL-series,whichisbasedonharmonicanalysis,althoughitwouldhave not paysufficientattentiontothem.Ididnot,however,includeTATE'sapproach Preface totheGermanEdition In chapterII,thespecialtreatment ofhenselianfieldsinsection6maybe Sections 11-14ofchapterImaymostlybedroppedfrom anintroductory From theverystartbookwasnotjustintendedasamodernsourcebook It wasadifficultdecisiontoexcludeIwasawaTheory,relativelyrecent .fl ,..
't7
p+.try (gyp CAD Ca. was .ti fit' ......
Ova a<< `"' 11.11- 'C7 't1 ;p, `.T 'L1 can xi Roch isanicesubjectforstudentseminarratherthananintroductory The subsequenttheoryconcerningthetheoremofGrothendieck-Riemann- they highlightanewapproachtoclassicalresultsofalgebraicnumbertheory. xii The readerisencouragedtoprovehisversatilityinlookinguptheliterature. much exercises,butadditionalremarkswhichdidnotfitwellintothemaintext. linked toclassicalproblems. is howeverunsatisfactorybecauseitremainsintheabstractrealm,andnever the Hilbertsymbol,andVI,78,stillleavesafully-fledgedtheory,which of higherramificationmaybeomitted.Cuttingoutevenmore,chapterV,3,on Sections 4-7ofchapterV,onformalgroups,Lubin-Tatetheoryandthe students arealreadyfamiliarwithprofinitegroupsandinfiniteGaloistheory. course. reaction tosuchapossibleslipoftheauthorbemitigatedbyGoethe'sdistich: case arises,itisforthereadertofindcorrectformulation.Mayreader's so thattheremightbeoccasionalmistakesinthewaytheyareposed.Ifsucha I shouldalsopointoutthathavenotactuallydonealltheexercisesmyself, proofreading andtheformattingoftext,nevertiring ofgoingintothe of Hecke'stheorythetaseriesandL-series.Ioweagreat debtofgratitude it wasC.DENINGERandU.JANNSENwhosuggestedthatI giveanewtreatment avoid mistakesandambiguities.TomyfriendsW-D.GEYER, G.TAMME,andK. or smallerparts,whichledtonumerousimprovementsand madeitpossibleto KAusz, B.KOOK,P.KoLCZE,TH.MosER,M.SPIESShavecriticallyexaminedlarger the SpringerVerlagforconsideringmywisheswithgenerosity.MystudentsI. * Erroriseverwithus.Yetsome angelicneed who didthetypesettingofmanuscriptinTEX.That bookcanappearis who arenotnamedhere.TremendousthanksduetoMrs. MARTINAHERTL minutest detail.Letmeheartilythankallthosewhoassisted me,andalsothose to Mrs.EvA-MARIASTROBEL.Shedrewthepicturesand helpedmewiththe WINGBERG Iowemuchvaluableadvicefromwhichthebook hasprofited,and Gently coaxesourstrivingmind upwards,towardstruth. The firstthreesectionsofchapterIIIshouldbepresentedinthelecturessince A wordontheexercisesatendofsections.Somethemarenotso Finally, inpresentingclassfieldtheory,itsavesconsiderabletimeifthe O0' During thewritingofthisbookIhavebeenhelpedinmanyways.thank 0-A "Irrtum verlaBtunsthe,dochzieheteinhoherBedurfnis Immer denstrebendenGeistleisezurWahrheithinan." (Translation suggestedbyBARRY MAZUR.) .-r ors .as
r-] Preface totheGermanEdition -,= Preface to the German Edition xiii essentially due to her competence, to the unfailing and kind willingness with which she worked through the long handwritten manuscript, and through the many modifications, additions, and corrections, always prepared to give her best.
Regensburg, February 1992 Jiirgen Neukirch
1 1 5 16 23 28 34 39 44 53 58 65 72 84 94 99 99 1.0 . . 116 123 106 134 143 160 166 176 183 183 194 . 208 224 ...... 152 ...... Table of Contents of Table ...... Chapter I: Algebraic Integers Chapter I: § 1. The Gaussian Integers § 1. The Gaussian § 2. Integrality § 3. Ideals § 4. Lattices § 5. Minkowski Theory § 6. The Class Number Theorem . § 7. Dirichlet's Unit Domains § 8. Extensions of Dedekind Theory § 9. Hilbert's Ramification § 10. Cyclotomic Fields § 11. Localization § 12. Orders § 13. One-dimensional Schemes § 14. Function Fields . Chapter H: The Theory of Valuations § 1. The p-adic Numbers § 2. The p-adic Absolute Value § 3. Valuations § 4. Completions § 5. Local Fields § 6. Henselian Fields . Extensions § 7. Unramified and Tamely Ramified c67 CO') § 8. Extensions of Valuations § 9. Galois Theory of Valuations § 10. Higher Ramification Groups .S'. Chapter III: Riemann-Roch Theory § 1. Primes § 2. Different and Discriminant § 3. Riemann-Roch § 4. Metrized o-Modules c07 Chapter IV:AbstractClassFieldTheory § 8.TheEuler-MinkowskiCharacteristic § 7.Grothendieck-Riemann-Roch § 6.TheChernCharacter § 5.GrothendieckGroups xvi § 1.TheLocalReciprocityLaw Chapter V:LocalClassFieldTheory § 7.TheHerbrandQuotient § 6.TheGeneralReciprocityLaw § 5.TheReciprocityMap § 4.AbstractValuationTheory § 3.AbstractGaloisTheory § 2.ProjectiveandInductiveLimits § 1.InfiniteGaloisTheory Chapter VII:ZetaFunctionsandL-series § 8.TheReciprocityLawofthePowerResidues § 7.TheIdeal-TheoreticVersionofClassFieldTheory § 6.GlobalClassFields § 5.TheGlobalReciprocityLaw § 4.TheClassFieldAxiom § 3.TheHerbrandQuotientoftheIdeaeClassGroup § 2.IdelesinFieldExtensions § 1.IdelesandIdeaeClasses Chapter VI:GlobalClassFieldTheory § 6.HigherRamificationGroups § 5.GeneralizedCyclotomicTheory § 4.FormalGroups § 3.TheHilbertSymbol § 2.TheNormResidueSymboloverQp § 4.TheHigher-dimensional GammaFunction § 3.ThetaSeries § 2.DirichletL-series § 1.TheRiemannZetaFunction e0' COD00-* ......
r-, ......
......
...... Table ofContents ...... N ...... 290 . . . . . 395 . 380 . 368 . . . . 317 . . . 317 310 299 284 275 265 261 261 255 246 243 233 419 414 405 385 373 357 357 352 346 341 333 327 453 443 434 419 '.0 Table ofContents Index Bibliography § 13.DensityTheorems § 12.TheFunctionalEquationofArtinL-series § 11.TheArtinConductor § 10.ArtinL-series § 9.ValuesofDirichletL-seriesatIntegerPoints § 8.HeckeL-series § 7.ThetaSeriesofAlgebraicNumberFields § 6.HeckeCharacters § 5.TheDedekindZetaFunction ...... 527 . . . 484 470 457 559 551 r-+ 542 535 (!t 517 504 493 xvii
It clearlysufficestofind y EZ[i]suchthat existence ofgaussianintegersy,psuchthat N U(0),aHIa12.So,fora,BE7Z[i],$ Proof :WeshowthatZ[i]iseuclideanwithrespecttothe functionZ[i]-+ rial. (1.2) Proposition.Thering7G[i]iseuclidean,therefore inparticularfacto- unique factorizationinZ[i]. in Z[i].Theanswertothisquestionisbasedonthefollowing resultabout so thattheproblemisnowwhenandhowaprimenumberpEZfactors In thisring,theequationp=x2+y2turnsintoproductdecomposition rational integersisfoundinthelargerdomainofgaussian prime numberoftheformp=a2-I-b2satisfies-1mod4,because (1.1) Theorem.Forallprimenumbersp is theremarkablefactthatconversealsoholds: perfect squaresare=0or-1mod4.Thisisobvious.Whatnotobvious Except for2,theyareall-1mod4,anditistrueingeneralthatanyodd show thefirstprimenumbersthatcanberepresentedasasumoftwosquares. 2=1+1, 5=1+4,13=4+9,17=1+16,29=4+25,37=1+36 The naturalexplanationofthisarithmeticlawconcerningtheringZ The equations p=a2+b2 (a,bEZ) w"+ Z[i]={a+biIa,beZ}, i= a=yp+p and Algebraic Integers § 1.TheGaussianIntegers p =(x+iy)(x-iy), Chapter I IpI2 p-1mod4. I 0, onehastoverifythe - yj yob SIR .^3 . Cry < 1.Now,the into twonon-unitsa,8of2L[i].Thenormz=x+iyisdefinedby there existsadecomposition not remainaprimeelementintheringZ[i].Indeed,ifthisisproved,then this: itissufficienttoshowthataprimenumberp-1mod4ofZdoes Therefore thereexistsanelementyEZ[i]with is notgreaterthanhalfthelengthofdiagonalmesh,i.e. lies insomemeshofthelatticeanditsdistancefromnearestpoint integer coordinateswithrespecttothebasis1,i).ThecomplexnumberE gaussian integersformalatticeinthecomplexplaneC(thepointswith 2 in thefactorialring7L[i]. not divideanyofthefactorsx+i,-andistherefore not aprimeelement Thus wehavepIx2+1=(xi)(x-i).Butsince fP2[i],pdoes by Wilson'stheorem,onehas admits asolution,namelyx=(2n)!.Indeed,since-1(p-1)!modp cannot beaprimeelementin7L[i],wenotethatthecongruence and thereforep=N(a)a2+b2,whereweputabi. i.e., byN(z)=1z12.Itismultiplicative,sothatonehas now lookatthisringabitmore closely. of algebraicintegerswitha concrete example.Forthesamereasonwewill introduced theringZ[i],but ratherinordertoprefacethegeneraltheory domains ofintegers.Butitwas notsomuchforthisequationthatwehave questions aboutrationalintegers mayleadtotheconsiderationofhigher Based onthisresultaboutthering2L[i},theorem(1.1)nowfollowslike The exampleoftheequationp=x2+y2showsthateven quite elementary Finally, inordertoprovethatarationalprimeoftheformp=1+4n -1(p-1)!=[1.2...(2n)][(p-1)(P-2)...(p-2n)] P [(2n)!] [(-l)2n(2n)!]=[(2n)!]2modp. are notunits,itfollowsthatN(a),N(P) N(x+iy)=(x+iy)(x-iy)=x2+y2, -1- x2modp 'CD p=a-,8 Chapter I.AlgebraicIntegers - y<2 1 (seeexercise1), < 1. .fl El 2. 3 G.1 J3, if they differ only by a unit factor, and J3, if they differ only by a unit factor, III ,8 into non-units a, 0 would imply that p2 = p = N(or) = N(a) N(p), with a2+b2=p,p=1mod4,a> Ibi >0, with a2+b2=p,p=1mod4,a> Ibi p=3mod4. N(a):=(a+ib)(a-ib)=a2+b21 7r = 1 +i, Numbers 7r = p, where p - 3 mod 4, are prime in Z [ i 1, because In order to answer the question for primes, i.e., irreducible elements of the question for primes, i.e., irreducible In order to answer When developing the theory of divisibility for a ring, two basic problems two basic for a ring, of divisibility the theory developing When in Z[i] implies an equation either N(a) = 1 or N($) = 1, so that with some prime number p. Hence either a or 8 is a unit. a decomposition p = a N(a) N(P), so that p = N(a) = N(a + bi) = a2 + b2, which according to (1.1) would yield p = 1 mod 4. of 7L[i]. n of Z[i], up to associated elements, (1.4) Theorem. The prime elements are given as follows. (1) (2) rr=a+bi (3) it=p, Here, p denotes a prime number of Z. because a decomposition 7r = a Proof: Numbers as in (1) or (2) are prime the ring Z[i], we first recall that two elements a, $ in a ring are called first recall that two elements a, $ in the ring Z[i], we associated, symbolically a - (exercise 1), i.e., if either a2 = 1, b2 = 0, or a2 = 0, b2 = 1. We thus obtain either a2 = 1, b2 = 0, or a2 = 0, b2 = (exercise 1), i.e., if the of the fourth group of units of the ring Z[iI consists (1.3) Proposition. The roots of unity, element 7r is also irreducible. that every element associated to an irreducible precise list of all prime numbers Using theorem (l.1) we obtain the following § 1. The Gaussian Integers Gaussian The § 1. ring in the units of the one hand, to determine prominent: on the are most question answer to the first elements. The on the other, its prime question, 7G[i] is a a = a + bi E easy. A number case is particularly in the present 1: only if its norm is unit if and Hence canddareintegers. follows that(2c)2=(2d)2- 0mod4,sincesquaresarealways-or=1. are integers,thenso2c and2d.From(2c)2+(2d)2=4b-0mod4it If canddarerationalintegers, thensoareaandb.Conversely,ifb Proof: Anelementa=c+idEQ(i)iszeroofthepolynomial with coefficientsa,bEZ. field Q(i)ofwhichsatisfyamonicpolynomialequation (1.5) Proposition.7G[i]consistspreciselyofthoseelements oftheextension gaussian integers,whichisindependentofachoicebasis. the basis1,i.However,wealsohavefollowingcharacterization ofthe "integers" inQ(i).Thisnotionofintegralityisrelativetothecoordinates as therationalintegersdoinfieldQ.Sotheyshouldbeviewed conjugate primefactors then 7r and theprimenumbersp=3mod4remaininZ[i]. shows that2 the squareofprimeelement1+i bers pEZdecomposein[i].Theprime2=(1+i)(1-isassociatedto proof. p =a2+b2(abi)-couldnotbeprime.Thiscompletesthe otherwise wewouldhavep=2or-1mod4andbecauseof(1.1) thus aunit.Moreover,p=3mod4hastoholdinthiscasebecause if p=2,itisassociatedto1+i.Ontheotherhand,N(7r)p2, N(or) =pweget7ra+biwitha2b2p,so7isoftype(2)or, N(7r) IN(p)=p2,sothateitherN(n)porp2.Inthecase with rationalprimespi,showsthatitIpforsome=pi.Thisgives of Z[i]isassociatedtoonethoselisted.Firstall,thedecomposition 4 The gaussianintegersplaythesameroleinfield The propositionalsosettlescompletelythequestionofhowprimenum- This beingsaid,wehavetocheckthatanarbitraryprimeelementrr is associatedtopsincep/zranintegerwithnormoneand x2 +axb (1 +i)2.Theprimenumbersp-1mod4splitintotwo ton Inn (i)={a+bila,bEQ} III Q [x] p =(a+bi)(a-bi), x2+ax+b =0 with . Indeed,theidentity1-i=-i =p, ...pr, a =-2c,bc2+d ..p Chapter I.AlgebraicIntegers 2. 'Cs (1 +i) in fullgenerality. bers, butoccursinmanydifferent contextsandthereforehastobetreated f (X)EZ[x].Thisnotionof integralityappliesnotonlytoalgebraicnum- integral, oranalgebraicinteger, ifitisazeroofmonicpolynomial ments ofKarecalledalgebraic numbers.Analgebraicnumberiscalled that itsunitsaregivenby±(l+/)",nEZ,anddetermine primeelements. Exercise 7.Showthatthering7L[-,f2]=Z+Z,/-2-iseuclidean. Showfurthermore integer d>1,hasinfinitelymanyunits. Exercise 6.Showthatthering7L[/]=Z+7L,/d-,foranysquarefreerational rational integerd>1,are±1. Exercise 5.Showthattheonlyunitsofring76[J]=7L+7L/,forany Exercise 4.Showthatthering76[i]cannotbeordered. a=u+ ivEZ[i]. Hint: Useexercise2toshowthatnecessarilyx+iy=saewithaunitEand where u,vEZ,u>0,(u,v)=1,notbothodd. possible permutationofxandy,bytheformula such thatx,y,z>0and(x,z)=1 Exercise 3.Showthattheintegersolutionsofequation prime numbersandsaunit,implies=E'4"PE"1J",withs',s"units. Exercise 2.Showthat,intheringZ[i],relationap=sy",fora,Prelatively Exercise 1.aE7Z[i]isunitifandonlyN(a)= case Z[i]arenottypical.Novelfeatureswillpresentthemselvesinstead. theory ofalgebraicintegers.Buttheanswerswefoundinspecial of primeelements,andtothequestionuniquefactorization. in thissectionacompleteanswertothequestionofunits, integer coefficients.Forthedomainofgaussianintegerswehaveobtained as beinganelementsatisfyingamonicpolynomialequationwithrational § 2.Integrality An algebraicnumberfieldisafiniteextensionK ofQ.Theele- These questionsindicatealreadythefundamentalproblemsingeneral The lastpropositionleadsustothegeneralnotionofanalgebraicinteger 'ti x = u2 - V2, § 2.Integrality x2 +yz=z2 y =2uv, ("pythagorean triples") z =u2+v2, are allgiven,upto CAD ..C 5 (2.1) Definition.LetACBbeanextensionofrings.AnelementbEis with 1. 6 of thenotionintegrality. integral. Thiswillbeaconsequenceofthefollowingabstractreinterpretation sum andtheproductoftwoelementswhichareintegraloverAagain b EBareintegraloverA. with coefficientsaiEA.TheringBiscalledintegraloverAifallelements called integraloverA,ifitsatisfiesamonicequation matrix, i.e.,a_ matrix withentriesinanarbitraryring,andletA*_(a!.)betheadjoint generated. over AifandonlytheringA[b1,...,b,]viewedasanA-moduleisfinitely (2.2) Proposition.Finitelymanyelementsb1, polynomial g(x)EA[x]wemaythenwrite a monicpolynomialofdegreen>1suchthatf(b)= 0. Foranarbitrary Proof ofproposition(2.2):LetbEBbeintegralover A andf(x)EA[x] this yieldstheimplication where Edenotestheunitmatrixofrankr.Foranyvector x=(xj,...,xr), by deletingthei-thcolumnandjrow.Thenonehas (2.3) Proposition(Row-ColumnExpansion).LetA=(aid)bean(rxr)- Thus A[b]isgeneratedasA-module by1,b, with q(x),r(x)EA[x]and deg(r(x)) In fact, if alb c K (a, b E A) is integral over A, i.e., (alb)" + at (alb)n-1 + ... + an = 0, with ai E A, then an + atban-t + ... + anb" = 0. Therefore each prime element 7r which divides b also divides a. Assuming alb to be reduced, this implies alb E A. We now turn to a more specialized situation. Let A be an integral domain which is integrally closed, K its field of fractions, LIK a finite field extension, and B the integral closure of A in L. According to (2.4), B is. automatically integrally closed. Each element ,8 E L is of the form b bEB, aEA, a because if an,Bn+...+a,P+ao=0,ai EA, an00, then b = a, 8 is integral over A, an integral equation (an,8)n+ +al(anp)+ao=0,ai EA, being obtained from the equation for P by multiplication by an-1. Further- more, the fact that A is integrally closed has the effect that an element P E L is integral over A if and only if its minimal polynomial p(x) takes its coef- ficients in A. In fact, let t3 be a zero of the monic polynomial g(x) E A[x]. Then p(x) divides g(x) in K[x], so that all zeroes f1,..., i3, of p(x) are integral over A, hence the same holds for all the coefficients, in other words p(x) E A[x]. The trace and the norm in the field extension L IK furnish important tools for the study of the integral elements in L. We recall the (2.5) Definition. The trace and norm of an element x E L are defined to be the trace and determinant, respectively, of the endomorphism Tx:L -+ L,TX(a)=xa, of the K-vector space L : TrLIK(x) =Tr(Tx),NLIK(x) =det(Tx). them equalto respect tothisbasishasobviouslyonlyblocksalongthe diagonal,eachof is abasisofLIK.Thematrixthelineartransformation TX:yHxywith basis ofLIK(x),then of x.Infact,1,x, of theminimalpolynomial Proof: Thecharacteristicpolynomialfx(t)isapower (iii) (ii) (i) over thedifferentK-embeddingsofLintoanalgebraicclosurekK,then (2.6) Proposition.IfLIKisaseparableextensionandor the followingGalois-theoreticinterpretation, In thecasewhereextensionLIKisseparable,traceandnormadmit Since TX+y=Tx+TyandTXyTsoTy,weobtainhomomorphisms we have of T,n=[L:K],werecognizethetraceandnormas § 2.Integrality In thecharacteristicpolynomial TrLIK (x)=F-ax, NLIK(X) =F1OrX. fx(t) =F1(t-ax), fx(t) =det(tid-Ti)t"-alti-1+ p(t) a1, a1x,..., TrLIK:L -KandNLIK:L*->K*. v Cr a1=TrLIK(x) ..., fx(t) =PX(t)d, -Cm 0 0 CAD xm-t isabasisofK(x)IK,andifa1, -Cm_1 =tm+C1tm-1 +...+Cm, 0 1 atxm-1 CI. ; .. and an=NLjK(x)- -Cm_2 d =[L:K(x)] 0 1 ; ad,adx,..., ... m= [K(x):K] ... + -C1 0 0 (-l)"a, EK[t] : L-+Kvaries adxm-1 ... ,ad is a 9 The correspondingcharacteristicpolynomialiseasilycheckedtobe proves (i),andthereforealso(ii)(iii),afterVieta. representatives, thenwefind into mequivalenceclassesofdelementseach.Ifvi,...,crmisasystem equivalence relation so thatfinallyfx(t)=px(t)d. 10 embeddings ofMispartitionedbytherelation (2.7) Corollary.InatoweroffinitefieldextensionsKCLM,onehas and Likewise forthenorm. tatives, thenHomK(L,k)=[QiILI into m=[L':K]equivalenceclasses.Ifcri, Proof :WeassumethatMIKisseparable.ThesetHomK(M,K)of- (see [143],vol.I,chap.II,§ 10). [M :MS]onehasK]i = [M:L]i[LK]iand subextension MS1K.Indeed, fortheinseparabledegree[M easily fromwhatwehaveshownabove,bypassingtothe maximalseparable The setHomK(L,K)ofallK-embeddingsLispartitionedbythe We willnotneedtheinseparablecaseforsequel.However itfollows . f x(t)=fl TrMIK(x) =[M:K]iTrMs1K(x), TrMIK(x) =F-axETrviMla,L(aix)_E07TrMIL(x) TrLIK OTrMIL=TrMIK =TrLIK(TrMIL(x)) 1(t - i=1 a^ai m tm ax)d or or -r46x=rx px(t)= COD -r = fm1!la_ai(t- .0. +citm-t +...+Cm i=1 m i=1 n: , i =1,...,m},andwefind . NLIK oNMIL=NMIK. CL) fl(t-o'ix), aIL=rIL NMIK(x) = ... ,am = px(t), Chapter I.AlgebraicIntegerg ax) is asystemofrepresen- i=1 m flat -ax).This : Nm,IK(x)[M:Kli K]i = 11 0n-1 ... , K. Because of K. Because extension L I K is L I K extension # 0. ... , an is a basis, then the 6n-1 one gets n 01- 02 -1 en-1 i E+' 01 02 on ... , an) = (x, Y) =TrLI K(xy) d(a1, ..., an) 1 1 j=1,..., n. It is nondegenerate because, for ei = Qi 0, j=1,..., n. It is nondegenerate because, d(a1, TrLjK(aiaj) = E(akai)(akaj), TrLjK(aiaj) = d (I, 0, ..., d(1, 0, ... , d(al, is an arbitrary basis of L I K, then the bilinear form (x, y) with det(M) (ei-'Oj-1))i, , ... , a, The discriminant of a basis a 1 , of a basis The discriminant cr. above it follows that d(al, ..., an) = det(M) # 0. (2.8) Proposition. If L I K is separable and a1, (2.8) Proposition. If L I K is separable discriminant and K -vector space L. is a nondegenerate bilinear form on the form (x, y) = Tr(xy) is nondegenerate. Proof: We first show that the bilinear i.e., L = K (0). Then 1, 0, Let 0 be a primitive element f o r L I K, form (x, y) is given by the matrix is a basis with respect to which the M = (TrLIK we have If a i respect to this basis is given by the matrix M = (TrLI K(ai aj)). From the In the special case of a basis of type 1, 0, In the special case each of the first is seen by successively multiplying where 0i = o-i 0. This the Vandermonde matrix (n - 1) columns in by 01 and subtracting it from the following. § 2. Integrality § 2. defined by L over the K -embeddings i = 1, ... , n, varies where ai, the relation (akai)' and (crkaj). is the product of the matrices the matrix (TrLIK(aiaj)) write Thus one may also rank [L:K].Inparticular,B admitsanintegralbasisoverA. (2.10) Proposition.IfLIK is separableandAaprincipalidealdomain, general If, however,Aisaprincipalidealdomain,thenonehas thefollowingmore of rankn=[L The existenceofanintegralbasissignifiesthatBis afreeA-module then everyfinitelygenerated B-submoduleM0ofLisafreeA-module of LIK,itslengthnalwaysequalsthedegree[L:K] the fieldextension. an A-basisofB).Sincesuchintegralbasisisautomatically abasis with coefficientsaiEA;iscalledanintegralbasis ofBoverA(or: written uniquelyasalinearcombination by thedeterminantdet(TrLIK(ajaj))=d.ThereforedaaeA,andthus and, asTrL1K(a;a)EA,theyaregiventhequotientofanelementA the systemoflinearequations Proof :Ifa=aleel+- discriminant d=(al, (2.9) Lemma.Leta1, discriminant isoftenusefulbecauseofthefollowing For ifaNLIK(x)=1,aEA,thenIfl,axyxforsomeyeB.The Furthermore, forthegroupofunitsBoverA,weobtainrelation account thatAisintegrallyclosed,i.e.,=BflK,(2.6)implies element ofL,thenallitsconjugatesaxarealsointegral.Takinginto closure BinthefiniteseparableextensionLIK.IfxEisanintegral closed integraldomainAwithfieldoffractionsK,andtoits 12 t-. A systemofelementscul,...,w,EBsuchthateach bEBcanbe After thisreviewfromthetheoryoffields,wereturntointegrally : K]. Ingeneral,suchanintegralbasisdoesnotexist. TrLI K(ala)_ x EB* - ... , ... , - +anonEB,aiK,thentheajareasolutionof TrLIK(x), cx be abasisofL(KwhichiscontainedinB, C3" Then onehas i NLIK(X) EA. NLIK (x)EA*. TrLIK(ata0aJ, Chapter I.AlgebraicIntegers UCH Putting We havetoshowthata;!EA.Putf3j=>ta;iw1. Let G(LL'IL')_ do formabasisofLL'IK.Nowletbeanintegralelement ofLL',and Proof: AsLflL'=K,wehave[LL':K]nn',sothe nn'productswiw' x, x'EA.Thencwico,,'isanintegralbasisofLL',discriminant d"'d'". d andd'arerelativelyprimeinthesensethatxd+x'd'=1,forsuitable resp. n',suchthatLflL'=K.Letwl,...,w",resp,co.....,wn,,bean proposition isinteresting.InsteadofintegralbasestheclosureB 0-] Jul, write integral basisofLIK,resp.L'withdiscriminantd,d'.Supposethat (2.11) Proposition.LetLIKandL'betwoGaloisextensionsofdegreen, of AinL,wewillnowsimplyspeakintegralbasestheextensionLIK. situations itcanalsobeanimportantone.Thisiswhythefollowing hence rank(M)=[L:K]. Finally, ideal domains,sinceMoisafreeA-module,soadM,andhencealsoM. According tothemaintheoremonfinitelygeneratedmodulesoverprincipal Then exists anaEA, Let A1,...,A,EMbeasystemofgeneratorstheB-moduleM.There also asystemofgeneratorstheK-moduleL,wehaverank(B)=[L:K]. rank(B) <[L:K],andsinceasystemofgeneratorstheA-moduleBis to lieinB.By(2.9),wethenhavedBCAal+ a basisofLIK.MultiplyingbyanelementA,wemayarrangeforthea; Proof :LetM;0beafinitelygeneratedB-submoduleofLand1,...,a" § 2.Integrality It isingeneraladifficultproblemtoproduceintegralbases.Inconcrete [L :K]=rank(B) Qj andG(LL'IL)=(al,...,Qn,).Thus mar' G(LL'IK)=(akcrrI k=1,...,n, f=1,...,n}. coo adMCdB , 0, suchthataµ;EB,i=1,...,r,soaMCB. a =(aria, aa,iw;w, 5.p i C'(D` atf EK. b =(fit, Min - lei - +Aa,,inparticular, , t3. coo 13 Therefore wiw;isindeedanintegralbasisofLL'IK.Wecomputethe of (wi)and(wl),onechecksinthesamemannerthatdai1EA,so entries inL,namelyd'fl=>id'aijwi.ThusEA.Swappingtheroles Since T*andahaveintegralentriesinLL',themultipled'bhas Write T*fortheadjointmatrixofT.Thenrow-columnexpansion(2.3)gives one findsdet(T)2=d'and 14 the (2,j)-entryismatrixQopwJ.whereQ=(akwi).Inotherwords, This matrixisitselfan(n'xn')-matrixwithentries(nn)-matricesofwhich (nn' xnn')-matrix discriminant Aofthisintegralbasis.SinceG(LL'IK)=fakoEIk proposition (2.10),everyfinitelygeneratedOK-submodule aofKadmits L flL'=Kthatandarelinearlydisjoint. separable extensions(notnecessarilyGalois),ifoneassumes insteadof Remark :Itfollowsfromtheproofthatpropositionis validforarbitrary matrix maybetransformedtolooklikethefirstone.Thisyields Here Edenotesthe(nxn)-unitmatrix.Bychangingindicessecond 1, The discriminant a 7L-basisai,...,a, integral closure0KC_KofZQinanalgebraicnumber fieldK.By ... ,n, The chiefapplicationofourconsiderationsonintegrality willconcernthe Q =1,...,n'}, M= d =det(M)2det(Q)2n'det((crw,)) Q 0 M =(akoewiwj)(akwiaaw'). d(at, ail =xdaij+ det(T) b=T*a. ..., an)= it Q 0 is thesquareofdeterminant a =Tb. Ed 1w' Eo wi x'd'ai1 det((o aj))2 ... n, E A...... Chapter I.AlgebraicIntegers . 2n Ea' n,wni Ecr ,wi = d"'dn ' . ... U 0.0 discriminant ofthealgebraicnumberfieldK, In thespecialcaseofanintegralbasiswt,...,wOKweobtain We maythereforewrite has integralentries.Itthereforedeterminantf1,sothatindeed then thebasechangematrixT=(ail),ai>iaijaj,aswellitsinverse, is independentofthechoicea7L-basis;ifai,...,ananotherbasis, § 2.Integrality polynomial ringA[t]. Exercise 2.Showthat,iftheintegraldomainAisintegrallyclosed,thenso generated Z-modules. to a7L-basisofa'.Thisproofispartthewell-knowntheoryfinitely of thedeterminantbasechangematrixpassingfroma7L-basis modules ofK,thentheindex(a':a)isfiniteandsatisfies (2.12) Proposition.Ifaca'aretwononzerofinitelygeneratedoK-sub- In general,onehasthe Exercise S.Showthat(1, and thatanintegralbasisofKisgivenby{1,,15}inthe secondcase,by the quadraticnumberfieldK=Q(,/D).Showthat Exercise 4.LetDbeasquarefreerationalinteger#0,1and d thediscriminantof p =(X2-Y3).Showthatisaprimeideal,butA/pnotintegrally closed. Exercise 3.InthepolynomialringA=Q[X,Y],considerprincipalideal Exercise 1.Is positive, resp.negativeterms, onefindsdK=(P-N)2+4PN. prefixed byapositiveornegative sign.WritingP,resp.N,forthesumof Hint: Thedeterminantdet(a1(o!) ofanintegralbasisu or =1mod4(Stickelberger'sdiscriminant relation). Exercise 7.ThediscriminantdK ofanalgebraicnumberfieldKisalways=0mod4 Exercise 6.Showthat(1,0,1(0+02)1isanintegralbasisof Q(0) {l, All wehavetoshowisthattheindex(a':a)equalsabsolutevalue i (1 + d(ai, o-'" 3+2,16- ..., in thefirstcase,andby(1, a') =det(T)2d(at, an algebraicinteger? d=4D, d=D, dK =d(oK)=d(wl,...,co.). His d(a) =d(ai,...,a,,). d(a) =(a':a)2d(a'). ,/ 2)isanintegralbasisofQ( if D=1mod4, if D-2or3mod4, Iv, ..., a,)=d(al,a,,) (d +a))inbothcases. is asumofterms,each ). '>' '0 -e-4=0. bow 3 w.. 15 bigger domainof"idealnumbers" whereuniquefactorizationinto"ideal idea wasthattheintegersof Kwouldhavetoadmitanembeddingintoa EDUARD KuMMER.Inspired bythediscoveryofcomplexnumbers,Kummer's grand eventsinthehistoryof numbertheory,thediscoveryofidealtheoryby different. or 1-2.Thetwoprimefactorizationsof21aretherefore essentially do notbelongtoOK,thenumbers3and7areassociated to1+2/ has nosolutionsinZ.Inthesamewayitisseenthat 7,1+2 imply NKIQ(a)=±3.Buttheequation decomposed intwoways, Example :TheringofintegersthefieldK=Q(,,f--5)isgivenby§2, uniqueness ofprimefactorizationdoesnotholdingeneral. 1 -2/areirreducible.Asthefractions instance, 3=afi,witha,Bnon-units,then9NKIQ(a)NKIQ(P)would exercise 4,asOK=Z+7G3.Inthisring,therationalinteger21canbe and y.However,contrarytowhathappensintheringsZZ[i], and theprimedecompositionofafollowsbyinductionfromthose5 of twonon-unitsa=0y.Thenby§2,onehas elements. Forifaisnotitselfirreducible,thenitcanbewrittenasproduct every non-unita-A0canbefactoredinOKintoproductofirreducible algebraic numberfieldKisatthecenterofallourconsiderations.AsinZ, 16 Realizing thefailureofunique factorizationingeneralhasledtooneofthe All factorsoccurringhereareirreducibleinOK.Forifonehad,for .--r Being ageneralizationoftheringZCQ,OKintegersan 1 spy 1+2/ 7 Chapter I.AlgebraicIntegers t". , and § 3. Ideals 17 prime numbers" would hold. For instance, in the example of 21 =3.7= (1+2J)(1-2/), the factors on the right would be composed of ideal prime numbers P1, p2, p3, p4, subject to the rules 3 = P1P2,7 = p3p4, P 2 - / - - - 5 This would resolve the above non-uniqueness into the wonderfully unique factorization 21 = (PIP2)(P3p4) _ (P1P3)(P2P4) Kummer's concept of "ideal numbers" was later replaced by that of ideals of the ring OK. The reason for this is easily seen: whatever an ideal number a should be defined to be, it ought to be linked to certain numbers a E OK by a divisibility relation a I a satisfying the following rules, for a, b, a. E OK, ala and alb = ala ± b;ala = al)a. And an ideal number a should be determined by the totality of its divisors in OK a= {aEOKIaIa}. But in view of the rules for divisibility, this set is an ideal of OK. This is the reason why RICHARD DEDEKIND re-introduced Kummer's "ideal numbers" as being the ideals of OK. Once this is done, the divisibility relation a I a can simply be defined by the inclusion a E a, and more generally the divisibility relation a I b between two ideals by b C a. In what follows, we will study this notion of divisibility more closely. The basic theorem here is the following. (3.1) Theorem. The ring OK is noetherian, integrally closed, and every prime ideal p ,-£ 0 is a maximal ideal. Proof: OK is noetherian because every ideal a is a finitely generated Z- module by (2.10), and therefore a fortiori a finitely generated OK-module. By § 2, OK is also integrally closed, being the integral closure of Z in K. It thus remains to show that each prime ideal p 0 0 is maximal. Now, p f Z is a nonzero prime ideal (p) in Z: the primality is clear, and if y E p, y # 0, and y,:+alyn-1+...+an=0 is an equation for y with a; E Z, an 0 0, then an E p fl Z. The integral domain o = OK /p arises from x = Z/pZ by adjoining algebraic elements and is therefore again a field (recall the fact that K[a] = rc(a), if a is algebraic). Therefore p is a maximal ideal. 0 ring 7G,theDedekinddomainsmaybeviewedasgeneralizedprincipalideal 'E. (3.2) Definition.Anoetherian,integrallyclosedintegraldomaininwhich developed byDedekind,whichsuggestedthefollowing foundation ofthewholetheorydivisibilityitsideals.Thiswas This isthesmallestidealcontainingaaswellb,inotherwords,it relation albisdefinedbyca,andthesumofideals ideals aandbofo(ormoregenerallyanarbitraryring),thedivisibility domain o,andwedenotebyKthefieldoffractionso.Giventwo we shallshowfurtheron. in general,notaprincipalidealdomain,butalwaysDedekindas and LIKisafinitefieldextension,thentheintegralclosureBofAinis, domains. Indeed,ifAisaprincipalidealdomainwithfieldoffractionsK, every nonzeroprimeidealismaximalcalledaDedekinddomain. 18 numbers". Still,thefactthat itholdsisremarkablebecauseitsprooffar factorization elements alonemayrefusetoprovide:theuniqueprime factorization. With respecttothismultiplicationtheidealsofowill grantuswhatthe the productofaandbby intersection aflbisthelcm(leastcommonmultiple)ofandb.Wedefine the greatestcommondivisorgcd(a,b)ofaandb.Bysametoken in o.Wepreparetheproofproper bytwolemmas. from straightforward,andunveils adeeperprinciplegoverningthearithmetic into nonzeroprimeidealspiofowhichisuniqueuptothe orderofthefactors. (3.3) Theorem.Everyidealaofodifferentfrom(0) and(1)admitsa Just astheringsofformOKmaybeviewedgeneralizations The threepropertiesoftheringOKwhichwehavejustprovenlay Instead oftheringoKwewillnowconsideranarbitraryDedekind This theoremisofcourseperfectly inlinewiththeinventionof"ideal .CD ab ={JaibilaiEa,b} a+b= {a+bIaEa,bEb}. i a=pi...p, Chapter I.AlgebraicIntegers . 'i~ elements b1ib2Eosuchthatblb2a,butb1,b20a.Puta1=(b1)+ thus admitsamaximalelementa.Thiscannotbeprimeideal,sothereexist stationary. ThereforefiIisinductivelyorderedwithrespecttoinclusionand is nonempty.Asonoetherian,everyascendingchainofidealsbecomes Proof: Supposetheset931ofthoseidealswhichdonotfulfillthiscondition P 1,P2, (3.4) Lemma.Foreveryideala § 3.Ideals prime.) SinceP2 for everyitherewouldexista1Ep;-,psuchthat..arp.Butis pl isamaximalideal.(Indeed,ifnoneofthepiwerecontainedinp,then possible. Thenoneofthep1,saypI,iscontainedinp,andsop1=pbecause Proof: LetaEp,;0,ndplp2. (3.5) Lemma.Letpbeaprimeidealofoanddefine products iscontainedina,acontradiction. both a1anda2containaproductofprimeideals,thethese a2 =(b2)+a.ThenaCa1,anda1Bythemaximalityofa, Proof of(3.3):I.Existence theprimeidealfactorization.LetWJIbe It followsthatxisintegralovero,beingazeroofthe monicpolynomial the determinantd=det(A)satisfiesdal. Writing Aforthematrix(x8;1-aii)weobtainA(al,..., an)t=0.By(2.3), Let usassumethatap-1=a.ThenforeveryxEp-1, thus a-lbEP-1.Itfollowsthatp-10o. i.e., a-1b¢o.OntheotherhandwehavebyC(a),abpco,and Then onehasap-1:_{>ta;x,IaiEa,x;p-1}0foreveryideala#0. decomposition. If9isnonempty, thenweargueasfor(3.4)thatthereexists set ofallidealsdifferentfrom (0)and(1)whichdonotadmitaprimeideal a contradiction. f (X)=det(XBti-a;1)Eo[X].ThereforeXo.Thismeans thatp-1=0, Now leta , pr such that k'1 0 beanidealofoandal, CAD Pr ¢(a),thereexistsbEP2 xa; _Ea,iai, p-1= IxEKIxpco}. v,' aDPiP2...pr. i 0 ofothereexistnonzeroprimeideals . 'IS . Prc(a)p,withrassmall a,i r=o. 000 ..., a,asystem . . =doe,0andthusd0. Pr suchthatb0ao, 'R3 of generators. Ono VII o-. C," 19 prime, wegeta=a1. Indeed, sinceaiIa,i=1, ... and foraEniaiwefindthat aiIa,andtherefore,thefactorsbeingrelatively an arbitraryringtakingintoaccountthat Theorem" fromelementarynumbertheory.Wemayformulate thisresultfor a1, ...,a,onehasananalogueofthewell-known"Chinese Remainder (a, b)=a+bo.Forproducta1...a,ofrelatively primeideals is writtenfortworelativelyprimeideals,insteadofthe correctformula Similarly, thenotationaIisoftenusedinstead of aI(a)and(a,b)=1 of "idealnumbers"-wewillwritewithaslightabuse notation then -followingthetraditionwhichtendstoattributeidealsrole that thepiarepairwisedistinct.Ifinparticularaisprincipalideal(a), In thesequelsuchanidentitywillbeautomaticallyunderstoodtosignify be twoprimeidealfactorizationsofa.Thenp,dividesafactorqi ideal factorizationofana pi=gi,forall i=1,...,r. Continuing likethisweseethatr=sand,possiblyafterrenumbering, pt and beingmaximalequalsQ1.Wemultiplybyp]1obtain,inviewof H. Uniquenessoftheprimeidealfactorization.Foraponehas: a p-1=p1 a it followsthatpp-1=o.Inviewofthemaximalityain971andsince By (3.5),onehasaCap-1andpp-1a.Sinceismaximalideal, inclusion oCp-1givesus a maximalelementinf9)1.Itiscontainedidealp,andthe 20 abcp=acporbcp,i.e.,plab=>pjaorpIb.Let cue Grouping togethertheoccurrencesofsameprimeidealsin p, i.e.,ap-10o,theidealadmitsaprimedecomposition -trpipit=o,that .fl pr, andsodoesa=ap-1pp1 .-. -I% . anIa,i.e.,aEa. a=pit ...p'r, a cap-1pp-1o. ... ,n,wefindon or- P2 a =pit...pr'. a =nai. a=PIP2...pr =gtg2...gs Pr =g2...gs, 0 ofo,givesaproductrepresentation i=1 n vi>0. ."1 the onehandthataCn=1 ai, °,a Chapter I.AlgebraicIntegers . prp,acontradiction. ... , sayqj, group JKofK.Theidentity elementis(1)=o,andtheinverseofa ideals ofK. as idealsino.Fordistinctionthelattermayhenceforth becalledintegral is anidealoftheringo.Fractionalidealsaremultiplied inthesameway a fractionalidealifandonlythereexistscE0, (3.8) Proposition.Thefractional idealsformanabeliangroup,theideal (a) =ao.Obviously,sinceoisnoetherian,ano-submodule a (3.7) Definition.AfractionalidealofKisafinitelygenerated 0-submod- of fractionalidealinthefieldfractionsK. may obtaininversesforthenonzeroidealsofobyintroducingnotion proves thesurjectivity. Putting x=xlyl+ and, bythesametoken,elementsY2,...,y,suchthat x -ximodai,i=1,2. may write1=a1+a2,aiEai,andputtingxx2a1x1a2weget For this,letximodaiEo/ai,i=1, has kernela=niai.Itthereforesufficestoshowthatitissurjective. ulea00ofK. Proof: Thecanonicalhomomorphism such thatai+aj=0fori;j.Then,ifa=n7 (3.6) ChineseRemainderTheorem.Letat, § 3.Ideals For instance,anelementaEK'definesthefractional"principal ideal" Now letobeagainaDedekinddomain.Justasfornonzeronumbers,we If n>2,wemayfindasbeforeanelementYlEosuchthat rye yi -lmodai, .`o o-®o/ai, a-®amodai, III yi=1mod a1, . +xnynwefindx-ximodai,i=1,...,n.This a-' =IXEKxaCc i=1 )? o/a yi -0modai yi =0modnai, 112 0/at. III CD. n III its ..., n,be i.1 11 ... ,an 1 ai, . i_2 for i0j. given. Ifn=2,we one has be idealsinaringo 0, suchthatcaCo 0 ofKis 21 product Finally, ifaisanarbitraryfractionalidealandcEo,#0,suchthat because pismaximal.Consequently,ifa=pl a primeidealp,(3.5)saysthatpCpp-1andtherefore Proof: Oneobviouslyhasassociativity,commutativityanda(1)=a.For 22 Proof: Everyfractionalidealaisquotient=b/coftwointegralidealsb group onthesetofnonzeroprimeidealspo. (3.9) Corollary.Everyfractionalidealaadmitsuniquerepresentationas ca co,then(ca)-1=c-1a1 if xaco,thenxabCb,soxEbbecauseab=o.Thuswehave then b=pi1 group ofidealsJK,whichwillbedenotedPK.Thequotient is integral,andthereforeclearlyalsoingeneral. decomposition ofthetypestatedincorollary.By(3.3),itisuniqueifa and c,whichby(3.3)haveaprimedecomposition.Thereforehas with v.E7Gandv,=0foralmostallp.Inotherwords,JKisthefreeabelian these resultscannotbehad for nothing.Theywillbeobtainedbyviewing in anumberfieldK,however, oneobtainsimportantfinitenesstheorems, may turnouttobecompletelyarbitrarygroups.Forthering OKofintegers groups o*andC1Kmorethoroughly.ForgeneralDedekind domainsthey same process.Thisimmediatelyraisestheproblemofunderstanding these to ideals,whereastheunitgroupo*measures contraction inthe CIK measurestheexpansionthattakesplacewhenwe pass fromnumbers where thearrowinmiddleisgivenbyaH(a).So theclassgroup the groupofunitso*o,itfitsintoexactsequence is calledtheidealclassgroup,orgroupforshort, ofK.Alongwith prepare thenecessaryconcepts, whichallcomefromlinearalgebra. the numbersgeometrically as latticepointsinspace.Forthiswewillnow which arefundamentalforthe furtherdevelopmentofnumbertheory.But The fractionalprincipalideals(a)=ao,EK*,formsubgroupofthe 'O- 1-) o*-+K*-) p,-1 isaninverse.ba=oimpliesthatbCa-.Conversely, CIK =JK/PK CAD is theinverseofca,soas1=o. a =fjYVp P CAD Chapter 1.AlgebraicIntegers pr isanintegralideal, pp-1 = o emu" a-1. ring ofintegers0(v').Showthat(p)=poisaprimeidealifandonly Exercise 3.Letdbesquarefreeandpaprimenumbernotdividing2d.othe Q (-47). are twoessentiallydifferentdecompositionsintoirreducibleintegralelementsof Exercise 2.Showthat Exercise 1.Decompose33+11 § 4.Lattices cam chain R2pipipe..... Hint: Showasin(3.4)that(0)isaproductp, descending chainofidealsai2a2) Exercise 7.InanoetherianringRinwhicheveryprimeidealismaximal,each Hint: Useexercise5. Exercise 6.EveryidealofaDedekinddomaincanbegeneratedbytwoelements. Choose rrEp Hint: Fora=p"theonlyproperidealsofo/aaregivenbyp/p" principal idealdomain. Exercise 5.Thequotientringo/aofaDedekinddomainbyanideal the Chineseremaindertheoremforcosetsn;modp;'". Hint: Ifa=pi'...p,':A0isanideal,thenchooseelementsn;Ep;gyp?andapply ideal domain. Exercise 4.ADedekinddomainwithafinitenumberofprimeidealsisprincipal the congruencex2=-dmodphasnosolution. homomorphism a-+Ncanbeliftedto h :a-+Msuchthat modules, i.e.,givenanysurjectivehomomorphismM Exercise 10.ThefractionalidealsaofDedekinddomain oareprojectiveo- factorization intoprimeideals.ShowthatoisaDedekinddomain. Exercise 9.Letobeanintegraldomaininwhichallnonzeroideals admitaunique in everyidealclassofC1K,thereexistsanintegralprimetom. Exercise 8.LetmbeanonzerointegralidealoftheDedekinddomaino.Showthat series. we usedatacrucialplacethe inclusion foh=g. In §1,whensolvingthebasic problemsconcerningthegaussianintegers, ti] C4" )40 p2 andshowthatp"=orr"+p". 54 =2.33 § 4.Lattices 13+,/---47 Z[i] C into irreducibleintegralelementsofQ( pr =(0)canberefinedintoacomposition °'n 2 becomes stationary. y-. .t'.. -°' rte, 13 - 44)i ors pr ofprimeidealsandthatthe 2 -47 N ofo-modules,each '00 p"-1/p" 0 isa 23 ). (4.2) Proposition.Asubgroup TcVisalatticeifandonlyitdiscrete. clearly issuchaneighbourhood. Thispropertyisindeedcharacteristic. then, extendingv1,...,vmtoabasis.. neighbourhood whichcontainsnootherpointsofP.In fact,if that everypointyETisanisolatedinthesense thatthereexistsa has thespecialpropertyofbeingadiscretesubgroup of V.Thisistosay instance Z+Z-CRisnot.Buteachlatticer= Zv1+ subgroup ofV.Butnoteveryfinitelygenerated isalattice-for of suchachoice.Notethat,firstall,latticeisfinitelygenerated vectors. Butwewillneedacharacterizationoflatticeswhichisindependent entire spaceV. the setofalltranslates(P+y,yEI",fundamentalmeshcovers structure ofV,ifm=n. is calledabasisandtheset develop Minkowski'stheorywefirsthavetointroducethegeneralnotionof essential partofthefoundationsalgebraicnumbertheory.Inorderto HERMANN MINKOWSKI(1864-1909)andhasledtoresultswhichmakeupan This pointofviewhasbeengeneralizedtoarbitrarynumberfieldsby and consideredtheintegersofQ(i)aslatticepointsincomplexplane. 24 a fundamentalmeshofthelattice.ThelatticeiscalledcompleteorZ- with linearlyindependentvectorsvl, is asubgroupoftheform (4.1) Definition.LetVbeann-dimensionalR-vectorspace.Alatticein lattice andstudysomeofitsbasicproperties. The abovedefinitionmakesuseofachoicelinearlyindependent The completenessofthelatticeisobviouslytantamounttofactthat {xlvl +- 0 _txlVI+- - . I xjER,Jai-xi , v ten Chapter 1.AlgebraicIntegers' of V,theset 11 ..., m} + zvm the entirespaceV. subset MCVsuchthatthe collection ofalltranslatesM+y,yEF,covers (4.3) Lemma.AlatticeTinV iscompleteifandonlyifthereexistsabounded given, say,asadiscretesubgroupTCV-iscomplete. m-dimensional spaceVO.ThisshowsthatTisalattice. By themaintheoremonfinitelygeneratedabeliangroups, Ttherefore vectors v1,...,yrarealsoR-linearlyindependentbecause theyspanthe admits aZ-basisv1,...,r Lbw Y i=l-ti+o, T CT0=Z(qIII) TO=ZU1+...+ZU"CT /J iE'Po, BCD-( +...+Z( 0.7 .,, Yoi ETOcVo U"). ... ,um e + Zv,..The of VOwhich can 25 The cubespannedbyanorthonormalbasisel, Then wehaveonVanotionofvolume-morepreciselyHaarmeasure. for eachvEN, show thatV=Vo.SoletvEV.SinceUYEF(M+y)wemaywrite, II° for ycF,coverV.LetVobethesubspacespannedbyP.Wehaveto fundamental mesh0 Proof :Ifr=Zv1+ 26 This doesnotdependonthechoiceofabasisv1 mesh ofr,andwewriteforshort v1, ...,vn,sothatvi=(Ekaikek.Since where A=(aik)isthematrixofbasechangefromel, has volume vectors vl,...,vn, and moregenerally,theparallelepipedspannedbynlinearlyindependent dimension nequippedwithasymmetric,positivedefinitebilinearform Since Misbounded, in X.Withthesedefinitions we have whole linesegment(ty+(1 -t)x10 v-r 00v (Dry v av convergestozero,andsinceVoisclosed, vol(O) =Idet((vi,vj))I kJ (xl v1+- t + Zv,,iscomplete,thenonemaytakeMtobethe a +lim (,):V xV--*R. ={xlvl+...+xnunIXi ER,0 ..., e Yv EVo. CIA . . . . . vn has volume1, of thelattice ..., ento 27 . 1 x1,x2 E X, SIN 1 ?N. vol(o n (Z X + y)) yEr = 2x2+y2, .SE vol(X) > 2° Vol(r). vol(X) > + y, y E r, were pairwise disjoint, then the same + y, y E r, were pairwise disjoint, then Y=Y1-Y2= 2x2- 2x1, Y=Y1-Y2= 2x2- AID (2x+y1) n(Zx+y2) oo. (2x+y1) n(Zx+y2) 2X 2x1 +Y1 vo1(O) > yEr vol(45) > > vol((.p - y) n 2 x) = vol(2 X) = 2n vol(X), vol(45) > > vol((.p - y) n 2 x) = vol(2 Now, if the sets is compact. Exercise 2. Show that Minkowski's lattice point theorem cannot be improved, by giving an example of a centrally symmetric convex set X C_ V such that vol(X) = 2° vol(r) which does not contain any nonzero point of the lattice P. If X is compact, however, then the statement (4.4) does remain true in the case of equality. Exercise 1. Show that a lattice r in 1' is complete if and only if the quotient 1W'/r Exercise 1. Show that a lattice r in 1' is complete But translation of o n (2 X + y) by -y creates the set (45 - y) n 1 X of But translation of o n (2 X + y) by cover the entire space V, therefore equal volume, and the 0 - y, y E r, obtain also the set i X. Consequently we would which contradicts the hypothesis. the euclidean vector space V and X a centrally symmetric, convex subset of V. convex subset and X a centrally symmetric, vector space V the euclidean y E F. nonzero lattice point at least one Then X contains yi, y2 E 17 such that point in this intersection, In fact, choosing a we obtain an element joining x2 and -x1, and therefore which is the center of the line segment belongs to x n r. (2 X +y) with a fundamental mesh 45 would be true of their intersections c n of r, i.e., we would have § 4. Lattices § 4. in lattice a complete Let r be Theorem. Point Lattice Minkowski's (4.4) Suppose that lattice points to show that there exist two distinct Proof: It is enough 28 Chapter I. Algebraic Integers' Exercise 3 (Minkowski's Theorem on Linear Forms). Let n L;(x1,.. i=1,..n, j=t be real linear forms such that det(a;j) 0 0, and let c1, ..., c,, be positive real numbers such that c1. c > I det(a;j) I. Show that there exist integers m1, ... , mn c Z such that IL1(mi, , mn)I § 5. Minkowski Theory The basic idea in Minkowski's treatment of an algebraic number field K IQ of degree n is to interpret its numbers as points in n -dimensional space. This explains why his theory has been called "Geometry of Numbers." It seems appropriate, however, to follow the current trend and call it "Minkowski Theory" instead, because in the meantime a geometric approach to number theory has been developed which is quite different in nature and much more comprehensive. We will explain this in § 13. In the present section, we consider the canonical mapping . j:K--)Kc:=fl C,ai) ja=(ra), which results from the n complex embeddings r : K -> C. The C-vector space Kc is equipped with the hermitian scalar product (*) (x,y) = F-xtyt t Let us recall that a hermitian scalar product is given by a form H(x, y) which is linear in the first variable and satisfies H (x, y) = H (y, x) as well as H (x, x) > 0 for x 0. In the sequel we always view KC as a hermitian space, with respect to the "standard metric" (*). The Galois group G(CIR) is generated by complex conjugation F : z v )Y. The notation F will be justified only later (see chap. III, §4). F acts on the one hand on the factors of the product 11t C, but on the other hand it also acts on the indexing set of r's; to each embedding r : K -+ C corresponds its complex conjugate i : K -- C. Altogether, this defines an involution F:KC --*KC Tr oF=wehaveon KRthek-linearmap the associatedHaarmeasure (see§4,p.26)thecanonicalmeasure.Since the Minkowskispace,itsscalar product( and, inanycase,(x,x)>0forx:0. view oftherelationsF(x,y)=(Fx,Fy)_(x,y), =(x,y)(y,x), on thek-vectorspaceK.Indeed,forx,yEKR,one has(x,y)EJRin scalar product The restrictionofthehermitianscalarproduct(,)from KCtoKRgivesa mapping given anon.Sinceis=fadforaEK,onehasF(ja)ja.Thisyields are thepoints(zt)suchthatzt=zt.AnexplicitdescriptionofKRwillbe consisting oftheG(CI1R)-invariant,i.e.,F-invariant,pointsKc.These gives theusualtraceofKIQ(see(2.6),(ii)), given asthesumofcoordinates.ItisalsoF-invariant.Thecomposite Finally, wehaveontheC-vectorspaceKc=ftClinearmap The scalarproduct(,}isequivariantunderF,that which, onthepointsz=(zt)EKv,isgivenby § 5.MinkowskiTheory We calltheeuclideanvectorspace We nowconcentrateonthek-vectorspace chi, (,):KRxKR- R (Fx, Fy)=F(x,y) TrKIQ(a) =Tr(ja). K-+K, Tr:KR -)R, Tr:Kc-+C, j :K--)KR. '7t -S' (Fz)t KR=Kc=Lll-J t if. t KR=[f[cC]+ , C ) thecanonicalmetric,and . 29 following manner.Someoftheembeddingsr:K->CCarerealinthatthey by theinclusionP-->C.FcorrespondstoF(a(9z)=a®Y. corresponds tothecanonicalmappingK®QP-->Cwhichisinduced Likewise, K®QC"Kr-.Inthisapproach,theinclusionKRKc product K®QR, the mappingj Remark: Wementioninpassing-itwillnotbeusedthesequelthat and itscompositewithj:K-iKRisagaintheusualtraceofIQ, 30 embeddings andaoverthefamilyofchosencomplexembeddings.SinceF pair somefixedcomplexembedding,andletpvaryoverthefamilyofreal of complexconjugateembeddings.Thusn=r+2s.Wechoosefromeach be therealembeddings.Thecomplexonescomeinpairs land alreadyinP.andothersarecomplex,i.e.,notreal.Let given bytherule(zr)i-a(xr)where This givesthe leaves thepinvariant,butexchangesa,F,wehave This isomorphismtransforms thecanonicalmetric(,)intoscalarproduct (5.1) Proposition.Thereisanisomorphism where ar=1,resp.2,if risreal,resp.complex. GET An explicitdescriptionoftheMinkowskispaceKRcanbegivenin : K-KRidentifiesthevectorspacewithtensor Xp =zp, KR =I(Zr)EHCI K ®QR--KR, at,Q1,...,cr ff,:K--)C P1... TrKIQ(a) =Tr(ja). (x,Y) _Earxryr, xa =Re(za), r Pr K r r Zp ER,zQ=Y, a ®xt-r(ja)x. f:Kp-*flR=Rr+2s >R xd =Im(za) Chapter I.AlgebraicIntegers' 31 and , + Zja,. in view of yQ = xQ in view of IdKI (oK : a). )) , and .f( : K - C, tl, ..., r, and _ teak) = AA'. 1 -xTYT = IdetAl = to ai I dK I (OK : a) . T aT u,xtyt transfers the canonical measure u,xtyt transfers the V y., Idet((jaf,jak))I112 e=i (x, Y) = vol(P) = It obviously differs from the standard Lebesgue meas- It obviously differs RT+2s. VOlcanonical(X) = 2 ((jai, jak)) 'mar vol(1') = zoza +zFF = z,f +z z' = 2Re(z-ZQ) = 2(xx' +xQxa). 2Re(z-ZQ) = = z,f +z z' = zoza +zFF +-' Using this proposition, Minkowski's lattice point theorem now gives the The mapping j : K -+ KR gives us the following lattices in Minkowski The mapping j : K -+ KR gives us the : a)2d(OK) = (OK : a)2dK, d(a) = d(al, ..., a,,) = (detA)2 = (OK The scalar product (x, y) The scalar product to number theory. form the matrix A = (reca; ). Then, according to (2.12), we have form the matrix A = (reca; ). Then, according and on the other hand This indeed yields following result, which is what we chiefly intend to use in our applications vol the corresponding canonical measure. space KR. of OK, then r = j a is a complete (5.2) Proposition. If a # 0 is an ideal volume lattice in KR. Its fundamental mesh has a, so that T = Z jal + Proof: Let al, ..., a, be a Z-basis of r We choose a numbering of the embeddings from KR to on R'+2, worked with the Lebesgue measure Minkowski himself on KR is the one suit. The corresponding measure most textbooks follow scalar product determined by the called the Minkowski metric on KP. This scalar product may therefore be the canonical metric, and denote by But we will systematically work with § 5. Minkowski Theory Minkowski § 5. iyt), = (xT + z = (zr) If isomorphism. an is clearly The map Proof: then zpzP = xPx' = (x= + iy') E KR, z' = (z'') x'Q , one gets and yQ = the scalar products. the claim concerning This proves ure by has 0 T . .vi 0, a E a ; in other Chapter I. Algebraic Integers I. Algebraic Chapter a a) = 2" vol(F). (C* =flC*. (z1)t---+(xr), p t E Hom(K,Q. I dx I (o I zT I < ct } is centrally symmetric and I zT I < ct } is centrally K* -L* C* r for all K N:K* z (2 )S r[c,>A(OK:a), K* f :Kn -- FIR, IdK I. Then there exists a E a, a 0 0, such that there exists a E IdK I. Then Ital < c7 S f (X) = ( xz) E FIR I Ixp I < Cp, xa + Xa < Ca } f (X) = ( xz) E FIR I Ixp I < Cp, xa vol(X) > 2T+Sn' vol(X) = 2S VOILebesgue(f (X)) = 2S rj(2cp) fl(7rca,) = 2+'7r' Fl c= . vol(X) = 2S VOILebesgue(f (X)) = 2S There is also a multiplicative version of Minkowski theory. It is based There is also a multiplicative version (5.3) Theorem. Let a 0 0 be an integral ideal of K, and let ct > 0, for let ct > 0, K, and ideal of integral 0 0 be an Let a Theorem. (5.3) cf and such that cr = C), be real numbers r E Hom(K, = (n) where A (5.1) vol(X) can be computed via the map convex. Its volume 32 = { (zz) E KR I Proof: The set X out to be 2' times xa = Re(za), xa = Im(za). It comes given by x, = zp, of the image the Lebesgue-volume This gives Now using (5.2), we obtain lattice point theorem is satisfied. So Thus the hypothesis of Minkowski's j a E X, a there does indeed exist a lattice point words IraI < cT. on the homomorphism homomorphism The multiplicative group K * admits the given by the product of the coordinates. The composite with Rbythemap(x,x)H 2x.Inthiswayweobtainanisomorphism The factor[RxIR]+now consists ofthepoints(x,x),andweidentifyit which isanalogoustotheonewesawabovefor[frC ]+ of complexconjugateones,91,vt,...,US,QS.Weobtain adecomposition before theembeddingsr diagram now passeverywheretothefixedmodulesunderG(CIR)andobtain i.e., thehomomorphismsofdiagramareG(CIR)-homomorphisms.We One clearlyhas on K*,K*asbefore,andthepointsx=(xr)EJrRby(Fx)rxT. and weobtainthecommutativediagram It inducesasurjectivehomomorphism logarithm topassfrommultiplicativeadditivegroups In ordertoproducealatticefromthemultiplicativetheory,weuse is theusualnormofKIQ, § 5.MinkowskiTheory The R-vectorspace[fl,R]+isexplicitlygivenasfollows. Separateas The involutionFEG((CIR)actsonallgroupsinthisdiagram,trivially Foj=j, Fot=rfoF,NoF=FoN,TroF=Tr, NKIQi K* Q* [FIR]+ =FIRxfl[RxR]+. f:C*_ r -- ) : K->Cintorealones,pl,...,p,-,andpairs i [flR]+ J NKIQ(a) =N(ja). f: KC* r + C*- R* P- K>Et p R, KC* III jN N _ zi v e e R'.+S R, loglzj. [flzR]+ R. 4-- jTr 33 is givenby homomorphism given bythesumofcoordinates.Identifying[ftR]+withI[8r+s, which againtransformsthemapTr 34 Exercise 3.Showthatineveryideala has volumevol(X)=2I7rsn"(seechap.III,(2.15)). Exercise 2.Showthattheconvex,centrallysymmetricset integral ideala;0ofKcontainsanelement Exercise 1.WritedownaconstantAwhichdependsonlyonKsuchthatevery where wewritexEKRCftC*as=(xt). In ordertocounttheidealsa0ofringOKweconsider theirabsolute ideal classgroupCIK=JK/PKofanalgebraicnumber fieldKisfinite. arithmetic andgeometricmeans, Hint: Useexercise2toproceedasin(5.3),andmakeuseoftheinequalitybetween where M=n°(n) excluded, whenitsconsideration wouldvisiblymakenosense.)Thisindex (Throughout thisbookthe caseofthezeroideala=0isoftentacitly norm As afirstapplicationofMinkowskitheory,wearegoing toshowthatthe ."t 2(x) =(logIxp1I Iral a =modp'+'isbasisoftheOK/p-vectorspacepi/pi+'.Sowehave `.° well asdet(A)=NKIQ(a)bydefinition. then, aswaspointedoutalreadyin§2,onehas Indeed, ifcol,...,coisaZ-basisofOK,thenco, ideal (a)ofOK,wherewehavetheidentity is finiteby(2.12),,andthenamejustifiedspecialcaseofaprincipal § 6.TheClassNumber pi/pi+l because otherwiseb'=by-'wouldbeaproperdivisorofppi+'p-iThus pi Npi+1andb=(a)+pi+l,thenp'D quotient p'/pi+'isanOK/p-vectorspaceofdimension1.Infact,ifae one haspi#p`+lbecauseoftheuniqueprimefactorization,andeach In thechain We arethusreducedtoconsideringthecasewhereaisprimepowerpV. Proof: BytheChineseremaindertheorem(3.6),onehas a 40,thenonehas (6.1) Proposition.Ifa=pi' (a) =aOK,andifA(aid)denotesthetransitionmatrix,aa)iEaidcvi, a consequenceof(5.3),iscrucialforthefiniteness idealclassgroup. defined onallfractionalidealsa=r[ppVp,vpEZ.The followinglemma, of theabsolutenorm.Itmaythereforebeextendedtoahomomorphism that (6.2) Lemma.Ineveryideal a The propositionimmediatelyimpliesthemultiplicativity fl(pV) =(OK:pV) 211 OK/p andtherefore ..B `+° dK/a =OK/plt®...®pK/prr 91(a) = NKIQ(a)I = (OK fl((a)) fl(ab) ='l(a)0'i(b) p Dp2...pV 01:JK : p)(pp2)... p'. istheprimefactorizationofanideal 0 ofOKthereexistsanaEa, a +-- = 2 I NKIQ(a) (pl)v1 (-)sIdKgt(a) ... R+ b.:D pi+1 ,c7 (pr)vr . `F" I det(A)_(OK:(a))as and consequentlypi=b, (pv-1 : pv) II} _ J(p)V. 0, such 35 The idealal=ab-1ay-1a E[a]thereforehastherequiredproperty. that For this,chooseanarbitraryrepresentativeaoftheclass, andayEOK, ideal atsatisfying prime idealspofboundedabsolutenorm.Sinceeveryintegral idealadmits because thismeansthatpI(p).Itfollowsthereareonlyfinitelymany is calledtheclassnumberofK. This beingtrueforalls>0andsinceINKIQ(a)Iisalwaysapositiveinteger, Hom(K, C),suchthatct=cTand Proof :Givens>0,wechoosepositiverealnumberscr,fort 36 y 540,suchthatb=a-tc OK.By(6.2),thereexistsaEb, absolute norm'Jt(a) - OHO plr where hK=(JK:PK) gy(p) =Pf. 0, suchthat < `r<1 IIct=(2)SIdKIM(a)+s. r 2 vi >Oand (7r2 ),,/, mud lS n 2 ls om.., 0, satisfyingltal 2 In thecaseoffieldp-throotsunity,questionwhether The theoremofthefinitenessclassnumberhKmeansthatpassing d =-1,-2,3,7,11,19,43,67,163. 'L1 "'"' Supposing erroneouslythat thiswasthecaseingeneral- f", d =2,3,5,6,7,11,13,14,17,19,21,22,23,29, ... y'. 62,67,69,71,73,77,83,86,89,93,94,97. 31, 33,37,38,41,43,46,47,53,57,59,61, cob '-' ... +-' ,4. r"' of integers X P+yP=Z hoc (OD C3, ",4 The equationyP=zP-xP were alwaysequal 0 1leadsto .U. '5`w 37 Following asurprisingdiscoverybythemathematicianGERHARDFREY,who numbers. We stilldonotknowtodaywhetherthereareinfinitelymanyregularprime the first25primenumbers<100onlythreeareirregular:37,59,and67. of theBernoullinumbersB2,B4,...,Bp-3arenotdivisiblebyp.Among irregular. Heevenshowedthatpisregularifandonlythenumerators instead ofhn=1.Inthiscasehecalledaprimenumberpregular,otherwise the argumentswehaveindicatedcanbesalvagedifoneonlyassumespth this way.KUMMER,however,didnotfallintotrap.Instead,heprovedthat to 1-someactuallythoughttheyhadproved"Fermat'sLastTheorem"in 38 important. and withahelpinghandfromRICHARDTAYLOR.See[144]. sufficient generalityin1994byANDREWWILES,aftermanyyearsofwork, conjecture, theTaniyama-Shimura-WeilConjecture.Thiswasprovedin who managedtoreduceFermat'sstatementanother,muchmoreimportant established alinkwiththetheoryofellipticcurves,itwasKENNETHRIBET, Hint: Usingexercise3,§5,proceedasintheproofof(6.3). there existsanintegralidealasuchthat Exercise 3.ShowthatineveryidealclassofanalgebraicnumberfieldKdegreen, -8, -11haveclassnumber1. Exercise 2.Showthatthequadraticfieldswithdiscriminant5,8,11,-3,4,-7, Exercise 1.Howmanyintegralidealsaaretherewiththegivennormfl(a)=n? Exercise 7.Showthat,forevery numberfieldK,thereexistsafiniteextensionL principal idealinthefieldL =K('i),inthesensethataoc(a). Exercise 6.Letabeanintegral idealofKandam=(a).Showthatabecomes the degreenoffield. Exercise 5.ShowthattheabsolutevalueofdiscriminantIdK Itendsto00with chap. III,(2.17)). algebraic numberfieldK Exercise 4.ShowthattheabsolutevalueofdiscriminantIdK I such thateveryidealofKbecomes aprincipalideal. The connectionwithFermat'slasttheoremhasatbecomeobsolete. The regularandirregularprimenumbersdohowevercontinuetobe -°O r-1 C/] N"1 01(a) IdKI "J' Chapter I.AlgebraicIntegers is >1forevery acv number sofpairsv,&:K-+Ccomplexconjugateembeddings.Inorder fact determinedbythenumberrofrealembeddingsp:K-*Rand roots ofunitythatlieinK,butgeneralisnotitselffinite.Itssize field K,thegroupofunitsoK.Itcontainsfiniteµ(K) main problemposedbytheringOKofintegersanalgebraicnumber In theupperpartofdiagramweconsidersubgroups to describe'thegroup,weusethediagramwhichwassetupin§5: § 7.Dirichlet'sUnit'Theorem Proof: Wehavetoshowthat µ(K)isthekernelofA.For is exact. (7.1) Proposition.Thesequence and weobtainthe and thecompositeA:=Eoj:oK-+H.Theimagewill bedenotedby that A(s)=£(js)0.This meansthat µ(K) Cker(.l).Conversely, letsEoKbeanelementinthekernel,so r :K-Canyembedding,we findlogIrkI=10,sothatcertainly After consideringtheidealclassgroupC1K,wenowturntosecond We obtainthehomomorphisms oK ={aEOKNKIQ(s)f1} H ={xE[HR]+lTr(x)01, S =fyEK*N(y)±11, NKIQ K* r *--> JR* § 7.Dirichlet'sUnitTheorem I 1 --3 --> KR-_[II (K) F =X(oK)CH, IN ) o* , low the "norm-onesurface", the groupofunits, the "trace-zerohyperplane". X) r I TE=1foreachembedding (IQ R. jTi t R] + o E µ(K)and 39 subset ofthelatticejoKin [ HrC]+(see(5.2)).ThereforeFisalattice. It containsonlyfinitelymany elementsofthesetjoKbecausethisisa bounded domain frR I `^b and itsufficestoshowthat,foranyc>0,thebounded domain{(xt)E subgroup. ThemappingXoK-+Harisesbyrestricting themapping Proof: Wefirstshowthatr=)t(0)isalatticein H, i.e.,adiscrete i.e., fiisassociatedtoa.Therefore,upmultiplicationbyunits,there one, then such thatIN(a)I=INKIQ(a)ia.Forif,8=a+ay,yEOK,isanother Proof: LetaEZ,>1.Ineveryoneofthefinitelymanycosets need thefollowing elements, andthus,beingafinitegroup,containsonlyrootsofunityinK*. (see (5.2)).ThereforethekernelofXcancontainonlyafinitenumber vector spaceKR.Ontheotherhand,jeisapointoflatticejoKKR 40 dimensional vectorspaceH,andisthereforeisomorphicto (7.3) Theorem.ThegroupPisacompletelatticeinthe(r+s-1)- are atmost(OK:aoK)elementsofnorm±a. because N(0)/flEoK,andbythesametoken oK/aoK thereexists,uptomultiplicationbyunits,atmostoneelementa elements aEOKofgivennormNKJQ(a)=a. (7.2) Lemma.Uptomultiplicationbyunitsthereareonlyfinitelymany r ((zt)) _(logIztl),thepreimageofthisdomainwithrespect to£isthe : K-->C,sothatjE=(re)liesinaboundeddomainoftheIR- Given thisproposition,itremainstodeterminethegroupP.Forthis,we Ixti <-c}containsonlyfinitelymanypointsofP=£(joK). Since l(zr)E[TC*I e-' icy and integersvi. = ^.1 y (K)->oK r 1 BCD 1 ESnXja-1CT,andthus =xja- s 1 (1, ..., =xj(ai s). t ..., st, t 1) E](8'+s ..., A(st)EH. 1 Chapter I.AlgebraicIntegers t =r+s-1,called r A 0 The vector 0, suchthat ... ,vt be El 11 43 10 7 ,lt(st) Ar+1(sr) 6 ... t(sr) r + s. We therefore get the r + s. We therefore l[8r+1. But this has volume l[8r+1. But this 2 ... xl+yl 5 X &O )t+1(st) x2 - dye = 4, x2 - dye = -4, E1 = a. (8,) in : 1(sl) .~i A01 Aot+1 3 ±det '`a 2 D 81 1+/ 2+V (1+-)/2 5+2J 8+3'7 3+,/-10 The importance of the regulator will only be demonstrated later (see The importance of the regulator will where R is the absolute value of the determinant of an arbitrary minor of rank value of the determinant of an arbitrary where R is the absolute of the field K. This absolute value R is called the regulator integer solutions - of the equation or - in case this equation has no rational Then for which x1, yl > 0 are as small as possible. equations x2 - dy2 = ±4 is called Pell's is a fundamental unit of K. (The pair of equation.) fundamental units el for Q(-,/-D): Exercise 2. Check the following table of except for the first entry, which equals except for the first of the unit lat- The volume of the fundamental mesh (7.5) Proposition. tice A(oK) in H is following matrix: t = r + s - 1 of the chap. VII, § 5). and d the discriminant of the real Exercise 1. Let D > 1 be a squarefree integer 2, exercise 4). Let x1, yl be the uniquely quadratic number field K = Q(VD) (see § equation determined rational integer solution of the § 7. Dirichlet's Unit tPheorem Unit Dirichlet's § 7. volume t-dimensional 1. The has length to H and orthogonal is obviously of the parallelepiped 1)-dimensional volume equals the (t + of 0 therefore Ao, L(et), ... , spanned by has only zeroes, a fixed one, say the i-th row, this row Adding all rows to integral closure0ofoinafinite extensionofitsfieldfractions. Dedekind domainoatthebase insteadofZ,andtakingOKthe We treatthisprobleminamore generalcontext,startingfromanarbitrary arises astohowaprimenumber pfactorsintoprimeidealsoftheringOK. cause anymisunderstanding. of K-animprecisemannerspeakingwhichis,however, notlikelyto the setofprimeidealsOK.Theyareoftenreferredto astheprimeideals wow §3, p.17)andisthereforea divisoroftheidealp0K.Hencequestion of integersanumberfieldK,wenowproposetomake afirstsurveyof Hint: ApplyMinkowski'slatticepointtheoremtotheunitintrace-zerospace. 0 Coo a°c § 8.ExtensionsofDedekindDomains ^_+ be aprimitivem-throotofunity,m>3.Showthatthenumbers be aprimitivep-tbrootofunity,panoddprimenumber.Show ... +Show that ..V .cam 0 ofoKcontainsarationalprimenumberp(see QC, CAD °"q .0+ _.b(1 III Chapter I.AlgebraicIntegers' amt ,-, Ti' 0 C1- fin" oco So. extension The exponente;iscalledthe ramificationindex,andthedegreeoffield This wealsodenoteforshort by in thesensethatonehasrelation Instead ofpOwewilloftenwritesimplyp.Theprimeideals X31occurringin ... the decompositionarepreciselythoseprimeidealsq3of 0whichlieoverp a productofprimeideals, s =rrxforsomexE0flKo,i.e.,p,acontradiction. If onehadpO=0,thenitwouldfollowthatsOspO c_rrO,sothat Writing 1=b+s,withbEpandsEa,wefinds0pand spcpa=7ro. In fact,let7rEPNp2(p in §12(see(12.8)). occasion. Infact,weshallgivetheproofinamoregeneralframework this case.Weshallcomebacktothegeneralcaseonamoreconvenient ask thereader'spermissiontocontentourselvesfortimebeingwith module. Thisshowsthat0isnoetherian,providedLIKseparable.We is itselfano-moduleoffinitetype,henceafortiorifinitelygeneratedO- ideal of0isalsocontainedinthisfinitelygeneratedo-module,andtherefore contained inthefinitelygeneratedo-moduleoat/d+ d =d(a1,..., very easy.Leta1,...,a,beabasisofLIKcontainedin0,discriminant chief interesttous,namely,ifLIKisaseparableextension,theproof ideal ino/p.Itremainstoshowthat0isnoetherian.Inthecaseof prime idealwhoseintersectionwitho/pwouldagainbeanonzero itself tobe,afield,becauseifitwerenot,thenwouldadmitnonzero the integraldomainO/q3isanextensionoffieldo/p,andthereforehas case o=Z(see(3.1)):pl3flisanonzeroprimeidealofo.Thus the nonzeroprimeidealsq3of0aremaximalisprovedsimilarlyasin Proof: Beingtheintegralclosureofo,0isintegrallyclosed.Thefactthat is againaDedekinddomain. L IKbeafiniteextensionofand0theintegralclosureoinL.Then (8.1) Proposition.LetobeaDedekinddomainwithfieldoffractionsK,let § 8.Extensionsof]iedekindDomains A primeidealp#0oftheringodecomposesina unique wayinto II. For aprimeidealpofoonealwayshas .-.°-w Then d fi =[O/Ti:o/p] 'L7 0), sothatfro=pawithp p p=g3no. = Tel. PO 0 by(2.8),and(2.9)tellsusthatis 3 Ip,andwecallTaprime divisor ofp. O. any . . s3,. e, . a, hencep+a=o. + Every 45 ... , wm over x, a contra- K] are connected by the K] are connected Chapter I. Algebraic Integers I. Algebraic Chapter w,,, E 0 be representatives : ... , is a basis of L I K. Assume the ... , aS is a system of generators of o and find a E a-' such that , for aij Ep. dim1e(O/q3i`) = eifi r i=1 . + amCJm = O. e,f1 =n. ..., co,,, and 211 a;jaj !=1 ..., am) a,,, E o not all zero such that O/pO = ®O/q3=' . O/pO = ®O/q3=' . + ocom and N = 0/M. Since 0 = M + p0, . + ocom and N = 0/M. Since 0 a1cv1 + a; ... , aa1w1 + ... + aa,,,com = 0 mod p ..., w,,, are therefore linearly independent over K. ..., w,,, are therefore linearly independent ... , w,,, of O/pO over K (we have seen in the proof of (8.1) ... , w,,, of O/pO over K (we have seen dimK(O/p0) = n In order to show that the cut are a basis of LIK, we consider the o- In order to show that the cut are a of N, then B be the adjoint matrix of A, whose entries are the minors of rank (s - 1) of a basis w 1, so certainly dimK (O/pO) < oo). that 0 is a finitely generated o-module, K, and hence also over o. Then col, ..., co,,, are linearly dependent over there are elements a1, Consider the ideal a = (al, aal, ..., aa,,, lie in o, but not a 0 a-1p, hence as ¢ p. Then the elements all belong to p. The congruence the (D1, thus gives us a linear dependence among diction. The l)1, modules M = co), + O, and hence also N, are finitely we have pN = N. As L I K is separable, generated o-modules (see p. 45). If c Let A be the matrix (a;j) - I, where I is the unit matrix of rank s, and let (8.2) Proposition. Let L I K be separable. Then we have the fundamental be separable. Then Let L I K (8.2) Proposition. identity based on the Chinese remainder theorem Proof: The proof is o/p, and it suffices vector spaces over the field x = O/pO and O/3 e` are to show that w1, In order to prove the first identity, let It is sufficient to show that col, 46 K is separable, L I the extension over p. If of j3; degree the inertia is called S,2 degree n = [L e;, ff and the the numbers law. following d ;0,becauseexpandingthedeterminant=det((a11)-I)wefind (see (2.3)).Hence of A.ThenonehasA(al,...,ar)`=0andBAdl,withddet(A), phenomena andastrikingsimplicity.Itisincompletein that afinitenumber element 0Ewithminimalpolynomial dim,, (T°/T"+1)=fiandtherefore has kernel4iandissurjectivebecause isomorphic toO/li,forifaET°-,T'+',thenthehomomorphism of K-vectorspaces.Thesuccessivequotientsq3'/T'+'inthischainare col, ...,w,,,isthereforeindeedabasisofLIK. (-1)s modpbecauseailEp.ItfollowsthatL=dLKcol+ and thereforedN=0,i.e.,dOCMowl+ § 8.ExtensionsofDedekindDomains ideals ofOwhichiscontainedino[B].Inotherwords the ringo[8]canbeconsidered.Thisconductorisdefined tobethebiggest of primeidealsareexcluded;onlythoserelatively totheconductorof decomposition ofpin0which,albeitnotcomplete,does show characteristic so thatL=K(B).Wemaythendeducearesultabout thenatureof (a) =aOsothatT°+ conductors ofo[O],andlet (8.3) Proposition.Letpbea prime idealofowhichisrelativelytothe Suppose nowthattheseparableextensionLIKisgivenbyaprimitive In ordertoprovethesecondidentity,letusconsiderdescendingchain a .N. finitely generatedo-module(seeproofof(8.1)),onehas s dim, ((9/3;`)=Edim,(q3;/q3°+1)eifi 0 =BA(al, /` ` O j5(X) =p1(X)e1...P'.(X)' ti, Lea ,/ ..., a,)'=(dal, p(X) EO[X], i / e;-1 v=o I` D...- °+1 93°+1. Since fi=[O/q3i a r-- j,-1/3j, D 3° is thegcdof13°+1and au, das)t (0) + OWm.Wehave , . : x],weobtain +Kmm. 47 0. so thatp01l is thepreimageof Then 53i,fori=1, be thepreimageof'3iwithrespecttocanonicalhomomorphism polynomial Ti(X),andwehave(0)=ni=113e' 0 correspondtotheprimeideals()5i),andtheyare principalideals pi (X)=pi(X)modpovertheresidueclassfieldo/p,withallEo[X] fi =[O/9,31:o/p]isthedegree ofthepolynomialpi(X).FurthermoreT" generated bythepi(0)modp0.Thedegree[O/Ti:a] is thedegreeof same situationholdsinthering0=O/pO.Thus prime idealsTiof In viewoftheisomorphismo`[X]/(p(X))=O/p0, f (X)r*(B),the that that thedegree[R/(pi):n]equalsofpolynomialpi(X),and principal ideals(pi)generatedbythep,(X)modp(X),fori=1, This showsthattheprimeidealsofringR=o[X]/(p(X))are the isomorphism o[X]/(P(X)), wehaveO'/pO' Its kernelistheidealgeneratedbypandp(X),inviewof0'=o[9]_ it followsthatp0fl0'=(p+a)(p0n0')cp0'. is surjective.Ithaskernelp0n0',whichequalsp0'.Since(p,ano)=1, The firstisomorphismfollowsfromtherelativeprimalityp0+=0.As Proof: Writing0'=o[9]andoo/p,wehaveacanonicalisomorphism monic. Then 48 degree ofpi(X),andonehas are thedifferentprimeidealsof0abovep.Theinertiadegreefi'43iis be thefactorizationofpolynomialp(X)=modintoirreducibles C 0',itfollowsthat0=p0+0',i.e.,thehomomorphism0'--0/pO Since p(X) The secondisomorphismisdeducedfromthesurjectivehomomorphism 1 o[x]l(n(x)) =®5[X]1(Pi(X))e' ale' andthereforep0=fi-1 X38'because)eifi=n. 13e' 0/pO =0'/p0'51X]l(P(X)). 3i=p0+pi(0)0, ..., r,variesovertheprime idealsof0abovep. (because ei=#(3"IvcN}), andPO:fi-1 o[x] --)o[X]/(p(X)) (X )el Flu (0) _(P)=f(Pi)e'- 0 -->O/pO. , theChinese 211 Tej... o[X)/(p(X)). l i_t i=t r i=1,...,r, }e- 211 'In r remainder theoremfinallygives . NowletJi= Chapter I.AlgebraicIntegers . p0+pi (0)0 ..., r, .fl i" 0 the discriminantsd(cot discriminant ofClo.Itisdefined tobetheideal1ofowhichisgeneratedby generated byB=6modTiandthereforeseparable.Hence pisunramified. of T(X)isnonzero.Theresidueclassfieldextensions o/pilo/pare has nomultipleroots.Butthisisthecasesincediscriminant d=modp to 1inthefactorizationofp(X)=modpo/p,so certainlyifp(X) In fact,by(8.3),theramificationindiceseiequal1assoon astheyareequal be thediscriminantofp(X)(see§2,p.11).Thenevery primeidealpofK which isrelativelyprimetodandtheconductor'of o[6]isunramified. p (X)Eo[X]beitsminimalpolynomial.Let Proof: Let6E0beaprimitiveelementforL,i.e.,L=K(6),andlet ideals ofKwhichareramifiedinL. (8.4) Proposition.IfLIKisseparable,thenthereareonlyfinitelymanyprime phenomenon. Infact,wehavethe itself iscalledunramifiedifallprimeidealspofKareinL. all ramified iffurthermorefi=1.Theprimeidealpiscalledunramified extension O/TiIo/pisseparable.Ifnot,itcalledramified,andtotally unramified overo(orK)ifei=1andtheresidueclassfield the moreidealpwillbetendtofactorintodifferentprimeideals. we nowunderstandthenameofinertiadegree:smallerthisdegreeis, prime idealofLoverp.Fromthefundamentalidentity called nonsplit,orindecomposed,ifr=1,i.e.,thereisonlyasingle one-has r=n=[L:K],sothatei=fi=Iforalli=1,...,r.pis if inthedecomposition § S.ExtensionsofDedekindDomains The precisedescriptionof theramifiedprimeidealsisgiven,by The casewhereaprimeidealpofKisramifiedinLanexceptional The primeidealJiinthedecompositionp=r[i_1ale' The primeidealpissaidtosplitcompletely(orbetotallysplit)inL, n+' ,3k areunramified,otherwiseitiscalledramified.TheextensionLIK fix. d=d(1,0,...,On-I)=F1 (6i-Bj)2E0 r"3 P.' CAD I?) , ..., o.)of ran Gr. p = i=1 r Tel eifi =n all basescol,... I ... mer i 0 r.. 49 i.e., whetherthecongruence p and(a,p)=1(exercise6).Letusassumethisforthesequel.Weare equation reducesimmediatelytothecasewherebisanoddprimenumber equation reciprocity law.Thelatterconcernstheproblemofintegersolutions number fieldQ(J)isintimatelyrelatedtoGauss'sfamousquadratic Example :Thelawofdecompositionprimenumberspinaquadratic the primeidealswhichramifyinL. in 0.WewillshowchapterIII,§2thattheprimedivisorsofctareexactly 50 and obtainthe a divisorof2(see§2,exercise4).Wemaythereforeapply proposition(8.3) for someaeZ.TheconductorofZ[/]inthering integers ofQ(,/)is relation withprimefactorization.(P)=1signifiesthat In thecaseofsquarefreea,Legendresymbol(P)bears thefollowing one alsohas squares hasindex2,i.e.,1F,/1Fp-Z/2Z.Since(p)=1 This isbecausethegroup1Fpcyclicoforderp-1andsubgroupF*2 have asolution.Thissymbolismultiplicative, defined symbol (a),which,foreveryrationalnumberarelativelyprimetop,is a solutioninthefield1Fpornot.ForthisoneintroducesLegendre the equationx2=d,foragivenelementmodpE1F,admits does ornothaveasolution.Inotherwords,wewanttoknowif then facingthequestionastowhetheraisquadraticresiduemodp, the simplestamongnontrivialdiophantineequations.Thetheoryofthis (8.5) Proposition.Forsquarefreea and(p,2a)=1,wehavetheequivalence G1. (p) (p .C. a = 1or-1,accordingasx2amodphasdoesnot x2-am(x-a)(x+a)mod p = 1 :-" `C7 x2+by=a, (p)-(p)(p)' (Pa x2 =amodp )--_a III p istotallysplitinQ(,). 211 F+" ..fi 't7 (a,bEZ), mod p. -y+ m Chapter I.AlgebraicIntegers cad :"'' a E]FP numbers £andp,thefollowingidentityholds: theory. like noneotherhasleftitsmarkonthedevelopmentofalgebraicnumber § 8.ExtensionsofDedekindDomains Proof :(pl) One alsohasthetwo"supplementarytheorems" (8.6) Theorem(Gauss'sReciprocityLaw).Fortwodistinctoddprime From this,aneasycomputationyields and since(1+i)n=-1imodp(p) Since (1+i)2=2i,wefind a primitive2-throotofunity.WeconsidertheGausssum deduce (F)_(-1)r. if and showthat For theLegendresymbol,onehasfollowingremarkablelaw,which In ordertodetermine(p),weworkintheringZ[i]ofgaussianintegers. In ordertoprovethefirstformula,weworkinring P-1 2 (P) ... is even,resp.odd.Since 2 (1 +i)P= - (-1)4modp, (2)(1+i)i21 (-1) ( P p nor l)=(-1) pz-1 p-1 eTL \P,)(D E i)((1+i)2) mod pimplies(pl)_(-1) aE(Z/IZ)* -1mod p. resp. = P'-' =p-1+iP+'P-1, (-1)Yz. 8 (p)=(-1) 2 P a (-) = (1+ 2 a 4 2'Y mod :)ice 24 2 z Ptl R IF modp, p, itfollowsthat since p02. 4 , where 2 we 51 is and L'IK,thenitisalsototally splitinthecompositeextension. Exercise 3.Ifaprimeidealpof KistotallysplitintwoseparableextensionsLI conductor Exercise 2.Foreveryintegral ideal2tof0,thereexistsa8E0suchthatthe Exercise 1.IfaandbareidealsofO,thenonehas a=ofl0and al b4=,aOlbO. apparent later(seeVII,§2and6). Gauss sumsdohaveahighertheoreticalsignificance,though, aswillbecome of decompositionprimesinthefield In §10,however,wewillrecognizethetruereasonwhy itholdsinthelaw Multiplying byranddividing±2yieldstheclaim. so that On theotherhandonehas This, togetherwiththecongruence(p)_£ from whichweindeedfindthat Now Ec(e)=0,asoneseesbymultiplyingthesumwithasymbol(E)_ For this,letaandbvaryoverthegroup(Z/Z)*,putc=ab-1deduce -1, andputting from theidentity(e)_(s)that 52 We haveprovedGauss'sreciprocitylawbyarathercontrived calculation. 1 (-')-r2 - (-1) rp - ' =(aE01aoCo[8])isprime to2tandsuchthatL=K(8). r-+ TP=t(i2) , implies (a ),aP a, b _ 361 t(Q) =t(-1)P2(-t)mod (-1 )r2 ( (C)b `-t gives e III = (-1)(-1)+ (Q) =r(-1)l2 >b i; r(a ( b(c-il ).aP P = +2... t of £-throotsunity.The P Chapter I.AlgebraicIntegers (1? ) mod pandtheidentity mod p. T r( mod p. P, e-t bc-b The ideals6q3,fororEG,are calledtheprimeidealsconjugateto43. so isvq3,foreacho-EG,because belongs to0,i.e.,Gactson 0.IfTisaprimeidealof0abovep,then ring 0ofintegralelementsL,theconjugatecra,for everyorEG,also duced intonumbertheorybyDAVIDHILBERT(1862-1943). Givenainthe The "ramificationtheory"thatarisesfromthisassumption hasbeenintro- group Galois extension.Theprimeidealsarethensubjecttothe action oftheGalois a particularlyinterestingandimportantturnonceweassume LKtobea 0 modp,where(a,p)=1,equals1+ Exercise U.Showthatthenumberofsolutionscongruenceax2+bxc= Exercise 10.Showthatthenumberofsolutionsx2-amodpequals1+(a). class ofpmod12. property of3beingaquadraticresidueornonresiduemodpdependsonlyonthe Exercise 9.StudytheLegendresymbol(p)asafunctionofp>3.Showthat Fibonacci number).Ifpisaprimenumber Exercise 8.Leta,,= the r,give'apermutationitofnumbers1,...,p-1.Showthatsgnn=(a). Exercise 7.Let(a,p)=1andavrmodp,v1,...,p-1,0 (b2 - mod p. P G"+ 1+,/5 4ac ) 2, 5,thenonehas b, resp.a-Imod8when81b. 2, 'ti s' = 1 2 III (an isthen-th ore P III P 53 Then thenormNLIK(x)=T1111 Proof: Let ideals q3of0lyingabovep,i.e.,theseprimeareallconjugateseach (9.1) Proposition.TheGaloisgroupGactstransitivelyonthesetofallprime 54 is calledthedecompositiongroupofq3overK.Thefixedfield H1EG axo hand, x0o-'3foranyaEG,henceaxq3orG.Consequently any aEG.BytheChineseremaindertheoremthereexistsx0suchthat other. index (G:Gp).Inparticular,onehas of thecosetsinG/Gq3,thena3variesoverdifferentprimeideals For ifq3isoneofthemandavariesoversystemrepresentatives of differentprimeidealsintowhichaidealpodecomposesin0. is calledthedecompositionfieldof93overK. (9.2) Definition.IfTisaprimeidealof0,thenthesubgroup In fact,forrc=G,onehastheequivalences conjugate subgroup The decompositiongroupofaprimeideal above p,eachoneoccurringpreciselyonce,i.e.,theirnumberequalsthe consider theequivalencerelation inGdefinedby in thecaseofanon-Galoisextension. ForsubgroupsUandVofagroupG, Remark: Thedecomposition groupregulatestheprimedecompositionalso The decompositiongroupencodesingroup-theoreticlanguagethenumber con x-0mod 3' r EGQ3 3 andT'betwoprimeidealsabovep.Assume Gq3 =G4ZcpK 3 flo=p,acontradiction. G p=1 or Zc3=IxELIax=x forallaEGG} bye a' =uav ^'t3 Gcp and x-1modag3 a-lra EGp3 G,p=aG93or- za 3=a'3 Zq3 =L II' r-.-, (a CG!a3=q3} EG ax belongs to43'flo=p.Ontheother III for uEU,V p isnonsplit. p istotallysplit, 1 r EaGg3a-I. Chapter I.AlgebraicIntegers a-irag3= 3 for allaCG. 3 conjugatetoq3isthe a0° . 3' a3 for § 9. Hilbert's Ramification Theory 55 The corresponding equivalence classes UaV={uovIuEU, VEV} are called the double cosets of G modd U, V. The set of these double cosets, which form a partition of G, is denoted U\G/V. Now let L I K be an arbitrary separable extension, and embed it into a Galois extension N I K with Galois group G. In G, consider the subgroup H = G(NIL). Let p be a prime ideal of K and Pp the set of prime ideals of L above p. If T is a prime ideal of N above p, then the rule H\G/GT -3 Pp,HuGT H a 93 n L, gives a well-defined bijection. The proof is left to the reader. In the Galois case, the inertia degrees fl,... ,fr and the ramification indices el, ..., er in the prime decomposition p = Tel...g3er 1 r of a prime ideal p of K are both independent of i, fi=...=fr=f,el=...=er=e. In fact, writing3 = TI, we find ',3t = o;3 for suitable oi E G, and the isomorphism a; : 0 - 0 induces an isomorphism O/ T -+ 0/a; T,a mod T Hal amod ai 13, so that ft = [ 0/04: o/p] = [ O/.p : 0/p], Furthermore, since o; (p0) = p0, we deduce from 3v I pO ai (3v) j ai (pQ) G = (aaT)" I p 0 the equality of the e;,i = 1,..., r.Thus the prime decomposition of p in 0 takes on the following simple form in the Galois case: (IIaq3)e, P =a where or varies over a system of representatives of G/GT. The decomposi- tion field Zp of 93 over K has the following significance for the decompo- sition of p and the invariants e and f. (9.3) Proposition. Let' 3Z = T n Zp be the prime ideal of Zp below q3. Then we have: (i) PZ is nonsplitin L, i.e.,' .3 is the only prime ideal of L above 543z. (ii)13 over Zq has ramification index e and inertia degree f. (iii) The ramification index and the inertia degree of 93Z over K both equal 1. 't7 K (p) Chapter I. Algebraic Integers I. Algebraic Chapter so that p = 3e"e'..., i.e., so that p = 3e"e'..., sae' in L, a mod q3 r-r as mod q3, n = efr, GT -) G(K(P)IK(p)). ZT] = e' f', i.e., we have e' f' = ef, and therefore ZT] = e' f', i.e., we 5 E G(K(`)IK(p)) = G(K(p)(6)IK(p)) . ... in Zq3 and 93z = ... in Zq3 and 93z v : O/93 -) O/T, The ramification index e and the inertia degree f admit a further index e and the inertia degree f The ramification Now let 9 be a primitive element for the maximal separable subextension 56 or T, are the above J3z ideals the prime ZT) = GT, G (L I (i) Since Proof: are and inertia degrees ramification indices in the Galois case, (ii) Since case reads identity in this the fundamental of the prime divisor, independent = [L : Zq] = ef. (G : Gds). We see therefore that #Gp where n :_ #G,)- = for or E G (L I Zp), and they are all equal to T. (L I Zp), and they for or E G of 13Z over K. ramification index of 93 over Zp, resp. Let e', resp. e", be the Then p = 3Z for the inertia obviously gets the analogous identity e = e'e". One also of Tz fundamental identity for the decomposition degrees f = ff". The in L then reads [L e' = e, f'=f,e"=f"=1. = O and a3 = q3, interpretation. Since a O interesting group-theoretic every or E Gtp induces an automorphism K(q3) = O/T and K(p) = o/p, we of the residue class field O/ 3. Putting obtain the IK (p) is normal and admits a surjective (9.4) Proposition. The extension K (q3) homomorphism K equals 1, i.e., ZT has the same Proof: The inertia degree of q3z over residue class field K(p) as K with respect to p. Therefore we may, and residue class field K(p) as K with respect = G. Let 9 E O be a representative do, assume that Z T = K, i.e., GT resp. g (X), the minimal polynomial of an element 9 E K (q3) and f (X), Then 9 = 9 mod q3 is a zero of the of 9 over K, resp. of 9 over K(p). g (X) divides f (X). Since L I K is polynomial j (X) = f (X) mod p, i.e., factors. Hence f (X) splits into linear normal, f (X) splits over O into linear of g (X). In other words, K factors over x (q3), and the same is true is a normal extension. of K(g3)IK(p) and to theextensionLITipthatG (K(T)IK(TT))=1,hence(TT)(T). inertia groupof93overTT,it followsfromanapplicationofproposition(9.4) follow fromK(P3T)=K(v3). AstheinertiagroupITof3overKisalso have toshowstatements(i)and (ii).Usingthefundamentalidentity,theyall Proof: Thefirsttwoclaims follow fromtheidentity#GT=ef.Soweonly In thiscaseonefindsfortheprimeidealTofTipbelow (ii) (i) If theresiduefieldextensionK(T)IK(p)isseparable,then onehas (9.6) Proposition.TheextensionTipIZrisnormal,andonehas Its propertiesareexpressedinthe and wehavetheexactsequence is calledtheinertiafieldof43overK. is calledtheinertiagroupof43overK.Thefixedfield (9.5) Definition.ThekernelITcGTofthehomomorphism mapped bythehomomorphisminquestiontoF.Thisprovessurjectivity. for someaEG(LIK).Since9-5'8modT,theautomorphismis of f(X)suchthat9'-QBmodq3.isaconjugate9,i.e.,6'=9 Then rrBisarootofk(X),andhenceI(X),i.e.,thereexistszero9' § 9.Hilbert'sRamificationTheory This inertiafieldTpappearsinthetoweroffields The ramificationindexof13overPTiseandtheinertiadegree is1. The ramificationindexoff3Tover43Zis1,andtheinertia degreeisf. #1T=[L:TT]=e, (GT:IT)=[Ti:ZT]=f. G(TgIZT) .=G(K(P)IK(p)) 1 -+IT-*GT-pG(K(T)IK(p))-f1. TT=Ix ELIax=xforallaElt} GT -*G(K(q3)IK(p)) In KCZTcTTCL, N , G(LITq) =IT. l: 57 0 aye `"30 extension maybeviewedasasubgroupofG=G(LIK). In thiscasetheGaloisgroupG(K(3)IK(p))-Gpofresidueclassfield residue fieldextensionK(q3)Ix(p)isseparablewefind on top,andtheinertiadegreesbottom.Inspecialcasewhere we haveindicatedtheramificationindicesofindividualfieldextensions 58 Hint: UsethefollowingresultofGALOIS(see[75],chap.II,§ 3):ifGisatransitive degree 1,thenitisalreadytotallysplit(theoremofF.K.SCHMIDT). Galois). IftheunramifiedprimeidealpinLhastwofactorsq3andq'3'of Exercise 3.LetLIKbeasolvableextensionofprimedegreep(notnecessarily sition groupGpiscyclicandcpageneratorofGp. where q=[K(q3):K(p)).ItiscalledtheFrobeniusautomorphism.Thedecompo- one andonlyautomorphismAPTEG(LIK)suchthat ideal whichisunramifiedoverK(i.e.,p=93f1inL),thenthere Exercise 2.IfLIKisaGaloisextensionofalgebraicnumberfields,andTprime Galois group,thenthereareatmostfinitelymanynonsplitprimeidealsofK. Exercise 1.IfLIKisaGaloisextensionofalgebraicnumberfieldswithnoncyclic in thenextchapter(seechap.II,§9). tions, belongsnaturallytothetheoryofvaluations,whichwewilldevelop G(NIL) andG,isthedecompositiongroupofaprimeideal43 overp. Hint: UsethedoublecosetdecompositionH\G/GT,where G=G(NIK),H split inLifandonlyitistotallyN. fields andNIKthenormalclosureofLK.Showthataprime idealpofKistotally Exercise 4.LetLIKbeafinite(notnecessarilyGalois)extension ofalgebraicnumber a EGwhichfixestwodistinctletters. solvable permutationgroupofprimedegreep,thenthereisnonontrivial permutation developed havereachedadegree ofabstractionwhichwewillnowbalance In thediagram Hilbert's ramificationtheory,withitsvariousrefinementsandgeneraliza- The conceptsandresults ofthetheoryasfar I p=1 cpTa=_aQmod g3forallaEO, § 10.CyclotomicFields K-3ZT-- TT--L III TT =L f <<< p isunramifiedinL. Chapter I.AlgebraicIntegers it hasnowbeen .0. principal ideal(A)intheringoofintegers does notonlyfurnishaworthwhileexamplebutinfactanessentialbuilding cyclotomic fieldQ(c),where to thetaskandmakethemmoreexplicitinexampleofn-th by somethingmoreconcrete.Wewillputtheinsightsofgeneraltheory Putting X=1,weobtaintheidentity nomial Proof: Theminimalpolynomialof Furthermore, thebasis1,c,..., and wehave (10.1) Lemma.LetnbeaprimepowerZ'andputA=1-.Thenthe field block forthefurthertheory. all numberfields,thisfieldoccupiesaspecial,centralplace.Sostudyingit § 10.CyclotomicFields But 1-g=Eg(l polynomial is4n(X)=rjdI (X-iii)and(see§2,p.11) fundamental identity(8.2)showsthat(A)isaprimeideal ofdegree1. the unite=Hgeg,henceto is integralaswell,i.e.,egaunit.Consequently£= e(1- 1 +- It willbeourfirstgoaltodetermineexplicitlytheringofintegers Let con ¢,(X) =(XeV- . +g-'Ifg'isanintegersuchthatgg'=1mod2",then = 1i...,dbetheconjugates of.Thenthecyclotomic For thisweneedthe d(1, j o 1-fig 1 - = ('X)d , 1 - f= 1-fig ), forthealgebraicintegereg=1-_ 7 where d=q?(f')_ gE(Z/nZ)* d-1 is aprimitiven-throotofunity.Among ±Ls, IZ 1) =XPv-1(P-1)+...+ over Qisthen-thcyclotomicpoly- 1 +g (1-i;g) S =Q°-1(v1-v1). i-1 i d'7 Since ... has thediscriminant is aprimeidealofdegree1, { (Mg/ Q]. Q} =o(t"),the Xe- + 1. 1-V with 59 Iterating thisprocedure,we find obtain Multiplying thisbyAandsubstitutingtheresulta,o= X20+ o =Z+),o,andafortiori Putting A=1-,lemma(10.1)tellsusthato/,lo = 7L/EZ,sothat power E".Sinced(1,,... Proof: Wefirstprovethepropositionincasewherenisaprime with Observing that Differentiating theequation 1, (10.2) Proposition.A7Z-basisoftheringointegers so that with theprimitiveE-throotofunity and substituting 60 The ringofintegersQ(c)isnowdetermined,forarbitraryn,asfollows. , ... d(1, .-. d-1, withd s=f'-1(vt-v-1). , ... , d-`) = for Xyields has norm±1weobtain = (p(n),inotherwords, (Xe°-` ( -1)01()= A20 + Eso c Ao +Z[f]=o. .d-1) =±Es,(2.9)givesus - 1)0n(X)= for all C o. -1)Ev' =fEE o. XE" fEveV t>1. But - 1 Chapter I.AlgebraicIntegers `(e-t)_P err is givenby _ +ES 1) =+E, we ,>~ 11 10. Cyclotoroic Fields 61 For t = s (p(2°) this implies, in view of £ o = JMt )o (see (10.1)), that o = tso+Z[x] = 'Iti, In the general case, let n = Ci' £v'. Then is a primitive £°' -th root of unity, and one has Q(O and ( ) (_ 1) fl Q Q. By what we have just seen, for each r, the elements 1, fit, ... , d`-1, where d; = rp(Q°'), form an integral basis of Since the discriminants d(1, ;,... ,d'-1) = ±Qi' are pairwise relatively prime, we conclude successively from (2.11) that the elements 1' .r', with ji = 0, ... , d1 - 1, form an integral basis of Q (c) IQ. But each one of these elements is a power of . Therefore every a n o may be written as a polynomial a = f with coefficients in Z. Since c has degree cp(n) over Q, the degree of the polynomial f may be reduced to rp(n) - 1. In this way one obtains a representation a= ao + alb+ ... + Thus 1, ,... ,gyp(")1 is indeed an integral basis. Knowing that is the ring of integers of the field we are now in a position to state explicitly the law of decomposition of prime numbers p into prime ideals of It is of the most beautiful simplicity. (10.3) Proposition. Let n = fp pvp be the prime factorization of n and, for every prime number p, let fp be the smallest positive integer such that pfp - 1 mod n/p"p. Then one has in Q the factorization P = (Pi ...pr)pVp), where p 1,... , pr aredistinct prime ideals, all of degree f. Proof: Since o = Z[], the conductor of Z[] equals 1, and we may apply proposition (8.3) to any prime number p. As a consequence, every p mod m, III because its is therefore the p fP , mod p. E 1F1,. Its degree is _,pm(X)°(P"P) mod p, Chapter I. Algebraic Integers I. Algebraic Chapter p,-(X))'P(P"P) mod p. &(X)=P1(X)...pr(X) 4.)'(P"') i,j (p1(X)... "P mod p, one has r7j = 1 mod p, for any prime "P mod p, one has is cyclic of order pfP-1. ]Fp fP 0. (X) = l (X - i 77j) 0,(X) = .On (X) mod p. factors into irreducible polynomials mod p. All we mod p. All irreducible polynomials factors into i On (X) = m (X) 10 'Pvp) 0n(X) = 0n(X) = fj(X - - 1 = (X - 1)P "P modulo p remains a primitive n-th root of unity. The smallest modulo p remains a primitive n-th Let us emphasize two special cases of the above law of decomposition: As the characteristic p of o/p does not divide n, the polynomials X' - 1 As the characteristic p of o/p does not unity it is the field lF extension field of lF p = Z / pZ containing cyclotomic polynomial field of decomposition of the reduced polynomial has no multiple roots, Being a divisor of Xn - 1 mod p, this and if Fp, then every pi (X) is the minimal is its factorization into irreducibles over polynomial of a primitive n-th root of unity therefore fp. This proves the proposition. where pl(X), ..., pr(X) are distinct irreducible polynomials over Z/p7L of polynomials over irreducible ..., pr(X) are distinct where pl(X), nj, varies over to see this, put n = p"Pm. As i, resp. degree fp. In order Since XP words, ideal p I p. In other This implies the congruence integer such that pfP Observing that fp is the smallest positive us to the case where p { n, and it is obvious that this congruence reduces hence cp(p"P) = Cp(l) = 1. in o/p. So Xn - 1 mod p has no and nXn-' have no common root passing to the quotient o -- o/p multiple roots. We therefore see that unity bijectively onto the group maps the group µn of n-th roots of the primitive n-th root of of n-th roots of unity of n/p. In particular, multiplicative group FP1 62 the minimal way as the same in exactly ideals into prime decomposes 0,(X) of polynomial is therefore that have to show i q j vary precisely of order m, resp. p"P, the products primitive roots of unity over o: roots of unity, i.e., one has the decomposition over the primitive n-th recently, Iwasawatheory. laws whicharethesubjectofatheoryhasbeendevelopedonly number theory.Atthesametimeoneencountershereplentyofastonishing context areinfactamongthemostdifficultproblemsposedbyalgebraic of thegroupunitsandidealclassgroup.Theproblemsarisinginthis decomposition ofprimesinthefield split in except inthecasewherep=2(4,n).Aprimenumber0istotally (10.4) Corollary.Aprimenumberpisramifiedin § 10.CyclotomicFields Proof: Thelittlecomputationin§8,p.51hasshownusthatQ*=r2with (10.5) Proposition.Let2andpbeoddprimenumbers,f*_(-1)Yf, following proper explanationofGauss'sreciprocitylaw(8.6).Thisisbasedonthe numbers tandp, of pflQ(2*).Thisimplies thatpistotallysplitinQ((,1F*). inertia degreeofpflZpoverQis1by(9.3),hencealso theinertiadegree is even.Since [Zp the indexofdecompositiongroupGp,orinother words,thedegree of into thesetofprimeidealsaboveP2.Thereforenumber ofprimeideals UPI =P2transformsthesetofallprimeidealslyingabove plbijectively r =G.aE(Z/tZ)* a primitiveC-throotofunity.Thenonehas: The completenessoftheseresultsconcerningtheintegralbasisand The lawofdecomposition(10.3)inthecyclotomicfieldprovides From thispropositionweobtain thereciprocitylawfortwooddprime : Q]ofthedecompositionfieldaprimeidealp p istotallysplitin above piseven.Nowassumeconverselythatthisthe case. Then say p=P1P2,thensomeautomorphismaof tot if andonly so thatQ(Q*)c is cyclic,itfollowsthatQ(e*)c_Z.The 2*) n0modp, °-° p1modn. p number ofprimeideals. p splitsinQ will notbematchedbyourstudy -2 t-1 P-I toy h,+ ..y 1.r If pistotallysplitin if andonly into aneven such that over p, 63 El number theoremisvalidmore generally forprimenumbersp=amodn,provided Let p10,,(xnP)forsuitablex. Deduce acontradictionfromthis.(Dirichlet'sprime n -thcyclotomicpolynomial0,,. Notallnumbers Hint: Assumethereareonlyfinitely many.LetPbetheirproductandconsiderthe are infinitelymanyprimenumbersp-1modn. Exercise 1.(Dirichlet'sPrimeNumberTheorem).Foreverynatural numbernthere reciprocity law,Artin'soneofthehighpoints inthehistory (a, n)=1(seeVII,(5.14)and §13)). of classfieldtheory,whichwewilldevelopinchapters IV-VI. of numbertheory,andcompellingcharm.Thislawis themaintheorem of thistheoryhasledtoatotallycomprehensivegeneralization ofGauss's successfully intheexampleofcyclotomicfields.Thefurther development developed thebasicsofthistheoryinprecedingsections andtestedit attempts tosolveitledhisseminaldiscoveryofideal theory.Wehave Since thisproblemrequiredworkingwiththen-thcyclotomicfield,KuM,MER's x" -amodp,withn>2,dominatednumbertheoryforlongtime. quest forsimilarlawsconcerninghigherpowerresidues,i.e.,thecongruences number theory.ItwasdiscoveredbyEULER,butfirstprovenGauss.The have (p) it belongstoFQ2,thelastcongruenceisequivalent( of By (10.3),thisnumberisr= an elementinthecyclicgroup]FEhasorderdividing such thatpf-1modf,i.e.,risevenifandonlyfadivisor in thefield By (8.5)and(10.5),weknowthat(P)=1ifonlypdecomposes then gives In fact,thecompletelyelementaryresult(p1)_(-1) as follows.Itsufficestoshowthat 64 Historically, Gauss'sreciprocitylawmarkedthebeginningofalgebraic 2 But thisistantamounttothecondition (p)=(p)=(p)L (i) asclaimed. III of £-throotsunityintoanevennumberprimeideals. 0.Y E f 1 , 2 where fisthesmallestpositiveinteger .(p)=(p)(-1) Chapter I.AlgebraicIntegers p(1-1)/2 for xE7L,canequal1. III pz ) =1.Sowedo 2 i 1 modf.Since f- (see §8,p.51) if andonly `. cue 65 IQ, in the case where n is odd. IQ, in the case ) gAP t-. .ti and 1_'2. 1 n A qt ), Q(/), Q(/) are the quadratic subfields of ), Q(/), Q(/) are the coy § 11. Localization K={b IaEA, bEA\(0}}. q H qS-1 AS-t={s a EKIaEA, sES}. a primitive n-th root of unity. a primitive for, n = 2Q, q ? 3. To "localize" means to form quotients, the most familiar case being the to form quotients, the most familiar To "localize" means § 11. Localization § 11. Exercise 2. For every finite abelian group A there exists a Galois extension L IQ extension a Galois there exists group A abelian every finite 2. For Exercise A. group G(LIQ) - with Galois exercise 1. Hint: Use cyclotomic contained in some field Q(J) is Every quadratic number Exercise 3. field subfields of Q Describe the quadratic Exercise 4. Q( Exercise 5. Show that A N {0} any nonempty S C_ A N (0) More generally, choosing instead of one again obtains a ring structure on which is closed under multiplication, passage from an integral domain A to its field of fractions passage from an integral the set such a multiplicative subset is the The most important special case of p of A. In this case one writes AP complement S = A N p of a prime ideal AP the localization of A at p. When instead of AS-1, and one calls the ring single prime ideal p of A at a time it is dealing with problems that involve a A.. This procedure forgets often expedient to replace A by the localization p, and brings out more clearly all the everything that has nothing to do with the mapping properties concerning p. For instance, the prime ideals q c p of A and the gives a 1-1-correspondence between any multiplicative set S, one has the prime ideals of Ap. More generally for (11.1) Proposition. The mappings are mutually inverse 1-1-correspondences between the prime ideals q C A-, S of A and the prime ideals £2 of AS-1. Forifa E£,then Furthermore onehas since to 0.Furthermoreonehas s" i.e., ss,=9, is aprimeidealofAS-1.Indeed,inobviousnotation,therelation Proof: IfqCA 66 from XsurviveinA(X). being eliminatedwhenpassingfromAtoA(X).Forinstance,ifXisfinite the primeidealsofAwhicharecontainedinUp. instead ofAS-1.TheprimeidealsA(X)correspondby(11.1)1-1to prime idealsofA.Inthiscaseonewrites which provestheproposition. mappings qi-+qS-1and to containansES,thenwewouldhave1=£becauseAS-1. is obviouslyaprimeidealofA,andonehasqCA maximal idealofA,thenone has identifying Ap/mpwiththe field offractionsA/p.Inparticular,ifpisa a uniquemaximalideal,namelymp=pAp.Thereiscanonical embedding (11.2) Corollary.IfpisaprimeidealofA,thenApalocal ring,i.e.,Aphas of Aatp.Herewehavethe the localization or omitsonlyfinitelymanyprimeidealsofA,then theprimeideals 5;. Usually SwillbethecomplementofaunionU,ExpoversetX Conversely, let12beanarbitraryprimeidealofAS-1.Thenq=flA In thecasethatXconsistsofonlyoneprimeidealp, ringA(X)is q, andhenceaora'belongtowhichshowsthat a = aE12flAimpliesqasq,whencer=becauses A(X)={f , impliesthats"aa'=qss'Eq.Thereforeaa'q,because . SisaprimeidealofA,then Ap=t 8 g A/p" =Ap/mp I f,gEA,g#OmodpforpEX} 12=qs-1={q H £flAarethereforeinversesofeachother, A/p I f,gEA,g0-0modp} q=onA, £. =qS-1. I Ap/mp, qEq, forn >1. SES} Chapter I.AlgebraicIntegers =X p,alltheothersare aim . S.Infact,ifqwere a or EqS-1. The s belong q. E field offractionsA/p.Letpbemaximalandn>1.ForeverysEA-, For n=1,fisinjectivebecausepmpflA.HenceAp/mpApbecomesthe homomorphism in p,theidealmp=pApisuniquemaximalideal.Letusconsider Proof: SincetheidealsofAVcorrespond1-1toAcontained § 11.Localization i.e., a/smodmpliesintheimageoff. exists ana'EAsuchthata-a'smodp'.Thereforea/sp"AV, s 0p.Thenas=bEp",sothatd A=p°-t+sA=p=pA=p(pn-l+sA)Cp"+sA=p"+sA=A. is clearfromthemaximalityofp,andforn>1itfollowsbyinduction: one hasp"+sA=A,i.e.,smodisaunitinA/p'.Forn1this element n.Sinceeverynotcontainedinpis aunit,itfollows (11.3) Definition.Adiscretevaluationringisaprincipalidealdomainowith The simplestlocalrings,exceptforfields,arediscretevaluationrings. ideal, ithastobethewholering.Sowehave Indeed, sincetheprincipalideal(a)isnotcontainedinanyothermaximal The valuationisafunction obviously characterizedbythe equation The exponentniscalledthevaluationofa.Itdenoted v(a),anditis may beuniquelywrittenas and n>0.Moregenerally,everyelementa0ofthefield offractionsK nonzero elementofomaythereforebewrittenass7r', forsomeE0*, that, uptoassociatedelements,rristheonlyprimeelement ofo.Every a uniquemaximalidealp#0. In alocalringwithmaximalidealm,everyelement0misunit. Surjectivity off:leta/sEAV,aA,s0p.Thenbytheabove,there Injectivity off:letaEAbesuchthatmp,i.e.,=b/swithbp", The maximalidealisoftheformp=(,r)iro,for someprime C/] ac' a=sir", v:K*-±7L. A*=A.m. (a) = EEO*, nEZ. amod p"Hamod pt(a) 0 inA/p",andhencea=A/p". Cal C3. '-r mp. 67 For if6Enpop,witha,b E 0,then valuation ring. m" =(Tr").Thusopisaprincipalidealdomain,and henceadiscrete choosing anyitEmNm2,onenecessarilyfinds(rr)= m, andfurthermore maximal idealm=popistheonlynonzeroprime ofop.Therefore, Proof: IfoisaDedekinddomain,thensoarethelocalizations op.The discrete valuationrings. domain ifandonlyif,forallprimeidealsp (11.5) Proposition.Letobeanoetherianintegraldomain.isDedekind therefore xEoS-1.ThisshowsthatoS-1isaDedekinddomain. of s=sl with coefficientsaEoS-1,thenmultiplyingitthen-thpower equation this holdsino.Finally,oS-1isintegrallyclosed,forifxEKsatisfiesthe It followsfrom(11.1)thateveryprimeidealof ifa Proof: LetQIbeanidealof plicative subset,thenoS-tisalsoaDedekinddomain. This isaconsequenceofthe This innocuouslookingfunctiongivesrisetoatheorywhichwilloccupyall Extending ittoKbytheconventionv(0)=oo,asimplecalculationshows a =seaS-1.Asisfinitelygenerated,so91,i.e.,oS_1noetherian. (11.4) Proposition.MoisaDedekinddomain,andSc'.(0)multi- of thenextchapter. that itsatisfiestheconditions 68 Letting pvaryoverallprimeideals (J1 The discretevaluationringsariseaslocalizationsofDedekinddomains. a EQandsES,thenonehas a.+ ... s v(ab) =v(a)+v(b), shows thatsxisintegralovero,whenceEoand xn+at xn-1+...+.at.0 4-1 a{xE 01xaEbo} S1 oS-t and o=no,. v(a +b)>mintv(a),v(b)}. p coo 'ti a =%no.Then2( 0.of o,wefindinanycasethat sn s E2tflo=a,sothat 0, thelocalizationsorare Chapter I.AlgebraicIntegers oS-' ismaximal,because "C3 S-1, because where x=p"albwithintegers a,brelativelyprimetop. x EQ*isgivenby is calledthep-adicvaluation ofQ.Thevaluationvp(x)anelement valuation associatedto7G(p), and thegroupofunitsconsistsallfractionsalbsatisfying ptab.The The maximalidealpZ(p)consistsofallfractionsalbsatisfying pIa,tb, called exponentialvaluations. Hence indeedvq(x)=vq.Inviewofthisrelation,thevaluationsvparealso In fact,forafixedprimeidealq;0ofo,thefirstequationaboveimplies relation totheprimeidealfactorization.IfXEK*and maximal becausethisissoinop.ThereforeoaDedekinddomain. hencea=1-aEbo,i.e., 6Eo. S Ea-\p.Asisnotcontainedinanymaximalideal,itfollowsthat=o, is anidealwhichcannotbecontainedinanyprimeofo.Infact,for ideal (p)=pZisgivenby (because poq=for0q)that is theprimefactorizationofprincipalideal(x),then,foreachp,onehas of thefieldfractions.Thesignificancethesevaluationsliesintheir valuation ringopandthecorresponding closed. Finally,from(11.1)itfollowsthateveryprimeidealp0ofois domains, theyareintegrallyclosed(see§2),soo=npopisalso any p,wemaywrite § 11.Localization The readershouldcheckthatthelocalizationofring Zattheprime For aDedekinddomaino,wehaveforeachprimeidealp;0thediscrete Suppose nowthattheoparediscretevaluationrings.Beingprincipalideal +-+ XOq =(11P"')OqgVgoqmq9 Z(p)=}b Ia,bEZ,pIb} E`7 cm, b = VP:Q -p- s p vp:K*) Z with cEo,s (x) =rl Vp(x) =if, VP =up(X). p U(oo}, .fl pVP p, sothatsa=bc,hence ado 69 following We denotebyCl(o),resp.CI(o(X)),theidealclassgroupsof0, same localizations pX =po(X),forpcX,anditiseasilycheckedthatoo(X)havethe prime idealsofO.By(11.1),the where Xisasetofprimeideals0owhichcontainsalmostall the ring 70 where vp:K*->ZistheexponentialvaluationofKassociated toop.Let is inducedbymapping proof of(11.5)).Thisshowstheexactnessato(X)*.Thearrow because op=o(X)px,henceaElp0*(seetheargumentin a r=o(X)*belongstothekernel,thenEopforp inclusion o(X)*-+K*,followedbytheprojectionsK* Proof: Thefirstarrowisinclusionandthesecondoneinducedby and onehasK*/o*=Z. (11.6) Proposition.Thereisacanonicalexactsequence O(X). They,aswellthegroupsofunitso*ando(X)*,arerelatedby in themiddle.Thearrow that aelp,o*=o(X)*and a=upmodo*.Thisshowsexactness vp (a)=0forpeX,and vp(ap)=(a)forpVX.Itfollows for someaEK*.Becauseofuniqueprimefactorization, thismeansthat ®pOX upmodopbeanelementinthekernel,i.e., `Y- To endthissection,wenowwanttocompareaDedekinddomaino o(X)={g If,gEO,g#0modpforpEX}, poX ® apmodopH11 pox fl poX pVP( P) ® K*/Op-aCl(o) Cl(o) -*CI(O(X)) pox ED K*10* (a9 Op =O(X)pX. = (a)_ 11` p¢X p C40) pvP(cf) pVp(c Chapter I.AlgebraicIntegers 0 ofo(X)aregivenas ) X, andalsoforpEX MOM) K*/o*. If 1, § 11. Localization 71 comes from mapping a i-3 ao(X). The classes of prime ideals p E X are mapped onto the classes of prime ideals of o(X). Since C1(o(X)) is generated by these classes, the arrow is surjective. For p 0 X we have po(X) = (1), and this means that the kernel consists of the classes of the ideals Ilex pp". This, however, is visibly the image of the preceding arrow. Therefore the whole sequence is exact. Finally, the valuation vp : K*, Z produces the isomorphism K*/op Z. For the ring of integers OK of an algebraic number field K, the proposition yields the following results. Let S denote a finite set of prime ideals of OK (not any more a multiplicative subset), and let X be the set of all prime ideals that do not belong to S. We put OK = OK (X) The units of this ring are called the S-units, and the group C1K = C1(4) the S-class group of K. (11.7) Corollary. For the group KS = (c4)* of S-units of K there is an isomorphism KS -a(K)x71,#S+r+s-1 where r ands are defined as in § 5, p. 30. Proof: The torsion subgroup of KS is the group µ(K) of roots of unity in K. Since Cl(o) is finite, we obtain the following identities from the exact sequence (11.6) and from (7.4): rank(KS) = rank(o*) + rank( ®Z) = #S -1- r + s- 1. PES This proves the corollary. (11.8) Corollary. The S-class group C1K = Cl(oK) is finite. Exercise 1. Let A be an arbitrary ring, not necessarily an integral domain, let M be an A-module and S a multiplicatively closed subset of A such that 0 0 S. In M x S consider the equivalence relation (m, s) (m', s') 4== 3 s" E S such that s" (s'm - sm') = 0. Show that the set Ms of equivalence classes (m, s) forms an A -module, and that M Ms, a i (a, 1), is a homomorphism. In particular, As is a ring. It is called the localization of A with respect to S. ring oKiscalledthemaximal orderofK. of KisasubringoOKwhich containsanintegralbasisoflengthn.The (12.1) Definition.LetKIQbean algebraicnumberheldofdegreen.Anorder `ti are notnecessarilyintegrallyclosed.Theseringsthe so-calledorders. theory ofringsalgebraicintegerswhich,likethering pushed todeviseatheoryofgreatergeneralitywhich comprises alsothe to importanttheoreticalaswellpracticalcircumstances, however,oneis interest becauseofitsexcellentpropertybeingaDedekind domain.Due then As=A. one hastheimplication: be anA-moduleandNCMasubmodulesuchthatM/Nisfinitelygenerated.Then Exercise 7(Nakayama'sLemma).LetAbealocalringwithmaximalidealm,letM Exercise 6.LetAbeanintegraldomain.IfthelocalizationAsisoverA, closed subsetsuchthatf(S)cB*.TheninducesahomomorphismAS->B. Exercise 5.Letf:A-+BbeahomomorphismofringsandSmultiplicatively in As.ThenonehasAST=(A5)T.. Exercise 4.LetSandTbetwomultiplicativesubsetsofA,T*theimage (iii) fm:Mm-->N.isinjective(surjective)foreverymaximalidealm. (ii) (i) fisinjective(surjective). conditions areequivalent: Exercise 3.Letf:M->NbeahomomorphismofA-modules.Thenthefollowing in thisway,thenAs/psisthelocalizationofA/p.withrespecttoimageS. to theprimeidealsofAwhicharedisjointfromS.IfpCandpsgAscorrespond Exercise 2.Showthat,intheabovesituation,primeidealsofAscorrespond1-1 72 The ringOKofintegersanalgebraicnumberfield Kisourchief fp :Mp-4Npisinjective(surjective)foreveryprimeidealp. 0 =Z+7G1'5-cQ(fs), M=N+mM =M=N. 112 X211 § 12.Orders Chapter I.AlgebraicIntegers stronger versionoftheChinese remaindertheorem. domain andKitsfieldoffractions. Wesetoutbyprovingthefollowing is amaximalideal. rank n.Thereforeo/pisafiniteintegraldomain,hence afield,andthusp ideal andaEpf1Z, generated o-module.Thisshowsthatoisnoetherian.If p every idealaisalsofinitelygenerated7L-module,and afortiorifinitely Proof: SinceoisafinitelygeneratedZ-moduleofrank n=[K The term"one-dimensional"referstothegeneraldefinitionofKrull domain. (12.2) Proposition.AnorderoofKisaone-dimensional noetherianintegral ideals poCpiC-..CPd. dimension ofaringasbeingthemaximallengthdchainprime integral domainsinwhicheveryprimeidealp one-dimensional noetherianintegraldomains.Thesearethe common generalizationofbothtypesringsletusconsiderfornowall of OKwhichareintegraloverZ,yetnolongerintegrallyclosed.Asa yet nolongerintegraloverZ.Nowwestudyorders.Theyaresubrings domain OK.TheyareextensionringsofOKwhichintegrallyclosed, what thismeansinthenextsection. domains withtheir"regular"localizationsopareconsidered.Wewillexplain that theyadmit"singularities",whichareexcludedaslongonlyDedekind is anorder.Thetheoreticalsignificanceoforders,however,liesinthefact M =7Lal+- practical applications.Forinstance,ifal, has lengthn.Ordersariseoftenasringsofmultipliers,andsuchhavetheir which, asQo=K,hastobeatthesametimeabasisofKIQ,andtherefore submodule ofthefree7L-moduleoK,odoescourseadmita7L-basis where al,...,a,areintegerssuchthatK=Q(al,a,).Beinga § 12.Orders In whatfollows,wealwayslet obeaone-dimensionalnoetherianintegral In theprecedingsectionwestudiedlocalizationsofaDedekind In concreteterms,ordersareobtainedasringsoftheform F"+ - - +7La,,then om 0, thenascpCo,i.e.,andohavethesame N(1 [iii 0 =7L1al,...,Cr,l pay ... ,a, is anybasisofKIQand 0 isamaximalideal. 0 isaprime : Q], 73 0 Indeed, foranyaEnpa,,theidealb=(xoIxaca)doesnotbelong Proof: Letap=otlaop.ForalmostallponehasAaandtherefore (12.3) Proposition.Ifa0isanidealofo,then 74 because itisthebiggestidealsuchthatas1c_o.Theinvertible idealsofo fractional ideal These formanabeliangroup,fortrivialreasons.Theinverse ofaisstillthe which thereexistsafractionalidealbsuchthat attention totheinvertibleideals,i.e.,thosefractionalidealsaofofor a group-unlessohappenstobeDedekind.Thewayoutisrestrict generated nonzeroo-submodulesofthefieldfractionsK,nolongerform ideal ofo/ap. and wehaveo/ap=op/aop,because now givestheisomorphism maximal ideal,whenceap+aq=o.TheChineseremaindertheorem(3.6) ... distinct primeidealspandqofo,theidealap+aqcannotbecontainedinany if pDa,thenistheonlyprimeidealcontainingap.Therefore,giventwo consequently, b=o,i.e.,a1 to anyofthemaximalidealsp(infact,onehasspaEaforsp aop =op,hencea,o.Furthermore,onehasanpapnpaap. is afractionalprincipalideal of op. every primeidealp (12.4) Proposition.Afractional idealaofoisinvertibleifandonlyif,for may becharacterizedasthosefractionalidealswhichare "locally"principal: For theringo,fractionalidealsofinotherwords,finitely cat I;' o/a =®op/aopop/aop. 0, s'= IXEKIxaco}, o/a =®o/ap, p ap =aop 211 ab=o. a Ea,asclaimed.(11.1)impliesthat, pJa S3. p2a p moddpistheonlymaximal Chapter I.AlgebraicIntegers p), 0 § 12. Orders 75 Ei=1 Proof: Let a be an invertible ideal and a b = o. Then 1 = ai b1 with ai E a, bi E b, and not all aib1 E op can lie in the maximal ideal pop. Suppose albs is a unit in op. Then ap = aloe because, for x E ap, xb1 E apb = op, hence x = xb1(biaj)-1a1 E aloe. Conversely, assume ap =aop is a principal ideal apop, ap E K*, for every p. Then we may and do assume that ap E a. We claim that the fractional ideal a-' = (x E K I xa C o} is an inverse for a. If this were not the case, then we would have a maximal ideal p such that as 1 C P C o. Let a1...... a be generators of a. As ai E ap op, we may write ai = aps, with bi E o, Si E o. p. Then siai E apo. Putting s = Si s,,, we have sai E apo for i = 1, ..., n, hence sap 1 a C o and therefore sap1 E a-1. Consequently, s = sap1ap E a la C p, a contradiction. We denote the group of invertible ideals of o by J(o). It contains the group P(o) of fractional principal ideals ao, a E K*. (12.5) Definition. The quotient group Pic(o) = J(o)/P(o) is called the Picard group of the ring o. In the case where o is a Dedekind domain, the Picard group is of course nothing but the ideal class group C1K. In general, we have the following description for J(o) and Pic(o). (12.6) Proposition. The correspondence a H (ap) _ (aop) yields an isomorphism J(o) = ®P(op). p Identifying the subgroup P (o) with its image in the direct sum one gets Pic(o) = ( ®P (op)) /P (o) . p Proof: For every a E J(o), ap = aop is a principal ideal by (12.4), and we have ap = op for almost all p because a lies in only finitely many maximal ideals p. We therefore obtain a homomorphism J(o) -) ® P(ap),a (ap) . p 76 Chapter I. Algebraic Integers- It is injective, for if ap = op for all p, then a C np op = o (see the proof of (11.5)), and one has to have a = o because otherwise there would exist a maximal ideal p such that a C p C O, i.e., ap c pop op. In order to prove surjectivity, let (apop) E (Dr P (op) be given. Then the o-submodule a = n apop p of K is a fractional ideal. Indeed, since apop = op for almost all p, there is some c E 0 such that cap E op for all p, i.e., ca c- n, op = o. We have to show that one has aop = apop for every p. The inclusion C is trivial. In order to show that apop C aop, let us choose c E o, c 0 0, such that cap'aq E o for the finitely many q which satisfy ap tag oq. By the Chinese remainder theorem (12.3), we may find a E o such that a-cmodp and aEcaplagoq forq¢p. Then s = ac-1 is a unit in op and apr E ng agoq = a, hence apop = (apE)Op C aOp. Passing from the ring a to its normalization 5, i.e., to the integral closure of o in K, one obtains a Dedekind domain. This is not all that easy to prove, however, because 6 is in general not a finitely generated o-module. But at any rate we have the (12.7) Lemma. Let o be a one-dimensional noetherian integral domain and o its normalization. Then, for each ideal a 0 0 ofo, the quotient o/ a& is a finitely generated o-module. Proof: Let a c a, a 0. Then 6/a6 is a quotient of 6/a6. It thus suffices to show that 61a6 is a finitely generated o-module. With this end, consider in o the descending chain of ideals containing ao an, = (a'(5 rlo,ao). This chain becomes stationary. In fact, the prime ideals of the ring o/ao are not only maximal but also minimal in the sense that o/ao is a zero- dimensional noetherian ring. In such a ring every descending chain of ideals becomes stationary (see § 3, exercise 7). If the chain a,,, = a,,, mod ao is stationary at n, then so is the chain a,,,. We show that, for this n, we have 6Ca-no+a6. extensions ofthefieldfractions ofaDedekinddomain. only inthecaseofaseparable extensionLIK,isvalidforgeneralfinite Remark :Theaboveproofis takenfromKAPz.ANsKY'sbook[82](seealso 2t/a0, andalsothe0-module2t. [101]). Itshowsatthesametime thatproposition(8.1),whichwehadproved finitely generatedo-module.Sincecisnoetherian,so is theo-submodule ideal of0andae2tflo,;0;thenbytheabove lemma0/aOisa one-dimensional andarethusreducedtothecaseL= K.Solet2tbean particular isnoetheriansincecnoetherian.Weargue asbeforethat0ois Then theringOo=o[w1, noetherian. Letw1, ideal ismaximal,arededucedasin(3.1).Itremains toshowthat0is Proof: Thefactsthat0isintegrallyclosedandevery nonzero,prime integral domainwithfieldoffractionsK.LetLIKbeafiniteextensionand0 (12.8) Proposition(KnuLL-AKizutu).Letobeaone-dimensionalnoetherian q.e.d. generated bya-"modao.Itisthereforeitselfafinitelyo-module, This contradictstheminimalityofh.Sowedohavenca-'o+ao. Let hbethesmallestpositiveintegersuchthat,8Ea-ho+ao.Itthen hence (1-xa)ahEcc,andtherefore Let ,B= the integralclosureofoinL.Then0isaDedekinddomain. u =alt+lu'+au',E0,it'&.Substitutingthisinto(*)gives we haveu=ah(p-au)Eahoflocaltah+1becauseh>n,hence (*) suffices toshowthath - xa)+Pxa= i.e., wefindsomexEosuchthatah=xah+1modcc, ... ,m = ah_1+a(il+W)Ea1-hn+a0. =alt +au CL, u' Ll be abasisofLIKwhichiscontainedin0. , cw,t] isafinitelygeneratedo-moduleandin ah with ueo,E6, (1 xa)a h E a_ho+ao. III 77 This conditionwillbeassumedforallthatfollows.Itavoidspathological the followinghypothesis: evident anddoesnotrequirethelengthyproofof(12.8)providedwemake main owithitsnormalizationn.Thefactthat6isaDedekinddomain 78 Proof: Ifpvariesovertheprimeidealsofa,weknowfrom(12.6)that In thesum,pvariesoverprimeideals#0ofoandopdenotesintegral (12.9) Proposition.Onehasthecanonicalexactsequence each otherbythefollowing in analgebraicnumberfield. situations andissatisfiedinallinterestingcases,particularfortheorders o-module. (*) oisanintegraldomainwhosenormalizationafinitelygenerated of (12.6),itfollowsthat ideals andisthereforeaprincipalidealdomain(see§3, exercise 4).Inview has tolieabovepop,thelocalizationoponlyafinite numberofprime for theintegralclosureo,ofop.Sinceeverynonzero prime idealofCep i.e., thereareonlyfinitelymanyprimeidealsofoabove p.Thesameholds product If pisaprimeidealofo,thenposplitsintheDedekind domainointoa closure ofopinK. and therefore Next wewanttocomparetheone-dimensionalnoetherianintegraldo- The groupsofunitsandthePicardoarecomparedwith -fl won sue., J(o) = 6*- "O9' P(op) =J(np)®POP) J(n) -®P(bp). `^J p p pap ® P(op)=®P(op). P C.' -p p p Chapter I.AlgebraicIntegers 1. fractions K,weobtainthecommutativeexactdiagram Observing thatP(R)-K*/R*foranyintegraldomainRwithfieldof This thenyieldstheexactsequence whereas lemma 3.3).Inourparticularcase,a,fi,andthereforealsoy,aresurjective, relating thekernelsandcokernelsofa,,y(see[23],chap.III,§3, lemma: thediagramgivesinacanonicalwayanexactsequence § 12.Orders If thisisthecase,then (12.10) Proposition.Forany primeidealp0ofoonehas Since 5isafinitelygeneratedo-module,wehavef the biggestidealofowhichiscontainedino,other words, ideals ofo,namelythedivisorsconductoro.This isbydefinition 6p/op in(12.9)aretrivial.Thereonlyfinitelymany non-regularprime thus adiscretevaluationring.Fortheregularprimeideals, thesummands For suchadiagramonehasincompletegeneralitythewell-knownsnake A primeidealp 1 -aker(a)--ker(fl)ker(y) 1 1 ->K*/o* 1 o* -6*-k®O*p/o*p ker(a) =n*/o* K*/o* --®K*/o*pPic(o)--*1 0 ofoiscalledregularifopintegrallyclosed,and s} °"t° f= {aEOlaoco}. p coker(a) -)coker(p)--coker(y)--*1 pn isaprimeidealof6andop =Cep. p f ®K*/o* -Pic(o)1. p p .A. and p isregular. P ker(p) _®n*/o* Pic(o) -fPic((5)-+1. p t 0. 79 (2) and it followsthatfap=fop.TheChineseremaindertheorem yields of owithrespecttothemultiplicativesubset prime idealsabovepoftheringo.Atsametime,op isthelocalization The integralclosurenpofoppossessesonlythefinitely manyprimeideals fortiori overop,wehaveocop.Letx1, that lieabovepop.Theygivethelocalizationsop,where pvariesoverthe (1) Proof :WeapplytheChineseremaindertheorem(12.3)repeatedly. We,have (12.11) Proposition.)poy/o=(n/f)"/(o/f)*. in (12.9). s Ef of theo-modulen.Wemaywritexi= ideal domain,itisintegrallyclosed,andsinceointegralovero,hencea m =pop(po)oppeepme,i.e.,e1andthusppn. hence p=q.Consequently,pope,withe>1,andfurthermore Therefore op=Op.Thus,by(11.5),isavaluationring,i.e.,pregular. aEo,sEoNp,then toEoandisEoNp,hence because pismaximal.Trivially,opCop,andifconverselya Proof: Assumepff,i.e., above p.Forifqisanotherone,thenop=Cpq,andtherefore o C 80 We nowobtainthefollowingsimpledescriptionforsum®pnp/o* Conversely, assumeopisadiscretevaluationring.Beingprincipal One hasfurthermorethatp=pa.Infact,istheonlyprimeidealofo m fla,pisaprimeidealofosuchthatco,hence N p.Itfollowsthatp i -I- o Cop.Ifm=popisthemaximalidealofopthen,putting p=5npopC5ngoq=q, np/fop =®o00 ®® op/fbp=®np/fnp- p p2P o/f =®op/fop f. f, andlettEf p p?p 21I al , witha;E0,SioNp. ... ,x,beasystemofgenerators p C43 p. Sincefisanidealofo, Chapter 1.AlgebraicIntegers p. Thentoco,hence S a = -Eo. EtQ p, is Setting n o for p This givestheclaim. sequence Proof: By(12.9)and(12.11), andsincePiC(OK)=C1K,wehavetheexact maximal order,andftheconductorofo. (12.12) Theorem.LetobeanorderinalgebraicnumberfieldK,oKthe the finitenessofclassgroup. following generalizationofDirichlet'sunittheoremandtheon by theconsiderationoforders.Forthem,(12.9)and(12.11)imply according to(12.10).Theclaimofthepropositionnowfollowsfrom(3). This remainstruealsoforp The kernelofVisasubgroupopwhichcontainedinop,and of op/fopgivepreciselyallthemaximalidealsop,sincefopcpop. not containedinanymaximalideal,andthepreimagesofideals This issobecausetheunitsinanyringarepreciselythoseelementsthat It issurjective.Infact,ifsmodfopaunitinop/fop,thenop. For fcpwenowconsiderthehomomorphism Passing tounitgroups,wegetfrom(1)and(2)that where hKistheclassnumberofK.Inparticular,onehas that contains o*.ItisthereforeequaltoWenowconcludethat (3) § 12.Orders Our studyofone-dimensionalnoetherianintegraldomainswasmotivated Then thegroupsoK/o*andPic(o)arefiniteandonehas 1 -oKlo*--j(oK/f)*/(o/f)* -Pic(o)--C1K--±1. (n/f)*/(o/f)* _®(op/fop)*/(op/fop)*. rank(o*) =r+s-1. #Pic(o) '-' opl op=(oplfnp)*I(op/fop)* A.. P :o (op/fap)*/(op/fop)* (oK f becausethenbothsidesareequalto1 p hK : O*)#(o/f)* #(oKlf)* 81 +-° won This groupiscalledthedivisorofo.Itselementsareformalsums on thesetofallmaximalidealspo(i.e.,prime;0). group ofo.Itsdefinitionstartsfromthefreeabelian prime decomposition.Thisgroupiscalledthedivisorclassgroup,orChow ideals ofo.Itisbasedonanartificialre-introductiontheuniqueness generalization oftheidealclassgroupwhichdoestakeintoaccountallprime the informationcarriedbynoninvertibles.Butthereisanotherimportant by restrictingattentiontotheinvertibleideals,andthusleavingaside domain oavoidstheproblemofuniquenessprimeidealdecomposition 82 Nevertheless, opdefinesahomomorphism valuation ringop.Ingeneral,opisnotanymoreadiscrete valuationring. where vp to guideus.Theretheprincipalideal(f)wasgivenby algebraic numbers(seechap.III,§3). where divisorsforanalyticfunctionsplaythesameroleasidealsdo The additivenotationandthenameofgroupstemfromfunctiontheory divisor groupDiv(o)andtheofidealsarecanonicallyisomorphic. Corollary (3.9)simplysaysthat,inthecaseofaDedekinddomain, with noEZand=0foralmostallp,calleddivisors(or0-cycles). by theequation with maximalidealin,the value v=vp(a)ofaEop,for of op-submodules.Inthespecial casewhereo,isadiscretevaluationring length ofastrictlydecreasingchain where £,,(M)denotesthelengthofanop-moduleM, i.e.,themaximal then weput which generalizesthevaluationfunction.Iff=albE K*,witha,bEo, f eK*a"principaldivisor"div(f).WeusethecaseofDedekinddomain The definitionofthePicardgroupaone-dimensionalnoetherianintegral In ordertodefinethedivisorclassgroupwehaveassociateevery : K*-+Zisthep-adicexponentialvaluationassociated tothe ordp(f) =£op(op/aop) M=MO M1#... .-j Div(o) C3. (.f) =lIp"pct) ordp:K*-,Z D =Enpp aop =tn". p p s;. p >p, Zp MP=0 Chapter I.AlgebraicIntegers c4- COQ 0, isgiven 83 Z, we can now associate to every Z, we can now associate ) Div(o). p p : K* - div : K* div : J(o) -- Div(o) div(f) = Fordp(f)p, CH' (o) = Div(o)/P(o) div(a) = E - ordp(ap)p. div : Pic(a) - CH' (o) div : Pic(o) - CH' (o). op/m° J m/m° J ... D m°/m° = (0). m/m° J ... D m°/m° op/m° J chi ."- The Chow group is related to the Picard group by a canonical homomor- The Chow group is related to the Picard Using the functions ordp Using the functions The property of the function ordp to be a homomorphism follows from follows ordp to be a homomorphism of the function The property of the ideal group J (o) which takes principal ideals into principal divisors, of the ideal group J (o) which takes principal and therefore induces a homomorphism In the special case of a Dedekind domain we obtain: (12.13) Definition. The quotient group phism invertible ideal, then, by (12.4), aop, which is defined as follows. If a is an ideal apop, ap E K*, and we put for any prime ideal p ; 0, is a principal This gives us a homomorphism element f E K* the divisor element f E K* the homomorphism and thus obtain a canonical form a subgroup are called principal divisors. They The elements div(f) D' which differ only by a principal 'P(o) of Div(o). Two divisors D and divisor are called rationally equivalent. group of o. is called the divisor class group or Chow § 12. Orders § 12. chain longest the op/m°, because of the op-module length to the It is equal is of submodules this case. valuation vp in with the exponential ordp agrees Thus the function is multiplicative proved) that the length function Q,,, the fact (which is easily of op-modules. on short exact sequences C"' possibility ofviewingnumbers asfunctionsonatopologicalspace. aspects inavarietyofways. Thisgeometricinterpretationhingesonthe fundamentally fromageometric pointofview,whichwillbringoutnovel by themethodsofarithmetic andalgebra.Butthetheorymayalsobetreated ideals 0arepreciselytheinvertibleprimeideals. Exercise 5.Inaone-dimensionalnoetherianintegraldomain theregularprime Hint: N(x+y')=x2-2y2,N(3+12-)=N(5+3V) =7. all integersolutionsof"Pell'sequation" Z +7GF2inQ(/).Showthats=1 Exercise 4.DeterminetheringofmultipliersocompletemoduleM= is anorderinK,butgeneralnotthemaximalorder. multipliers where a,,._a,aarelinearlyindependentelementsofK.Showthatthering in Kisasubgroupoftheform Exercise 3.LetKbeanumberfieldofdegreen=[K:Q].Acompletemodule unit oftheorderZ+Z.1a fundamental unitoftheorderZ+Z/fieldQ(,)isalso Exercise 2.Letaandbbepositiveintegersthatarenotperfectsquares.Showthe phism C[X,Y]-C[t],Xi-*t21,y--+t(t21),haskernel(Y2XZX3). Hint: Forinstanceinthelastexample,putt=X/Yandshowthathomomor- their normalizations. are one-dimensionalnoetherianrings.Whichonesintegraldomains?Determine Exercise 1.Showthat is anisomorphism. (12.14) Proposition.IfoisaDedekinddomain,then 84 The firstapproachtothetheory ofalgebraicnumberfieldsisdominated C[X, Y]/(X2-Y3), C[X,Y]/(XY -X), § 13.One-dimensionalSchemes div :Pic(o)--)CHt(o) o={uEKI aMcM} x2-2y2=7. + Z/a/inthefieldQ(,/, C[X, Y]/(Y2-X2X3) m C[X,Y]/(XY -1), is afundamentalunitof0.Determine Alt coo ''C Chapter I.AlgebraicIntegers ). of o. usually notHausdorff.Theclosed pointscorrespondtothemaximalideals a topologicalspace(observe thatV(a)U(b)=(ab))which,however,is be theclosedsets,whereavaries overtheidealsofo.ThisdoesmakeXinto defined bystipulatingthatthesets as beingthesetofallprimeidealspo.TheZariski topologyonXis ring o,oneintroducesthespectrum lowing geometricperceptionofcompletelygeneralrings. Foranarbitrary in theresidueclassfieldK(p)=C[x]/pas"value"offatpoint residue class subsets. Inthenewformulationtheyaresets (i.e., onlythefinitesetsandMitself).Thesearedefinedtobeclosed of theform All thatcanbesalvagedalgebraicallyarethepointsetsdefinedbyequations p EM.ThetopologyonCcannotbetransferredtoMbyalgebraicmeans. which sendstheresidueclassf(x)modpto(a).Wemaythusviewthis of Mwehavethecanonicalisomorphism of theringC[x]asfunctionsonMfollows.Foreverypointp=(x-a) We mayviewMasanewkindofspaceandinterprettheelementsf(x) We denotethesetofallthesemaximalidealsby points ofthecomplexplanecorrespond1-1tomaximalidealsC[x]. the pointaformsmaximalidealp=(x-a)ofC[x].Inthisway The setofallfunctionsf(x)inthepolynomialringC[x]whichvanishat purely algebraicwayasfollows.LetaECbepointinthecomplexplane. as functionsonthecomplexplane.Thispropertymaybeformulatedina with complexcoefficientsa1EC,whichmaybeimmediatelyinterpreted § 13.One-dimensionalSchemes The algebraicinterpretationoffunctionsgivenabove leads tothefol- In ordertoexplainthis,letusstartfrompolynomials V(f)={pEMI f(p)=0} p'' (°D .f (p) V(a)=IpIpDa} a", M =Max(C[x]). C[x]/p-4C, X =Spec(o) `"y .f (x)modpEK(p) f(x)=0 ={pEMI p2(f(x))}. t,, 85 be everywhereornowhereatall.Thisintuitioncomplieswiththefactthat is f(x)itself,viewedasanelementofC(x).Thisshouldbethought field ic(p)=C(x).The"value"ofapolynomialfEC[x]atthispoint in thecaseofringo=C[x],pointp(0)hasresidueclass out tobeextremelyuseful-andhasanintuitivereasonaswell.Forinstance of o/p.Sothevaluesfdonotingenerallieasinglefield. and isanelementoftheresidueclassfieldK(p),i.e.,infractions space X:the"value"offatpointpisdefinedtobe 86 The integersaEZareviewedasfunctionsonXbydefiningthevalueof point (0),theclosureofwhichistotalspaceX.Thenonemptyopensets For everyprimenumberonehasaclosedpoint,andthereisalsothegeneric Example: ThespaceX=Spec(Z)mayberepresentedbyaline. space X.Thisiswhypalsocalledthegenericpointof the closureofpointp=(0)inZariskitopologyXistotal of asthevaluefatunknownplacex-whichonemayimagineto the ringoasfunctionsonspaceX=Spec(o)isobtained byforming Thus everyprimefieldoccursexactlyonce. The fieldsofvaluesarethen at thepoint(p)toberesidueclass in Xareobtainedbythrowingoutfinitelymanyprimenumbersp1,..., "regular functions"onUis given by subset ofX.Ifoisaone-dimensional integraldomain,thentheringof the structuresheafox.This meansthefollowing.LetU#0beanopen Admitting alsothenon-maximalprimeidealsasnon-closedpoints,turns The elementsfEonowplaytheroleoffunctionsontopological An importantrefinementofthegeometricinterpretation ofelements Z/2Z, Z/3Z,Z/5Z,Z/7Z,Z/11Z, `M' 2 o(U)={f (p' a(p) =amodpEZ/pZ. 3 9 f (p):=modp Ig(p)¢0 5 °'w z,. 7 11 for allpEU Chapter I.AlgebraicIntegers generic point ... ,Q. } 11 , p,t. .7' § 13. One-dimensional Schemes 87 in other words, it is the localization of o with respect to the multiplicative set S = o N UpEU p (see § 11). In the general case, o(U) is defined to consist of all elements s = (sp) E fl op PEU which locally are quotients of two elements of o. More precisely, this means that for every p c= U, there exists a neighbourhood V C U of p, and elements f, g E osuch that, for each q E V, one has g(q) 0 0 and sq = f/g in oq. These quotients have to be understood in the more general sense of commutative algebra (see § 11, exercise 1). We leave it to the reader to check that one gets back the above definition in the case of a one-dimensional integral domain o. If V C U are two open sets of X, then the projection 117 op - fI op pEU pEV induces a homomorphism PUv : o(U) --) OW), called the restriction from U to V. The system of rings o(U) and mappings PU v is a sheaf on X. This notion means the following. (13.1) Definition. Let X be a topological space. A presheaf .F of abelian groups (rings, etc.) consists of the following data. (1) For every open set U, an abelian group (a ring, etc.) .F(U) is given. (2) For every inclusion U C V, a homomorphism PUV :.F(U) --> .F(V) is given, which is called restriction. These data are subject to the following conditions: (a).F(0) = 0, (b) pUU is the identity id :.F(U) -> F(U), (c) PUw = Pvw o puv, for open sets W C V C U. The elements s E .F(U) are called the sections of the presheaf .F over U. If V C U, then one usually writes puv(s) = s I V. The definition of a presheaf can be reformulated most concisely in the language of categories. The open sets of the topological space X form a category Xtop in which only inclusions are admitted as morphisms. A presheaf of abelian groups (rings) is then simply a contravariant functor .F : Xtop -+ (ab), (rings) into the category of abelian groups (resp. rings) such that .F(0) = 0. where U'=f and, foreveryopensubsetU ofX,ahomomorphism induces acontinuousmap be ahomomorphismofringsandX=Spec(o),X' Spec(o').Thencp structure sheafoxisdroppedfromthenotation.Nowlet The couple(X,ox)iscalledanaffinescheme.Usually, however,the OX, P=Op. pu V,formasheafonX.Itisdenotedbyoxandcalledthestructure that suIw=svw.Theequivalenceclassesarecalledgermsofsections union Uu3T.F(U)ifthereexistsaneighbourhoodWcuflvofxsuch sections sucF(U)andsvEF(V)arecalledequivalentinthedisjoint on X.ThestalkofoxatthepointpCXislocalizationop,i.e., (13.3) Proposition.Theringso(U),togetherwiththerestrictionmappings at x.TheyaretheelementsofJ where Uvariesoverallopenneighbourhoodsofx.Inotherwords,two limit (seechap.IV,§2) for alli,j,thenthereexistsasectionsEF(U)suchthatsIU,=sii. (ii) s= s'. (i) if, forallopencoverings(Ut)ofthesetsU,onehas: 88 (13.2) Definition.ApresheafFonthetopologicalspaceXiscalledasheaf The proofofthispropositionfollowsimmediatelyfrom thedefinitions. We nowreturntothespectrumX=Spec((9)ofaringoandobtain 0.i The stalkofthesheafFatpointxEXisdefinedtobedirect Ifs, s'EF(U)aretwosectionssuchthatsIu,=foralli,then 211 If SiE.F(Ui)isafamilyofsectionssuchthat (U). Themapsffhavethefollowing twoproperties. fU :o(U)--*o(U'), f :X' .ti si Iu,nuj=SiIU,nUj Fx = (p X, : 0 . Uax .f (p') ) 0' (U) s > s ofIu', Chapter I.AlgebraicIntegers -tom." .ti (fl w., 89 : o -* o'. .Lo fin PUNT I a(U') Ti nl r > o(V') x f > I JJJ--- o(U) o(V) PUV X: a(f (p')) = 0 = fu(a) (p') = 0. a(f (p')) = 0 = fu(a) `ti U' C X' and a E o(U) one has U' C X' and a E o(U) F." : o(U) - o(U') which satisfy conditions a) and b) is called a morphism satisfy conditions a) and b) is called : o(U) - o(U') which If V C U are open sets, then the diagram then the sets, C U are open If V A continuous map f : X' -* X together with a family of homomorphisms A continuous map although some of them are a bit The proofs of the above claims are easy, noetherian integral do- We will now confine ourselves to considering § 13. One-dimensional Schemes One-dimensional § 13. a) fu* is commutative. b) For such a morphism, to the scheme X. When referring to from the scheme X' show that every not written explicitly. One can the maps ff are usually and X = Spec(o) is two affine schemes X' = Spec(o') morphism between (p described above by a ring homomorphism induced in the way of a very extensive theory which lengthy. The notion of scheme is the basis As introductions into this important occupies a central place in mathematics. [51] and [104]. discipline let us recommend the books to illustrate geometrically, via the mains o of dimension < 1, and propose of the facts treated in previous sec- scheme-theoretic interpretation, some tions. Spec(K) consists of a single point 1. Fields. If K is a field, then the scheme as the structure sheaf. One must (0) on top of which the field itself sits are all the same because they differ not think that these one-point schemes essentially in their structure sheaves. valuation ring with maximal ideal p, 2. Valuation rings. If o is a discrete of two points, the closed point x = p then the scheme X = Spec(o) consists and the generic point Y7 = (0) with with residue class field K(p) = o/p, of fractions of o. One should think residue class field K(77) = K, the field neighbourhood described by the of X as a point x with an infinitesimal generic point 77: This intuition is justified by the following observation. 90 Chapter I. Algebraic Integers The discrete valuation rings arise as localizations l op = { g I f, g E 0, g(p) 01 of Dedekind domains o. There is no neighbourhood of p in X = Spec(o) on which all functions E op are defined because, if o is not a local ring, we find by the Chineseg remainder theorem for every point q # p, q 0, an element g E o satisfying g= 0 mod q and g° 1 mod p. Then I E o, as a function is not defined at q. But every element E op is defined on a sufficiently small neighbourhood; hence one may sayg that all elements of the discrete valuation ring o, are like functions defined on a "germ" ofs neighbourhoods of p. Thus Spec(op) may be thought of as such a "germ of neighbourhoods" of p. We want to point out a small discrepancy of intuitions. Considering the spectrum of the one-dimensional ring C[x], the points of which constitute the complex plane, we will not want to visualize the infinitesimal neighbourhood Xp = Spec(C[x]p) of a point p = (x -a) as a small line segment, but rather as a little disc: This two-dimensional nature is actually inherent in all discrete valuation rings with algebraically closed residue field. But the algebraic justification of this intuition is provided only by the introduction of a new topology, the etale topology, which is much finer than the Zariski topology (see [103], [132]). 3. Dedekind rings. The spectrum X = Spec(o) of a Dedekind domain o is visualized as a smooth curve. At each point p one may consider the localization op. The inclusion o op induces a morphism f : Xp = Spec(op) _+ X, which extracts the scheme Xp from X as an "infinitesimal neighbourhood" of p: Xp X = Spec(o) smooth, butwillhavesingularitiesatcertainpoints. scheme X=Spec(o)asacurve.Butnowthecurvewillnotbeeverywhere number fieldwhichisdifferentfromthemaximalorder.Againweview domain owhichisnotaDedekinddomain,e.g.,anorderinalgebraic 4. Singularities.Wenowconsideraone-dimensionalnoetherianintegral § 13.One-dimensionalSchemes The onlysingularpointistheorigin.Itcorrespondsto the maximal-ideal where (a,b)variesoverthepointsofC2whichsatisfyequation prime ideals o =C[x,y]/(y2-x3),theclosedpointsofschemeXaregivenby not generatedbyasingleelement.Forexample,intheone-dimensionalring no longeradiscretevaluationring,thatistosay,themaximalidealpop These arepreciselythenongenericpointspforwhichlocalizationopis of thesingularitiesthatwere justdiscussed.Indeed,ifX=Spec(o)and noetherian integraldomaino means,ingeometricterms,takingtheresolution 5. Normalization.Passing to thenormalization5ofaone-dimensional cannot begeneratedbyasingle element. maximal idealpoopoofthelocalringisgeneratedbyelements 5F,57,and po =(x,y),wherexmod(y2-x3),y -x3)Eo.The p =(x-a,y-b)mod(y2-x3) 1 - Xp=Spec(op) Y b2-a3=0. x X =Spec(o) 91 '-O in thecasewhereresidue classfieldsofoarealgebraicallyclosed(like It isgraphicallyrepresentedbythefollowingpicture: of Xwhicharemappedtopbyf,Themorphismfisa"ramified covering." the primedecompositionofpin0,then431 and the morphisminducedbyinclusionoyO.Ifpisamaximalidealof in L.LetY=Spec(o),XSpec(O),and Let LIKbeafiniteseparableextension,andOtheintegralclosureofo if andonlyr=1,e11f1=(o/pl:o/p)=1. is aregularpointofX-inthesensethato,discretevaluationring the differentpointsofkthataremappedtopbyf.Onecanshow If po=pi' 6. Extensions.LetobeaDedekinddomainwithfieldoffractionsK. Since oisaDedekinddomain,theschemeXtobeconsideredassmooth. X =Spec(o),thentheinclusiono-inducesamorphismf:-*X. 92 This picture,however,isafair renderingofthealgebraicsituationonly CAD [an pej is theprimefactorizationofpino,thenpt,... T ramified points p0 = f:X-+Y 1 t , ... GCS r r3. Chapter I.AlgebraicIntegers 3, arethedifferentpoints X Y , prare between boththeories. algebraic curvesoverafinite fieldF,,.Infact,averycloseanalogyexists section iscorroboratedveryconvincinglybythetheory of functionfields This establishesafirstlinkofGaloistheorywithclassical topology.This In chap.II,§7,wewillseethatthecompositeoftwounramifiedextensions link ispursuedmuchfurtherinetaletopology. our presentcontextthefundamentalgroupofscheme Yby cally isomorphictothefundamentalgrouprrt(X).Therefore wedefinein topology. TherethegroupofcoveringtransformationsAutX (X)iscanoni- that theuniversalcoveringspaceX-->ofatopological spaceplaysin is calledtheuniversalcoveringofY.Itplayssameroleforschemes ideal p54-0ofo.TheschemeY=Spec(o)withthemorphism rule, therewillbeinfinitelymanyprimeidealslyingaboveagiven dimensional integraldomain,butingeneralnolongernoetherian,and,asa unramified extensionofK.TheintegralclosurenoinKisstillaone- closure KofK,allunramifiedextensionsLIiscalledthemaximal of Kisagainunramifled.ThecompositeK,takeninsidesomealgebraic denoted byAuty(X).Wethushaveacanonicalisomorphism of theramifiedcoveringX/Y.Thegrouptransformationsis is commutative.Suchanautomorphismcalledacoveringtransformation or morphism orEGinducesvia ideals that"ramify." of thepointsX31,..., of Y,exceptwhenpisramifiedino.Atapointramification,several are exactlyn=[L:K]pointsq31,..., for theringC[x]).Then,fromfundamentalidentity>teiff=n,there § 13.One-dimensionalSchemes The geometricpointofviewalgebraicnumberfields explained inthis : X-*X.Sincetheringoisfixed,diagram If LIKisGaloiswithgroupG=(L),theneveryauto- rr1(Y) =Auty(Y)G(KIK). l,t coalesce.Thisalsoexplainstheterminologyof G(LIK) =Auty(X). f :YY X ate, f---A : 0->anautomorphismofschemes Yf IoM j3, ofXlyingaboveeachpointp 93 The primenumberscorrespondtothemonicirreduciblepolynomialsp(t)E important modelforthetheoryofalgebraicnumberfields. since theyareimmediatelyrelatedtogeometry,actuallyserveasan fields. Theyrepresentastrikinganalogywithalgebraicnumberfields,and 94 ]Fpd ontheother,residesoneofmostprofoundmysteriesarithmetic. and thegrowingofcardinalitiesp,p2,p3,p4,...residuefields by f(a)E]Fpd,ifa1Fpdiszeroofp(t).Thisduetotheisomorphism actually givenbythevaluef(a)EFp,ifp(t)=t-a,ormoregenerally value offatapointp=(p(t))theaffineschemeXSpec(Fp[t])is becomes muchmoreapparentinthat,foranelementf=(t)E1Fp[t],the fields havethesamecharacteristic.ThegeometriccharacterofringIFp[t] as theirresidueclassrings.Thedifferenceis,however,thatnowallthese ]Fp[t]. Liketheprimenumberstheyhavefinitefields1Fpd,d=deg(p(t)), and itsfieldoffractions1Fp(t).LikeZ,1Fp[tIisalsoaprincipalidealdomain. analogies withthepolynomialringlFp[t]overfieldlFpelements different primeideals.Itisthereforecrucialtobuild atheorywhichis where itisduetothefactthatchoiceofunknown twhichdetermines the theory. of o,whichmustbetakenintoaccountinafully-fledged developmentof hides awayafinitenumberoffurtherprimeideals,besides theprimeideals difference withthenumberfieldcaseisseeninthat functionfieldK for arbitraryone-dimensionalnoetherianintegraldomains. Butthecrucial as foralgebraicnumberfields.Thisisclearfromwhatwehavedeveloped tween, ontheonehand,progressionofprimenumbers2,3,5,7,..., which takestheresidueclassf(p)=modpto(a).Inanalogybe- Let Kbeafiniteextension of F(t)andotheintegralclosure1F[t]inK. valuations, orschemetheoretically, i.e.,inageometricway. independent ofsuchchoices. Thismaybedoneeitherviathetheoryof t' =1/t,determinesacompletelydifferentring1Fp[1/t],and thuscompletely the ringofintegralityFp[t] The ringZofintegerswithitsfieldfractionsQexhibitsobvious We concludethischapterwithabriefsketchofthetheoryfunction This phenomenonappearsalreadyfortherationalfunction field]Fp(t), One obtainsthesamearithmetictheoryforfiniteextensionsKofFP(t) Let usfirstsketchthemore naivemethod,viathetheoryofvaluations. ... ((DD § 14.FunctionFields 1Fp[t]/p +]Fpd ..r is totallyarbitrary.Adifferentchoice,say ,-r (30 °C3. Chapter I.AlgebraicIntegers BCD '"' in' y." :'' 95 E lFp(t), g 4d4 p div(f) _ Evp(f)p, Y P S3. 0 of o there is an associated normalized an associated o there is 0 of X3. = deg(g) - deg(f). (a) = flpuP(a) (a) 4r" Div(K), vp:K -- ZUfool vp:K -- (L) npp I np E 96, np = 0 for almost all p} . npp I np E 96, np = 0 for almost all p} CI (K) = Div(K)/P(K). i'1 v00 p C.' div : K* Div(K) 1Fp[t], it is defined by vp(a + b) > min [v, (a), vp (b) 1. vp(a + b) > min [v, vp(ab) = vp(a) + vp(b), VP (0) = 00, For an arbitrary finite extension K of Fp(t), instead of restricting attention For an arbitrary finite extension K of Fp(t), , g E an analogue of the ideal group we form the "divisor group", i.e., the free an analogue of the ideal group we form abelian group generated by these symbols, We consider the mapping the image of which is written P(K), and we define the divisor class group of K by finitely many other discrete valuations of K. In the case of the field FP(t) finitely many other discrete valuations ones associated to the prime ideals there is one more valuation, besides the valuation vim. For p = (p(t)) of Fp[t], namely, the degree f = yIFp[y] of the ring lFp[y], where It is associated to the prime ideal p all normalized valuations of the y = 1/t. One can show that this exhausts field FP(t). normalized discrete valuations vp of K to prime ideals, one now considers all p has kept only a symbolic value. As in the above sense, where the index § 14. Function Fields Function § 14. ideal p prime 11, for every By § function i.e., a surjective discrete valuation, the properties satisfying (i) (ii) (iii) in the the valuations and the prime decomposition The relation between o is given by Dedekind domain the subring o to discrete valuation of K does not require The definition of a from o, there are and in fact, aside from those arising be given in advance, isomorphisms and ov.RemovingfromX thepointper,resp.po,onegetscanonical obvious howtoobtainasheaf ofringsoxonXfromthetwosheavesou We nowidentifyintheunionUVpointsof - {per}withthose isomorphism f:Fp[u,u-1]--Fp[v,v-1],uHv-1,yields abijection Removing fromUthepointpo=(u),and p,,,_(v)fromV, of U-{po}bymeansgyp,andobtainatopologicalspace X.Itisimmediately one hasU-(po}=Spec(Fp[u,u-1]),V-{p.)Spec(1Fp[v, v-1]),andthe more preciselythetwoaffineschemesU=Spec(A) andV=Spec(B). is obtainedbygluingthetworingsA=F,[u]and B=IFp[v],or field K.Itshowsallprimeidealsatonce,andmissesnone. This generalizationofaffineschemesisthecorrectnotionforafunction the sheafcXtoU,isisomorphicanaffineschemeinsenseof§13. there existsaneighbourhoodUwhich,togetherwiththerestrictionouof topological spaceXwithasheafofringsoxsuchthat,foreverypointX, Spec(o) andasheafofringsaxonX.Moregenerally,schemeis fields isprovidedbythenotionofscheme.Inlastsectionweintroduced will beexplainedinchap.III,§1. correction. Theadequateappreciationofthissituationanditsamendment number fieldcasethatshouldberegardedasadeficiencywhichcallsfor be consideredasstrange.Onthecontrary,itisratherfinitenessin deg isfinite.Theinfinitudeoftheclassgroupfunctionfieldsmustnot deg(div(f)) =0,sothatthemappingdegisindeedwell-defined.Asan For aprincipaldivisordiv(f),fcK*,wefindbyaneasycalculationthat residue classfieldofthevaluationringp,andwhichassociatesto Unlike theidealclassgroupofanalgebraicnumberfield,thisisnot affine schemesaspairs(X,ox)consistingofatopologicalspaceX= one obtainsherethefactthat,ifnotCI(K)itself,kernelCI°(K)of analogue ofthefinitenessclassnumberanalgebraicfield, class ofanarbitrarydivisora=EPnppthesum which associatestotheclassofpdegreedeg(p)=[x(p):1Fp] finite. Rather,onehasthecanonicalhomomorphism 96 In thecaseK=]Fj,(t)forinstance,corresponding scheme(X,ox) The ideal,completelysatisfactoryframeworkforthetheoryoffunction (X-{per}, °X_(poo)) 't7 `r3 0 :V-(p.) S3. 211 ."1 deg(a) _Enpdeg(p). deg:Cl(K)- > (U, OU), ) U-{po},p'f-1(p) p (X _(po},OX-(p0})2- Chapter I.AlgebraicIntegers (V, Ov) .'' mow,,, § 14. Function Fields 97 The pair (X, ax) is the scheme corresponding to the field lFp(t). It is called the projective line over IF,, and denoted P. Po More generally, one may similarly associate a scheme (X, ox) to an arbitrary extension K TIFF (t). For the precise description of this procedure we refer the reader to [51]. admits ap-adicexpansion leads ustothenotionofp-adicnumber.First,everypositive integerfEN in Z,foranyrationalnumberfEQaslongitlies thelocalring such anexpansioncanalsobewrittendown,relativeto aprimenumberp provided thereisnopoleatz=a,i.e.,aslong(z-a) t h(z).Thefactthat g, hEC[z],theyaredefinedbytheTaylorexpansion and moregenerally,forrationalfunctionsf(z)=h(z)EC(z),with derivatives atthepointz=aaregivenbycoefficientsofexpansion reasonably defined.Inthecaseofpolynomialsf(z)EC[z],higher "value" oftheintegerfE7Latp,butalsohigherderivativescanbe in theresidueclassfieldK(p)=7L/pZ. point pEX,i.e.,theelement space XofprimenumbersinZ,associatingtothemtheir"value"atthe be viewedinanalogywiththepolynomialsf(z)EQz]asfunctionson from theobservationmadeinlastchapterthatnumbersfE7Lmay which playssuchapredominantroleinfunctiontheory.Theideaoriginated introduce intonumbertheorythepowerfulmethodofpowerseriesexpansion century bythemathematicianKURTHENSEL(1861-1941)withaviewto This pointofviewsuggeststhefurtherquestion:whethernotonly The p-adicnumberswereinventedatthebeginningoftwentieth The TheoryofValuations f(z) 7L(p)={hI g,hEz,pth}, § 1.Thep-adicNumbers f f(z) =Eav(z-a)', L`100 f(p):= fmodp Chapter II v=0 00 =ao+a,(z-a)+...+a,.(z -a)', In concretecases,onesometimeswritesthenumberfsimplyassequence Here aiE(0,1,...,p-1}denotestherepresentativeoffimodZ/pZ. following systemofequations: unique. Itiscomputedexplicitlybysuccessivelydividingp,formingthe atives ofthe"fieldvalues"K(p)=Fp.Thisrepresentationisclearly with coefficientsaiin(0,1, 100 This notationshouldatfirstbeunderstoodinapurelyformalsense,i.e., integers, letaloneforfractions,oneisforcedtoallowinfiniteseries As soonasonetriestowritedownsuchp-adicexpansionsalsofornegative of digitsao,aIa2 represented intheform following propositionabouttheresidueclassesinZ/pnZ. denoted byZ,,. series (1.1) Definition.Fixaprimenumberp.Ap-adicinteger isaformalinfinite (1.2) Proposition.Theresidue classesamodpnEZ/p'Zcanbeuniquely where 0 a CAD ... an,forinstance v=o E avPv=ao 00 sna,pv, n=1,2,... 216 =1,331 216 =0,0022 216=0,0011011 ti. ao+alp+a2p2+... n-1 ... ,p- v=o fn_I =an_1+Pfn, ..., n-1. f, =an fi=a1+Pf2, f =ao+Pfl, + 1}, i.e.,inafixedsystemofrepresent- ..: =ao+alp+a2p2+...... Theset Chapter II.TheTheoryofValuations (5-adic). (3-adic), (2-adic), fib, +an_]pn-1 of allp-adicintegersis mod pn +alp+a2p2+.... 40. proved forn-1.Thenwehaveauniquerepresentation Proof (inductiononn):Thisisclearforn=1.Assumethestatement § 1.Thep-adicNumbers for whichwefind,bytheprecedingproposition, classes denominator ofwhichisnotdivisiblebyp,definesasequenceresidue is uniquelydeterminedbya,andthecongruenceofpropositionholds. for someintegerg.Ifg=an_,modpsuchthat0 where g,hEZ, moo v=0 E avPvZ.p. 00 an_2p"-2 +gpr-1 n=1,2,..., pn-1 ao+alp+a2p2+... coo (gh, p)=1, _n= etc., n =1,2,..., av(z -a)v,wenow 101 integers p-adic numbersf=F° into complications.Theydisappear onceweuseanotherrepresentationofthe as onedoesitwhendealingwithrealnumbersdecimal fractions,leads approach, definingsumandproductviatheusualcarry-over rulesfordigits, turn 7LPintoaring,andQpitsfieldoffractions.However, thedirect In fact, b)11p=1+p+p2+... In fact,wehave Examples: a)-1=(p-1)+1)p1)p2 expansion forwhichwewerelooking. This establishesthearithmeticanalogueoffunction-theoreticpowerseries 7L C7LP.Thus,foreveryrationalnumberfEQ,weobtainanidentity now identifyQwithitsimageinQp,sothatwemaywriteCQpand expansion, thena-bisdivisiblebypnforeveryn,andhence=b.We which takesZinto7LPandisinjective.Forifa,bEhavethesamep-adic as itsp-adicexpansion. is thep-adicexpansionofh 102 One candefineadditionandmultiplicationofp-adic numbers which hence hence -1 In thiswayweobtainacanonicalmapping -1 =(p-1)+(pI)p+.--+1)pn-l-Pn w>' aop 1-p (p -1)+(p-1)p++(p -M 1-+ 1 +a1p 1=(1+p+...+pn 1)(1-p)+pn, =-I+ D+---+ III -m+1 001 sn=7, r-. , f = then weattachtofthep-adicnumber +...+a,7 Q --)QP, n-1 v=O viewing themnotassequences ofsums 00 cad D'I-I Chapter II.TheTheoryofValuations Mod Dn. l-, 1)pn-1 +an,+1p+...EQP mod pn. . yields abijection the sequence(YO,=0ofresidueclasses (1.3) Proposition.Associatingtoeveryp-adicinteger follows fromthe The modifiedrepresentationofthep-adicnumbersalludedtoabovenow by This setiscalledtheprojectivelimitofrings7L/p'Zanddenoted we nowconsiderallelements(xn)nENwiththepropertythat In thedirectproduct The termsofsuchasequencelieindifferentrings7L/pn7L,buttheseare and wefind related bythecanonicalprojections but ratherassequencesofresidueclasses § 1.Thep-adicNumbers n 7L/P'Z = 7L/pn7L. Inotherwords,wehave '~7 C1, 7L/p7L n=I l f JZ/Pn7 Xn(xn+l) =xn 00 (xn)nEN EflZ/P'ZI sn =F_ E-I i-+ s,: =snmodpnEZ/p'Z. n-1 7L p=) v=0 n=1 Z/p2Z EX2 00 n(Sn+I) =Sn f =EavpV { (xn)nENIxnE7L/pn7Gl mod pnE7L/pn7L, for all v=0 n 00 71, /pn7L. ) 7L/p37L 4X3 n =1,2,... (xn+1) =xn,n 103 'L1 becomes thefieldoffractionsZ. with gc7LP,additionandmultiplicationextendfromZPtoQp ring ofp-adicintegers7Lp. are definedcomponentwise.WeidentifyZPwithjmZ/p'Zandobtainthe subring ofthedirectproductFlo"17G/p"7Lwhereadditionandmultiplication limit Im7L/p"Zofferstheadvantageofbeingclearlyaring.Infact,itis 104 heart, theDiophantineequations.Suchanequation up totheirdestinyentirelywithinarithmetic,moreprecisely atitsclassical field Qpofp-adicnumbersinthesameway. and therebyisrealizedasasubringofZ.WeobtainQsubfieldthe the subsetZistakentosetoftuples 0 (a modp,ap2,p3,...)EflZ/pnZ a =ao+alp+...+an-1P"-1modp", (3. III 0.w F(xt,...,xn)-0mod p" F(x1, ...,xn)=0modm. ,-, F(xt,...,xn)=0 em. 7Lp = f = n p-mg ..., xn], 7G/ p"7L ooh Chapter II.TheTheoryofValuations and thequestioniswhether n=1 00 Proof: Asestablishedabove,weviewtheringZasprojectivelimit is solvableinp-adicintegers. is solvableforarbitraryv>1ifandonlytheequation cients, andfixaprimenumberp.Thecongruence (1.4) Proposition.LetF(xl,...,x,)beapolynomialwithintegercoeffi- § 1.Thep-adicNumbers be givenforeveryv>1.Iftheelements(X!"))10, E11-,Z/PvZare with If now components overtheindividualringsZ/p"7G,namely,congruences Viewed overtheringonright,equationF=0splitsupinto n =1,writingx" fits ourneeds.Forsimplicityofnotationweonlycarry this outinthecase therefore extractasubsequencefromthesequence(x("), ...,xn"))which of theequationF=0.Butthisisnotautomatically the case.Wewill already in Conversely, letasolution(x,("),...,x,,,"))ofthecongruence F (xl, such that element ylEZ/pZ.Hencewe maychooseasubsequence{x,1,11)of{x"} there areinfinitelymanyterms x"whichmodparecongruenttothesame In whatfollows,weview(x") asasequenceinZ.SinceZ/pZisfinite, ... , xn) =0,thenthecongruencesaresolvedby tom. F x5 =ylmodp e 7Lp= Z/p"7G, foralli=1, (XI, - III x,("). Thegeneralcasefollowsexactlythesamepattern. AID" 7IP = . . F(xl, ...,xn)-0modp". . .., F(xi,...,xn)=0mod-p" X . x n) =( xnl F(xl,...,xn)=0 V Z/p"7G, isap-adicsolutionoftheequation ) Z/p"7L CfJZ/pt'Z. and 1 0 modp", ..., n, F(x,1,1)) =0modp. cam, v=1 x(")E 7Gn Ill n then wehaveap-adicsolution )vEN p" v =t,2, P .. . 105 (1) numbers reflectedintherepresentationbyp-adicdigits? Exercise 6.Howistheaddition,subtraction,multiplicationand divisionofrational identity. Exercise 5.ThefieldQpofp-adicnumbershasnoautomorphisms exceptthe Exercise 4.Writethenumbers3and-as5-adicnumbers. Exercise 3.Showthattheequationx2=2hasasolutionin7L7. and onlyifas00. Exercise 2.Ap-adicintegera=ao+a,pa2p2...isunitinthering7Lpif Hint: Writep"a=b+c117P0 The representationofap-adic integer .4.. 0,' x(,2) my2modp2and xvk) ao +alpa2P2..., § 2.Thep-adicAbsolute Value = Ykmodpk NIA F(Yk) m0modpk Yk =Yk-1mod e III Nip and fl, F(xl,21) =0modp2, F(xvk)) .-r Chapter 11.TheTheoryofValuations 0 (x(vk-t)} the termsof see § 2. The p-adic Absolute Value 107 resembles very much the decimal fraction representation ao+a1(10)+a2(10)2+...0 Let a =b, b, c EZ be a nonzero rational number. We extract from b and from c as high a power of the prime number p as possible, (2) a = Pmb' (b'c', p) = 1, and we put 1 . lalp=PM Thus the p-adic value no longer measures the size of a number a E N. Instead it becomes small if the number is divisible by a high power of p. This elaborates on the idea suggested in (1.4) that an integer has to be 0 if it is infinitely divisible by p. In particular, the summands of a p-adic series ao + a1 p + a2 p2 + form a sequence converging to 0 with respect to IIp. The exponent m in the representation (2) of the number a is denoted by u p (a), and one puts formally up (0) = oo. This gives the function vp:Q--±ZU(cc), which is easily checked to satisfy the properties 1)vp(a) = oo a = 0, 2) vp(ab) = up (a) + up (b), 3)vp(a + b) mintvp (a), up (b) 1, where x + oo = co, co + oc = oo and oo > x, for all x E Z. The function vp is called the p-adic exponential valuation of Q. The p-adic absolute value is given by I Ip:Q R, aH lalp=p-°Pl°l. In view of 1), 2), 3), it satisfies the conditions of a norm on Q: product formula: conjunction withtheabsolutevaluesI context byI number s>0(see(3.7)).Theusualabsolutevalue all normsonQ:anyfurthernormisapowerI 3) 2) 108 polynomial p(t), We extractfromg(t)andh(t) thehighestpossiblepowerofirreducible which aregivenbythemonicirreduciblepolynomialsp(t) Ek[t].Forevery of 7L,wehaveinsidek(t)thepolynomialringk[t],prime idealsp k(t) overafinitefieldk,withwhichwestartedourconsiderations. Instead analogy ofthefieldrationalnumbersQwith functionfield so thatonehasindeedfpIalp=1. the signequals of a,theexponentvppispreciselyexponentialvaluationvp(a)and Proof: Intheprimefactorization where pvariesoverallprimenumbersaswellthesymboloo. (2.1) Proposition.Foreveryrationalnumbera;0,onehas 1) as follows.Letf(t)= such p,onedefinesanabsolutevalue One canshowthattheabsolutevalues The notation IabIP =Ialplblp, la +blp a g(t), h(t)Ek[t]beanonzero rationalfunction. w.. _ I +., .-. a.+ a=± flpUP lp fl IaIP=1, P IaI. : k(t)--lib a h (t) g(t) Plc PO Ip, itsatisfiesthefollowingimportant j , and Chapter H.TheTheoryofValuations (g h,P)=1, Ialp I 1 lp andI Ip or I I I I essentially exhaust is denotedinthis IS, forsomereal 0 of 0 order ofthezero,resp.pole,functionf=(t)atta. above. Inthecasep=(t-a)foraEk,valuationvp(f)isclearly and obtainforvpI and qafixedrealnumber>1.Furthermoreweputvp(0)=oo101p0, where qp=qdp,dpbeingthedegreeofresidueclassfieldpoverk and put § 2.Thep-adicAbsoluteValue valuations vpareassociatedtotheprimeidealspofk[t].Putting p =(1/t)oftheringk[l/t]ck(t)insamewayasexponential the functionf(1/t)atpointt=0.Itisassociatedtoprimeideal of f(t)atthepointinfinityoo,i.e.,orderzero,resp.pole, where f=h namely is notderivedfromanyexponentialvaluationvpattached toaprimeideal. and amazingmanner.Thedecisivedifferencebetween theabsolutevalue perspective, anditwillfulfilltheexpectationsthusraised inanevergrowing obeys ourconstantleitmotivtostudynumbersasfunctions fromageometric to avirtualpointatinfinity.Thisofviewjustifies thenotationI the ordinaryabsolutevalueI now denotesthepointatinfinity(seechap.I,§14,p.95). where pvariesovertheprimeidealsofk[t]aswellsymboloo,which the uniquefactorizationink(t)yields,as(2.1)above,formula motivated byfunctiontheory. new, analyticconstructiondoes agreewithHensel'sdefinition,whichwas construction ofthefieldreal numbers.Wewillverifyafterwardsthatthis now giveanewdefinitionof thefieldQofp-adicnumbers,imitating I I,, ofQandtheabsolutevalueI 'CJ But forthefunctionfieldk(t),thereisonemoreexponentialvaluation In viewoftheproductformula(2.1),aboveconsideration showsthat Having introducedthep-adic absolutevalue CAD .-O 0, g,hEk[t].Itdescribestheorderofzero,resp.pole, vp(f)=m, v. (f)=deg(h)-deg(g), Ip thesameconditions1),2),3)asforupandI v,,, :k(t)-- I f1.=q`-W, I ofQshouldbethoughtasbeingassociated Ilif lp=l, p I,,, ofk(t)is,however,thattheformer ZU{oo), IfIp=q,v.(.f) I Ip onthefieldQ,letus 'SC 8'G 109 I", Ip Ip mod m m 1 P n- 00 n,m>n0. Ip is by definition a sequence by definition Ip is , a,,Pv I Chapter H. The Theory of Valuations Theory H. The Chapter n= for all coIXnIp ER. n-I V=0 ... n Qp := R/m. := Urn Xn= vp:Q -- ZU(oo). 00 n-I v=m >avpv, 0 ~'' Q2, Q3,Q5,Q7,Q11,...,Q=R. Ix +YI 1 p,i.e.,everyCauchysequenceinQ.convergeswith ZP:={xEQpIIxIp<1} n vp(xn) =-logyIx.IP vp(xn) = p max{ IxIp,IYIp} IxIp=P-vP(X). for n> I I pofthering7Lin no. I Ip liesin 111 0, 112 Chapter H. The Theory of Valuations Proof: That 7Lp is closed under addition and multiplication follows from Ix +ylp If (xn } is a Cauchy sequence in Z and x = lim x,,then IIp < 1 implies n->co also x [ , < 1, hence x E Z. Conversely, let x = lim Xn E 7L p , for a n-ioo Cauchy sequence in Q. We saw above that one has Ix lp = Ixn Ip for n > no, i.e., xn = anb,with an, b e Z, (bn, p) = 1. Choosing for each n > no a solution y,, E Z of the congruence b y - an mod pn yields I xn - y, l p < p and hence x = lim yn , so that x belongs to the closure n-roo of Z. 0 The group of units of Zpis obviously 7LP=Ix EZpIlxlp=l} Every element x E Qp admits a unique representation x=p'u withmEZanduEZp. For if vp(x) = m E Z, then vp(xp-m) = 0, hence Ixp-n' I p = 1, i.e., u = xp'n' E Z. Furthermore we have the (2.4) Proposition. The nonzero ideals of the ring Zpare the principal ideals Pn7Lp={xEQvp(x)>n{ with n > 0, and one has zpl Pnz'p = 7L/p' Z. Proof: Let a ; (0) be an ideal of 7Lp and x = pn'u, u E Zr,, an element of a with smallest possible m (sinceI x I p < 1, one has m > 0). Then a = pniZp because y = pnu' E a, u' E 7Lp, implies n > m, hence y =(pn-mu/)pn'E p'ZPI The homomorphism Z--+7Lp/pnZp,ai) amodpnZp, has kernel pn7L and is surjective. Indeed, for every x E Zp, there exists by (2.3) an a E Z such that Ix-alp <-t1 , P we obtainanisomorphism i.e., vp(x-a)>n,thereforexaEp'Zandhencemodp. These sequencesconstitutedtheprojectivelimit where snwasthepartialsum which weidentifiedwithsequences as formalSeries and thefieldQpwhichwasgivenin§1.Therewedefinedp-adicintegers § 2.Thep-adicAbsoluteValue It isnowpossibletoidentifybothdefinitionsgivenfor Zp(andtherefore It isclearthatthefamilyofthesehomomorphismsyields ahomomorphism we obtain,foreveryn>1,asurjectivehomomorphism We viewedthep-adicintegersaselementsofthisring.Since is anisomorphism. (2.5) Proposition.Thehomomorphism also forQp)viathe We nowwanttoestablishthelinkwithHensel'sdefinitionofringZ E--- 13. iIT1 n sn =s,modpnEZ/p'Z, Z/P0Z v=0 00 71.p/pnzp =Z/pnz, 7p/Pn7p =Z/p'Z. 7Gp) jZ/p'Z. Zp --+m_Z/pnZ (xn)fEF1 EX avP Zp --kZ/p'Z. sn=r- avpv. V n-I v=0 n 0 00 C14 n 10M8 n=1,2,... 0 where qEZp[[X]],andr7Lp[X] isapolynomialofdegree ,.CC the wholekernel, +an_ipn_1). +a,-, p'-'itsp-adic a p-adicinteger, o ap°=0. P which p-1' .I1 IT -S 115 app let ". Show that U(X) _ that U(X) ". Show is the trivial valuation is a monic polynomial I o r(g) P ), E' b,X Chapter H. The Theory of Valuations Theory H. The Chapter E(-1)'p'(r o I:K -+ IR 00 x = 0, I d(x,Y) = Ix -Yl § 3. Valuations cad "angle inequality". aunit in Z[[X]]andwrite f(X) = pP(X)+X"U(X) f(X) aunit in Z[[X]]andwrite f(X) = p'P(X)U(X), U(X) `-' .f (X) _ q(X)=- unique representation a Ix + y I < Ix l + I Y I IxYl = IxIIYI, The procedure we performed in the previous section with the field Q in The procedure we performed in the previous where I We tacitly exclude in the sequel the case enjoying the properties (ii) 0 0. Defining the distance between of K which satisfies lxI = 1 for all x two points x, y E K by makes K into a metric space, and hence in particular a topological space. (3.2) Definition. Two valuations of K are called equivalent if they define the same topology on K. 116 in 76, [[X ]] and P (X) E 7L [X ] where U (X) is a unit mod p. satisfying P (X) = X" be generalized to arbitrary fields order to obtain the p-adic numbers can using the concept of (multiplicative) valuation. K is a function (3.1) Definition. A valuation of a field (i) Ix1>0,andlxI=0 (iii) Hint: Let r be the operator x(0 the operator Let r be Hint: < n - 1. Show that P (X) of degree with a polynomial r(qf) = r(g). in Z [[X]] such that power series is a well-defined nonzero power series Preparation Theorem). Every Exercise 9 (p-adic Weierstrass Therefore if111and1 For anarbitraryvaluationI Proof: If1 for allxEK. only ifthereexistsarealnumbers>0suchthatonehas (3.3) Proposition.TwovaluationsI § 3,Valuations so thatIx12 numbers (withn;>0)whichconvergestoafromabove.Thenwehave Then IxIi=yforsomeaEIR.Letm;/nibesequenceofrational Now letyEKbeafixedelementsatisfyingIYII>1.LetXK,x00. the conditionthat(xn)nENconvergestozerointopologydefinedby1 may beconsideredavariantoftheChineseremaindertheorem. We usethisfortheproofoffollowingapproximation theorem,which to thecondition hence Ixlt=IxI'.ButIYIi>1implies1Y121, s>0. IX 12=lyz.ForallxEK,x;0,wethereforeget which convergestoafrombelow(*)tellsusthatIx12>IyI2 ixil =1y11 3I? IX ll=Ixlz foralIi =1,...,n. Fog IY12 I, and1 11, I 1and Ix12 <1. Ix12<1. yne, zi .. , I2 VIA a=[aEZIIlall<1} If I = max[ laol, ..., lanI } . If I = max[ laol, ..., m IInII` < r Ilaill be a nonarchimedean valuation of Q. Then be a nonarchimedean valuation of IXI:AIYI = Ix+yI=max}IxI,Iyl}. IXI:AIYI I %I is the same as the proof of Gauss's lemma for polynomials I %I is the same as the II I,, (0, 1, ..., n - 1) II be archimedean. Then one has, for every two natural numbers II be archimedean. Then one has, for every II Ilai 11 Ip. As a matter of fact: E I p or I - + 111 < 1, and there is a prime number p such that II p II < 1 because, - + 111 < 1, and there is a prime number For the field Q, we have the usual absolute value I For the field Q, we Now let 11 Ilmll § 3. Valuations § 3. +...+ant°, f(t) =a0+alt 11 1 + . II b II = 1 and hence Remark : The strong triangle inequality immediately implies that implies immediately inequality strong triangle : The Remark valuation extend the nonarchimedean One may setting, for a polynomial canonical way by field K(t) in a the function The proof If + gI < maxflf I, IgI) is immediate. The triangle inequality that I fg I = I f I this lemma by the once we replace the content off in over factorial rings absolute value If 1. nonarchimedean and for each prime number p the the archimedean valuation, valuation I Q is equivalent to one of the valua- (3.7) Proposition. Every valuation of tions I Proof: Let imply IlxJI = 1 for all x E Q*. The if not, unique prime factorization would set is an ideal of Z satisfying p7L C a = bpm with p f b, so that b V a, then we have a = p7L. If now a E Z and a 11 where s = - log 11pII/ log p. Consequently (*) n,m>1, In fact, we may write where ai r < log m / log n and II ai ll = Ill inequality 120 Chapter II. The Theory of Valuations Substituting here mk for m, taking k-th roots on both sides, and letting k tend to oo, one finally obtains Ilnlllogm/logn or Ilmlll/togn: IInII1/t°gn. IImII Swapping m with n gives the identity (*). Putting c = IInlll/tog" we have IInII =Clogn and puttingc = e' yields, for every positive rational number x = a/b, Ilxll =eslogx = IxIS. Therefore IIII is equivalent to the usual absolute value II on Q. Let II be a nonarchimedean valuation of the field K. Putting v(x) _ -log IxIforx # 0,and v(0) = oo, we obtain a function v:K-+JRU{oo) verifying the properties (i) (ii)v(xy) = u(x) + v(y), (iii)v(x + y) > min{v(x), v(y)}, where we fix the following conventions regarding elements a E JR and the symbol oc: a < oo, a + oo = oo, oc + oo = oo. A function v on K with these properties is called an exponential valuation of K. We exclude the case of the trivial function v(x) = 0 for x 0 0, v(O) = oo. Two exponential valuations vl and v2 of K are called equivalent if vl = sv2, for some real number s > 0. For every exponential valuation v we obtain a valuation in the sense of (3.1) by putting IxI for some fixed real number q > 1. To distinguish it from v, we callI I an associated multiplicative valuation, or absolute value. Replacing v by an equivalent valuation sv (i.e., replacing q by q' = qs) changes I I into the equivalent multiplicative valuation I S. The conditions (i),(ii),(iii) immediately imply the (3.8) Proposition. The subset o = { x E K I v(x)>0} ={xEKI IxI <1} is a ring with group of units o*=Ix EKI v(x)=0} ={xEKI IxI=1} and the unique maximal ideal p={xEKI v(x)>0} =IX EKI IxI <1}. § 3. Valuations 121 o is an integral domain with field of fractions K and has the property that, for every x E K, either x E o or x-1 E o. Such a ring is called a valuation ring. Its only maximal ideal is p = {x E o I x-1 0 o). The field a/p is called the residue class field of o. A valuation ring is always integrally closed. For if x E K is integral over o, then there is an equation xn+a1xn-1+...+an= 0 with ai E o and the hypothesis x o, so that x-1 E o, would imply the contradiction x = -a1 - a2x-1- - an(x-1)n-1E o An exponential valuation v is called discrete if it admits a smallest positive value s. In this case, one finds v(K*) =sZ. It is called normalized if s = 1. Dividing by s we may always pass to a normalized valuation without changing the invariants o, o*, p. Having done so, an element 7r E osuch thatv(7r) = 1 is a prime element, and every element x E K* admits a unique representation X = u 7rm with m E Z and u e o*. For if v(x) = m, then v(x n-m) = 0, hence u = x7r-" E 0*. (3.9) Proposition. If v is a discrete exponential valuation of K, then o = Ix cKIv(x)>0} is a principal ideal domain, hence a discrete valuation ring (see I, (11.3)). Suppose v is normalized. Then the nonzero ideals of o are given by pn=nno={xEKI v(x)>n},n>0, where n is a prime element, i.e., v(7r) = 1. One has pn/pn+1 a/p Proof : Let a ; 0 be an ideal of o and x 54 0 an element in a with smallest possible value v(x) = n. Then x = u 7rn, u E o*, so that 7r1o C_ a. If y = E n' E a is arbitrary with s E o*, then m = v(y) > n, hence y = (E nm-n)nn E 7rno, so that a = 7r"o. The isomorphism pn/pn+t = o/p results from the correspondence an" H a mod p. which haskernelU(1+1) choose aprimeelement7r,bythesurjectivehomomorphism the kernelofwhichisU(').Thesecondisomorphism isgiven,oncewe surjective homomorphism Proof: Thefirstisomorphismisinducedbythecanonical andobviously n>1. have the Regarding thesuccessivequotientsofchainhigherunitgroups,we is calledthen-thhigherunitgroupandU(1)ofprincipalunits. of subgroups valuation andt neighbourhoods ofthezeroelement.Indeed,ifvisanormalizedexponential (3.10) Proposition.o*/U(n)=(o/pn)*andU(n)/U(n+1) -o/p,for then soisx-1because1-=Jx-1jx11Ilx '13 u Hmodpn, 1 +7rnaH Chapter II.TheTheoryofValuations qR_1 1 q n-i } 1 .. . . } a modp, n>0 , ,, .) U(n) (] of § 4. Completions 123 Exercise 1. Show that i z _ (zz )'/2 =Vf-NC1R (z) Iis the only valuation of C which extends the absolute value I of R. Exercise 2. What is the relation between the Chinese remainder theorem and the approximation theorem (3.4)? Exercise 3. Let k be a field and K = k(t) the function field in one variable. Show that the valuations vp associated to the prime ideals p = (p(t)) of k[t], together with the degree valuation vm, are the only valuations of K, up to equivalence. What are the residue class fields? Exercise 4. Let o be an arbitrary valuation ring with field of fractions K, and let F = K*/o*. Then P becomes a totally ordered group if we define x mod O* > y mod o* to mean x/y E O. Write r additively and show that the function v:K-) FU(oo), v(0) = oo, v(x) = x mod 0* for X E K, satisfies the conditions 1) v(X) = 00 X = 0, 2) v(xy) = v(x) +v()'), 3) v(x + y) > miri(v(x),v(y)}. v is called a Krull valuation. § 4. Completions (4.1) Definition. A valued field (K, 11) is called complete if every Cauchy sequence {an}fEty in K converges to an element a E K, i.e., nicclimI a,= - a I = 0. Here, as usual, we call {an}fEN a Cauchy sequence if for every s > 0 there exists N E N such that Ian-aml I 0.. In order to prove that K = JR or = C we show that each The fields JR and C are the most familiar examples of complete fields. The fields JR and C are the most familiar to an archimedean valuation of I isomorphism a : uniqueness of completions, there is an IaI = laal as required. a quadratic equation over R. For this, consider the continuous function f : C -+ JR defined by 124 value (an ), N the absolute sequence of real numbers, real numbers. As in the case of the field Cauchy sequence of I complete with respect to the extended one proves that K is 1 of the completion (K, contains (K, 1 a : K -+ K' such that IaI gives a K -isomorphism archimedean valuation. Amazingly They are complete with respect to an More precisely we have the enough, there are no others of this type. a field which is complete with respect (4.2) Theorem (OsTRowsxr). Let K be onto JR or C satisfying for some fixed s E (0, 1]. of generality that JR c K and that the Proof: We may assume without loss valuation 1 replacing I restriction of I with respect to the restriction closure Q in K we find that Q is complete One embeds the field K into K by sending every a e K to the class of the class of e K to the every a sending into K by field K embeds the One I a, a, ... ). The valuation sequence (a, constant Cauchy Cauchy represented by the a E K which is giving the element K to K by the uniqueness asequence (an) in K. Finally one proves a E K is a limit of the equation2-(zo+io)zozo=0. is thereforenonempty,bounded,andclosed,thereazoESsuchthat minimum m.Theset Note herethatz+i,zzERCK.Sincelimf(z)=oo,hasa § 4.Completions From thisandtheinequality may substitute For fixednEN,considerontheotherhandrealpolynomial hence Iz1>Izolandthus where 0 elf g(x) =x2-(zo+io)xzoioE, G(P)I 2= Fog S={zECCI f(z)=m} Pal) i=1 + zoioInI 2n 1) i=1 11 f(ai)> f(z1)>m. < - (ai+ai)x (I +( i=1 2n m .-. In = z-+co )n)2. f (zO)n+En=mnEn, aia;) (al)m2n-1 i=1 2n7 N 125 126 Chapter H. The Theory of Valuations be an exponential valuation of the field K. It is canonically continued to an exponential valuation v of the completion k by setting v (a) = lim n- oo where a = lim an E k' an E K. Observe here that the sequence v(a0) n-+00 has to become stationary (provided a 0) because, for n > no, one has v(a - an) > i (a), so that it follows from the remark on p. 119 v(an) = v(an - a + a) = min{v(a - a),v"(a)j = v(a). Therefore it follows that v(K*) = v(K*), and if v is discrete and normalized, then so is the extension U. In the nonarchimedean case, for a sequence {an}oOty to be a Cauchy sequence, itsuffices that an+1 - an be a nullsequence. In fact, v(an - a,,,) minor (4.3) Proposition. If o c K, resp. o c K, is the valuation ring of v, resp. of v, and p, resp. p, is the maximal ideal, then one has and, if v is discrete, one has furthermore a/ n = a/pnforn>1. Generalizing the p-adic expansion to the case of an arbitrary discrete valuation v of the field K, we have the (4.4) Proposition. Let R c_ o be a system of representatives for K = o% such that 0 E R, and let 7r E o be a prime element. Then every x 0 0 in K admits a unique representation as a convergent series x =7rn2(ao+a17r +a27r2+...) whereat E R,ao:A0,rn EZ. have foreveryn>1thecanonical homomorphisms valuation. Letobethevaluation ringwiththemaximalidealp.Wethen valuation theory.Toexplain this, letKbecompletewithrespecttoadiscrete limit special instancesofthesameconcretemathematicalsituation. chapter, betweenpowerseriesandp-adicnumbers,thus appearsastwo with aiEkandmeZ.Themotivatinganalogyofthe beginningofthis where 0 In §1weidentifiedtheringZofp-adicintegersasbeing theprojective In thecaseofrationalfunctionfieldk(r)andvaluationvpattached In thecaseoffieldrationalnumbersQandp-adicvaluationvp im 7L/pn7L.Weobtainasimilarresultinthegeneral settingof f(t) =(t-a)m(ao+a1(t-a)+a2(t-a)2+...), u =ao+a17r CD. q., u =ao x =pm(ao+a1p+a2p2+...), coo + a17r... +-0 i.. + .ay 0 o/pn w°) o an_17rn-1 ;(o ... ,an-1E °,3 + an,rn ..., p- which isuniquelydetermined an-17tH-1 +7rnb,: + Trn+lbn+1 1 formasystemof R havebeenfound, , ,L+ for somebn+1E ZIP ,.7 127 128 Chapter II. The Theory of Valuations and O/p 1 O/p24? O/p3 This gives us a homomorphism o -l m o/pn n into the projective limit O/Pn = { (xn) E 1 1 O/p1 I n (xn+1) = Xn } n n=1 Considering the rings o/p" as topological rings, for the discrete topology, gives us the product topology on fl1 o/p", and the projective limit o/pn becomes a topological ring in a canonical way, being a closed n subset of the product (see chap. IV, § 2). (4.5) Proposition. The canonical mapping O - O/pn n is an isomorphism and a homeomorphism. The same is true for the mapping O* O*/U(n). n Proof: The map is injective since its kernel is n,°Op" = (0). To prove 1 surjectivity, let p = rro and let R C O, R 3 0, be a system of representatives of o/p. We saw in the proof of (4.4) (and in fact already in (1.2)) that the elements a mod p" E O/p" can be given uniquely in the form a = ao + al lr + ... + an_ 1 n n-1 mod p', where a, E R. Each element s E o/p" is therefore given by a sequence of sums n sn =ao+a1rt+...+an-1n-1,n = 1,2, ..., with fixed coefficients a1 E R, and it is thus the image of the element x = lim sn = E n. n- co The sets P, = o/p" form a basis of neighbourhoods of the zero element of flt o/p°. Under the bijection O - 1 m_ O/p° V §4. Completions 129 the basis of neighbourhoods pn of zero in o is mapped onto the basis of neighbourhoods P, n o/p° of zero in lam o/p°. Thus the bijection is a homeomorphism. It induces an isomorphism and homeomorphism on the group of units O*_ ( O/pn)* (O/pn)* = One of our chief concerns will be to study the finite extensions L I K of a complete valued field K. This means that we have to turn to the question of factoring algebraic equations f (x) = anxn + an-lxn-i + ... + ao = 0 over complete valued fields. For this, Hensel's seminal "lemma" is of fundamental importance. Let K again be a field which is, complete with respect to a nonarchimedean valuation I I. Let o be the corresponding valuation ring with maximal ideal p and residue class field Ic = o/p. We call a polynomial f (x) = ao + aix + ... + anxn E o[x] primitive if f (x) # 0 mod p, i.e., if IfI =max{Iaol, ..., IanI} = 1. (4.6) Hensel's Lemma. If a primitive polynomial f (x) E o[x] admits modulo p a factorization f (x) - g (x)h (x) mod p into relatively prime polynomials g, h E ic[x], then f (x) admits a factoriza- Lion f(x) = g(x)h(x) into polynomials g, h E o[x] such that deg(g) = deg(g) and g(x)_-g(x)modpandh (x) h(x) mod p. Proof: Let d = deg(f), m = deg(g), hence d - m > deg(h). Let go, ho c= o[x] be polynomials such that go _- g mod p, ho = h mod p and deg(go) = m, deg(ho) < d - m. Since (g, h) = 1, there exist polynomials a(x), b(x) E o[x] satisfying ago + bho - 1 mod p. Among the coefficients of the two polynomials f - goho and ago + bho - 1 E p[x] we pick one with minimum value and call it .7r. degree f -gn-ihn-1=(g,,-Iqn+hn-IPn)7tnmod7Cn+1 gn-1qn +hn-lPn=goqnhoPnfnmodit, gn =gn-1+ CID hn-1 =ho+q1n'...Qn-i7rn-1, go(af, +hoq)hop,°f,mod7r. gn-i =go+pig... b(x)fn(x) =q(x)go(x)+p,(x), h=ho+qi g =go+p17r+p27r2+ goafn +hobf,,=fnmod7r. f =gn-lhn-imod7tn. III pnjrn, can III hn =hn-1+gnirn, III III CAI Chapter II.TheTheoryofValuations pn-i7rn-1 , f". 1 modit,one § 4. Completions 131 Example : The polynomial x P-1-1 E Zp [x] splits over the residue class field Z p / p7L p = F p into distinct linear factors. Applying (repeatedly) Hensel's lemma, we see that it also splits into linear factors over Z,,. We thus obtain the astonishing result that the field Qp of p-adic numbers contains the (p -1)-th roots of unity. These, together with 0, even form a system of representatives for the residue class field, which is closed under multiplication. (4.7) Corollary. Let the field K be complete with respect to the nonar- chimedean valuation 1 1. Then, for every irreducible polynomial f (x)_ ao + a1x + + a"x" E K [x] such that aoa, : 0, one has IfI =max{laol,Ia,:l}. In particular, a = 1 and ao E o imply that f E o[x]. Proof: After multiplying by a suitable element of K we may assume that f c o[x] andI f I = 1. Let ar be the first one among the coefficients a 0 .... . a"such that l ar I = 1. In other words, we have f (x) = x' (ar + ar+tx + ... + a"x"-r)mod p. If one had max(I ao I,I a" 1) < 1, then 0 < r < n and the congruence would contradict Hensel's lemma. From this corollary we can now deduce the following theorem on extensions of valuations. (4.8) Theorem. Let K be complete with respect to the valuation 1 1. Then II may be extended in a unique way to a valuation of any given algebraic extension L I K. This extension is given by the formula n lal = INLIK(01)1, when L I K has finite degree n. In this case L is again complete. Proof: If the valuation I I is archimedean, then by Ostrowski's theorem, K = IR or C. We have NCIR(z) = zz = Iz12 and the theorem is part of classical analysis. So let I I be nonarchimedean. Since every algebraic extension L IK is the union of its finite subextensions, we may assume that the degree n = [L : K] is finite. be theminimalpolynomialofaoverK.ThenNLIK(a)=±a'Eo,sothat Conversely, letaEL*andNLIK00o.Let The implicationae0=>NLIK(a)Eoisevident(seechap.I,§2,p.12). its integralclosureinL.Thenonehas Existence oftheextendedvaluation:letobevaluationringKand0 a-1 E be theminimalpolynomialofaoverK.Thenonehas a1,...,adEoand valuation ring0.LetT,resp.T',bethemaximalidealof0,We Uniqueness oftheextendedvaluation:let as itsvaluationring. to K,itclearlygivesbackthegivenvaluation.Equallyobviouslyhas0 the formulalal=nINLIK(a)IdoesdefineavaluationofLand,restricted and then,by(*),toaE0=+10,whichistriviallytrue.Thus reduces, afterdividingbyaor$,totheimplication and IupI=a laol <1,i.e.,aoEo.By(4.7)thisgivesf(x)o[x],a0. (*) 132 let Vbeann-dimensionalnonmed vectorspaceoverK.Then,foranybasis is deducedfromthefollowinggeneralresult. would allowustofindanaELsuchthatl<1 are equivalent.Foriftheywerenot,thentheapproximation theorem(3.4) Jul <1=>Jul'andthisimpliesthatthevaluations This showsthe show that0c0.LetaEN0'andlet v1, ...,v,ofVthemaximum norm (4.9) Proposition.LetKbe complete withrespecttothevaluationI I' areequalbecausetheyagreeonK. For thefunctionlal="iNLIK(a)I,conditions The factthatLisagaincompletewithrespecttotheextended valuation 3', hence1= I fi IIxiv1 f (x)=xd+ inclusion 0c0.Inotherwords,wehavethat are obvious.Thestrongtriangleinequality f (x)=xd+alxd-1...+ad 0 ={aELINLIK(a)E01 lal <1 la+PI <-max{lal,Ifl} -a1a-1 - 11 =max{Ixi1,...,Ix,,I} ad_lxd-1 la +ll<1, - + ...aor=K[x] Chapter H.TheTheoryofValuations ad(a-1)d ET',acontradiction. I I' beanotherextensionwith . la l'>1.ThusI I al=0 I I and a =0 I and p'. I and I I' proved byinductiononn.Forn=1wemaytakepIviI.Supposethat x 1v+ Then thenorm constants p,p'>0suchthat norm onV.'Itsufficestoshowthat,foreveryI Proof: Letv1,...,vnbeabasisandII is ahomeomorphism. isomorphism is equivalenttothegivennormonV.Inparticular,Vcompleteand § 4.Completions For x=x1v1+ from U'=,(Vi+vi),i.e.,thereexistsp>0suchthat Since 0U1(V1+vi),thereexistsaneighbourhoodofwhichisdisjoint of I everything isprovedfor(n-1)-dimensionalvectorspaces.Let and weobtainthetopologicalisomorphismK'-V,(xi,...,xn) given bytheformula uniquely toLisatrivialconsequenceoftheorem(4.8). The extensionwis so thatonehasIxI>plxrl=pllxll. if n=[L:K] coefficients inK. then theelementsofKturn out tobe,by(4.4),theLaurentseriesintwith over Kintolinearfactors.We maythereforeviewKasasubfieldofK,and the minimalpolynomialofa overIF,,,then,byHensel'slemma,p(X)splits hence Fp((t))cK.Infact, ifK=Fp(a)andp(X)EFp[X]cK[X]is In thiscasewefindK=K((t)),foraprimeelement t ofK(seep.127), hand thecharacteristicofKisnotequaltozero,then ithastoequalp. fact thatKIQpisoffinitedegreeresultsfromthelocal compactness ofthe chap. I,§2,n°4,th.3),butitalsofollowsfrom(6.8)below. Ifontheother vector spaceK,byageneraltheoremoftopologicallinear algebra(see[18], is thecompletionofQwithrespecttovp,inotherwords QpCK.The because v(p)>0.InviewofthecompletenessK, closureofQinK then therestrictionofvtoQisequivalentp-adic valuationvpofQ and pthecharacteristicofitsresidueclassfieldK.IfKhas characteristic0, ring ofkwith1asi-thcoefficient,fromwhichweobtainanontrivialrelation absolute value,yieldsalinearcombinationwithcoefficientsinthevaluation relation ),lxl+ is linearlyindependentoverk.Indeed,dividinganynontrivialk-linear are linearlyindependent,thenanychoiceofpreimagesxi,...,x"EK so istheresidueclass which itselfisobviouslyagaindiscrete.SinceK1koffinitedegree, by (4.8),withrespecttotheextendedvaluation Proof: AfiniteextensionKofk=Q.orlFp((t))isagaincomplete, neighbourhood, sothatKislocallycompact. For everyaEK,theset+oisanopen,andatsametimecompact hence compact.Beingaclosedsubsetofthecompactproductno"Io/p", fields QpandlFp((t)). (5.2) Proposition.Thelocalfieldsarepreciselythefiniteextensionsof of QandlFp(t),wenowobtainthefollowingcharacterizationlocalfields. it followsthattheprojectivelimitm_o/p",andthuso,isalsocompact. logically. Sincep°/p°}i Proof: By(4.5)wehaveo=lamo/p",bothalgebraicallyandtopo- is compact. (5.1) Proposition.AlocalfieldKislocallycompact.Itsvaluationringo §5. LocalFields ix i+.A, Conversely, letKbealocalfield,vitsdiscreteexponential valuation, e.. In happyconcordwiththedefinitionofglobalfieldsasfiniteextensions BCD = 0byreducingtox.ThereforeKisalocalfield. +.k"x" =0,.kiek,bythecoefficientXjwithbiggest .,, .t", field extensionKIIFp,forifY1, .,, '-c ,.. = o/p(see(3.9)),theringsc)/p'arefinite, .a) tip can .n, coo I a="NKJk(a) cpr ... ,x,EK 135 , series fieldsFq((t)),withq=pf.Thelocalofcharacteristic0,i.e., extension ofQporIFp((t)),i.e.,toalocalfield(see[137],chap.1,§3). to anondiscretetopologyisisomorphiceitherI[8orC,finite Remark :OnecanshowthatafieldKwhichislocallycompactwithrespect kernel UMandmapsµq_1bijectivelyontoK*.Henceo* =µq_1xU0). Proof: ForeveryaEK*,onehasuniquerepresentation=i"uwith the followingfact. multiplicative groupK*.Forthedefinitionoflogarithmwemakeuse former isnotdefinedonallofK,whereasthelattergivenwhole logarithm function.Incontrasttotherealandcomplexcase,however, p-adic numberfields.Forthemonehasanexponentialfunctionanda the finiteextensionsKIQoffieldsp-adicnumbersQp,arecalled 136 such thatlogp=0whichonprincipalunits(1+x)EUM isgivenbythe determined continuoushomomorphism (5.4) Proposition.Forap-adicnumberfieldKthere is (q -l)-throotsofunity.Thehomomorphismo*--)-K*, u>modp,has into linearfactorsoverKbyHensel'slemma,o*containsthegroupµq_1of n EZ,ato*sothatK*=(zr)xo*.SincethepolynomialXq-1-1splits of principalunits. elements intheresidueclassfieldK=o/p,andUM1+pisgroup Here rrisaprimeelement,(7r)=(irkIkEZ},q#Kthenumberof decomposition (5.3) Proposition.ThemultiplicativegroupofalocalfieldKadmitsthe to K.Observingthatvp(x) >0,sothatc=p"P(x)1,andp"P(v) x2 2 ,}, Chapter H.TheTheoryofValuations + x3 3 a uniquely 0.i III K 137 - , 3 Z3 + 2 Z2 In p ln(c°/v) E µq_1. It follows that In v K by log(l+z) =z- In c U(n) ) ---> K such that X(p) = 0, then we and log exp U(1) ring be a p -adic number field with valuation pn : a =ntip(a)0) (a)(a), mutually inverse isomorphisms (and homeomor- x3 ?.(4q-1) = 0 for each log a = vp (a) log n + log (a). 1 1, two q-1 xz e VP(n>=vvp(x)-vp(v)>vlnp-lnp= P log((1 + x)(1 + y)) = log(' + x) + log(' + y) log((1 + x)(1 + y)) x" L.' We prepare the proof by the following elementary lemma. For every a E K*, choosing a prime element n, we have a unique choosing a prime element n, we For every a E K*, (5.5) Proposition. Let K IQ o and maximal ideal p, and let po = pe. Then the power series o and maximal ideal p, and let po = pe. exp(x)=1+x+2 +3i + yield, for n > l+x, l+y E U(1). E µq_1, (a) E U(1). the normalized valuation of K, 0v(a) where vp = evp is we define loge = -e log (p) As suggested by the equation p = new(p)(p), : K* and thus obtain the homomorphism log property that log p = 0. If ?.: K * It is obviously continuous and has the is any continuation of log find that ?.(') = (p), so that ?.(jr) = loge, and thus 0 = e.(rr) + k((p)) = e?.(rr) + log +log (a) = logo, for all a E K*. X(a) = vp(a)A(n)+,L((a)) = vp(a) logrr and independent of the choice of jr. log is therefore uniquely determined phisms) § 5. Local Fields Local § 5. of valuation compute the we logarithm), the usual In pp (with vp (v) < giving x°/v of the series, the terms logarithm series converges. i.e., the that x°/v is a nullsequence, This shows because It defines a homomorphism converge provided power series and all series in it is an identity of formal representation a,) 138 Chapter II. The Theory of Valuations (5.6) Lemma. Let v = E'=o ai p', 0 < ai < p, be the p-adic expansion of the natural number v E N. Then Vp(V!)_P-1i=oai(P`-1). Proof : Let [c] signify the biggest integer < c. Then we have [VIP] = at + a2P + ... + a,, pr-1 +...+a,-p'-2 [viP2] = a2 [V/Pr] = ar. Now we count how many numbers 1, 2, ... , v are divisible by p, and then by p2, etc. We find vp(v!) = [vIP]+...+[vIPr] =at+(P+1)a2+...+(P'-t +...+1)ar and hence (P-1)vp(v!) = (P-1)al+(p2-1)a2+...+(Pr-1)ar= >ai(P` -1). i=O Proof of (5.5): We again think of the p-adic valuation vp of Qp as being extended to K. Then vp = evp is the normalized valuation of K. For every natural number v > 1, one has the estimate vp(v) 1 v-1 - p-1 for if v = pavo, with (vo, p) = 1 and a > 0, then vp(v) a a 1 a 1 v-1pavo-1 - pa-1p-1 pa-t+...+p+lp-1 For vp (z) >p-11 , z ; 0, i.e., vp (z) >p-e this yields zv VP (V) vp vp(Z) _ (V - 1)vp(Z) - Vp(V) >(v - 1)( 1 > 0, (v p - 1 v-1) and thus vp(log(1 + z)) = vp(z). For n > e log therefore maps Ut11 into pn. P For the exponential series E °x/v!, we compute the valuations vp(x°/v!) as follows. Writing, for v > 0, v=ao+alp+...+arp' 0 locally compactmultiplicative groupK*ofalocalfieldK. of ordinarypowers(1+x)zi, ziEZ,ifz=limz,. expressed asthelimit to theneighbourhood(1+x)zU(n+l)off(z).Inparticular, (1+x)zmaybe is continuousbecausethecongruencez=z'modgn7Lp implies(1+x)z This obviouslyextendsthe7L-modulestructureofU(1).The function (1 +x)"modU('+'),sothattheneighbourhoodzq"7L pofzismapped is a7L/q°Z-module)andoftheformulas reason forthisisthatU(1)/U(i+1)=o/p,by(3.10),soU(1)/U(n+1) of thefactthatU(1)/U(n+l)hasorderqnforalln(whereq=#o/p- and everyzE7Lp,onehasthepower(1+x)ZU(1).Thisisaconsequence module (wherep=char(K))inacanonicalway,i.e.,forevery1+xEU) This provestheproposition. for theseareidentitiesofformalpowerseriesandalltheconverge. into U(").Furthermore,onehasforvp(x),vp(z)> Therefore vexp(x)-1 For vp(x)> Putting s=ao+- exponential series.Iffurthermorex we getfrom(5.6)that § 5.LocalFields After thisdiscussionwecan nowdetermineexplicitlythestructureof Gtr For anarbitrarylocalfieldK,thegroupofprincipalunitsU(1)isa7L Vp(XV )-vp(x)=(v-1)vp(x)- vp(v!)= V,( V! U(1) _ x" p( v. ' p exp log(1+z)=1z P-1i-o e = vV,(X)- 1 1 , i.e., vp(x)> n - +a,thisbecomes -Cl ) =vp(x),i.e.,forn> ai(P`-1)= U(1)/U(n+1) (1 +x)z=limX)z` V-SV f(z) =(1+x)Z p CD'' = UVp(x)- 1 0 andv>1,thenonehas P- and , this and p-1 v-1 1 1-> 00 (v-(ao+a1+...+a,.)) 7Lp = implies theconvergenceof log expx=, p e p-1 p- 1 p ^U3 ,exp mapsthegroupp" n mo` 1 e I 7L/q"7L. that I } sv- p-1 - p- S >0. 139 D - can Proof: By(5.3)wehave(bothalgebraicallyandtopologically) (ii) topologically) (i) elements intheresidueclassfield.Thenfollowinghold. (5.7) Proposition.LetKbealocalfieldandq=pfthenumberof 140 on modulesoverprincipalidealdomains,thereexistsin U(1) afree,finitely the groupµpaofrootsunityinKp-powerorder.Bymaintheorem generated Zp-moduleofrankd,soisUM.ThetorsionsubgroupU° U(n) =Zd.Sincetheindex(Utl) al, ...,adover7Gp,i.e.,o=Zpa1®...®7Gpad isomorphism ofZp-modules.Bychap.I,(2.9),oadmitsanintegralbasis Since log,exp,andf(z)=(1+x)Zarecontinuous,thisisatopological isomorphism (i) Assumechar(K)=0.Fornsufficientlybig,(5.5)givesusthe This reducesustothecomputationofZp-moduleU(1). topologically) where a>Oandd=[K:Q]. prime topweconsiderthecontinuous homomorphism The followingargumentistakenfromthebook[79]ofK.IWASAWA. (ii) both algebraicallyandtopologically. generated, andthereforeclosed,Zp-submoduleVofrank dsuchthat Let a)1, If Khascharacteristic0, If Khascharacteristicp,thenone(bothalgebraicallyand If char(K)=p,wehaveKFq((t))(seep.127)and 112 K* =(7C)X/tq-1x0)ZED gn ..., cvfbeabasisofIFqI]Fp. Foreverynatural Zp K* =Z®Z/(q-1)Z®7L/p°7L®7Gp, UM log:U(°)-, pn=7rno-o. K* =Z U(n), U(I) _ I.LpaXV=Z/p°7G = 1+Pl+t]F'q[[t]] gn(al, ...,af)_fl(1+witn)°`. 211 Z/(q -1)76®Z. then onehas(bothalgebraicallyand .C. 211 : U(n))isfiniteandU01)afinitely a)> Chapter II.TheTheoryofValuations ®Zd 211 i=1 f - 1)Z 211 (3. number nrelatively tin ®01) Z d.Therefore CDR ..fl ments ofF.,andweget(1).Furthermoreonehasg(psa)=1modpm+1 As avariesovertheelementsofZ and hence,sinceweareincharacteristicp, Indeed, forco=F1biro;ElFq,biZ,aimodp,wehave (2) and, fora=(a1,...,af)E7Gp (1) This functionhasthefollowingproperties.Ifm=nps,s>0,then § 5.LocalFields surjective. is denseinU(1).SinceAcompactandgcontinuous,actually of U(m)/U(n,+1)isrepresentedbyanelementg(A).Thismeansthatg(A) natural number.As converges becausegn(an)EU(n).Letm=nps,with(n,p)1,beany each factorbeingacopyofZ where theproductfT(np)=1Z i =1,...,f w =0 Consequently all n' satisfies m(an) ICI U(m+1) (m+1) bra f wps 0. Thenonehas,for , 0, i.e.,an ai =_0modp,for varies overtheele- fl gn(an) 0, which 0, and 0 for 141 is aHaarmeasureonthelocally compactgroupK*. that fodx=1.Thenonehas u (a)=facdx.Furthermore, of K,anddxtheHaarmeasure onthelocallycompactadditivegroupK,scaledso Exercise 6.LetKbeap-adicnumberfield,vpthenormalized exponentialvaluation basis ofneighbourhoods1inK*. Exercise 5.IfKisap-adicnumberfield,thenthegroupsK*", fornEN,forma both openandclosed. Sao Exercise 4.Forap-adicnumberfieldK,everysubgroupof finite indexinK*is Exercise 3.Undertheabovehypothesesonehas The seriesconvergesevenforxEKsuchthatvp(x)> has Exercise 2.LetKIQbeap-adicnumberfield.For1+xeU(1)andZE7Z extension ofthenormalizedvaluationonQp. exp :pr- log Exercise 1.Thelogarithmfunctioncanbecontinuedtoacontinuoushomomorphism because (n,p)=1,i.e.,n7LP7LP. and whenchar(K)=ponegetssimply(U:U")#µ(K)#A,/InI one has when char(K)=0,resp.p>0.Fromtheexactsequence have Proof: ThefirstequalityisaconsequenceofK*=(ir)xU.By(5.7),we powers K*"andU"inthemultiplicativegroupK*unitU: of K,thenonefindsthefollowingindicesforsubgroupsn-th (5.8) Corollary.Ifthenaturalnumbernisnotdivisiblebycharacteristic 142 (U :U1)=#IL(K)#(Zp/n7ZP)d 1 (1 +x)Z=exp(zlog(1x)) .}_ ) l-tn(K) 4A° U =µ(K)xZd P' and theexponentialfunctiontoa,continuoushomomorphism (K* :K*")=n(UU") where Pr=(xeQ #.t(K)/µ(K)". Whenchar(K)=0,thisgives: 112 .L' , g(K) , n A(K) resp. and u=0 co I log(l +x)==zlog(1+x). Chapter H.TheTheoryofValuations U =µ(K)x7LP vP(x) > Ixip d . Inly#µ,:(K). #A,(K)pduP(n) P- I e P } andv,,istheunique = #,u,(K)/InIp, A(K)lg(K)' 1, one of (4.6).Onealsocallsthevaluation vorthevaluationringohenselian. (6.1) Definition.Ahenselian field category ofalgebraicextensions.Letusdefineingeneral: henselization ofaninfinitealgebraicextensionLIKis possible withinthe valuation vwhose ringosatisfiesHensel'slemmainthesense to thecompletion(see§9,exercise4).Theconsequence isthattakingthe also beobtainedinapurelyalgebraicmanner,withoutthe analyticrecourse offers theadvantageofbeingitselfanalgebraicextension ofKwhichcan to v.Itenjoysalltherelevantalgebraicpropertiesof completionK,but algebraic overK. in ov,becausethecoefficientsoff,andthereforealsothose ofgandhare already takesplaceoverooncethehighestcoefficientofgischosentobe such thatg factorization ino relatively primefactorsg(x),hthenwehavebyHensel'slemma(4.6)a and ifaprimitivepolynomialf(x)E we have (otherwise itfollowsfrom(6.6)and§6,exercise3below).Indeed,by(4.3) in K-henceparticularchar(K)=0thisisimmediatelyobvious though Kwillnot,asarule,becomplete.Whenisalgebraicallyclosed Then Hensel'slemmaholdsintheringoaswellneven with maximalidealp,,,whichisassociatedtotherestrictionofvK,,, consider theseparableclosureKofinK,andvaluationringo,C its completion.Leto,resp.bethevaluationringsofK,K.Wethen ones. Forexample,let(K,v)beanonarchimedeanvaluedfieldand0) in amuchbiggerclassofnonarchimedeanvaluedfieldsthanthecomplete lemma alone,withoutthefullstrengthofcompleteness.Thisisvalid § 6.HenselianFields The valuedfieldKiscalledthehenselizationof Kwithrespect Most resultsoncompletevaluedfieldscanbederivedfromHensel's U.e mod p,h Sam' KcKUcK, § 6.HenselianFields o/p =oa/pv=,66, f (x)=g(x)h(x) mod li,deg(g)=deg(g).Butthisfactorization la= is afieldwithnonarchimedean Ova off 'O. splits over ... c)' into 143 E"" `ti E"" 'i7 if LIKhasfinitedegreen.Inanycase,thevaluationringofextended extension LIK.Itisgivenby tion 1 (6.2) Theorem.LetKbeahenselianfieldwithrespecttothevalua- follows. the coefficients.ItreliesonnotionofNewtonpolygon,whicharisesas the valuationsofrootsapolynomialinterms In ordertoprovethis,weappealamethodwhichallowsusexpress that, conversely,theuniqueextendabilityalsocharacterizeshenselianfields. complete field(see(4.8)).Whatisremarkableaboutourcurrentsetting valuation istheintegralclosureofringKinL. 144 This producesapolygonalchainwhichiscalledtheNewton polygonof the lowerconvexenvelopeofsetpoints (i, v(a;))ER2,ignoringhoweverthepointoo)ifa1=0.Wenowtake be apolynomialsatisfyingaoa;0.Toeachtermaix'weassociatepoint slopes arestrictlyincreasing, andwhicharesubjecttothefollowing f W. The proofofthistheoremisverbatimthesameasincasea Let vbeanarbitraryexponentialvaluationofthefieldKandlet The polygonconsistsofasequence oflinesegmentsS1,S2, 1. ThenI I admitsoneandonlyextensiontoanygivenalgebraic { (0,v(ao)),(1,v(a1)), ""p f (x) r-. IaI = (i, v(ai)) NLIK(a)1, ... ,(n, Chapter II.TheTheoryofValuations v(an))} E K[x] . ... whose '7r .ti Newton polygonoff,thenf(x)haspreciselys-rrootsal,...,ar-,. polynomial overtheheldK,vanexponentialvaluationofandw way that assume thatan=1.Wenumbertherootsal,...,ELoffinsucha Proof: Dividingbyanonlyshiftsthepolygonupordown.Thuswemay value extension tothesplittingfieldLoff. (6.3) Proposition.Letf(x)=ap+a1x § 6.Henselian.Fields symmetric functionsoftherootsaj,weimmediatelyfind where m1 every aEGinducesrc-automorphism 5ofQ/43,andthezeroesa`a=Qa would implythattheconstant coefficientaocouldnotbelongtoo.Thus as EOforallcrG.Indeed, ifa that o0=0,o,q3T.Ifa isazerooff(x)andititsmultiplicity,then because I means thatf(x)=ao. K-automorphism aEG=(LIK),wehavevaI= Ia I of theuniqueextension the Newtonpolygonisasegmentwhichdoesnotlieon thex-axisandthis if I=max{Iaoi,Ian}.Wemayassumethata"isaunit, becauseotherwise for someirreduciblepolynomialip(x)EK[x]andaconstant a. we havedeg(f)=0ordeg(f),andfind polynomial suchthataoa"00,andletj(x)=fmodpEic[x].Then extension toanygivenalgebraicextension.Wefirstshow: with in(6.2).Letusassumeconverselythat Proof: Thefactthatahenselianvaluation only ifthevaluationI (6.6) Theorem.Anonarchimedeanvaluedfield(K,I We nowproceedtodeduceHensel'slemmafromtheuniqueextendability. it, bycontrast,asaconsequenceoftheuniquenessextendedvaluation. and thusobtainedtheuniquenessofextendedvaluation.Hereweobtain polynomial witha,, The valuesofallcoefficientslieonorabovethislinesegmentandwegetthe with auniqueextensiontothesplittingfield,onehas (6.5) Corollary.Letf(x)=ao+arx+-- there isonlyoneslope,i.e.,theNewtonpolygonconsistsofasinglesegment. § 6.HenselianFields Let LIKbethesplittingfieldoff(x)overand0 valuationring As fisirreducible,theNewtonpolygonasinglelinesegment andthus Let f(x)=ao+alx In (4.7)wededucedthisresultforcompletefieldsfromHensel'slemma If thepolynomialfisirreducible,then,byabovefactorizationresult, I andthecomposite °v' orb CT' ICS I canbeuniquelyextendedtoanyalgebraicextension. 0. Then,ifI I Ifl=max{loot,Ia"I}. I to L,withmaximalidealT.Foranarbitrary + I oorextendthesamevaluation. Thisshows ' 0,thenHv I is anonarchimedeanvaluationofK I I I extends uniquelywasdealt I admits oneandonly E K[x]beanirreducible a primitive,irreducible Il' =IFooa">1 I) ishenselianifand for allaEL, 0'00 Old ,.o `-C 147 I fi 1, ICs I fi I = 1. The f1 (x) are I fi I = 1. The f1 (x) U3, amp JEJ h(x) - h(x) mod p. Chapter II. The Theory of Valuations Theory II. The Chapter f1(x)...fr(x), mod p and f = gh. III and iEI iEI f(x) = g(x)h(x) f (x) = fi (x) ... fr (x) f (x) = fi (23 f (x) - g(x)h(x) mod p f(x) = ICs g=aflfi, h=bfl fj, g=aflfi, h=bflfj III g(x) - g(x) mod p We have introduced henselian fields by a condition of which the reader We have introduced henselian fields Let now f (x) E o[x] be an arbitrary primitive polynomial, and let primitive polynomial, (x) E o[x] be an arbitrary Let now f with relatively prime monic factors g (X), h (x) E K [x], admits itself a splitting with relatively prime monic factors g (X), (6.7) Proposition. A nonarchimedean field (K, v) is henselian if any monic (6.7) Proposition. A nonarchimedean into monic factors g(x), h(x) E a[x] such that wherea,b EKand(1,...,r)=I UJanddeg(f1)=deg(fi)fori E1. wherea,b EKand(1,...,r)=I UJanddeg(f1)=deg(fi)fori We now put for a, b E o* such that a - a, b restricted to manic polynomials will find weaker versions in the literature, by the following only. Both are equivalent as is shown over the residue class field K = o/p as polynomial f (x) E o[x] which splits 148 1 = If I = r[ into irreducibles over K. Since be its factorization by suitable constants yields multiplying the f1 that irreducible polynomials in o[x]. It follows therefore primitive, a constant factor, or deg(f 1) = deg(f1), and f i is, up to where deg(f 1) = 0 factorization into polynomial. If f = g h is a the power of an irreducible have g,h E K[x], then we must relatively prime polynomials of f (x) are all conjugate over K. It follows that f (x) = dip(x)m, if irp(x) is if irp(x) = dip(x)m, that f (x) It follows over K. conjugate are all of f (x) E o*, we have furthermore a over K. Since a polynomial of the minimal = deg(f). that deg(f) § 6. Henselian Fields 149 Proof (E. NART): We have just seen that the property of K to be henselian follows from the condition that the Newton polygon of every irreducible polynomial f (x) = ao + aix + + a"x" E K[x] is a single line segment. It is therefore sufficient to show this. We may assume that a" = 1. Let L I K be the splitting field of f. Then there is always an extension w of v to L. It is obtained for example by taking the completion k of K, extending the valuation of K in a unique way to a valuation v of the algebraic closure K of k, embedding L into k, and restricting v to L. It is also possible to get the extension w directly, without passing through the completion. For this we refer to [93], chap. XII, § 4, th. 1. Assume now that the Newton polygon of f consists of more than one segment: m e Let the last segment be given by the points (m, e) and (n, 0). If e = 0, we immediately have a contradiction. Because then we have v (a;) > 0, so that f (x) E o[x], and ao - - am_t = 0 mod p, a", # 0 mod p. Therefore f (x) _ (X"-' + + a",)X' mod p, with m > 0 because there is more than one segment. In view of the condition of the proposition this contradicts the irreducibility of f. We will now reduce to e = 0 by a transformation. Let a E L be a root of f (x) of minimum value w(a) and let a E K such that v(a) = e. We consider the characteristic polynomial g(x) of a-la' E K(a), r = n - m. If f (x) = ft I(x - a-), then g(x) = fn I(x - aD2a-1). Proposition (6.3) shows that the Newton polygon of g(x) also has more than one segment, the last one of slope -w(a-1ar) = v(a)- rw(a) = e - re = 0. Since g(x) is a power of the minimal polynomial of a-la', hence of an irreducible polynomial, this produces the same contradiction as before. Let K be a field which is henselian with respect to the exponential valuation v. If L IK is a finite extension of degree n, then v extends uniquely to an exponential valuation w of L, namely w(a) = n v(NLJK(a)) . I K and let K C X. 0 we find w(si) E v(K*). In Chapter U. The Theory of Valuations Theory U. The Chapter i=0,.. ,e-l, and p = 43e n = slle, [L:K]=ef, F E aijWjri = 0 ;-S e-1 f i=Oj=1 e = (w(II)Z : v(7r)Z) I=l,.. ,f, v(K*) C w(L*) e = e(w I v) _ (w(L*) : v(K*)) e = e(w I f=f(wlv)=[;.:x] Wjn'i, 150 the degree index of the extension L I K and is called the ramification NLIK, is discrete degree. If v, and hence w = nv o is called the inertia ideal and a 0, q3, Fl, are the valuation ring, the maximal and if o, p, 'r, resp. and we find so that v(7r) = ew(l1), deduces the familiar (see chap. I) for some unit s E 0*. From this one p 0 = it 0 = RIO = c3e, or interpretation of the ramification index: of and the fundamental identity (6.8) Proposition. One has [L : K] > if v is discrete and L I K is separable. of a basis of Proof : Let cot, ..., w f be representatives of which represent the various 7ro, ..., zre-1 E L* be elements the values of e will be a consequence of what cosets in w(L*)/v(K*) (the finiteness in the discrete case form even a basis are linearly independent over K, and This follows from (6.2) by taking the logarithm. For the value groups and groups the value For the logarithm. by taking (6.2) follows from This w, one gets the inclusions fields of v and residue class The index resp. L, then one has prime element of K, for instance 3r; = II' . We show that follows). If v is discrete, we may choose the elements of LIK. Let with ai j E K. Assume that not all ai j = 0. Then there exist nonzero sums si = rj laijwj, and each time that si § 6. Henselian Fields 151 fact, dividing si by the coefficient ai of minimum value, we get a linear combination of the wi,... , cwfwith coefficients in the valuation ring o c K one of which equals 1. This linear combination is # 0 mod 93, hence a unit, so that w(si) = E v(K*). In the sum si7ri, two nonzero summands must have the same value, say w (si 7ri) = w (sgrj) ,i 0 j, because otherwise it could not be zero (observe that w(x) w(y) = w(x+y) = min(w(x), w(y)}). It follows that w(7ri) = w(7r,) + w(sj) - w(si) = w(7rj) mod v(K*), a contradiction. This shows the linear independence of the wJ 7ri. In particular, we have of < [L : K]. Assume now that v, and thus also w, is discrete and let II be a prime element in the valuation ring 0 of w. We consider the o-module e-1 f M = Y > ocoj7ri i=o,j=1 where ni = rJi and show that M = 0, i.e., [w.j7ri } is even an integral basis of 0 over o. We put f N=Locvj, j=1 so that We find that O=N+rlo, because, for a E O, we have a = a, wl + + a f wf mod 110, ai E o. This implies 0 =M + ]3e = M + pO. Since L I K is separable, 0 is a finitely generated o-module (see chap. I, (2.11)), and we conclude 0 = M from Nakayama's lemma (chap. I, § 11, exercise 7). Remark: We had already proved the identity [L : K] = of in a somewhat different way in chap. I, (8.2), also in the case where v was discrete and L I K separable. Both hypotheses are actually needed. But, strangely enough, the separability condition can be dropped once K is complete with respect to the discrete valuation. In this case, one deduces the equality 0 = M in the above proof from 0 = M + pO, not by means of Nakayama's lemma, but rather like this: as pi M C M, we get successively 0=M+p(M+pO)=M+p2O=...=M+pvO for all v > 1, and since is a basis of neighbourhoods of zero in 0, M is dense in 0. Since o is closed in K, (4.9) implies that M is closed in 0, so that M=0. nonarchimedean valuation. Exercise Exercise 3.Afieldwhichishenselianwithrespecttotwoinequivalentvaluations then onehasK(a)S-=K a1, ...,a,,beitsconjugatesoverK.IffEKissuchthat Exercise 2(Krasner'sLemma).LetaEKbeseparableoverandlet= given precision. then ithasrrootsP1,...,$,.whichapproximatetheatat polynomial g(x)ofdegreenhasallcoefficientssufficientlyclosetothosef(x), its decompositionintolinearfactors,withm;>1,a;¢ajfori of degreenand of itscoefficients.Moreprecisely,onehas:letf(x)EK[.x]beamonicpolynomial Exercise 1.Inahenselianfieldthezeroesofpolynomialarecontinuousfunctions in algebraicgeometryisthefollowing: in thesenseof(6.7)holdsforit.Acharacterizationtheseringswhichisimportant Remark: AlocalringowithmaximalidealpiscalledhenselianifHensel'slemma Hint: TheNewtonpolygon. zero aEP. and pthemaximalideal.Kis Exercise S.LetKbeanonarchimedeanvaluedfield,othevaluationring, Hint: IfaP=aEK,oneisforcedtoputw(a) inseparable extensionLIK. separably closed(TheoremofF.K.SCHMIDT). 152 follows. extensions isplayedbythe unramifiedextensions,whicharedefinedas labelled w,0,3,A,respectively. Anespeciallyimportantroleamongthese If LIKisanalgebraicextension, thenthecorrespondinginvariantsare ring, themaximalidealand theresidueclassfieldbyn,p,x,respectively. a nonarchimedeanvaluationvor splits intoadirectproductA=r[;=tA;oflocalringsA;. f (x)=x"+a,_1x"-' The proofisnotstraightforward,wereferto[103],chap.I,§4, th.4.2. More generally,everyvaluationofKadmitsauniqueextensiontoanypurely In thissectionwefixabasefieldKwhichishenselian withrespectto A localringoishenselianifandonlyeveryfinitecommutative o-algebraA --, r-. § 7.UnramifiedandTamelyRamifiedExtensions 4. Aseparablyclosed la-fl wok Chapter H.TheTheoryofValuations if andonlyeverypolynomial i=2,...,n, henselian withrespect P r-+ v(a). , ar toanypreviously j. Ifthemonic N0) p hasa to any all caseswherevextendsuniquelytoL. Remark: ThisdefinitiondoesnotrequireKtobehenselian;itappliesin finite unramifiedsubextensions. An arbitraryalgebraicextensionLIKiscalledunramifiedifitaunionof extension a.IKoftheresidueclassfieldisseparableandonehas (7.1) Definition.AfiniteextensionLIKis § 7.UnramifiedandTamelyRamifiedExtensions and f(x)=modpEK[x].Since A =K(a).LetaE0belifting,f(x)o[x]theminimalpolynomialof and, beingseparable,isthereforegeneratedbyaprimitiveelementa, explanatory. WemayassumethatLIKisfinite.ThenAJKalsofinite Proof :Thenotations Each subextensionofanunramifiedextensionisunramified. closure KIKandletL'=LK'.Thenonehas (7.2) Proposition.LetLIKandK'Jbetwoextensionsinsideanalgebraic the formulafordegree. from whatwehavejustprovedthatL'ILisunramified.Hence soisLIK,by This implies[L':K']= We obtain irreducible overK',becauseotherwiseg(x)isreducible by Hensel'slemma. g(x) modp'EK'[x].Beingafactoroff(x),isseparable andhence let g(x)eo'[x]betheminimalpolynomialofaover K'andg(x)= one hasL=K(a)andI(x)istheminimalpolynomialof unramified. (7.3) Corollary.Thecomposite oftwounramifiedextensionsKisagain We thushaveL'=K'(a).InordertoprovethatL'IK' isunramified, If LIKisasubextensionoftheunramifiedextensionL' K, thenitfollows [A' :K']<[L'=deg(g)[K'(a) :K']. [X:K]:5 deg(.f)=deg(f)=[K(a):K]<[L:K]K], (f1 L IKunramified=L'K'unramified. o, p,x; : K'],i.e.,L'IK'isunramified. CAD [L:K]=[A:K]. o', p',K;0,T,A; own loo called unramifiedifthe 'C7 0', q3',A' over K. are a., Iv" Iii self- "r7 153 f =modp.Then(x)isirreducibleandbyHensel'slemmahasaroot if theextensionAIKof residueclassfieldsisseparableandonehas (7.6) Definition.Analgebraic extensionLIKiscalledtamelyramified extension. is evengeneratedbytheserootsofunitybecausetheygenerate a,IK. that thedegreeofeachfinite subextensionofLITisprimetop. ([L one hasthefollowingweakernotionaccompanyingthat ofanunramified over K,,,.,byHensel'slemma.IfKisafinitefield,thenthe extensionK,,,K of Kbecausetheseparablepolynomialx°'-1splitsover a,.andhencealso K,,,- containsallrootsofunityordermnotdivisibleby thecharacteristic (nr =`nonramifiee').Itsresidueclassfieldistheseparable closureKSIK. of Kissimplycalledthemaximalunramifiedextension K,,,IKofK claim thenfollowsfrom that K(a)Iisunramified,soCT,andthusaEAo. in Lsuchthati=amod93,i.e.,[K(a):K]K].Thisimplies minimal polynomialofaandf(x)Eo[x]monicsuchthat separable overK.WehavetoshowthataE Proof: LetAobetheresidueclassfieldofTandassumea`EXis group ofTequalsthatK. (7.5) Proposition.TheresidueclassfieldofTistheseparableclosure) subextension ofLIK. T )Kofallunramifiedsubextensionsiscalledthemaximal (7.4) Definition.LetLKbeanalgebraicextension.Thenthecomposite (and residuefield)extensionsaremultiplicative. unramified aswellbecauseseparabilityistransitiveandthedegreesoffield L IKisunramified,hencesoL'L',by(7.2).Thisimpliesthat Proof: ItsufficestoshowthisfortwofiniteextensionsLIKandL'IK. of Kintheresidueclassfieldextensiona,IKLIK,whereasvalue 154 If thecharacteristicp=char(K)ofresidueclassfield ispositive,then The compositeofallunramifiedextensionsinsidethealgebraicclosureK In ordertoprovew(T*)=v(K*)wemaysupposeLIKbefinite.The [T :K]?(w(T*)v(K*))[).o=K]. : TI,p)=1.Intheinfinitecase thislatterconditionistakentomean y.. '-h ..y Chapter U.TheTheoryofValuations Let I(X)EK[x]bethe 1t3 [L The inequality[L class field.So,bywhatweproved first,wehave construction, bothfieldshavethesamevaluegroupand thesameresidue where ai=cibieK,i.e.,wehaveK(m1a1,... ,"Va,)CL.By solution LeiEL.Puttingai=yipi1L,wefindw(ai) =cotand which tendsto1inA.ByHensel'slemmatheequation xmi-ui=0hasa As ),=xwemaywritesibiui,wherebieKand uiisaunitinL W(yim') =v(ci),withciEK,sothatyinT'cisifor some unitsiinL. that (mi,p)=1.LetyiEL*beanelementsuch w(yi)=wi.Then w(L*) =nv(NLJK(L*))Cv(K*),where[L:K], wehavemiIn,so quotient w(L*)/v(K*)andmi and thusm-0modp,whichcontradictspt[L:K]IK]. the sameresidueclassessiink,becauseA=is.Hence0-i=1Eims, a units=p8/bELNKwithtrace the element Writing a=a1,...,a,nfortheconjugatesandTr(a)_Fi_1ai, ramified ifandonlyLITistamelyramified,thisthecase,then Proof: WemayassumethatK=TbecauseLIisobviouslytamely resp. (e,p)=1. v(K*) =w(L*),wemaychooseabEK*suchthatv(b)w(p)andobtain ([L :K],p)=1.Wefirstshowthate1impliesLK.LetaEN [T :K]_[X=f.LetLIKbetamelyramified,sothatK,land such that(mi,p)=1.Inthiscasethefundamentalidentityalwaysholds: the extensionLITisgeneratedbyradicals (7.7) Proposition.AfiniteextensionLIKistamelyramifiedifandonly is unramified,resp.tamelyramified,simplyamountstosayingthate=1, identity ef=[L:K]holdsandXIKisseparable,tosaythattheextension the valuationvofKhasauniqueextensiontoL.Whenfundamental §7. UnramifiedandTamelyRamifiedExtensions Now letcot,...,co,Ew(L*)beasystemofrepresentativesforthe As before,inthisdefinitionKneednotbehenselian.Weapplyitwhenever : K]=e,nowfollowsby inductiononr.IfLI=K("`'al 'z, 'c7 = a--1-"ELNKhastraceTr(,B) : K] III hill [L:K]=ef. a1,.. F" the 1 si=0.Buttheconjugateshave CD, order of , m'a,) nsra, r, . cwt mod v(K*).Since --, III ,.: A.. Pi =0.Since , then 155 classes of&)2,..., Also e(LIL1)>[L:L1],becausew(L*)/w(L1)isgeneratedbytheresidue cot Ew(L*)yields ramified, thenA1Irc1isseparable;henceAl=Klandp{[L1 following diagram separable closureKi=iZ by passingtothemaximalunramifiedextensionK1=K,,,.,whichhas assume withoutlossofgeneralitythatKisseparablyclosed.Thisseen where (m,p)=1.Thegeneralcasethenfollowsbyinduction.Wemay ramified, In ordertoprovethatanextensionL=K('at 156 tamely ramified. in otherwordsf=1,andsoAKptne.This shows thatLIKis hence m=n.Thus for somenewbEK*.Sincexm-aisirreducible,we haved=1,and factors, hencealsooverKbyHensel'slemma.Therefore aD1=bda xd thus all=aEbdforsomeunitsinK.As(d,p) = 1,theequation m =dn.Consequentlyw(a")v(b),bEK*,andv(bd) =w(am)v(a); let n=ord(w(a)modv(K*)).Sincemw(a)v(a)Ev(K*),wehave a' EK*,andifm'=m/d,thena=°:a'and[K("` In fact,ifdisthegreatestdivisorofmsuchthata=a'dforsome [L :K]=T],i.e.,LIKisalsotamelyramified. where LflKt=TKand1K1(,).IfnowIK1istamely Every subextensionofatamely ramifiedextensionistamelyramified. closure KIK,andL'=LK'. Thenwehave: (7.8) Corollary.LetLIKand K'jKbetwoextensionsinsidethealgebraic old Let a - s=0splitsovertheseparablyclosedresiduefieldxinto distinctlinear `r4 'Z3 it e =e(LIL1)e(L1IK)>[L:L1][L1K]=K]. L IKtamelyramified;L' I K'tamelyramified. Za-. Wemayassumethat[L suffices tolookatthecaser=1, e(L1IK) =(w(L*):v(K*))>ml[L1K]. cv,.. Thus e>n=[L:K]>ef >e, I.t of xasitsresidueclassfield.Weobtainthe Chapter H.TheTheoryofValuations : K]=[K(°)m. C1. , ..., i.e., L=K(/), fl/21) K]=m'.Now is tamely : `i' K1] _ §7. Unramified and Tamely Ramified Extensions 157 Proof: We may assume without loss of generality that LIK is finite and consider the diagram L L' I I T T' I I K K'. The inclusion T C al_..T' follows from (7.2). If LIK is tamely ramified, then L = T (°'' (m; , p) = 1; hence L' = LK' = LT' _ T'(`'1-5-,, ... ,DVa,. ), so that L'IK' is also tamely ramified, by (7.7). The claim concerning the subextensions follows exactly as in the unramified case. (7.9) Corollary. The composite of tamely ramified extensions is tamely ramified. Proof: This follows from (7.8), exactly as (7.3) followed from (7.2) in the unramified case. (7.10) Definition. Let L I K be an algebraic extension. Then the compos- ite V I K of all tamely ramified subextensions is called the maximal tamely ramified subextension of L I K. Let w(L*)(P) denote the subgroup of all elements co E w(L*) such that mco E v(K*) for some m satisfying (m, p) = 1. The quotient group w(L*)(p)/v(K*) then consists of all elements of w(L*)/v(K*) whose order is prime to p. (7.11) Proposition. The maximal tamely ramified subextension VI K of LIK has value group w(V*) = w(L*)(P) and residue class field equal to the separable closure As of K in ),IK. Proof: We may restrict to the case of a finite extension LIK. By passing from K to the maximal unramified subextension, we may assume by (7.5) that k = K. As p {' e(V I K) = #w(V*)/v(K*), we certainly have w(V*) c w(L*)(P). Conversely we find, as in the proof of (7.7), for every w E w(L*)(P) a radical a = E L such that a E K, (m, p) = I and w(a)=co,so that one has a E V,and wE w(V*). Nakayama's lemma. represents abasisofAlK, wehave0= (iii) SinceLIKisunramified, wehavepO from this. that µ"clF*f of unityandisgeneratedbyit.Consequentlyfthesmallest numbersuch distinct linearfactors,sothatA=lFqfcontainsthegroup g"ofn-throots polynomial X"-1splitsover0andthus(because(n, p) =1)overAinto [L :K]=[K()[A lemma cannotsplitintofactors.4,andhavethesame degree,sothat Indeed, beingadivisorofX"-1,4,(X)isseparable andbyHensel's reduction 4,(X)istheminimalpolynomialof Proof: (i)If4,(X)istheminimalpolynomialof (iii) O=a[0 generated bytheautomorphismq: natural numbersuchthatof-1modn. n-th rootofunity.Inthetwocases(n,p)=1andnp',thisextension (ii) (i) Then onehas: of valuationrings,resp.residueclassfields,LIK.Supposethat(n,p)=1. (7.12) Proposition.LetL= K withfiniteresidueclassfield=1Fq,q=p'. choose asourbasefield,insteadofQ.,anydiscretelyvaluedcompletefield behaves completelydifferently.Letusfirstlookatthecase(n,p)=1and Important Example: ramified ifitisnottamelyramified,i.e.,V0L. extension LIKiscalledtotally(orpurely)ramifiedifT=K,andwildly picture: 158 If LIKisfiniteande=e'p°where(e',p)1,then[V:T]e'.The The resultsobtainedinthissectionmaybesummarizedthefollowing The extensionLIKisunramifiedofdegreef,wherefthesmallest The GaloisgroupG(LIK)iscanonicallyisomorphictoG(AWK)and , i.e.,suchthatnIq owe v(K*) =w(T*)cw(L*)(P)w(L*). K .fl Consider theextensionQ C : K] r.. III T CC{ III f -1.Thisshows(i).(ii)resultstrivially C and letOIo,resp.ASK,betheextension H q. f. LIKisthereforeunramified.The A3 V Chapter H.TheTheoryofValuations C C and since1, pO, and0=o[ff]by = L A over K,thenthe mod for aprimitive , ... 3 overK. f coo 0 p. 159 1)P'-'(p-J) mod = 0. is therefore bijective, since is therefore bijective, - 1 is a root of the equation - 1 is a root of the r5' + 1 ... + 1 = 0, hence p-2 + fl(1 - Qc) _ (1) = p. I K correspond 1-1 to the subextensions of the I K correspond 1-1 to the subextensions of + p-' be a primitive p"' -th root of unity. Then one has: Then one of unity. p"' -th root primitive be a (Z/p`Z)*. Qp] _ V(p'). The canonical primitive p-th root of unity, i.e., primitive p-th root is a f-, ' (p-t )p`-' + (p-2) pm-' + (p-t )p`-' + (p-2) pm-' in the proof of (6.8), - 1 is a prime element of Qp(c). As is totally ramified of degree co(p":) = (p - 1) p' ramified of degree is totally NQp(C)IQ (1 - Q(). Qp c T = Qp(cn') c V = T(cp) C is the valuation ring ofQp is the valuation is a prime element of Zp[f ] with norm p. element of Zp[f ] is a prime = pn - 1 - (7L/pm7G)*, Q H n(v), where ac (7L/pm7G)*, Q H If n is a primitive n-th root of unity and n = npm, with (n', p) = 1, If n is a primitive n-th root of unity §7. Unramified and Tamely Ramified Extensions Ramified Tamely and Unramified §7. Let Proposition. (7.13) (i) (ii) (iii) (iv) Proof: polynomial on the left, Denoting by 0 the criterion: is irreducible because it satisfies Eisenstein's ¢ (X -I-1) = 0. But this = (XP' - 1)/(Xp"'-' - 1) _ (X - 0(1) = p and O(X) It follows that [Q both groups have order gp(pm). Thus valuation vp of Qp, we find Writing w for the extension of the normalized vp(p) = 1, i.e., Qp(c)IQp is totally furthermore that rp(pm)w(c - 1) = ramified and is the valuation ring of Qp(c). This it follows that Zp[c - 1] = 7Gp[c] concludes the proof. the following result for the maximal then propositions (7.12) and (7.13) yield extension: unramified and the maximal tamely ramified of Qp is obtained by adjoining all Exercise 1. The maximal unramified extension Show that the subextensions of K,, roots of unity of order prime to p. I K the maximal unramified extension. Exercise 2. Let K be henselian and K,,, separable closure KSIK. Exercise 3. Let L I K be totally and tamely ramified, and let A, resp, 1', be the value group of L, resp. K. Show that the intermediate fields of LIK correspond 1-1 to the subgroups of 6 /1'. by 1 valuations, todenotethecorrespondingmultiplicativevaluationinbothcases with currentusage,toemploythelettervsimultaneouslyforbothtypesof exponential valuations.Inspiteofthis,itwillproveadvantageous,andagrees absolute valueswhilethelettervhashithertobeenusedfornonarchimedean notation here,becausearchimedeanvaluationsmanifestthemselvesonlyas archimedean ornonarchimedeanvaluation.Thereisalittlediscrepancyin field Kextendstoanalgebraicextensioningeneral.Soletvbearbitrary extensions wewillnowstudythequestionofhowavaluationv 160 values 1 we obtainbyrestrictionofvtorLanextension denoted byvandtheuniqueextensionofthislattervaluationtok,U. algebraic closurek,ofK.ThecanonicalextensionvtoKisagain clarifying remarks. will behenceforthcalledthe localizationofLwithrespecttow.When completions Li,,,ofallfinite subextensionsLi1KofLIK.Thisunion completion ofLwithrespecttow,buttheunionL. =UiLiwofthe where, inthecaseofaninfiniteextensionLIK,L,, does notmeanthe valuation. ItextendsinauniquewaytocontinuousK-embedding The mappingr absolute value1 of thevaluationvtoL.Inotherwords,ifv,resp.U,aregivenbyabsolute [L :K] K isobviouslycontinuouswithrespecttothis CID r:L,,,-) K, r.L -)K,,, IxI. =Irxli w=llor Chapter R.TheTheoryofValuations n- 00 Iv extendspreciselythe v.. .3' because ifLIKisfinite,thenthefieldLKCL,,,completeby(4.8), have unique extensionofthevaluationvfromKtoL,,,IK,,.We The canonicalextensionofthevaluationwfromLtoL,,,isprecisely (*) complete extensionrLKofK,,.Weconsiderthediagramfields sequence ink,Noteherethattherxconvergesfinite where § 8.ExtensionsofValuations power seriesexpansions,i.e.,thelocalstudyoffunctions.Thediagram(*) hence onaglobalobject,whereaspassingtoKandL,,,signifieslookingat elements oftheextensionLarealgebraicfunctionsonaRiemannsurface, arises fromthecaseofafunctionfieldK,forexampleK=C(t),where number theory,theso-calledlocal-to-globalprinciple.Thisterminology L,u IKandthusrepresentsoneofthemostimportantmethodsalgebraic shows thepassagefrom"globalextension"LIKto"local The fielddiagram(*)isofcentralimportanceforalgebraicnumbertheory.It satisfy therelation degree n ray .-' .x. is aw-CauchysequenceinL,andhence 1X'Iw = L -K K ?D' L Lw =LKv ".3 n ... I NLwIKv(x)I -a .fl : L-Kgaveusanextension ) K Lw Kv , boa 't7 K over '-n a v-Cauchy , we 161 r'L K E G(K Chapter II. The Theory of Valuations Theory II. The Chapter n-.oo n->oo r'L be the K-isomorphism a = r' o r-I. r'L be the K-isomorphism a = r' o x = lim rx ax := lim arxn : L -> K be two K-embeddings such that : L -> K be two : rL .°; f(X) = fl (X)" ... fr(X)mr K,,. Clearly the correspondence x -> ax does not depend on the K,,. Clearly the correspondence x -> Let r and a o r, with a E G (K I K,), be two embeddings of L conjugate a E G (K I K,), be two embeddings Let r and a o r, with Conversely, let r, r' Conversely, let r, Those who prefer to be given an extension L I K by an algebraic equation Those who prefer to be given an extension Let L = K (a) be generated by the zero a of an irreducible polynomial 162 (i) Every extension w of the valuation v arises as the composite w = v o r w = composite as the v arises the valuation w of extension (i) Every and r' are if and only if r v o r' are equal extensions v o r and (ii) Two over K,,. conjugate This is the unique which is again denoted by w. the canonical valuation, w. The restriction v o r has to coincide with r : L. -+ K,,, the valuation Indeed, rL is dense in rL K,,, so every element x E rL K can be written Indeed, rL is dense in rL K,,, so every as a limit for some K -embedding r : L -> K,,. K -embedding r : L for some localization of an extension of v to L and Lw the Proof: (i) Let w be any v from K to L,,,. Choosing extension of the valuation that w = v o r. a K-embedding r : L -+ K such of r to L is therefore (ii) v from K, to 'E", the only extension of the valuation over K,,. Since v is induced to L and thus v o r = U o (a o r). The extensions one has u = v o or, therefore the same. by r and by or o r are v o r = v o r'. Let or We can extend or to a K -isomorphism to a finite subextension of L. As for some sequence xn which belongs in r'L an isomorphism rL K choice of a sequence (xn}, and yields a which leaves K fixed. Extending a to v o r = U o r', the sequence r'xn = arxn converges to an element v o r = U o r', the sequence r'xn = arxn conjugate over K,,. gives r' = v o r, so that r and r' are indeed concrete variant of the above f (X) = 0 will appreciate the following extension theorem. f (X) E K [X] and let of the and Visahomomorphismof Kv-algebras.Thishomomorphismisthesubject is infactaKv-algebra,with themultiplication(a0b)(a'®b')=aa'bb', K -vectorspaceLisliftedto a Kv-vectorspaceL®KK.Thislatter,however, To beginwith,thetensorproduct istakeninthesenseofvectorspaces,i.e., hence acanonicalhomomorphism indicate thatwisanextensionofthevaluationvKto L.Theinclusions L on thecompletionL,,,iofLwithrespecttowi. be thecorrespondingK-embeddingofLintoK.Thenonehas follows: letaiEKvbeazerooffiand ri extendstoanisomorphism of foverthecompletionK. irreducible factorsfl, of theirreduciblepolynomialf(X)EK[X]. (8.2) Proposition.SupposetheextensionLIKisgeneratedbyzeroa factor fi.With(8.1),thisgivesthe in k,: embeddings r:L the completionK.Ofcourse,miareoneiffisseparable.TheK- be thedecompositionoff(X)intoirreduciblefactorsfl(X), and r'(a)areconjugateoverK,i.e.,iftheyzeroesofthesameirreducible § 8.ExtensionsofValuations Let L1Kbeagainanarbitraryfiniteextension.Wewill writewIvto The extendedvaluationwiisexplicitlyobtainedfromthefactorfias Then thevaluationswt, Lw inducehomomorphismsL®KKv-*Lv,viaa®b r and r'areconjugateoverKvifonlythezeroesr(a) Kv arethengivenbythezeroes8off(X)whichlie ... ,frinthedecomposition f (X)=fl(X)II'...f,.(X r:L-*Kv, (p:L®KKv-'* fLw ri:L -)-Kv, ri ..., w,-extendingvtoLcorrespond1-1the : L, coo wi=Uori. Kv(ai) aHai, wIv P''>4 )mr ... , f,. (X)over ab, and 163 i-1 1.w (a) . IK, : K] = dimK(L) _ wIv I flLw, wIv Chapter H. The Theory of Valuation's Theory H. The Chapter wIv F1 Kv[X]l(fw) wIv TrLIK(a) = E wIv wIv of f(X), and in view of the separability, in view of the of f(X), and (a) = fl char. polynomialLW LOK Kv = F1 fJ L,, LIK [L:K]=E[Lw:Kv] I wIv L®KKv K,[X]l(f) char. polynomial NLIK(a) _ fl K-vector space L. Therefore where the top arrow is an isomorphism by the Chinese remainder theorem. is an isomorphism by the Chinese remainder where the top arrow the right is induced by X --* aw K[X]/(f) = K (u) = L. The arrow on = Kv(aw) = L. Hence the and is an isomorphism because Kv[X]/(fw) bottom arrow is an isomorphism as well. then one has (8.4) Corollary. If L I K is separable, and (8.3) since [L Proof: The first equation results from isomorphism dimK, (L ®K Kv). On both sides of the by a. The characteristic let us consider the endomorphism: multiplication L ®K K is the same as that on the polynomial of a on the K -vector space This implies immediately the identities for the norm and the trace. 164 Lw. K - f,,,1, then L ®K IK is separable, If L Proposition. (8.3) and let so that L = K (a), element for LIK, a be a primitive Proof: Let an w v, there corresponds polynomial. To every [X] be its minimal f (X) E K factor f. (X) E irreducible into an Lw as embedded Consider all the f (X) = f.1, f,,, (X). we have of a under L -+ L. of K and denote by aw the image algebraic closure K polynomial of aw = Kv(aw) and fw(X) is the minimal Then we find Lw a commutative diagram over K. We now get a ®1 and is an isomorphism because The arrow on the left is induced by X r-+ § 8. Extensions of Valuations 165 If v is a nonarchimedean valuation, then we define, as in the henselian case, the ramification index of an extension w I v by ew = (w(L*) : v(K*)) and the inertia degree by fw = [Xw : K], where X, resp. K, is the residue class field of w, resp. v. From (8.4) and (6.8), we obtain the fundamental identity of valuation theory: (8.5) Proposition. If v is discrete and L I K separable, then Tewf=[L:K]. wv This proposition repeats what we have already seen in chap. I, (8.2), working with the prime decomposition. If K is the field of fractions of a Dedekind domain o, then to every nonzero prime ideal p of o is associated the p-adic valuation vp of K, defined by vp(a) = vp, where (a) = fp p°' (see chap. I, § 11, p. 67). The valuation ring of vp is the localization op. If 0 is the integral closure of o in L and if PO= Tel ... Te. I I' is the prime decomposition of p in L, then the valuations wi = -Ti, i = 1, ... , r, are precisely the extensions of v = vp to L, e; aree the corresponding ramification indices and fi = [0/3i:o/p] the inertia degrees. The fundamental identity r eifi=[L:K] i=1 has thus been established in two different ways. The raison d'etre of valuation theory, however, is not to reformulate ideal-theoretic knowledge, but rather, as has been stressed earlier, to provide the possibility of passing from the extension L I K to the various completions LI K where much simpler arithmetic laws apply. Let us also emphasize once more that completions may always be replaced with henselizations. Exercise 1. Up to equivalence, the valuations of the field Q(om) are given as follows. 1) Ia + b,I1 = ja + bJI and ja + b/I2 = ja - b/1 are the archimedean valuations. of ordinaryGaloistheory, concerning the1-1correspondencebetween nutshell likethis: Galois theoryalsoinchap. IV,§1below.Itsmainresultcanbeputina extensions inthissectionto befinite.Ontheotherhand,wetreatinfinite reader whohappensnottoknowthistheoryshouldfeel freetoassumeall for extensionsthatarenotnecessarilyfinite,usinginfinite Galoistheory.The so wemayberatherbriefhere.However,formulateand proveallresults Galois extensionsofalgebraicnumberfields.Thearguments staythe of valuationsv,theprimeidealspandtheirdecomposition p=Ti'- "Hilbert's ramificationtheory"-seechap.I,§9,where westudied,instead action ontheextendedvaluationswIv.Thisleadstoadirect generalizationof is thevaluationringofw. Exercise Exercise 4.LetLIKbeafiniteseparableextension,othevaluationringof of vinL,thenthelocalizationOp0atprimeideal valuation, andwanextensiontoL.If0istheintegralclosureofvaluationringo Exercise 6.LetLIKbeafinitefieldextension,vnonarchimedeanexponential chap. I,(8.3)? then onehas of vtoLando,resp.Ow,arethevaluationringscompletionsK,,,Lw, discrete valuationvand0itsintegralclosureinL.IfwIvariesovertheextensions of Qadmit? Exercise 3.Howmanyextensionsto Exercise 2.DeterminethevaluationsoffieldQ(i)Gaussiannumbers. where yisasolutionofx2-5=0inQP. extensions ofI exactly oneextensionofI 166 In thecaseofaGaloisextension LIKofinfinitedegree,themaintheorem We nowconsiderGaloisextensionsLIKandstudytheeffect oftheGalois 3) Ifpisaprimenumber#2,5suchthat(5)=1,thentherearetwo tai 2) Ifp=2or5aprimenumber#2,suchthat(5)-1,thenthereis Ia+b-Ip, 5. Howdoes IP toQ(/),namely § 9.GaloisTheoryofValuations y = Ia+bylp, proposition IP toQ(/3),namely j a+b,/5-1 O ®oOu=fOw () doesthearchimedeanabsolutevalueI resp. " (8.2) lag w1u - Ia+bJIpz nil Chapter II.TheTheoryofValuations relate to Dedekind's proposition, = Ia-byIP, 3 ={aEOIw(a)>0} - - mer in ?same, defined by (9.2) Definition.Thedecomposition groupofanextensionwvtoLis because ifM=Ml. a relationfl1XM1_0withfinitelymanyMt,andthis isacontradiction We havenetXM the wholeopenneighbourhoodaG(LIM)liesincomplement ofXM. nonempty, aswehavejustseen,andalsoclosedbecause, foraEGNXM, sions andconsiderthesetsXM={aEG(woorI w'JM1.Theyare for allaEG.Thenonewouldhavethenorm=NLI K(x)=FlaEGax If wandw'arenotconjugate,thenthesets The correspondencecanbesalvaged,however,byconsideringacanonical that Iav=JJoaxw<1andlikewisev>1,contradiction. find anxELsuchthat would bedisjoint.Bytheapproximationtheorem(3.4),weableto Proof: Letwandw'betwoextensionsofvtoL.SupposeLIKisfinite. sions wIv,i.e.,everytwoextensionsareconjugate. (9.1) Proposition.ThegroupGactstransitivelyonthesetofexten- so thatthegroupGactsonsetofextensionswIv. of KandwanextensiontoL,then,foreveryaEG,oalsoextendsv, group G=G(LIK).Ifvisan(archimedeanornonarchimedean)valuation homomorphisms ofG(LIK). tacitly restrictsattentiontoclosedsubgroups,andaccordinglycontinuous of G(LIK).Otherwise,everythinggoesthroughasinthefinitecase.Soone that theintermediatefieldsofLIKcorrespond1-1toclosedsubgroups of Galoistheorythenhastobemodifiedintheinfinitecasebycondition thus turnedintoacompact,Hausdorfftopologicalgroup.Themaintheorem where MIKvariesoverthefiniteGaloissubextensionsofLK.G(LK)is for everyaEG(LIK),asbasisofneighbourhoodsthecosetsM), topology onthegroupG(LIK),Krulltopology.Itisgivenbydefining, G (LIK)ceasestohold;therearemoresubgroupsthanintermediatefields. the intermediatefieldsofLIKandsubgroupsGaloisgroup § 9.GaloisTheoryofValuations If LIKisinfinite,thenweletMvaryoverallfinite Galois subexten- }°, So letLIKbeanarbitrary,finiteorinfinite,Galoisextensionwith Gw=GW(LIK)={a EG(LIK)I woo=w}. {woolaEG} 0, becauseotherwisethecompactnessofGwouldyield . . M,,thenXM=ni-1XM; Iaxlw <1 and and Iaxlw' >1 ,G? Z-4 '.1°' in' 167 model oftheargumentforexampledecompositiongroup. topology. Theproofofthisisroutineinallcases.Letusjustillustratethe class fieldofv,resp.w. resp. q3,themaximalideal,andletx=o/p,X0/93,beresidue which aredefinedasfollows.Leto,resp.0,bethevaluationring,p, two furthercanonicalsubgroups diagram into this,wewilltreatthefunctorialpropertiesofgroups Gw,Iw,R. behaviour ofthevaluationvKasitisextendedtoL. Butbeforegoing in G. w oa=w,sothator that woo-IM=woaMIMM.AsListheunion of alltheM,weget Since aMEaG(LIM),wehaveaMIM=aIM,andw o aM=wimplies am ofG,HereMIKvariesoverallfiniteGaloissubextensionsLK. This meansthat,ineveryneighbourhoodaG(LIM),thereissomeelement subgroups wewillencounterinthesequel,areallclosedKrull that onealwayshaso-0=0andax/xE0,forallxL*. and theramificationgroupby (9.3) Definition.TheinertiagroupofwIvisdefinedby 168 If visanonarchimedeanvaluation,thenthedecompositiongroupcontains Consider twoGaloisextensionsLIKandL'IK' acommutative The groupsGw,I,,Rwcarryverysignificantinformation aboutthe Let aEG=G(LIK)beanelementwhichbelongstotheclosureofGw. The subgroupsGw,Iw,RwofG=G(LIK),andinfactallcanonical Observe inthisdefinitionthat,forOrEG Rw=Rw(LIK)={aEGwI A Iw=I,,(LIK)={aEG,,,Iax=xmod93 forallxE0} (=- Gw.ThisshowsthatthesubgroupGwisclosed G.QI.DRw, K L T ACA ."S r )L' K' T 1 mod3 Chapter H.TheTheoryofValuations w , the identitywoa=implies for allxEL*}. .fl 169 > 0, 4,, 1) : K -+ K' are r-1Rwr, - sense. = Rwor w'(-rxx r-1 makes r*(o.') = G(L IK), and we trivially get the L l)) L -a L' and r - r-11wr, : = L T;. Iwor = I.,(L'IK')-a I.(LIK), R.,(L'IK') -+ RW(LIK). w(r-1rxx Gw,(L'IK') -+ GW(LIK), = - 1) r* : G(L'IK') --* G(LIK), r* : G(L'IK') x Crx Gwar = r-1Gwr, w(- IxIwoo = Irxlw = Ir-loa'rxlw = Ior'rxlw' = I rxlw' = Ixlw, IxIwoo = Irxlw = Ir-loa'rxlw = Ior'rxlw' w(Qx -x) = w(r-1(a'rx - rx)) = w'(a'(rx) - (rx)) > 0, w(Qx -x) = w(r-1(a'rx - rx)) = the decomposition, inertia, and ramification groups of conjugate the decomposition, inertia, and ramification If the two homomorphisms r field M of L I K by the Another special case arises from an intermediate Now let w' be a valuation of L', v'=W'IK'and to =w'or,v=WIK valuation of L', v'=W'IK'and to Now let w' be a r* then becomes the inclusion G(L I M) and or E I. (L I K). If Q' E Rw' (L' I K') and X E L*, then and or E I. (L I K). If Q' E Rw' (L' I K') so that Q E Rw (L I K). (9.4) are of course isomorphisms. isomorphisms, then the homomorphisms L', we find for each r E G(LIK): In particular, in the case K = K', L = i.e., valuations are conjugate. diagram § 9. Galois Theory of Valuations Theory Galois § 9. : G(L' I K') -* G(L I K) induces homomorphisms (9.4) Proposition. r* v is assumed to be nonarchimedean. In the latter two cases, r* (v). If x E L, then one has Proof : Let Q' E G,,,, (L' I K') and v = K') and x E 0, then so that or E GW (L IK). If Q' E I,,,(L'I with homomorphisms r which will typically be inclusions. They induce a They induce be inclusions. typically which will r homomorphisms with homomorphism and thus is true of rL I rK, normal, the same that, L I K being Observe here with c rL, so that composing one has a'rL Then we have the 170 Chapter IL The Theory of Valuations (9.5) Proposition. For the extensions K C M C L, one has GW(LIM) = Gw(LIK)nG(LIM), Iw(LIM) = Iw(LIK)nG(LIM), RW(LIM) = Rw(LIK) (1 G(LIM). A particularly important special case of (9.4) occurs with the diagram L Lw i K which can be associated to any extension of valuations w I v of L I K. If L I K is infinite, then Lu, has to be read as the localization in the sense of § 8, p. 160. (This distinction is rendered superfluous if we consider, as we may perfectly well do, the henselization of L I K.) Since in the local extension L,,, I K the extension of the valuation is unique, we denote the decomposition, inertia, and ramification groups simply by G (L,, I K,), I (L,,, I K,), R (LwI KU) . In this case, the homomorphism t* is the restriction map G(LwIK1) -) G(LIK),a i )orIL, and we have the (9.6) Proposition. GW(LIK) = G(LwIK0), Iw(LIK) = I(LwIK1), Rw(LIK) = R(LwIK1) Proof: The proposition derives from the fact that the decomposition group G,,, (L I K) consists precisely of those automorphisms a E G(L I K) which are continuous with respect to the valuation w. Indeed, the continuity of the a E Gw (L I K) is clear. For an arbitrary continuous automorphism or, one has IxIw < 1 Iaxlw = IxI < 1, because Ixl,,, < 1 means that x' and hence also ax" is a w-nullsequence, i.e., (axles < 1. By §3, p. 117, this implies that w and w o a are equivalent, and hence in fact equal because w I K = w o a 1 K , so that a e G,,, (L IK). Since L is dense in L,,,, every a E GW(LIK) extends uniquely to a continuous of Lw and it is clear that & E I (Lw I Kv), resp. & E if or c= I,,,(LIK), resp. or E Rw(LIK). This proves the bijectivity of the mappings in question in all three cases. (iii) Zw=LnKv(theintersection istakeninsideLw). same valuegroupasv. (ii) (i) (9.8) Proposition. the followingproposition. is calledthedecompositionfieldofwoverK. (9.7) Definition.ThefixedfieldofGu,, decomposition groupalsodescribesthewayavaluationvextendstoan mentioned alreadyinchap.I,§9-andleftforthereadertoverify In particular,thenumber#Wofextensionsequalsindex(G:Gw).As extensions ofvtoLandchoosingafixedextensionw,weobtainbijection to thevaluationw.ItcontrolsextensionofvLinagroup-theoretic H =G(NIL),G.Gw(NIK),togetabijection extension NIK,chooseanwofvtoN,andputG=(NK), arbitrary separableextensionLIK.Forthis,weembedKintoaGalois manner. DenotingbyG\GthesetofallrightcosetsG,,Q,W of (9.6)-allautomorphismsvEGthatarecontinuouswithrespect G =(LIK)forthefieldextensionLK. and similarlyIt,(LIK)=I(LwIK,)RI(LIK)R(LWIKI) with theGaloisgroupofL,,IKandwrite of Ktothelocalsituation.WeidentifydecompositiongroupGU,(LIK) § 9.GaloisTheoryofValuations The roleofthedecompositionfieldinextensionLIK isdescribedby The decompositiongroupG,uconsists-aswasshownintheproof We nowexplaintheconcretemeaningofsubgroupsGw,I,,,,Rw The abovepropositionreducestheproblemsconcerningasinglevaluation The restrictionwzofwtoZw extendsuniquelytoL. If visnonarchimedean,wZ hasthesameresidueclassfieldand Zw=Zw(LIK)=IX ELIQx=xforallerEGw}, Gm\G/H -W,GwQHf--*woOrIL 'CS Gw\G -ZW,,, a) Gw(LIK) =G(LWIKV), G,,,o- i-->wo. 0.. 171 (iii) TheidentityZw=LflKfollowsimmediatelyfromGw(LIK) thus w'=woor,forsomeorEG(LIZ,)=Gw,i.e.,w. Proof: (i)Anarbitraryextensionw'ofwztoLisconjugatewoverZ"; 172 Qq3 =3,everyvEGwinducesaK-automorphism the valuationringofwand of L.ItisthekernelacanonicalhomomorphismG.Forif0 the sameholdstrueforZw=LflK. (ii) SinceKhasthesameresidueclassfieldandvaluegroupasK, G(LW IK.). have anexactsequence with kernel1w. of theresidueclassfield have toshowthatthisneighbourhoodcontainsanelement oftheimagef(t), neighbourhood ofF,wherettIKisafiniteGaloissubextension ofAIK.We set, iscompactandhenceclosed.LetdEG(,lIK) QG(,lit)bea dense inG(AIK)becausef(Gw),beingthecontinuous image ofacompact the surjectivityoff:Gw also AIK,istheunionoffinitenormalsubextensions. Inordertoprove in chap.I,(9.4).IntheinfinitecaseAIKisnormalsince LIK,andhence Proof :InthecaseofafiniteGaloisextension,wehave proved thisalready (9.9) Proposition.TheresidueclassfieldextensionAIKisnormal,andwe is alsosurjective,andifrEGw ismappedtovIN,,thenf(r)E57G(XIp), p. AsG(MIZw)-*G(MIK)issurjective,thecomposite subextension MIZwofLwhoseresidueclassfield M containsthefield r EGw.SinceZ.hastheresidueclassfieldK,thereexists afiniteGalois as required. The inertiagroupIwisdefinedonlyifwanonarchimedeanvaluation Gw =G(LIZW)-+G(MIZW) -G(MIK)-*G(liJK) 'v'0 v: O/3--O/3, 1--> 1w-)Gw) and weobtainahomomorphism Gw -)AutK(A) 3 themaximalideal,then,sinceQO=(9and G (A.IK)allonehastoshowisthatf(Gw) xmod TF)axmodg3, (DD Chapter II.TheTheoryofValuations ... 1. ...' It hasthefollowingsignificanceforextensionLIK. is calledtheinertiafieldofwoverK. (9.10) Definition.ThefixedfieldofI., § 9.GaloisTheoryofValuations identity onX,s,i.e.,X,ifandonlyitbelongstoG(LIT).Consequently, residue classfieldextension.AnelementaEG(LIK)thereforeinducesthe is unramified:everyfiniteGaloissubextensionhasthesamedegreeasits we haveanisomorphism extensions arealsounramified.By(7.5),ThastheresidueclassfieldXs,and maximal unramifiedsubextensionofLIK.ItisGalois,sincetheconjugate Proof: By(9.6),wemayassumethatK=Zwishenselian.LetTIbethe (9.11) Proposition.T.IZ.isthemaximalunramifiedsubextensionofLZ,,,. extensions ofK. and hastheseparableclosureK,sIKasitsresidueclassfield. Theisomorphism the inertiafieldofthisextensionismaximalunramified extensionT(K G Surjectivity followsfrom(9.9)andtheinjectivityfactthatTIK where canonical homomorphism shows thattheunramifiedextensionsofKcorrespond1-1 totheseparable For theinertiafield,(9.9)givesusisomorphism Like theinertiagroup,ramificationgroupRwis thekernelofa If inparticularKisahenselianfieldandK,sIitsseparable closure,then cam. y-' 't" Tw=Tw(LIK)={xELI cx=xforallaEIw}, X(LIK) =Hom(AIr,X*), I~- G(TWIZW) =G(),IK). G(TIK) -;G(X3Irc). G(TIK) =G(KsIK) 1w -+X(LIK), 211 173 Rw isindeedtheuniquep-Sylow subgroup. order primetop.Thisalsoapplies tothegroupIw/RwCX(LIK),sothat they areoffiniteorder.The groupX(LIK)=Hom(aIF,s*)thereforehas as/a =1mod3,acontradiction. ThisprovesthatR.isap-group. primitive 2-throotofunity form L=K'(a)witha have indeedK'=a,,sothatLIistamelyramified,and by(7.7)isofthe over K',whichhasdegreeeither1orI,asthisisso for f(x).Thuswe v`" then f(x) a liftingofa,andf(X)EK'[x]istheminimalpolynomial ofaoverK', other hand,thedegreehastobeapoweroff,forif E sandifaLis s,. _cK',sothat.XJK'ispurelyinseparableandofp-power degree.Onthe field. WeshowthatK'=s.SinceRwCIw,wehave thatTCK.Thus prime orderI follows formallyfromthis. the casewhereLIKisafiniteextension.Theinfiniteofproposition Proof of(9.12):By(9.6),wemayassumethatKishenselian.Werestrictto p-power order,resp.anorderprimetop.Inunderstandthisbetter, finite quotientgroupsofR,,,,resp.Iw/R,,,,byclosednormalsubgroupshave infinite caseithastobeunderstoodinthesenseofprofinitegroups,i.e.,all Remark :IfLIKisafiniteextension,thenitclearwhatthismeans.Inthe x EL*.Forifx'L*isanelementsuchthatw(x')-w(x)modT, we referthereadertochap.IV,§2,exercise3-5. (9.12) Proposition.Rwistheuniquep-SylowsubgroupofI. Iw -*X(LIK)withkernelR. au/u then w(x')=w(xa),aEK*.Thenx'xau,uO*,andsince This definitionisindependentofthechoicerepresentativeSEand w(x) =3andput is givenasfollows:for homomorphism where A=w(L*),andI'v(K*).IfaEIw,thentheassociated 174 Since p=char(s),theelements ins*haveorderprimetop,provided If R.werenotap-group,thenwewouldfindanelementorERwof One seesimmediatelythatmappingor III 1 modq3(becauseorcIp),onegetsax'/x'-ax/xq3. = g(x)n,,whereg(x)EK'[x]istheminimalpolynomial ofa III u'' p. LetK'bethefixedfieldoforanditsresidueclass Xe(3) = E K'.SinceorRw,wehaveontheotherhand Va-, aEK.Itfollowsthatas 3 modI'Ea/I',chooseanxL*suchthat Xa:d/F---> s* ..4. ax x mod Chapter H.TheTheoryofValuations try r-- XQisahomomorphism f. 't7 = a,witha I,' that Since A*hasnoelementsof orderdivisiblebyp,wehaveontheotherhand C/) consisting ofallelementsorderprimetop,wherep= char(K).Thus Furthermore, by(7.5),wehavew(Tw)=v(K*) =: F,andputting inertia degree1.Thus,by(7.7), d =w(L*),weseethat As TwIKisthemaximalunramifiedsubextensionofVw IK,VwTwhas suffices toshowthat already seenthatRwisthekernelofarrowonright.Ittherefore extension LIK.Intheinfinitecaseprooffollowsasin(9.9).Wehave before, thatKishenselian.Werestricttoconsideringthecaseofafinite Proof: By(9.6)wemayassume,ashavealreadydoneseveraltimes (9.15) Corollary.Wehavetheexactsequence the proofof(9.12)).ThereforeVwIVistamelyramified,and=V. by p,theresiduefieldextensionofMIVisseparable(seeargumentin contains themaximaltamelyramifiedextensionVofT(andthusZ). Since thedegreeofeachfinitesubextensionMIVwIisnotdivisible all finiteGaloissubextensionsofLITdegreeprimetop.ThereforeVw field ofRw.SinceRwisthep-Sylowsubgroup1w,Vwunion not change,wemayassumethatK=Zwishenselian.LetVwbethefixed Proof :By(9.6)andthefactthatvaluegroupresidueclassfielddo ofLIZw. (9.14) Proposition.VwIZwisthemaximaltamelyramifiedsubextension is calledtheramificationfieldofwoverK. (9.13) Definition.ThefixedfieldofRw, § 9.GaloisTheoryofValuations Vw=Vw(LIK)=txELlorx=x forallQERwJ, X(LIK) =Hom(A/P,X*) Hom(A(P)/fX*) 1 (Iw:Rw)=[Vw:T.]=#X(LIK) [Vw :TwI=#(w(Vw)lw(Tw*)) ) Rw--)Iw-)X(LIK)--1. [Vw :T.]=#(d(P)lr) II. is thesubgroupd(P)/1'ofa/1' coo 175 W-5 ramified extensionLIK,suchthatGisthesemi-directproductofx(LK)with furthermore thateverytamelyramifiedextensioncanbeembeddedintoa a F-modulebylettingv=alEFoperateon1viarHara-1. G =G(LIK),I(LIK)andFG/IG(AIK).Then1isabelianbecomes Exercise 1.LetKbeahenselianfield,LtamelyramifiedGaloisextension, the Pontryagindualofgroupd(P)IFsothatindeed generated byradicalsT,,,(Va-)IT,yofdegreem.ThisshowsthatX(LIK)is contains anelementoforderm,becausethenthereisaGaloisextension But (7.7)impliesthatXcontainsthem-throotsofunitywhenever6(P)/T 176 p. 143. over Kisisomorphictothehenselizationofwithrespectv,insense§6, extension totheseparableclosureKofK.ThendecompositionfieldZrv Exercise 4.LetvbeanonarchimedeanvaluationofthefieldKandletan the maximaltamelyramifiedextensionbyT K =1Fp((t))isgivenbyTF,((t)),whereIF,,thealgebraicclosureoflFp,and Exercise 3.Showthatthemaximalunramifiedextensionofpowerseriesfield the maximalunramifiedabelianextensionTofQ, Exercise 2.ThemaximaltamelyramifiedabelianextensionVofQPisfiniteover Hint: Use(7.7). G(Ak): G-x(LIK)>1G(XIK). ramification groupofLIK by by vL=ewtheassociatednormalizedvaluationofL. field characteristicp,whichadmitsauniqueextension wtoL.Wedenote and thatvKisadiscretenormalizedvaluationofK,with positiveresidue are nowgoingtostudy.WeassumethatLIKisafinite Galoisextension valued fieldsareonlythefirsttermsinawholeseriesof subgroupsthatwe (10.1) Definition.Forevery realnumbers>-1wedefinethes-th Show thatthereisacanonicalisomorphism1=x(LIK)ofF-modules. The inertiagroupandtheramificationinside Galoisgroupof G,s=Gs(LIK)=IaEGI ago § 10.HigherRamificationGroups [Vm :Tw]=#(A(P)/F)#X(LIK). CAD "ti ,-, Chapter H.TheTheoryofValuations I MEN,(m,p)=1)). 211 II[ As ramification groupR=(LIK)whichhavealreadybeendefinedin(9.3). Here UL(S)denotesthes-thgroupofprincipalunitsL,i.e.,U(o)=0* s >0,themapping (10.2) Proposition.Let7CLE0beaprimeelementofL.Foreveryinteger of normalsubgroupsG.Thequotientsthischainsatisfythe and rO=0,theramificationgroupsformachain § 10.HigherRamificationGroups of (9.5). from thedefinitionoframificationgroupsfollowing generalization change offields.IfonlythebasefieldKischanged,then wegetdirectly the uniquep-SylowsubgroupininertiagroupI=Go. for s=0.Inparticular,wefindagainthattheramificationgroupRG1is therefore abeliangroupsofexponentp,fors>1,andorderprimeto UL)/U(1) =X*and and UL)=l+7rLO,fors>1. is aninjectivehomomorphismwhichindependentoftheprimeelement7rL. extension need somepreparation.We will assumeforthesequelthatresiduefield but theindexingchanges.For theprecisedescriptionofsituationwe mapped underG(LIK)--) (L'IK)intotheramificationgroupsofL'K, Galois subextensionL'IK.It istruethattheramificationgroupsofLIKare alls>-1, that (10.3) Proposition.IfK'isanintermediatefieldofLIK, thenonehas,for Clearly, G_1=G,Goistheinertiagroup1I(LK),andG1 We nowstudythebehaviourofhigherramification groupsunder We leavetheelementaryprooftoreader.Observethatonehas Matters becomemuchmorecomplicatedwhenwepass fromLIKtoa VL(r-laza I KofL1Kisseparable. GslGs+1 -- - a)= G=G_12Go_DG1 DG212... GS(LIK') =GS(LIK)nG(LIK'). VL(r-1(Qra UL)/ULS+1) = LL +'c1 U)/U+1) X, fors>1.ThefactorsGS/G,s+1are - ra))=VL(a(ra)ra) 011 -> U7rL 7rL \'G 177 generate thesameidealin0. r EH=G(LIL').Ittherefore sufficestoshowthattheelements elements ofGwithimage a'inG'=G(L'IK)arethengivenbyar We chooseafixedEG=(LIK)suchthatOr L'=a'.Theother 0' =o[y],withthevaluationringofL'.Bydefinition, wehave Proof: Fora'=1bothsidesareinfinite.Let01,and let0=o[x]and (10.5) Proposition.Ife'=eLIL'istheramificationindex ofLIL',then the followingrule. Passing toaGaloissubextensionL'IKofL generator xandwemaywrite proof of(6.8),wesawthatthe where 0=o[x].Thisdefinitiondoesnotdependonthechoiceof form anintegralbasisof0overo.Henceindeed=o[x]. we obtainvL(f(x+7rL))=1sincef'(x)E0*,because1'(Y)=A0.Inthe from Taylor'sformula required, i.e.,ifvL(f(x))>2,therepresentativex+7rLwilldo.Infact, then wecertainlyhavevL(f(x))>1becausef(X)=0.Ifxitselfisnotas 7r =f(x)isaprimeelementof0.Indeed,ifxanarbitraryrepresentative, polynomial j(X)of5F.ThenthereisarepresentativexE0suchthat admits aprimitiveelementY.Letf(X)Eo[X]beliftingoftheminimal Proof :AstheresiduefieldextensionXJKisseparablebyassumption,it an xE0suchthat=o[x]. (10.4) Lemma.Theringextension0ofoismonogenous,i.e.,thereexists 178 For everyaeGwenowput xfnr=xJf(x)`, e iL'IK(a')=vL(a'Y f (x+70_.f(x)'nLb7r2, G,(LIK)={cr EGIiLIK(a)>s+1}. a=ay-y iL'IK(a') = iLIK(a) =VL(vx-x), j=0,.. and - Y), 1 e aIL,-v' b= fl(arx-x) ,.. , F- f -1, iLIK(a) =VL(ax-x). Chapter II.TheTheoryofValuations r EH iLIK(a). , the numbersiLIK(a)obey i=0,.. ,e-1, b ED, '"' cad , cad § 10. Higher Ramification Groups 179 Let f (X) E O'[X] be the minimal polynomial of x over L'. Then f (X) _ IJVEH (X - ix). Letting u act on the coefficients of f , we get the polynomial (Q f) (X) _ fl 1(X - v rx). The coefficients of ov f - f are all divisible by a = cry - y. Hence a divides (6 f)(x) - f (x) = ±b. To show that conversely b is a divisor of a, we write y as a polynomial in x with coefficients in o, y = g(x). As x is a zero of the polynomial g(X) - y c= 0'[X], we have g(X)-y=f(X)h(X),h(X)EO'[X] Letting or operate on the coefficients of both sides and then substituting X = x yields y - cry = (o-f)(x)(Qh)(x) = ±b(uh)(x), i.e., b divides a. We now want to show that the ramification group G, (L I K) is mapped onto the ramification group Gt (L' I K) by the projection G(LIK) -} G(L'IK), where t is given by the function 71LIK : [-1, oo) -+ 1-1, cc), s dx t=r1LIK(s)= Jo(Go : Gx) Here (Go : Gx) is meant to denote the inverse (Gx : Go)-1 when -1 71LIK(s)=1 gi =#Gi. The function 71LIK can be expressed in terms of the numbers iLIK(a) as follows : (10.6) Proposition. 77LIK (s)=go EoEG min{ iLJK (Q), s + 1 }- 1. Proof: Let 9(s) be the function on the right-hand side. It is continuous and piecewise linear. One has 0 (0) = 77L I K (0) = 0, and if m > -1 is an integer and m 9'(s) =1 #{QE G Ii L I K(9) > m+2} = 1 = 17LIK(s) go (Go : G, +1) Hence 0 = 71 L I K L' I K , LL' , . . #HS t = IILIL'(S). 1 s = 7(lLIK(t). eLIL' CIO r)LIL'(1LIK(Q) - 1) > 17LIL'(S) where Chapter II. The Theory of Valuations Theory II. The Chapter 1/ILIK = lkLIL' 0 ,bL'IK t = 77LIL'(S). where #(G/H)i and 1 1 F min{iLIK(r),m} eL'IK e reH LIK satisfy the following transitivity condition: II' #GS = iLIK(Q) - I > S iL'IK(6') - 1 > 71LIL'(S) a-' E G:(L'IK), iL'IK(Or') - 1 = 1ILIL'(iLIK(a) - 1) . iL'IK(Or') - 1 = 1ILIL'(iLIK(a) iL'IK(Or') = .A. eLIK G`(LIK) := GS(LIK) GS(LIK)H/H =G,(L'IK) GS(LIK)H/H t1LIK = 1IL'IK o 1ILIL' m, and iLIK(Or) > m, so that iLIK(0r) = m. If t 0 Hm-i, > m, so that iLIK(0r) = m. If t 0 m, and iLIK(Or) '>, CSC The function 11 L I K is by definition strictly increasing. Let the inverse The function 11 L I K is by definition U' E GSH/H (10.8) Proposition. If L' IK is a Galois subextension of L I K, then (10.8) Proposition. If L' IK is a Galois of the extensions L I K Proof: For the ramification indices we have eLIK = eL'IKeLIL'. From (10.7), we obtain GS/HS = (G/H)r, t ='ILIL'(S); thus 180 (*) the formula (*), and e' = eLIL' = #Ho, (10.6) gives Since iLIK (r) = iLIL'(r) which in turn yields One defines the upper numbering function be 1LIK : [-1, co) -+ [-1, oo). of the ramification groups by The functions 11LIK and (10.7) Theorem (HERBRAND). Let L' I K be a Galois subextension of L I K of L subextension be a Galois L' I K Let (HERBRAND). Theorem (10.7) has (L I L'). Then one and H = G choose an every v' E G', we = G (L' I K). For G = G (L I K), G' Proof : Let maximal value iLIK (a) and show that preimage g E G of (L I L'), then If r E H belongs to Hm_t = Gm-i Let m = iLIK (Q). iLIK(t) we therefore and iLIK(ur) = iLIK(r). In both cases then iLIK(t) < m this gives = min{iLIK (r), m}. Applying (10.5), find that iLIK (o t) § 10. Higher Ramification Groups 181 This equation is equivalent to 71LIK(s) = 77 L'IK(t)nLIL'(s) = (nL'IK o 71LIL')'(s). As I1LIK(0) = (nL'IK c 71LIL')(0), it follows that r)LIK = 77L'IK o 71LIL' and the formula for i(r follows. The advantage of the upper numbering of the ramification groups is that it is invariant when passing from L I K to a Galois subextension. (10.9) Proposition. Let L' I K be a Galois subextension of L I K and H = G (L I L'). Then one has Gt (LIK)H/H = Gt (L'IK). Proof: We put s = 1IIL'IK(t), G' = G(L'IK), apply (10.7) and (10.8), and get GrH/H= G*LIK(t)H/H = GnLIL'(>kLlx(t)) = LIL'(s)) =G's=G'`. Exercise 1. Let K = Qp and K" = whereis a primitive p"-th root of unity. Show that the ramification groups of K,, IK are given as follows: G,, = G(K"IK) for s = 0, Gi=G(K,,IK1)for I GS = 1 for p"-1 < s. Exercise 2. Let K' be an intermediate field of LIK. Describe the relation between the ramification groups of L I K and of L I K' in the upper numbering. also servesasitsown"residue field,"i.e.,weput prime p.Foraninfiniteweadopttheconvention-that thecompletionKp We writepIifisthecharacteristicofresiduefield K(p)ofthefinite the completion.However,thiswillnowherecreateany riskofconfusion. ideal oftheassociatedvaluationring,orevenfor maximalidealof stands fortheprimeidealofringointegersK, orforthemaximal notation pIoo,thefiniteonesbytoo. isomorphic toRorC.Theinfiniteprimeswillbereferred tobytheformal K -*C.pisrealorcomplexdependingwhetherthe completion Kpis which areinducedbythepairsofcomplexconjugatenon-realembeddings which aregivenbyembeddingsr embeddings r (1.1) Definition.Aprime(orplace)pofanalgebraicnumberfieldKisa function fieldsisachieved.Thisthetasktowhichwenowturn. lacking, asthe"primesatinfinity".Inthiswayaperfectanalogywith in chapterVI.Theotheradvantageliesthefactthatarchimedean thereby enactingtheimportant"local-to-globalprinciple".Thiswillbedone chapter. Thefirstoneresultsfromthepossibilityofpassingtocompletions, advantages comparedtotheideal-theoretictreatmentgiveninfirst the valuation-theoreticpointofview.Thisapproachhastwoprominent are calledfiniteprimesandthearchimedeanonesinfiniteprimes. class ofequivalentvaluationsK.Thenonarchimedeanequivalenceclasses valuations bringintothepicturepointsatinfinity,whichwerehitherto algebraic numberfields.Wewanttodeveloptheirbasictheoryfrom CF' In thecaseofafiniteprime,letterphasmultiple meaning: italso The infiniteprimespareobtained,accordingtochap.II,(8.1),fromthe !T! coo Having setupthegeneraltheoryofvaluedfields,wenowreturnto : K-aC.Therearetwosortsofthese:therealprimes, "7L Riemann-Roch Theory K(p) :=Kp, CZ. Chapter III cad § 1.Primes : K-+R,andthecomplexprimes, when ploo. ti' vii 4U. w w fora E K fora e L*, fora E L. Chapter III. Riemann-Roch Theory III. Riemann-Roch Chapter Ialp = lral if p{co, ifp 100. resp.eglp=1. resp. (p prime number or p = oo) we put (p prime number pfp (p)fq3lp, efp { Ip K -+ R by vp:K*--*R lalp = 9Z(p)-v1(a) vp(a)=-log Iral, fp=[K(p):K(p)], gy(p) = 0. If p is the infinite prime associated to the embedding 0. If p is the infinite prime associated to lalp = Iral, fpp=[LT:Kp], (ii) M(93) = (v) IaIT = INL,IKp(a)lp (iv) vp(NLIKp(a)) = fpjpvp(a) (iii) v,,p(a) = e,,p,pvp(a) C is an embedding which defines p. C is an embedding If L I K is a finite extension of K, then we denote the primes of L by 93 If L I K is a finite extension of K, then To each prime p of K we now associate a canonical homomorphism canonical a now associate p of K we prime To each p I p For an arbitrary prime (1.2) Proposition. (i) F pjp eTjp fqijp = ETIp [LT : Kp] = [L : K], (1.2) Proposition. (i) F pjp eTjp fqijp r : K -+ C, then one finds in the class 93, when restricted and write TI p to signify that the valuations an infinite prime 93, we define the to K, give those of p. In the case of index, by inertia degree, resp. the ramification the For arbitrary primes 931p we then have furthermore if p I oo, and so that fp = [Kp : RI e as being an infinite prime This convention suggests that we consider unramifced with inertia degree 2. number, and the extension C IR as being We define the absolute value I for a 0 0 and 101 if p is real, resp. complex. 184 the p-adic finite, then vp is K* of K. If p is multiplicative group from the = Z. If p condition vp(K*) is normalized by the valuation which exponential then vp is given by is infinite, where r : K -,2 p. 69).ThisthereforeholdsforalmostallFrom(1.2)andformula(8.4) which demonstrateshowimportantitistoincludetheinfiniteprimes. § 1.Primes subgroup offractionalprincipalideals,andby chap.. II,(2.1): for KfromtheproductformulaQ,whichwasprovedalreadyin (which includesthecasep=oo,QpR),weobtainproductformula of chap.II, occur intheprimedecompositionofprincipalideal(a)(seechap.I,§11, Proof :Wehavevp(a)=0andthereforeJalp1forall{oowhichdonot and (1.3) Proposition.GivenanyaEK*,onehasIalp=1foralmostallp, number vER, (1.4) Definition.ArepleteidealofKisanelement the group account alsotheinfiniteprimes. the idealclassgroupC1KofK. where 118+denotesthemultiplicativegroupofpositivereal numbers. The normalizedvaluations In ordertounifynotation,we put,foranyinfiniteprimepandreal Let usnowextendthenotionoffractionalidealK bytakinginto We denotebyJ(o)thegroupoffractionalidealsK,P(o) f jIalp=IIFT P y..+ P PIP = C-0 NKIQ(a) =flNK,IQ,(a) JO) :=J(o)xr[1R+, 1 jTT r P PIP Pic(o) =J(o)/P(n) 11I NKvlQp(a)I p°:=e°E1R+. I fi lalp=1. p I psatisfythefollowingproductformula, PIP Pl- III p =T 1 P lI NKIQ(a)I i p = 1. !-+ 185 , , pv,(a). pica P fl p°p PI- 1 Ja I p i= 1. fl 1 plop p Chapter M. Riemann-Roch Theory M. Riemann-Roch Chapter Her =: F1 pvp a. pv. `n(p)vp p resp. (1) x ate. evp,...)E flR evp,...)E pica and x fl .... pvp .-. l(p)vp(a) = 17 p pl- 91 : J(o) --> R. i 0 : Pic(a) ---> R+ m(a) p Pic(a) = J(o)/P(o) pica pfca of x fl 1, a,° a = fl [a] = 1 1 pvp(a) = (a) x of = r[ pvp ip°p = PI- 9q([a]) = E K* we associate the replete principal ideal E K* we associate the replete principal a The absolute norm is multiplicative and induces a surjective homomor- The absolute norm is multiplicative and 186 Given a system of real numbers vp, pIoo, let flp,, pvp always denote the denote pvp always let flp,, vp, pIoo, numbers of real a system Given vector ideal R. Then every replete quantities eVp in product of the and not the representation the unique product a E J(o) admits fi oo, and vp E R for p I oo. Put where vp E Z for p is an element of of is a fractional ideal of K, and a00 so that a = of x a,. ideals time, we view a1, resp. ate, as replete r[pl,, R. At the same of J (5) the decomposition Thus for all elements P (5) of J (5). The factor group These replete ideals form a subgroup or replete Picard group. is called the replete ideal class group, a replete ideal a = fp p"p is defined (1.5) Definition. The absolute norm of to be the positive real number phism ideal [a] is equal to 1 in view of the The absolute norm of a replete principal product formula (1.3), We therefore obtain a surjective homomorphism on the replete Picard group. are conjugatesof3andthus havethesameinertiadegreef.Therefore of Lasp=(31. (iv) Theprimeidealplyingbelow and thereforeforallrepleteidealsofL. of L,by(1.2): ETIm fplpeT,p=[L index. Proof: (i)isbasedonthetransitivityofinertiadegree andramification NLIK(a) (vi) (v) Foranyrepleteprincipalideal[a]ofK,resp.L,onehas of OL,onehasNLIK(g3)OL= (iv) (iii) (ii) (i) (1.6) Proposition. These homomorphismssatisfythe Here thevariousproductsignshavetobereadaccordingourconvention. which aredefinedby an extensionfieldLareaffordedbythetwohomomorphisms § 1.Primes iMIK =iMILoiLIK. The relationsbetweentherepleteidealsofanumberfieldKandthose For achainoffieldsKCLM,onehasNMIK=NLIKoNMILand NLIK(1LIKa) = NLIK(2l) =NLIK(L)fistheidealofKgeneratedbynorms If LIKisGaloiswithgroupG,thenforeveryprimeideal43 (NLIK()) = NLIK (q3)d=pf0F1 (ii) of allaE2[f. follows from(1.2) 9Z(NLIK(3)) =Ot(pfq'I°)=(p)fTIpMP), iLIK([a]) =[a], . 93r)e,withprimeideals'3i =QiT,QiEG/Gp,which a[L:K1 NLIK(H'I3 : iLIK(l K]. (iii)holdsforanyreplete"primeideal"43 J(5K) '-J(5L), R(t)- ' .p for aEJ(OK). for 2tEJ(OL). i=1 pm resp. NLIK([a])=[NLIK(a)] i LaCG0'93' Tif in viewofthefundamentalidentity ) = NLIK ILIK 3 decomposesinthering0ofintegers = fl 1 11 - p7 7Tip PT IP i=1 rEGqj la. Pfd' II q3e3IPva o-i z3=rI6T. . or EG 187 But theargumentwhichyielded(v)appliesequallywelltolocalizations If 2.tfisaprincipalideal(a),thenof=(NLIK(a))NLIK(2if),by(v). (vi) LetofbetheidealKwhichisgeneratedbyallNLIK(a),withaE21f. v (NLIK(a))=F_s,111p If, ontheotherhand,aEL*,then(1.2)andchap.II,(8.4)implythat (v) Foranyelement 188 Div(o) toconsistofallformalsums diagrams arecommutative: (1.7) Proposition.ForeveryfiniteextensionLIK,thefollowingtwo groups ofKandL,weobtainthe replete principalideals,theyinducehomomorphismsofthePicard for allprimeidealspofo,andconsequently=NLIK(2tf) § 3,exercise4).Wethusget number ofprimeideals,andisthereforeaprincipalidealdomain(seechap.I, Cep Io,oftheextension01omaximalordersLK.0,hasonlyafinite to definethedivisorclassgroup group P(o)ofprincipaldivisors div(f)=rpf,,.vp(f)p,whichallowedus where vpEZ,and=0for almostallp.Containedinthisgroupisthe theoretic languageofdivisors.Inchap.I,§12,wedefined thedivisorgroup Since thehomomorphismsiLIKandNLIKmaprepleteprincipalidealsto Let usnowtranslatethenotionswehaveintroducedinto thefunction- iLIK ([a])=( fig NLIK([a]) = ago °G° lay coo 4.' NLIK(Hg3uP(a)) =flpvP(NLIK(a)) (af) p=NLIK((2Lf)p)NLIK(2tf)p p 1 'LIK ftNLIK K*, (1.2)impliesthatvp(a)=eq3lpvp(a).Hence CH' (o)=Div(o)/P(o). Pic(oK) Pic(15L) T pva(a)) Dvpp, fTIpv1}(a). Hence C31 = Fj PT IP pf- 1 T 93eTj,v,(a) p [L:K] > Chapter III.Riemann-RochTheory I[8+. I[8+ 1 I id `-C = 17 = [NLIK(a)1. 1 ?' mvrp(a) = [a] § 1. Primes 189 It follows from the main theorem of ideal theory, chap. I, (3.9), that this group is isomorphic to the ideal class group CIK, which is a finite group (see chap. I,(12.14)). We now extend these concepts as follows. (1.8) Definition. A replete divisor (or Arakelov divisor) of K is a formal sum D=Evpp, P where v, E Z for p { 00, vp EFfor p I oo, and vp = 0 for almost all p. The replete divisors form a group, which is denoted by Div(o). It admits a decomposition Div(5) - Div(o) x ® lRp. pica On the right-hand side, the second factor is endowed with the canonical topology, the first one with the discrete topology. On the product we have the product topology, which makes Div(5) into a locally compact topological group. We now study the canonical homomorphism div : K* Div(o), div(f) _ vp (f)p. p The elements of the form div(f) are called replete principal divisors. Remark: The composite of the mapping div: K* -> Div(as) with the mapping Div(5) -* jl R, F vpp h--3 (vpfp)ploo, plc p is equal, up to sign, to the logarithm map A:K*) 11 R,)(f)_(...,logIfIp,...), pica of Minkowski theory (see chap. I, §7, p. 39, and chap. III, §3, p. 211). By chap. I, (7.3), it maps the unit group o* onto a complete lattice T = ,l(o*) in trace-zero space H = {(xp) E j-jpl,,. R I Ipl. Xp = 0}. (1.9) Proposition. The kernel of div : K* -> Div(a) is the group µ(K) of roots of unity in K, and its image P(o) is a discrete subgroup ofDiv(?). ideal classgroupCIK=CH' (o)ofKasfollows. The kernelCH'(o)°ofthis map ismadeupfromtheunitgroupo*and Thus weobtainawell-defined continuoushomomorphism div(f) EP(o)that From theproductformula(1.3),wefindforanyreplete principaldivisor which sendsarepletedivisorD=E.vpptotherealnumber group R.Itisinducedbythecontinuoushomomorphism map ontothegroupZ; of afunctionfield(seechap.I,§14).Forthelatterweintroducedthe*degree to thequotienttopology.Itiscorrectanalogueofdivisorclassgroup CHI (o)becomesalocallycompactHausdorfftopologicalgroupwithrespect is calledtherepletedivisorclassgroup(orArakelovgroup)ofK. except 0.ThisshowsthatP(o)=div(K*)liesdiscretelyinDiv(o.). is aneighbourhoodof0inDiv(a)whichcontainsnorepleteprincipaldivisor of Pexcept0.Consideringtheisomorphisma: Therefore µ(K)is where Pis Proof: Bytheaboveremark,compositeofdivwithmapDiv(o)-* (1.10) Definition.Thefactorgroup (vp)plco f-->EplO,f"p,theset(0)xaUCDiv(o)®ploollp= exists aneighbourhoodUof0in A: K* fpI00R, Epvpp 190 As P(o)isdiscreteinDiv((5),andthereforeparticularclosed, deg(div(f)) _rlog'l(p)vl(f)=log(rjIfIpI) CWT flploo a completelattice deg(D) =T_vploggy(p)log(j-j'7t(p)1P). N . By r-* 1 the kernelofdiv. (vp fp)pl,,yields,uptosign,thehomomorphism chap. I,(7.1),thelatterfitsintoexactsequence CH' (o)=Div(o)/P(o) for CHI(o)weobtainadegreemapontothe deg :CH'(o)-->R. p p deg :Div(o)--)R (K) in trace-zerospaceH_c l lploo R whichcontainsnoelement Chapter III.Riemann-RochTheory Since Pis p a.+ p Fjploo R-->®plooRp, 0, a lattice, = 0. 1 lploo coo, there `BCD R. 191 Oil 0. 1 > P(o) - -) 0 xp = 0}. There is an exact xp = 0}. > Div(o) -* 0. rpl 1 'P (5) Div(o) - Div(o) - 0, Div(o) - Div(o) > ) Div(o)° a fp p. Restricting to Div(o)° yields the exact fp p. Restricting a p l- 1 0 -- fl 0-fHIP -kCH'(n)°-) CH1(o)-- 0. -kCH'(n)°-) 0-fHIP 0 -) H/P --) CH'(o)° -- p CH1(o) -- 0 -) H/P --) CH'(o)° -- p CH1(o) 0 --) H/J,.(o*) -- CH1(5)° --a CH1(o) --p 0. 0 --) H/J,.(o*) -- CH1(5)° --a CH1(o) 0 -) ;.(o*) 0 -f H The two fundamental facts of algebraic number theory, the finiteness of The two fundamental facts of algebraic § 1. Primes § 1. in of units lattice the complete denote T = X(o*) Let Proposition. (1.11) [1p,OOR I space H = {(xp) E trace-zero sequence Proof: Let Div(o)° be the kernel of deg : Div(o) -* R. Consider the exact be the kernel of deg : Div(o) -* R. Proof: Let Div(o)° sequence where a((vp)) commutative diagram III, §3, (3.3)), this gives rise to the Via the snake lemma (see [23], chap. exact sequence the class number and Dirichlet's unit theorem, now merge into (and are the class number and Dirichlet's unit that the kernel CH' (n)° of the in fact equivalent to) the simple statement degree map deg : CH1(o) -* R is compact. is compact. (1.12) Theorem. The group CH' (5)0 Proof: This follows immediately from the exact sequence Proof: This follows immediately from As T is a complete lattice in the R-vector space H, the quotient H/P is a compact torus. In view of the finiteness of CH1(o), we obtain CH 1 (6)0 as the union of the finitely many compact cosets of HIP in CH1(5)°. Thus CH1(oo)° itself is compact. These aretopologicalisomorphismsonceweequip by thetwomutuallyinversemappings We obtainthefollowingextensionofchap.1,(12.14). Pic(5) =J((5)/P(n)isalocallycompactHausdorfftopologicalgroup. a waythatP(a)becomesdiscretesubgroupofJ(o)by(1.9),and Replete principalidealscorrespondtorepletedivisorsinsuch The minussignhereismotivatedbyclassicalusageinfunctiontheory. as topology onfPI,,,,R.TheimageofadivisorD=Pvppisalsowritten with theproducttopologyofdiscreteonJ(o)andcanonical 192 and wegetacommutativediagram obviously relatedbytheformula on thegroupDiv((5)thereisdegreemapdeg:Div(o) -+R.Theyare topological isomorphism (1.13) Proposition.Themappingdiv with exactrows.(1.12)now yields the The correspondencebetweenrepleteidealsanddivisorsisgiven On thegroupJ(o)wehavehomomorphism91: J(n) ->R*,and div :J(o)-)Div(d), 0 0 - Div(5) --J(o), Picl (o)° CH1(5)° deg(div(a)) =-log97(a), div :Pic(o) J(a)=J(o)X r1R+ o(D) =fjp-",. > Picl(o) divl = p CH,,,lll1(5) + CHl(o). div(fl p"P) : PI- 'ti Chapter III.Riemann-RochTheory J(o) -*Div(o)inducesa P p vPp 1--flp-", deg -logl ;JJJ][8 R* F -vpp, P P > ) 0 0 different innature.Thecontroversyhasfinallygivenwaytotherealization the valuation-theoreticnotionofprimes.Boththeoriesareequivalentin controversy persistedbetweenthefollowersofDedekind'sideal-theoretic Throughout thehistoricaldevelopmentofalgebraicnumbertheory,a is compact. (1.14) Corollary.Thegroup § 1.Primes ideals. Theirreduciblesubschemesandtheirformallinearcombinations, the subjectofChowtheory.Vectorbundlesarebydefinitionlocallyfree K-theory. FromthesecondisconstructedaseriesofgroupsCH'(X), constructs aseriesofgroupsKi(X)whichconstitutethesubjectalgebraic Z-scheme X,onestudiesonthehandtotalityofallvectorbundles, in higherdimensionalarithmeticalgebraicgeometry.There,onan an importantmathematicalprinciple.However,allthisbecomesevidentonly own properworld,andthattherelationbetweentheseworldsisexpressedby that neitherapproachisdominant,eachoneinsteademanatesfromits sense oftheaboveisomorphismresult,buttheyarealsofundamentally approach, andthedivisor-theoreticmethodofbuildinguptheoryfrom fort numbers, forpES.Theneverye>0,thereexistsanx Ksuchthat let pobeanotherprimeofKwhichdoesnotbelongtoS.Let apEKbegiven Exercise 1(StrongApproximationTheorem).LetSbeafinite setofprimesand into aseminalmathematicaltheory(forfurtherreading,see [13]). groups CH`(X)andKi(X).Thustheinitialcontroversy hasbeenresolved group extendstothegeneralsettingasahomomorphicrelationbetween and divisors.Theisomorphismbetweendivisorclassgroupideal i.e., thecyclesofX,aretobeconsideredasgeneralizationsprimes ox-modules. InthespecialcaseX=Spec(o)thisincludesfractional and ontheother,thatofallirreduciblesubschemesX.Fromfirst,one the rule Exercise 4.LetKandLbenumber fields,andletr:K->Lbeahomomorphism. Exercise 3.Showthattheabsolute norm07:Pic(7L)-R+isanisomorphism. nonempty subsetofHom(K,R).Thenthereexistsaunits Ksatisfyingre>1 Exercise 2.LetKbetotallyreal,i.e.,K.=IRforallplop. LetTbeaproper The precedingisomorphismresult(1.13)invitesaphilosophicalreflection. Given anyrepletedivisorD= S and0 fix -app t,(D) =1(Fvxt.fsip)p, o.' S. P v TofL,definearepletedivisor ofKby TIP 17' .C. -t, '.1 193 This impliesdax1Eoandthus daxE0. satisfy thesystemoflinear equationsY' a egino.Indeed,ifx=x1a1 +- and d=det(Tr(a;aj))itsdiscriminant, thenad*9tc_0foreverynonzero It isagainafractionalideal. Forifa1, the dual0-module 'i+ (see chap.I,§2).Itallowsustoassociateeveryfractional ideal2lofL form ontheK-vectorspaceL,viz.,trace from thefactthatwearegivenacanonicalnondegenerate symmetricbilinear extensions Throughout thissectionweassumesystematicallythat theresiduefield with fieldoffractionsK,andlet0c_LbeitsintegralclosureinL. an extensionofvaluedfields. fields. Thisgoalleadsusnaturallytothenotionsofdifferentanddiscriminant, number fieldswhichshowsthecompleteanalogywiththeoryoffunction homomorphism where eq,denotestheramificationindexofToverK.Showthatr*inducesa L bytherule Exercise 5.GivenanyrepletedivisorD=Y_pvppofK,definea that rinducesahomomorphism where fTlpistheinertiadegreeofq3overrKandg3Ipsignifiesrp=g3JLK.Show as weshallexplainin§3and7.Theycontroltheramificationbehaviourof Exercise 6.Showthatr.or*=[L:K]and 194 Let LIKbeafiniteseparablefieldextension,ocDedekinddomain It isourintentiontodevelopaframeworkforthetheoryofalgebraic deg(r*D) =deg(D), of 01oareseparable.Thetheorythedifferentoriginates § 2.DifferentandDiscriminant *2[ _{xEL r* :CH'(oK)-->CH'(oL). r.: CH'(oL)-CH1(oK)- r*(D) _EY'vpepT, T (x,y)=Tr(xy) p Tip deg(r*D) =[L:K]deg(D). -I- xnaE*91,withxtK,then theaxi Tr(x 9[)Col. j=1 axeTr(ajaj)=Tr(xaaj) Eo. ... ,a,e Chapter Ill.Riemann-RochTheory 0 isabasisofLIK CID 5'C) 195 21 -a o extends uniquely to a uniquely to 21 -a o extends Tr(xy)). The module dual to 0, Tr(xy)). The module 1 Olo' x H (y H Tr(xy)) . x H (y H (y I Jolo = K in view of 2(K = L, we may consider of 2(K = L, we K in view CCIA = C-CIBCB IA *O = Horn, (0,o), olo=*O={x ELI Tr(xO)Co} olo=*O={x ELI HomK (L, K), x *21 -* Homo(2(, o), *21 -* Homo(2(, °F' o°. (i) Let A = o C K, and let B C L, C CM be the integral closure If 3I p are prime ideals of 0, resp. o, and OT I o, are the associated If 3I p are prime ideals of 0, resp. o, The notion of duality is justified by the isomorphism justified duality is of The notion an integral ideal of L. We As Col, D 0, the ideal Oolo C 0 is actually § 2. Different and Discriminant and Different § 2. f : every o-homomorphism Indeed, since f : L -- K-homomorphism *2( with the image of of HomK (L, K), namely, o) as a submodule Homo (2t, respect to L a distinguished place in this theory. will obviously occupy fractional ideal (2.1) Definition. The different. Its complementary module, or the inverse is called Dedekind's inverse, is called the different of 010. and O. (2.2) Proposition. I~' Proof : of o in L, resp. M. It then suffices to show that will frequently denote it by 3'7. LIK, provided the intended subrings o, 0 'are will frequently denote it by 3'7. LIK, provided way, we write CLIK instead of Lrolo. evident from the context. In the same mariner under change of rings o The different behaves in the following has OMIK = OMILT)LIK (i) For a tower of fields K C L C M, one o, one has 3JS-iols-io = S-132, olo. (ii) For any multiplicative subset S of (iii) completions, then The inclusionDfollowsfrom Therefore TrLIK,(xy)Eop.Thisshowsthat(EoloCCos,iop. then belongstoop,sinceTrLIK(xr7)EoCbutthesameistrueof to 0withrespectvip',for93'1p,q3'#',T.Theleft-handsideoftheequation (ii) istrivial. This doesindeedimplyLEBI1ATCIAcTCIB,andso so thatTr In viewofBC=C,theinclusionCisderivedasfollows: 196 time asanidealof0(i.e., theidealon2T),then(2.2),(iii)givesus toloOsp =(EopIop,andso37, ol0O'= these elementsarecloseto0.ThereforeTrLIK(y)E opnK=o,i.e., since TrL,,IKp(xy)Eop.LikewiseTrLIKp(y)opfor 93'193,because for q3' to xwithrespectvip,andsufficientlyclose0 respecttovqy, elements TrLIKp(xrj)becausetheyareclosetozero with respecttovp. allows ustofindanrlinLwhichiscloseywithrespectvj,and (see chap.II,(8.4)).LetXEL"oloandyOp.Theapproximationtheorem (tolo isdensein(to,4,Iop.Inordertodothis,weusetheformula (iii) By(ii)wemayassumethatoisadiscretevaluationring.Weshow (2.3) Corollary.0=FITOT. E (tolo.ThisshowsthatCoioisdenseinCO.,(,,, otherwords, If weput2=OLIKandOT = If ontheotherhandxECo,4,iop,andif ,--L T,then jL( CIAC)C(EBIA,andthus TrMIK(('-CIAC) =TrLIK(BTrMIL((-CIAC))CA, TrLIK(xr7) =TrL,,IKp(xil)+F_TrL.IKp(x>)) E Colo.Infact,ify0,thenTrLS,IKp(y)op, qtr °.o TrLIK =ETrL,IK, °'i Q'-B11ATrMIL(1rCIAC) CB. CDT ... Tip Chapter III.Riemann-RochTheory TrM and considerDTatthesame °o' E Lissufficientlyclose CIA CT-CIBCBIA .., I B C LOO A. mss, In thespecialcasewhere0=o[a]wethenobtain: of theelementaby f (X)Eo[X]betheminimalpolynomialofa.Wedefinedifferent which wasDedekind'soriginalwaytodefine § 2.DifferentandDiscriminant For ifa1, The dualbasisof1,a,...,0-1withrespecttoTr(xy)isthengivenby and Proof :Letf(X)=ao+-alX+ (2.4) Proposition.If0=o[a],thenthedifferentisprincipalideal Considering nowthecoefficientofeachpowers X,weobtain form roots al,...,an,soisidenticallyzero.Wemaywritethisequationinthe as thedifferenceoftwosidesisapolynomialdegree I ofKpdefinestheprime93. an_ 4 b-lx0Cb-to[a]=*o[a] r EGp=G(KplKp). ai-2 + Chapter III.Riemann-RochTheory -fl .. +an-i+t, .4: . o., ago TrLIK(xo[aJ) C0), Q.,x EbOLI'K. zero of1M=f(X)modp, sothatf'(a)EO*andthuss=0e-1. s =vp(f'(a)).AssumeLIK is unramified.Thena=mod9,3simple a suitableEO.Letf(X)be theminimalpolynomialofa.(2.4)saysthat ring withmaximalidealp. Then, bychap.II,(10.4),wehave0=o[a]for Proof: By(2.2),(iii),wemay assumethatoisacompletediscretevaluation ramification indexof9,3overK.Thenonehas TIOLIK (2.6) Theorem.Aprimeideal',3ofLisramifiedover Kifandonly L IKasfollows. Therefore vT(SLIK(a))=vT(]Jr#1(a-ra))vcp(SL1,1K,(a)),asrequired. since Ia-viaI elements inGp.Butnow where vrunsovertheK-embeddingsdifferentfrom1,andrijarecertain Furthermore, where rrunsthroughtheKp-embeddingsLT-Kpdifferentfrom1. we haveOp=o,[a].Now one thenfindsK(a+r°y)=K(a+iri`y)K(y)).Sinceaiscloseto modify bya+ir°y i =2,...,r,areverysmall.WemayevenassumethatLK(a)(ifnot, theorem, wenowpickanaE0suchthat which areconjugateoverop/pzeroornot.)UsingtheChineseremainder (Choose a=1,resp.0,accordingastheresidueclassesrfmod'j3 § 2_DifferentandDiscriminant Let 9,33bethemaximalpowerof9,3dividingOLIK, andletebethe The differentcharacterizestheramificationbehaviourofextension r~7 Ia -rijvial=Irij'a-vial=Irijla-a+a-vial=1, SLIK(a) = s =e-1, e r361 if 9,3iswildlyramified. if q3istamelyramified, I a-, i=2 j `'J and I uia-1,for 199 0 Fori =0,...,e-1,wefind is anEisensteinpolynomial.Letuslookatthederivative be chosentoaprimeelementof0.Inthiscasetheminimalpolynomial unramified extensionandassumethatLIKistotallyramified.Thenamay 200 of commutativerings,considerthehomomorphism characterization, whichisduetoE.KAHLER.ForanarbitraryextensionBIA into higherdimensionalalgebraicgeometry,isbroughtoutbythefollowing e PO' e -1, yi dxi.Thelinkwiththedifferent isnowthis. da=0 foraEA. d(xy) =xdy+ydx, F1. d:B-) SBBIA (7.5), wemaynowpasstothemaximal 01,wededucethat l)alae-2 + ... +ae-I. ao =1, i)aiae-i-I)} 201 different and the .-, the annihilator of the 0- of the the annihilator forall yc0}. ..O S,i. On the other hand, (fol, is a the principal ideal generated by S-'bolo. We may therefore assume that o is is 0olo={xE0Ixdy=0 DLIK = NLIK (ILIK) d(pai, -,Pan) =NLIK(,8)2d(al, ...,a,,). (2.10)). So we have ZLIK = (d(al, ... , an)). Dedekind's (2.10)). So we have ZLIK = (d(al, We will frequently write ZLIK instead of aolo. If al, ._a, a is an We will frequently write ZLIK instead A most intimate connection holds between the A most intimate (2.9) Theorem. The following relation exists between the discriminant and (2.9) Theorem. The following relation the different: of o, then clearly Ds-iols-'o = Proof: If S is a multiplicative subset S-1Zol0 and 2s-,ols-io = o is a principal ideal domain, so is 0 a discrete valuation ring. Then, since it admits an integral basis al, ...,an (see chap. I, § 3, exercise 4), and (see chap. I, complementary module bolo is generated by the dual basis a', ... , an which is characterized by TrL I K (ai principal ideal (P) and admits the o-basis Pat, ..., Pan of discriminant (2.8) Definition. The discriminant c0olr, is the ideal of 0 which is generated (2.8) Definition. The discriminant c0olr, all the bases a , ... , an of L I K which by the discriminants d (a 1 , ... , a,) of are contained in 0. integral basis of 01o, then tLIK bases contained in 0 are transforms d(a1, ... , an) = dLIK, because all other in o. The discriminant is obtained of the given one by matrices with entries I K (see § 1). from the different by taking the norm NL is any commutative A-algebra and B' = B ®A A', then it is easy to see A-algebra and B' = B ®A A', then is any commutative is preserved ®A A. Thus the module of differentials that SlB,IA' = S2BIA II, (10.4), that valuation ring. Then we find by chap. a complete discrete of x, then S2BtA E A[X] is the minimal polynomial B = A[x], and if f (X) f'(x). On the other (exercise 3). The annihilator of dx is is generated by dx claim. OBIA = (f'(x)). This proves the hand, by (2.4) we have o. The latter is defined as follows. discriminant of O I § 2. Different and Discriminant and Different § 2. is 001o different The Proposition. (2.7) i.e., module S2nlr,, A. If A' put 0 = B and o = clarity, let us greater notational Proof: For assume that A is and completion, and we may therefore under localization 202 Chapter III. Riemann-Roch Theory But (NLIK(B)) = NLIK(Qoly) = NLIK(DLI'K) = NLIK(DLIK)-1, and (d(al, ...,a,:)) = cZLIK. One hasd(al, ...,a,) =det((ataj))2,. d(a...... an) = det((a,aj' ))2, for 0,; E HomK(L,K), and Tr(a;ai') = 3jj. Then d(a1,... , d(a',... ,an) = 1. Combining these yields LIIK =(d(a1, (d(ai, ...,a')) = (d(fiai, ...,Pan)) = NLIKPLIK)_2OLIK and hence NLIK(OLIK) ='OLIK (2.10) Corollary. For a tower of fields K C L C M, one has DMIK = 0 1NLIK K Proof: Applying to 3ZMIK = JMILDLIK the norm N,MIK = NLIK 0 NMIL, (1.6) gives MK I. '0MIK = NLIK((MIL)NLIK(T i) = NLIK(OMIL)tJLIK Putting 0 = 1LIK and DT _ -OLIK, and viewing Zp also as the ideal D fl o of K, the product formula (2.3) for the different, together with theorem (2.9), yields: (2.11) Corollary.-0 = fp ZDp. The extension L I K is called unramified if all prime ideals p of K are unramified. This amounts to requiring that all primes of K be unramified. In fact,the infinite primes are always to be regarded as unramified because ep1p = 1. (2.12) Corollary. A prime ideal p of K is ramified in L if and only if p 10. In particular, the extension L IK is unraniified if the discriminan t 0 = (1). Combining this result with Minkowski theory leads to two important theorems on unramified extensions of number fields which belong to the classical body of algebraic number theory. The first of these results is the following. only dependondandn,the coefficientsoftheminimalpolynomiala r differentfromro,contradicting roa because ifQ(a)CKthenthe restrictionroIQ(a)wouldadmitanextension r tAro,io,onehasroa the conjugatesr0aandboaofahavetobedistinct. Since IraI<1for This aisprimitiveelementofK,i.e.,onehasK =Q(a).Indeed, INKIQ(a)I =HrIral?1impliesIroal>1;thusIm(roa) (*) (4.4)), wethusfindaEOK,00,suchthatja=(ra) EX,thatis, in KR-seechap.I,(5.2).ByMinkowski'slatticepoint theorem(chap.I, where vol(OK)isthevolumeofafundamentalmesh ofthelatticej0K convenient choiceofC,thevolumewillsatisfy where Cisanarbitrarilybigconstantwhichdependsonlyonn.Fora (see chap.I,§5)considertheconvex,centrallysymmetricsubset Choose oneofthem:ro.IntheMinkowskispace discriminant d.SuchafieldKhasonlycomplexembeddingsr:-->C. only finitelymanyfieldsKIQofdegreencontaining/--Iwithagiven only byaconstantfactor.Sowearereducedtoprovingthatthereexist Finally, thediscriminantofL( extension ofdegreem=n[K:Q]withdiscriminant(d)DKIQNKIQ(-O). if LIKisanextensionofdegreenwithdiscriminant0,thenIQ discriminant. WemayassumewithoutlossofgeneralitythatK=Q.For that thereareonlyfinitelymanyextensionsLIKofdegreenwithgiven Proof: IfLIKisanextensionofdegreenwhichunramifiedoutsideS, of divisorstheideala-fvEsp"(t+'=).Itthereforesufficestoshow then, by(2.12)and(2.6),itsdiscriminantc1LIKisoneofthefinitenumber given degreenwhichareunramifiedoutsideofS. set ofprimesK.ThenthereexistonlyfinitelymanyextensionsLIK (2.13) Theorem.LetKbeanalgebraicnumberfieldandletSafinite § 2.DifferentandDiscriminant Since theconjugates-iceof aaresubjecttotheconditions(*),which IIm(roa)I ) IQdiffersfromthediscriminantofL Idl =2'vol(OK), r ra. Izrl <1forr ro. ThisimpliesK=Q(a), Iral <1 for l-, ro,To} r 0 sothat , ro,io. 203 '[1 It followsfromthefollowingboundondiscriminant. degree anddiscriminant. 'L7 bounded regionX.Thereforethereareonlyfinitelymanyfieldswithgiven discriminant disgeneratedbyoneofthefinitelymanylatticepointsain are boundedoncedandnfixed.ThuseveryfieldKIQofdegreewith 204 lattice pointtheorem(chap.I,(4.4))asfollows.ConsiderinKRthe was exercise2ofchap.I,§5),wededucethepropositionfromMinkowski's Leaving asidetheproofofthisformulaformoment(whichincidentally Its volumeis convex, centrallysymmetricsubset Proof: InMinkowskispaceKR=[]lrC]+,rEHom(K,C),considerthe degree nsatisfies (2.14) Proposition.ThediscriminantofanalgebraicnumberfieldK inequality betweenarithmeticandgeometricmeans, that jaEXt.Asthisholdsforalls>0,andsince Xt containsonly for somes>0.Ifthisisthecase,thereexistsan a Eo,#0,such therefore holdsifandonly2'"nsn is vol(f')=.Theinequality r =jodefinedbyo.Bychap.I,(5.2),thevolumeofafundamentalmesh Given this,itremainstoprove thefollowinglemma. we obtainthedesiredresult: finitely manylatticepoints,itcontinuestoholdfors=0. Applyingnowthe The secondtheoremalludedtoaboveisinfactastrengtheningofthefirst. I CIA IdKI 0 § 2. Different and Discriminant 205 (2.15) Lemma. In Minkowski space KR = { f, C ] +, the domain Xr = {(zr) E KRI IzrI = -Ir-t,s(1)n 206 Chapter III. Riemann-Roch Thecry By induction, this implies that n(n - 1) 1(n- r -F- 1)I0's(1) In the same way, one gets Io,s(1) wt(1 - wt)2s-2dw1Io.s-1(1), 0 and, doing the integration, induction shows that 1 1 Io,s(1) _(2s) !10,0(1) = () l- Together, this gives lr,s(1) =1and therefore indeed n. 2rn.s vol(X1) = 2SVol(f (X,)) = 2s2r4-s(2Z)st"Ir,s(1)= tn. n ! If we combine Stirling's formula, nn e n!= /27rn(-) elzn,0<0<1, e with the inequality (2.14), we obtain the inequality (7r JdKI > ) 1 e2n- 6n 4 2zrn This shows that the absolute value of the discriminant of an algebraic number field tends to infinity with the degree. In the proof of (2.13) we saw that there are only finitely many number fields with bounded degree and discriminant. So now, since the degree is bounded if the discriminant is, we may strengthen (2.13), obtaining (2.16) Hermite's Theorem. There exist only finitely many number fields with bounded discriminant. 7r n/2 The expression an =n" satisfies n!( 4) an+1 >1, ()1/2(1+n)n4 i.e., an+1 > an. Since a2 = 2 > 1, (2.14) yields phenomenon is:thereare"auf3erwesentliche Diskriminantenteiler".) (See [60],chap.III,§25,p.443. TheuntranslatableGermancatchphraseforthis to holdingeneral.Counterexample: K=Q,LQ(a),a3-a22a80. words, (7LIKequalsthegcdof alldiscriminantsofindividualelements.Thisfails discrete valuationringandtheresidue fieldextension.ti.MK Show that0LIKisgeneratedby alldiscriminantsofelementsd(a)ifoisacomplete Exercise 1.Letd(a)=d(1,a,... numbers p(see[39]). in 1985bytheFrenchmathematicianJ.-M.FONTAINE:over thefieldQ,there are nosmoothpropercurveswithgoodreductionmod pforallprime in K. of genusg>1withcoefficientsinKadmitsonlyfinitely manysolutions basis forFALTINGS'sproofofthefamousMordellConjecture: of thisresultcanbegaugedbythenon-expertfromfactthatitwas proved in1983bythemathematicianGERDFALTINGS(see[35]).Theimpact good reductionoutsideafixedfinitesetofprimesS.Thisconjecturewas exist onlyfinitelymanysmoothcompletecurvesofgenusgoverKwith the RussianmathematicianI.S.JrtFAREvI(formulatedconjecturethatthere to anextensionLIKbeingunramified.InanalogyHermite'stheorem, prime pifitsreductionmodisagainasmoothcurve.Thiscorresponds over theresidueclassfieldofp.OnesaysthatXhasgoodreductionat curve X,onemaydefinethe"reductionmodp".Thisisadefined number fieldKbyallsmoothcomplete(i.e.,proper)algebraiccurvesdefined Their significanceisseenespeciallyclearlyinthelightofhigherdimensional over Kofafixedgenusg.IfpisprimeidealK,thenforanysuch analogues. Forinstance,letusreplacethefinitefieldextensionsLIKofa (2.18) Theorem.ThefieldQdoesnotadmitanyunramifiedextensions. different fromQis0A-1. (2.17) Minkowski'sTheorem.ThediscriminantofanumberfieldK § 2.DifferentandDiscriminant A 1-dimensionalanalogueofMinkowski'stheorem(2.18) wasproved Every algebraicequation These lasttheoremsareoffundamentalimportancefornumbertheory. Combining thisresultwithcorollary(2.12),weobtainthe CAA C1. ,b, CV' f(x,y)=0 for anelementaE0suchthatL=K(a). c'' is separable.Inother CIO 'C3 c°r I 207 justice toourgoalalsocouchthenumber-theoreticresultsinageometric Exercise 4.ForatowerM2LKofalgebraicnumberfieldsthereisanexact x E0,andthereisanexactsequenceofO-modules Exercise 3.ThemoduleofdifferentialsSloloisgeneratedbyasingleelementdx, Hint: If0=o[x](seechap.II,(10.4)),thensVL(SLIK(x))_E OLIK =`y,onehas field extensionXlx,andletG;,i>0,bethei-thramificationgroup.Then,if Exercise 2.LetLIKbeaGaloisextensionofhenselianfieldswithseparableresidue 208 Roch theoremstandsoutasthemostimportantrepresentative. Ifnumber the classicaltheoremsoffunctiontheory.Amonglatter, theRiemann- function-theoretic fashion,andconverselytogivearithmeticsignificance We nowhavetoaskthequestionwhatextentthispointofviewdoes theory in§1isaterminologyreminiscentofthefunction-theoreticmodel. Exercise 5.If sequence ofoM-modules The Riemann-Rochproblem istocalculatethedimension space o(D)(U)ofsectionsthesheafo(D)overUis given as of XandK(U)istheringmeromorphicfunctionson U,thenthevector i.e., anox-modulewhichislocallyfreeofrank1.IfU isanopensubset To eachdivisorD=EpEXvpPonXcorrespondsa linebundleo(D), a compactRiemannsurfaceXwiththesheafoxofholomorphic functions. going totackle. adequate waytoincorporatethisresultaswell.Thisisthe taskwearenow theory istoproceedinageometricmanner,itmustwork towardsfindingan The notionofrepletedivisorintroducedintoourdevelopmentnumber First recalltheclassicalsituationinfunctiontheory.There thebasicdatais o(D)(U) _{fEK(U)ordp(f) >-vpforallPEU}. iii is aprimitivep"-throotofunity,then .N. 0 0-rDole) -'O--),Sl1In--*O. S2LIK (9OM £(D) =dimH°(X,o(D)) § 3.Riemann-Roch v s =E(#G;-1). r.7 r =o 00 -+ S'MIK"''MIL0. Chapter M.Riemann-RochTheory ca' QE !CD y~, VL (x-vx). replaced by The formulareads H°(X, o(D))itself,butfortheEuler-Poincarecharacteristic In itsfirstversiontheRiemann-Rochtheoremdoesnotprovideaformulafor of thevectorspaceglobalsections § 3.Riemann-Roch an algebraicnumberfieldKwiththeroleofpointsaspaceXwhich explanations ofchap.I,§14andIII,1.Weendowtheplacesp module ofX,andIC=div(w)istheassociateddivisor(seeforinstance dual toH°(X,w(9of-D)),where=Q1istheso-calledcanonical is thenobtainedbyusingSERREduality,whichstatesthatH1(X,o(D)) The classicalRiemann-Rochformula and deg(D)=0,sothatX(ox)1-g;thenthisequationmayalsobe where gisthegenusofX.FordivisorD=0,onehaso(D5'=OX replete ideal replete divisor number vp(f) for p{oo,isdefinedtobetheintegervp(f),and plooitisthereal this spaceX.Theorderofthepole,resp.zerofat thepointpEX, elements fEK*willbegiventheroleof"meromorphic functions"on should beconceivedofastheanalogueacompactRiemann surface.The [51], chap.III,7.12.1andIV,1.1.3). and theset In ordertomimicthisstateofaffairsinnumbertheory,letusrecallthe More precisely,foragivendivisorD=pvpp,weare interested inthe H°(o(D)) X(o(D)) =dimH°(X,o(D)) £(D)-f()C-D)=deg(D)+1-g _ {fEo(D)f10 log IrfI.InthiswayweassociatetoeachEK*the X(o(D)) =deg(D)+1-g, div(f) _F_vp(f)pEDiv(n). X (o(D))=deg(D)+X(ox). H°(X, o(D))=o(D)(X). f EK*I o(D) =LI p div(f) >-D} p coo If lp- ..y 209 000 +, FP] Io'pX fj[CxIC]+ 2T11 fjKp, p complex x Chapter III. Riemann-Roch Theory III. Riemann-Roch Chapter lpp , 1 R flKp x till p real a ®x l ((ra)x) a ®x F-- ((tpa)x) plc., plo p complex pl- = real prime, p = pp : Kp -i-* real prime, p = pp = up : Kp -- (C. complex prime, a xzyT on KR is then transformed into µ = H !rp KR = j-j K. p xpyp + E (xpyp +xpyp). KR, ply j-jKp = KR Z plop p real = = C i p (x, Y) _ K ®, R K ®Q R -- fl Kp, K®R K®R p: K --> IR a, v : K We have the correspondences t E Hom(K, C), and the product jZplo. K. The reader will allow us to the product jZplo. K. The reader t E Hom(K, C), and 210 vpp D = 57p vp and D' = Yp divisors D between D' > the relation where no longer that H°(o(D)) is > vp for all p. Note defined to mean vp is simply Instead completely missing. of Hl (X, o(D)) is An analogue a vector space. we the size of HO(O(D)), of measuring directly the problem of attacking at the "Euler-Poincare model by looking in the function-theoretic proceed as we want defining this, ideal o(D). Before of the replete characteristic" KR = [ 11.1 between the Minkowski space to establish the relation situation in the following sketch. explain this simple isomorphisms There are the following -> Kp (see chap. II, (8.3)). They fit tp being the canonical embedding K into the commutative diagram where the arrow on the right is given by a I-± (or a, va). Via this isomorphism, where the arrow on the right is given by we identify KR with rjpl... Kp: The scalar product (x, y) The Haar measure u on KR which is determined by (x, y) becomes the product measure § 3. Riemann-Roch 211 where µp = Lebesgue measure on K. =R, if p real, AP = 2 Lebesgue measure on Kp = C, if p complex. Indeed, the system 1/,/2-, i// is an orthonormal basis with respect to the scalar product xy + l y on K. = C. Hence the square Q = {z = x + iy 0 < x, y < 1/,v/'2-) has volume A, (Q) = 1, but Lebesgue volume 1/2. Finally, the logarithm map .£:[ j l (C' ] + [ II R] +, x i__* (log I x r I) r r studied in Minkowski theory is transformed into the mapping £: flKp -* FIR,xF-> logIxPIp), PIE PI- for one has the commutative diagram K* [ H, R ]+ lIplooKp FjPlooR, where the arrow on the right, [FIR]+ = fjR x f[R x R1+ -- II R, r p o plop is defined by x H x for p p, and by (x, x) H 2x for a H p. This isomorphism takes the trace map x H El x, on [flrR] + into the trace map x H Epl,,.xp on fploo R, and hence the trace-zero space H = { x E [ TIr R] + I Fr xr = 0} into the trace-zero space H={xE IIRI E xp=01. Ploo plo In this way we have translated all necessary invariants of the Minkowski space KR to the product jlPloo Kp. To a given replete ideal a = of - a., = IZ pUPx lIpUP Ply plo0 we now associate the following complete lattice ja in KR. The fractional ideal of 9 K is mapped by the embedding j: K -- KR onto a 212 Chapter III. Riemann-Roch Theory complete lattice jaf of KR = j-jpl,, Kp. By componentwise multiplication, aoo =1 Ipl p°P = (..., evP, ...)pl. yields an isomorphism a. : KR -> KR, (xp)pl00 H (evPxp)ploo, with determinant (*) det(a0) = fl e°PfP = f `y (p)vP = 91(ac,). pl00 PI- The image of the lattice jaf under this map is a complete lattice ja := a00 jaf. Let vol(a) denote the volume of a fundamental mesh of ja with respect to the canonical measure. By (*), we then have vol(a) = 0t(a00) vol(af). (3.1) Definition. If a is a replete ideal of K, then the real number X (a) = - log vol(a) will be called the Euler-Minkowski characteristic of a. The reason for this terminology will become clear in § 8. (3.2) Proposition. The Euler-Minkowski characteristic X (a) only depends on the class of a in Pic(a) = J(n)/P(5). Proof: Let [a] _ [a]f - [a]te _ (a) x [a],, be a replete principal ideal. Then one has [a]a = aaf x [a]ooaoo. The lattice j (aaf) is the image of the lattice jaf under the linear map ja : KR --> KR, (xp)pl00 E-> (axp)pl00. The absolute value of the determinant of this mapping is obviously given by Idet(ja)l= fl Jalp = fJ 01(p)-Up(°) PI- PI- For the canonical measure, we therefore have vol(aaf) =([a]o.)-1 vol(af). Taken together with (*), this yields vol([ala) ='1([a]00a00)vol(aaf) ='1(a00)vol(af) = vol(a), so that X([a]a) = X(a). The explicit evaluation of the Euler-Minkowski characteristic results from a result of Minkowski theory, viz., proposition (5.2) of chap. I. Hence chap. 1,(5.2)thevolumeofafundamentalmeshisgivenby vol([a]a) =vol(a)and97([a]a)97(a),thatofisanintegralidealK.By Proof: Multiplyingbyarepleteprincipalideal[a]wemayassume,as (3.3) Proposition.ForeveryrepleteidealaofKonehas § 3.Riemann-Roch theorem: we deducefromproposition(3.3)thefirstversionofRiemann-Roch Observing that the degreeofrepleteidealatoberealnumber This isafinitesetbecausejH°(a)liesinthepartof latticejafcKR number fieldK,wedefine is ananalogueofH°(X,ox).Foreachrepleteideal a =r{ppvpofthe There isnoimmediateanalogueofH'(X,ox)inarithmetic. However,there Poincare characteristicandthegenusgofRiemannsurface Xinquestion: (3.4) Proposition.ForeveryrepleteidealofKwehavetheformula the dimension,weput2(a) = 0ifH°(a)0,andinallothercases which isboundedbytheconditions If 'L7 In viewofthecommutativediagramin§1,p.192,wewillnowintroduce In functiontheorythereisthefollowingrelationshipbetween theEuler- vol(a) =Xt(a,)vol(af)fl(a)\/IdKm(af) -ti v'. X (o)=dim '., H°(a)={f EK*Ivp(f)>v,forallp). deg(a) =-log9l(a)deg(div(a)) VOl(af) = x (a) vol(a) = X(o) =-log t?(a) :=log H°(X, (CD ox) -dimH1(X IdK I(o:af). I dK19't(a). V #H°(a) f10 vol(W) IdKI, e-vpfp ( c?) , ox)= . , pIoo.Astheanalogue . 1 -g. IdK 192(a) . 213 of b!) 'ZS 2r(271)S I dKI 21' (27r)s Chapter III. Riemann-Roch Theory III. Riemann-Roch Chapter #u (K) f(o) = log r 9q(a) + 0 (9t(a)1) s so that I dK I 2 (2n) vol(W) = 2r(27r)S, vol(W) = X (a) = deg(a) + E(o) - g. E(D)=deg(D)+1-g+E(1C-D), g = 40) - X (o) -log W=I(zz)EKR=[fc]+I Izrl <1}. W=I(zz)EKR=[fc]+I #H°(at) = H°(o) = µ(K), Observe that the genus of the field of rational numbers Q is 0. Using this Observe that the genus of the field of The analogue of the strong Riemann-Roch formula The analogue of the strong Riemann-Roch 214 of the set volume is the vol(W) factor the normalizing where (see the primes of K of real, resp. complex, s, is the number where r, resp. In particular, one has proof of chap. I, (5.3)). Ip = 1 for all p, so for all p, and jlp I f Ip = 1 implies If because If ip _< 1 of all roots of subgroup of K* and thus must consist that H°(o) is a finite definition of leads us necessarily to the following unity. This normalization ad hoc by the field, which had already been proposed the genus of a number ANDRE WELL in 1939 (see [138]). French mathematician field K is defined to be the real (3.5) Definition. The genus of a number number (3.4) takes the following shape: definition, the Riemann-Roch formula a of K one has (3.6) Proposition. For every replete ideal arithmetic analogue of Sen-e duality. to SERGE LANG and which reflects an This volume is given explicitly by is given explicitly This volume of Minkowski theory, which is due hinges on the following deep theorem of real, resp. complex, primes, As usual, let r, resp. s, denote the number and n = [K : Q]. ideals a = Flp p"o E J(o ), one has (3.7) Theorem (S. LANG). For replete if T(a) --> oo. Here, as usual, 0(t) denotes a function such that O(t)/t remains bounded as t -+ oo. 2ti P(o),wherec=encl. so that9t(p)"p f3. For theproofoftheoremweneedfollowing fpnp =fp(vp "'I a =11p"PI i P n gipp (5). ItthereforesufficestoshowthatBic_2tiP(5) p app =app(apo)"j1 Ifpvp p l00 of =at,iJ2(p)"P n 1 log0"t(ap))+cp = np log`i(a)+nci, p l- HR plop fpco p", n 1 log91(a)+cl .fl Eplpp xp=01. rya 40i 215 We thereforehavetocountthelatticepointsinr=ja-1cKRwhichfall over theset%i. preceding lemmaJ(a)=Uh12t1P(O),itsufficestoshow(*)foraranging #H°(a-1) and01(a)thusdependonlyontheclassamodP(5).Asby mapped bijectivelyviaxHaxontothesetH°([ala1).Thenumbers for allaEJ(a)satisfying01(a)>C.ForEK*,thesetH°(a-')is We havetoshowthatthereareconstantsC,C'such Proof of(3.7):AsO(t)=-1,wemayreplaceHO(a1)by 216 This implies as well one has As YcI'nPa=H°(a1)XandasUYEY(F+y)C P. c We considerthesets where Dp={xEKpI into thedomain with itscanonicalmeasure.Sinceof=atforaFjpp"PE91j,wehave (*) For this,weshallusetheidentificationofMinkowskispace H°(a-1)=H°(a-')U{0}={f Eaf1IIfIp<01(p) H°(a')={f Ea,'IIfIp::07(p)""forpIoo}. #H°(a') - won I#H°(a 1) X NY={yE #Y vol(F) O h'1 o}. C01(a)'_n for ploo}. .a' § 3. Riemann-Roch 217 For the set Pa = fl ,Dp, we now have vol(Pa) = fl 2171(p)"P FT 2,r fl(p)"P = 2r(2n')sO1(aoo) p real p complex (observe here that, under the identification Kp = C, one has the equation lx I p = Ix12). For the fundamental mesh F, (3.3) yields vol(F) = I dK Ifl(af' ) From this we get #H0(a 1) - 2r(27r)' IR(a) < #(X N Y). IdKI Having obtained thisinequality,itsufficesto show that there exist constants C, C' such that #(XNY)=#{yEFI (F+y)naPa54 O} 1 - Dp, ti ) 2ap(t - 2), if p real, 12-fDp,(p, 0) H ap (p cos 2 rO, p sin 27r6),if p complex, where up = 91(p)"P. We bound the norm II II of the derivative dcp(x) : R" -* lR' (x E I"). If dcp(x) = (a,k), then Ildv(x)II < nmax Iatkl. Every partial derivative of v is now bounded by 2ap, resp. 2nap. Since a r= ?t;, we have that up ='31(p)"P < cO1(a)fP/", for all ploo. It follows that II dcp(x) II < 27rn max aplf' < ci Y1(a)1/". The mean value theorem implies that cp(x) (**) jI - w(y)II < cl1R(a)'1" IIx- y II, where IIII is the euclidean norm. The boundary of Pa, aPo=U[dDpx f Dq], p q9-`p is parametrized by a finite number of boundary cubes I"-l of P. We subdivide every edge ofP'-1 intom = [' 1(a)1/"] > C' = 1 segments of and x(D)=1(D)-i(D).Putting a1=o(D),wefindby(3.7)that Proof: Theformulafor1(D) followsfromX(D)=deg(D)+1(o)-g in particular,i(D) have theformula We callthenumber The indexofspecialtyi(D)satisfies (3.9) Theorem(Riemann-Roch).Foreveryrepletedivisor DEDiv(o)we the indexofspecialtyDandget divisors. LetD=FpvppbearepletedivisorofK, of theRiemann-Rochtheorem.Wewanttoformulateitinlanguage independent ofaE2t1,asrequired. for allaE2t;with01(a)>1,whereC=2nc3isconstantwhich parts p(In-1),wedoindeedfindthat translates, andsincetheboundary8PQiscoveredbyatmost2nsuch o(I"-'), weseethatcp(1"-1)meetsatmost translates F+y.Sincetherearepreciselyma-1= mesh F.Theimageofasmallcubeunderrp is boundedbyaconstantc3whichdependsonlyonc2andthefundamental The numberoftranslatesF+y,yEF,meetingadomaindiameter #{yErI (F+y)n8PQ00} may=' #H°(a ') 2''(27r)S (,-1)112 1(D)=deg(D)+1(o)-g+i(D). 0 fordeg(D)--+oc. c1 Tt(a)'1"< i(D) =O(e° i (D)=1(D)-X g CS' 01(a) 1dKI vol(W) (1 +m(a)01(a)-1/") (n-1)1/2ct - deg(D)) and Chapter III.Riemann-RochTheory To conclude this section, let us study the variation of the Euler-Minkowski characteristic and of the genus when we change the field K. Let L I K be a finite extension and o, resp. 0, the ring of integers of K, resp. L. In §2 we considered Dedekind's complementary module (LIK = {x ELI Tr(xO) C o} = Homo(O,o). It is a fractional ideal in L whose inverse is the different',i)LIK. From (2.6), it is divisible only by the prime ideals of L which are ramified over K. (3.10) Definition. The fractional ideal CuK = (EKIQ = Homz(o,7L) is called the canonical module of the number field K. By (2.2) we have the (3.11) Proposition. The canonical modules of L and K satisfy the relation COL = (ELIKWK . The canonical module WK is related to the Euler-Minkowski character- istic X (o) and the genus g of K in the following way, by formula (3.3): vol(o) = IdK 1- (3.12) Proposition. deg (OK = -2X (o) = 2g - 2Q (o). Proof: By (2.9) we know that NKIQ(ZKI(Q) is the discriminant ideal DKIQ = (dK), and therefore by (1.6), (iii): (DKIQ)-1 Yt(wK) _ = IdK I-1, so that, as vol(o) = IdK 1, we have indeed deg (OK = -109M(a)K) = log I dK I = 2 log vol(o) _ -2x(o)=2g-2(o). genus ofL,resp.K.Thenonehas (3.13) Proposition.LetLIKbeafiniteextensionand9L,resp.9K,the Hurwitz formulaoffunctiontheory. 220 primes TofLunramified,sincetheydonotoccurintheideal QLIK phenomenon innumbertheoryweareforcedtodeclare alltheinfinite Thus thepropositionfollowsfrom(3.12). so that Proof: Sinceo)L=(tLIK1K,onehas In particular,inthecaseofanunramifiedextensionLIK: Replacing it,forinstance,by the"Minkowskimetric" with the"canonicalmetric" made. Infact,inchap.I,§5,weequippedtheMinkowski space be thecase.Itisratheraconsequenceofwell-hiddeninitial choicethatwe Investigating thematteralittlemoreclosely,however, this turnsoutnotto the coveringofRiemannsurfacesinquestion.Inordertoobtainsame analogy totheideal(ELIKtakesaccountofpreciselybranchpoints as unramified.Infact,infunctiontheorythemodulecorrespondingby we tookin§1,reallyhadnochoicebuttoconsidertheextensionCLR As forthegenus,wenowobtainfollowinganalogueofRiemann- The Riemann-Hurwitzformulatellsusinparticularthat,thedecision Thus thefactthatCIIRisunramifiedappearstobeforced bynatureitself. t((OL) _(iLIK(DK)((tLIK)= gL -f(oL)=[L: C:. deg wL=[L:K]coK+(ELIK X (°L)_[L:K](OK) (x, Y)_ (x,Y) _ExY, KR K](gK = [111c]+ 'C7 ',.y - f(oK))+ T T aTXTYT, ',bb Chapter III.Riemann-RochTheory O(WK)IL:K] tN(ELIK), deg (ELIK +-+ . NIA-' . Minkowski measurebyatilde,wegettherelations Distinguishing theinvariantsofRiemann-Rochtheorywithrespectto measures onKubelongingto(,)and(,)arerelatedasfollows: ar =1ifrT,atz § 3.Riemann-Roch The absolutenormaswellthedegreeofarepleteidealremainunaltered: to define The followingmodificationsensuefromthis.Foraninfiniteprimeponehas ramified, topute"p1p=[Lp:Kp],fsl3jp1,andinparticular such thatLT only ifoneenriches'tLIKintoarepleteidealinwhichallinfiniteprimesj3 Substituting thisintotheRiemann-Hurwitzformula(3.13)preservesitsshape (the latterincasethatH°(a)00),whereasthegenusremainsunchanged. to hold.Bythesametoken,idealcLIKhasbereplaced bythereplete in orderfortheequation The canonicalmodulewKhoweverhastobechanged: In thesamewayasin(3.13),thisyieldsRiemann-Hurwitz formula so that ideal the mathematicianUwEJANNSEN proposesasanaloguesofthefunctionfields In viewofthissensitivityto thechosenmetriconMinkowskispaceKu, tap gL -f(cL)=[L: vp(a) =-eplogItal, ,o, T(a) =IR(a), 0.w k (a)=X+log2S, K. occur.ThisforcesustoconsidertheextensionCRas VOlcanonical (X)=2sVOlMinkowski deg&K =-2y(o)2g-21(o) LIK =1 ep =[Kp:R], C3. WL = COK =WK1 .... deg(a) K](gK ''t T, changesthewholepicture.TheHaar p complex ,v, p'=e'14, e iLIK(WK). - 3Ioo !(a) =2(a)+log2' F1 1?(OK)) +2degLIK p#1 p2log2 log 91(a)=deg(a). fp =1. g321og2 `n(p) =e. . 221 where coKistheinverseof different2K The canonicalmoduleofkisdefinedtobetherepleteideal ^r7 gives theabsolutenormfl(a)_f57t(p)°P,anddegree every repleteidealaofK,thereisuniquerepresentation a=flPIP,which so thatagainlap=Ira Further, weput which weinterpretastherepleteideal(1)x(1, number and wedefinethevaluationvpofK*associatedtopby J(o) xr[R.Weput up =ar.Atthesametime,wealsouseletterpforpositivereal Let p=prbetheinfiniteprimecorrespondingtor:K-+C.Wethenput them thefollowinginvariants.Let theory. Wedenotemetrizednumberfields(K,(, precise manner,anditisoffundamentalimportanceforalgebraicnumber fields. Thisideadoesindeeddojusticetothesituationinquestionavery ar >0,a7=az,onKR.Letthesenewobjectsbecalledmetrizednumber metric ofthetype not justnumberfieldsKbythemselvesbutequippedwitha 222 Lay' pl- Y(p) =efP ep =l/ap moo =(ap degK (a)_-log'Y if pisreal,andJalp=IraI2complex.For vp(a) =-eplogIrai. (X,Y)K =>atXryr. (X,Y)K p =e°`PER* and and p )p E lalp = r fp =ap[Kp:R], (yrXryr, pica IQ ofKIQ,and Chapter III.Riemann-RochTheory , R+. . pl- fl R ) K)asKandattachto , fl(p)-up(a), 1, eaP, a`^ , p - , 1)E `"D §2, exercise4).Hencen= 1, sothatK=Q. But thereisnorealquadratic fieldwithdiscriminantIdKI=4(seechap.I, This isnotcompatiblewithI dK to ODLYZKO(see[111],table2): this canonlyhappenifn<6.Butforcaseonehassharper bounds,due bound (2.14)onthediscriminant #µ(K) =2sothatIdKI4r-1,wherenr[K:Q]. Inviewofthe Since Jr Proof. Wehave number fieldsofgenus0. (3.14) Proposition.Themetrizedfields(Q,axy)aretheonly all ofwhichhavegenusg=0.Oneevenhasthe is replacedbythecontinuousfamilyofmetrizedfields(Q,axy),aER+, does notdependonthechoiceofmetric. The genus with determinantfr because Ta:(KR,(,)K)--+(K1ei( Distinguishing theirinvariantsbythesuffixkyieldsrelations using thedefinitionsgivenabove,tometrizednumberfieldsk=(K,(,)K). § 3.Riemann-Roch Just asinfunctiontheory,thereisthennolongeronesmallestfield,butQ The Riemann-Rochtheorymaybetransferredwithoutanyproblem, gg =tk(°K)-XK(°K)Q(oK)X(OK)log g =log is transcendental,onehass=0,i.e.,Ktotallyreal. Thus Xg(°K) =-logvolk(OK) Qf(oK) = #/L(K) IdK Il/n 2 (27r) n ar, andtherefore vo1K(X) _farvol(X), IdK I log IdK > = aid i>~ #H°(OK) vo1K(W) = I1/2 3,09 1/1: =4nn 0 3 > r nn (yr n!4/ , )), (xr)H( 4,21 = £(OK)-logfar. = X(OK)-log/a-T, 4 , sowemay ln/2, 1-+ #A (K) 5,30 'ti 5 6,35 ar xr),isanisometry I dK 6 conclude thatn<2. r #/k (K) 2r (2,r),' 2r (2n)S IdK 223 Thus thefundamentalidentity of LIK,q3belongingtoaandpr=wedefinetheramification satisfy therelationaT>fiawheneverr=CrIKK.If931pareinfiniteprimes L =(L, 224 Finally wedefinethedifferentofL(Ktoberepleteideal ideals" p=e1r,93A,weput is preserved.Also93unramifiedifandonlyaT index andinertiadegreeby present book. then LET;Kpisalwaysramified.Inthiswaytheconvention foundin convention, weobtainthegeneralRiemann-Hurwitzformula where /3 where OLIKisthedifferentofLand in thecaseofrepleteideals isembeddedinamuchmorefar-reaching many textbooksisnolongerincompatiblewiththecustoms introducedinthe An extensionofmetrizednumberfieldsis If weconsideronlynumberfieldsendowedwiththeMinkowski metric, The Riemann-Rochtheorywhich waspresentedintheprecedingsection p = gL -PL(oL)=[L:K](gk-ik(OK)) )L), suchthatKCLandthemetrics OLIK =DLIK2,EJ(5L)J(nL)XfJR+, ,8 epIp =aT/fo (x,Y)K = and ap=at(93belongstoorpraIK)-Withthis iLIK(p) _flePIP, oho 0. =(f'p/ap)`p10oE111+, § 4.Metrized0-Modules T TIP EeTIpf`ii =[L:K] TIP arxrYT, and .fTIp =fiolaT[LT:Kp]. (x,Y)L NLIK(93) _pfMlc. Chapter III.Riemann-RochTheory T1- a pairk=(K,(,)K), Q Naxoya 2 1 TI- degDLlK. For "repleteprime -c1 however, proofswerenotgivendirectly,aswewilldohere,butusually Theory andGrothendieck-Riemann-Roch"taughtbyGÜNTERTAMME.There, places. Theeffectisthataleadingroleclaimedbylinearalgebra-for compactification whichisaccomplishedbytakingintoaccounttheinfinite doing so,ourprincipalattentionwillbefocusedasbeforeonthekindof algebraic varieties,andwhichwewillnowdevelopfornumberfields.In has beenconstructedbyAL.EXAI'IDERGROTHENDIECKforhigherdimensional comprehensive generalization.Thistheoryissubjecttoaformalismwhich that thetheorydisplaysitstruenature,andbecomessusceptibletomost theory whichtreatsfinitelygeneratedc-modules.Itisonlyinthissetting § 4.Metrized0-Modules b°> induces acanonicaldecompositionofrings freely gobackandforthinthesequelwithoutfurtherexplanation.Theset It admitsthefollowingtwofurtherinterpretations,betweenwhichwewill the passagefromKtoRandC,westartbyconsideringring deduced asspecialcasesfromthegeneralabstracttheory. which wereferto[151.Ourtreatmentisbasedonacourse"Arakelov and weidentifyitwithitsimage.Intheinterpretation(2), theimageofaeK The fieldKisembeddedinvia (3) Hom(X(C),C) ofallfunctionsx:X(C)-C,i.e., Alternatively, theright-handsidemaybeviewedas thesetCA' (2) (1) view theextensionCRasunramified. FinducesaninvolutiononK automorphism FEG(FIlF), inaccordancewithourdecisionof§1to by F.Thisunderlinesthefact thatithasapositionanalogoustotheFrobenius as thefunctionx(o)=aa. appears asthetuple Let Kbeanalgebraicnumberfieldand0theringofintegersK.For tin We denotethegeneratorof theGaloisgroupG(CIR)byF,orsimply .-. K °.' coo K-K®QC, a-+a®1, aa ofconjugatesa,andintheinterpretation(3) OEX(C) Kc X(C) =Hom(K,C) KC=K®QC. gyp' C, Hom(X(C),C). atzI- oX(C) zaa. 225 = 226 Chapter III. Riemann-Roch Theory which, in the representation Kr = Hom(X ((C), (C) for x : X (C) -+ C, is given by (Fx) (a) = x(o). F is an automorphism of the R-algebra Kc. It is called the Frobenius correspondence. Sometimes we also consider, besides F, the involution z i-+ z on KC which is given by z(a) = z(a). We call it the conjugation. Finally, we call an element x E KC, that is, a function x : X (C) -} C, positive (written x > 0) if it takes real values, and ifx(a)> 0for all a EX(C). By convention every o-module considered in the sequel will be supposed to be finitely generated. For every such o-module M, we put Mc=M®z C. This is a module over the ring Kr = o ®z C, and viewing o as a subring of Kr - as we agreed above - we may also write MC=M®oKc as M ®Z C = M ®o (o ®z P. The involution x i-* Fx on Kr induces the involution F(a(9 x)=a®Fx on Mc. In the representation MC = M ®z C, one clearly has F(a(9 z)=a®z. (4.1) Definition. A hermitian metric on the Kc -module Mn is a sesqui- linear mapping (,)M:M(c xMe --) Kc, i.e., a K(C -linear form (x, y)M in the first variable satisfying (x,Y)M = (Y,x)M, such that one has (x, x) M > 0 for x 0. The metric (,) M is called F-invariant if we have furthermore F(x,Y)M = (Fx, FY)M § 4. Metrized o-Modules 227 This notion may be immediately reduced to the usual notion of a hermitian metric if we view the Kc -module MC, by means of the decomposition Kc = ®a C, as a direct sum Mc=M(&oKc= ® Ma aEX(C) of cC -vector spaces Ma = M ®o.Q C. The hermitian metric (, )M then splits into the direct sum (X,Y)M = ® (XO.,ya)MQ aEX(C) of hermitian scalar products (, )Mv on the C-vector spaces Ma. In this interpretation, the F -invariance of (x, y) M amounts to the commutativity of the diagrams (') u Ma X Ma (C FxF I F Ma xMQ- .(C. (?mQ (4.2) Definition. A metrized o-module is a finitely generated o-module M with an F -invariant hermitian metric on MC. Example 1: Every fractional ideal a C K of o, in particular o itself, may be equipped with the trivial metric (x, y) = xY = ® XaYa aEX (C) on a ®z C = K ®Q C = Kc. All the F -invariant hermitian metrics on a are obtained as a(x, Y) = axY = ®a(o,)xaYa , a where a E K(c varies over the functions a : X (C) R+ such that a(a) = a(v). Example 2: Let L 1 K be a finite extension and 21 a fractional ideal of L, which we view as an o-module M. If Y(C) = Hom(L, C), we have the restriction map Y (C) -* X (C), t H t I K, and we write t l or if o= t I K For the complexification MC _ 2t ®z C = Lc, we obtain the decomposition Mc= ®CC= ® Ma, rEY(C) aEX(C) power ARM.Infact,wehavethat the tensorproductM®M',dual=Homo(M,o)andn-thexterior on the[L:K]-dimensionalC-vectorspacesMa. standard metrics where Ma=®rIaC.Misturnedintoametrizedo-modulebyfixingthe 228 standard textbooksofcommutativealgebra(seeforinstance [90],chap.IV, two, becauseoisaDedekinddomain).Fortheproof,we referthereaderto is alsoexact.Thisequivalenttoanyofthefollowing conditions(thelast F' -+F-aF"thesequence are definedbytheconditionthatforeveryexactsequenceofo-modules x=(,x)M:Mc--* Kc. Here .x, and themetricsontheseKc-modulesaregivenby (v) M=a®onforsomeideal aofoandsomeintegern>0. nonzero aE0, (iv) Mistorsionfree,i.e., themapM-+M,xHax,isinjectiveforall ideal p, (iii) Mislocallyfree,i.e.,®oopafreeor-module foranyprime (ii) Misadirectsummandoffinitelygeneratedfreeo -module, (i) Misprojective, conditions areequivalent: (4.3) Proposition.Foranyfinitelygeneratedo-module Mthefollowing §3, or[16],chap.7,§4). Among allo-modulesMtheprojectiveonesplayaspecialrole.They If MandM'aremetrizedo-modules,thensoistheirdirectsum®M', (M (DM')c=McED in thecaseofmoduleMc,denoteshomomorphism (x1 A...Axn,YtA...AYn)AnM=det((xi,Yi)M). (x,Y)M =(x,Y)M, (x ®x',Y®Y')Mom,=(x,Y)M (x ®x',Y(DY')mom,=(x,Y)M+(x',Y')M', Homo (M,F')--kF)-F") Mc = HomK,, (Mc, Kc), (x,Y)MQ =LxrYt (M(DoM')c=Mc®KcM,c, rIa (nnM)c Chapter III.Riemann-RochTheory (x',Y')M', = nnKCMc, resp. resp. 'LS 't3 resp. "CS 229 e2upo the replete ideals a such that a(or) = QEX(C) i.e., plco R+, 1 pv" 11 p" = afa. f:M-3M' 17 ptoo afc =of®zC=KC a = rk(M) = dimK(M (&o K). rk(M) = dimK(M (x,y)a=axy= ® e2vp°xvyv a : X (C) L-*L®oK=K(a®1), xH f(x)(a(9 1), L-*L®oK=K(a®1), The ordinary fractional ideals, To see the connection with the Riemann-Roch theory of the last section, with the Riemann-Roch theory of To see the connection In order to distinguish them from projective o-modules, we will henceforth we will o-modules, projective from them to distinguish In order (4.4) Definition. Two metrized o-modules M and M' are called isometric (4.4) Definition. Two metrized o-modules of a-modules which induces an isometry fc : Mc -+ MC'. a,, = flpi,,. p"p yields the function the F -invariant hermitian metric the metrized o-module thus obtained (see example 1, p. 227). We denote by L (a). are thus equipped with the trivial a,,, = 1, and in particular o itself, metric (x, y)Q = (x, y) = xy. if there exists an isomorphism gives an injective o-module homomorphism L -)- K, x H f (x), onto a o-module homomorphism L -)- K, gives an injective K. fractional ideal a c replete ideal to generalize, we observe that every which we are about o-module. In fact, the identity of K defines an invertible, metrized place defined by a : K -* C. Since where pa denotes as before the infinite obtain on the complexification pQ = p, , one has a (Q) = a (o-), and we § 4. Metrized o-Modules Metrized § 4. coherent The rank of a c-modules coherent. finitely generated call arbitrary the dimension M is defined to be o-module 910.5 invertible o-modules, L of rank 1 are called o-modules The projective The is an isomorphism. -> a, a ®a i--* a(a), them L ®o L because for or isomorphic to a are either fractional ideals of K, invertible o-modules rank 1 and a E L, Indeed, if L is projective of fractional ideal as o-modules. (iv), mapping a 0 0, then, by (4.3), a(a) =e2vpa,withvp,ER, makesofwiththemetrichintometrized for somefunctiona:X(C) -R+suchthata(Q)=(a).Puttingnow h(x, y)=(gc'(x),g(l(y))L on Kc.Itisoftheform for theunderlyingo-moduleontoafractionalidealaf. Theisomorphism because up,(a) gE an isomorphism (ii) LetLbeaninvertiblemetrizedo-module.Asmentioned before,wehave of =bf(a),thisyieldsab[a]. then impliesthata=,By,sovpµp+vp(a)for allpkoo.Inview identity is givenasmultiplicationbysomeelementaEbf1af=Homo(bf,af).The homomorphism Therefore bf-kaf,xHax,givesanisometryL(a)-(b). o-module isomorphismbf If a=b[a],thenvpµp+vp(a);thus/3y,andofbf(a)The Proof: (i)Leta=fl,pvp,b (iii) (ii) Everyinvertiblemetrizedo-moduleisisometrictoanL(a). and L(b)ifonlytheydifferbyarepleteprincipalideal[a]:=b[a]. (i) Tworepleteidealsaandbdefineisometricmetrizedo-modulesL(a) (4.5) Proposition. 230 (, )a.Indeed,viewingaasembeddedinKE,wefind=®,and Conversely, letg:L(b)-*L(a)beanisometry.Thentheo-module : L(ab) =L(a)®,L(b),L(at)L(a). Lc -*afc=KEgivesustheF-invarianthermitian metric ,3(x, y)=(x,y)b(g(x),g(y))aa(ax,ay)ay-1(x, y) (ax, ay),,=a(ax,ay) a(a) =e2 log IaaI,sothat as =®e-2vpa , P(a) = h(x, y)=axy af, xHax,takestheform(,)bto or g:L -oaf g:bf*af =ay_1(x,y) ^S' Hp p/`p, [a]=rjppup(a),andlet III e2upo (a) =Y-i , Chapter III.Riemann-RochTheory Y(a) = coo (x, y)=(x,y)b- 112 e2UVo(a) exterior powerofM,i.e., detM, aninvertiblemetrized o-module.Thedeterminantisthehighest are calledanadmissiblemonomorphism,resp.epimorphism. (aCM,)-L ismappedisometricallyontoM,'C'. M ismappedisometricallyonto (4.6) Definition.Ashortexactsequence Thus ggivesanisometryL(a)-L(a1). underlying o-moduleswhichsplitsisometrically,i.e.,in thesequence of metrizedo-modulesisbydefinitionashortexact sequenceofthe so thatgC(x)=a-1X,andthus we have to thefractionalidealof1.ForinducedKC-isomorphism isomorphism L(a) ®aL(b)=L(ab). between theo-modulesunderlyingL(a)®,,(b)and(ab)thenyields,as isomorphism (iii) Leta= isometric toL(a). (ab,a'b')ab =aPaba'b'a(a,a'),B(b,b')(a,a')a(b,b')b,anisometry o-module L(a)associatedtotherepleteideala=offl § 4.Metrizedo-Modules To eachprojectivemetrized o-moduleMisassociateditsdeterminant The homomorphismsa,,Binashortexactsequenceofmetrized o-modules The o-moduleHomy,(af,o)underlyingL(a)isisomorphic,viathe (gc(x),gC(Y))L l gC (X)(Y)=xYa-laxYa-1(Y,x)L(a), , pUP, b=fppAP,a(v)e2"Pa,(a)e2µpa.The 0-) M"-MC--)0, of ®,bf._afbf, 0-) M' gC :KC det M=ARM, N.. (a) = = aIxY(x,Y)L(aI) a-2X ,Y)L(a)=a-2(x,Y)L(a) HomKc (KC,Kc) ai n =rk(M). a(& bi and theorthogonalcomplement ) ) ab, ax), p°P, andLis 231 Proof: Letn'=rk(M')andn"rk(M").Weobtainanisomorphism sequence ofprojectivemetrizedo-modules,wehaveacanonicalisometry (4.7) Proposition.If0-+M' 232 Thus Kisanisometry. Then wehave j =1, of Kc-modules.Letx induces anisomorphism morphism Kisanisometry.Accordingtotherulesofmultilinearalgebrait since amiA m +am',+t,whereM',+1EM',isanotherpreimageofithen This mappingdoesnotdependonthechoiceofpreimages,forif,say, where mi of projectiveo-modulesrank1bymapping (K (X'(9,8x),K(y'0$y))detM=(ax'Ax,ay'Ay)detM am' A ..., , x'=AX', y'=Ay; n", andfurthermore ..., ... A ... A mn arepreimagesofmi K :detMc®KcM,c'-> amn, Aam',+,=0.Weshowthattheo-moduleiso- am',A(mi +am/n/+0Am'''' K :detM'®oM"--fM det M'®oM"-M. , y, EMc,i=1, =am' A...Aam',niniA...Ain11 = det((xi,yk)M,)det((,8xj,fyt)M") = det = (x',Y')detM'(fix,,Y)detM" M 4M"-->0isashortexact x =Ajxj, (9 fix,Y'0$Y)detM'®detM" , ... Chapter III.Riemann-RochTheory A...Aam',nmin...mmn,, 0 Yk)M' .. . m",, underfM-+W. , ... n' , Y =Ajyj. mm and xi,y;Ea2 (fixi,Yyt)M" fee ' § 5. Grothendieck Groups 233 Exercise 1. If M, N, L are metrized o-modules, then one has canonical isometrics M®oN - N®o M,(M (Do N)®oL = M®o(N(Do L), M®o(N(D L) = (M ®o N)®(M (go L). Exercise 2. For any two projective metrized o-modules M, N, one has A(Ms (DN) =SAMOAN.I I+j=n Exercise 3. For any two projective metrized o-modules M, N, one has det(M ®o N) - (detM)®rk(N) ®o (detN)®`". Exercise 4. If M is a projective metrized o-module of rank n, and p > 0, then there is a canonical isometry det(A M) (det M)®(P-!) . § S. Grothendieck Groups We will now manufacture two abelian groups from the collection of all metrized o-modules, resp. the collection of all projective metrized o- modules. We denote by (M) the isometry class of a metrized n-module M and form the free abelian group Fo(o) = ® Z{M),resp.F°(o) = ® Z[M), (M) (M) on the isometry classes of projective, resp. coherent, metrized o-modules. In this group, we consider the subgroup Ro(o) c Fo(n),resp.R°(o) S; F°(5), generated by all elements (M') - {M} + {M"} which arise from a short exact sequence 0-) M' -aM -) M"-3.0 of projective, resp. coherent, metrized o-modules. (5.1) Definition. The quotient groups Ko(o) = Fo(o)/Ro(o), resp. K°(n)= F°(5)/R°(o) are called the replete (or compactified) Grothendieck groups of o. If M is a metrized o-module, then [M] denotes the class it defines in Ko(o), resp. K°(n). 0 . M ®N and (M (D N) ®L M ®N and (M (D Chapter III. Riemann-Roch Theory III. Riemann-Roch Chapter )I. fl. K°(o) -a K°(5). [M] = [MI] + [M"]. [M] = [MI] ) M' -k M --) M"-+0 ) M' -k (m)[0) := IM 011 M') (m)[0) := IM 011 [M' ® M"] = [M'J + [M"] [M' ® M"] = [M'J moo 0-fE-->F--) M-) 0 0 0 --a M' - ) M ---f M" 0-;N®M'- ) NOM (N)((M'}-{M}+(M")) ={N®M'}-{N®M}+{N(9 M") (N)((M'}-{M}+(M")) ={N®M'}-{N®M}+{N(9 The construction of the Grothendieck groups is such that a short exact that a is such groups Grothendieck of the construction The o-module M in K°(o) its Associating to the class [M] of a projective sequence group: in the an additive decomposition o-modules becomes of metrized extending the product becomes a K°(o)-module: that N ®M by linearity, and observing ring and F°(o) right away that F°(o) is a commutative M® (N ®L), we find a generator {M'} - {M) + (M"}, the element 234 In particular, one has and K°(o) then even induces a ring structure on K°(5), The tensor product and R°(o) S F°(5) the subgroups R°(o) c F°(n) is a subring. Furthermore For if turn out to be F°(o)-submodules. metrized o-modules, and N is a is a short exact sequence of coherent is clear that projective metrized o-module, then it o-modules as well, so that, along with is a short exact sequence of metrized This is why K°(o) = F°(5)IR°(a) will also belong to R°(5), resp. R°(5). a KO (5) -module. is a ring and KO (5) = F°(o)/R°(n) is by [M]), defines a homomorphism class in K°(o) (which again is denoted We will show next that the It is called the Poincare homomorphism. The proof is based on the Poincard homomorphism is an isomorphism. following two lemmas. o-modules M admit a "metrized (5.2) Lemma. All coherent metrized of metrized o-modules in which E and F are projective. projective resolution", i.e., a short exact sequence § 5. Grothendieck Groups 235 Proof: If a t, ..., a,is a system of generators of M, and F is the free o-module F = o", then F-->M,(xi, ..., x,,) i )Ext(xt, i=l is a surjective o-module homomorphism. Its kernel E is torsion free, and hence a projective o-module by (4.3). In the exact sequence 0 ) Er ) Fc1 Mo ) 0, we choose a sections : MC --> Fo of f , so that Fc = Ec ® sMc. We obtain a metric on Fc by transferring the metric of Mo to sMc, and by choosing any metric on EC. This makes 0 -* E F - M -± 0 into a short exact sequence of metrized o-modules in which E and F are projective. In a diagram of metrized projective resolutions of M 0 -a E -k F -) M -- 0 0--) E' -k F' -* M -+0 the resolution in the top line will be called dominant if the vertical arrows are admissible epimorphisms. (5.3) Lemma. Let 0 ) E') F' M -- 0, 0 -- E" F" f M 0 be two metrized projective resolutions of the metrized o-module M. Then, taking the o-module F = F'XMF"{(x',x")EF'xF"I f'(x')=f"(x")} and the mapping f : F -- M, (x',x") H f'(x') = f"(x"), one obtains a third metrized projective resolution 0 ) E= E' x E" which dominates both given ones. 236 Chapter III. Riemann-Roch Theory Proof: Since F' ® F" is projective, so is F, being the kernel of the homomorphism F' (D F" ff M. Thus E is also projective, being the kernel of F -+ M. We consider the commutative diagram 0 - E' - F(C' f MC - 0 s T 1 0-) Ec- FFf_ Mc --a 0 S I I f 1 0 -+ E" FF =_MC 0, ,.,, where the vertical arrows are induced by the surjective projections F'T) F',F - F". The canonical isometries s':M(C ) s'Mc,s":M(C -) s"M(C give a section s:MC-) FF,sx=(s'x,s"x), of F which transfers the metric on Me to a metric on sM(C. EC = E'C x E"(C carries the sum of the metrics of E', E", so that F(c = Ec ® sMc also receives a metric, and 0) E-) F - M -) 0 becomes a metrized projective resolution of M. It is trivial that the projections E E', and E --> E" are admissible epimorphisms, and it remains to show this for the projections 7r': F -* F', ir" : F -+ F". But we clearly have the exact sequence of o-modules 0,E" `)F=F'xMF" F' 0, where ix" = (0,x"). As the restriction of the metric of F to E = E' x E" is the sum of the metrics on E' and E", we see that i: E"C -* M'(C is an isometry. The orthogonal complement of i E" in FF is the space F' x s"M r` s"a E F' x s"M '(x') = a Indeed, on the one hand it is clearly mapped bijectively onto FF,and on the other hand it is orthogonal to iE" E. For if we write x' = s'a+e', with e' E Ec, then (x', s"a) = sa + (e', 0), and then weget (y', s"b)EFC'xMcs"MMandx'=s'a+e',y's'bd',withd'Ec, Finally, theprojectionF.xMcs"Mc-i-FC'isanisometry,forif(x',s"a), where (e',0)EECandwefindthat,forallx"EC", § 5.GrothendieckGroups and associatingtotheclass{M}inF°(o)difference [F]-[E]ofthe resolution by choosing,foreverycoherentmetrizedo-moduleM,aprojective Proof: Wedefineamapping is anisomorphism. (5.4) Theorem.ThePoincarehomomorphism have thefollowingidentityin Ko(n): bottom one.ThenE-*Finduces anisometryker(a)-Zker($),sothatwe of twometrizedprojectiveresolutions ofM,withthetoponedominating first consideracommutativediagram classes [F]and[E]inKo(o).Toseethatthismappingis well-definedletus ((x', s"a),(y's"b)) [F] -[E]=[F']+[ker(l)] -[E'][ker(a)]=[F'][E']. (ix", (x',s"a)) .-n (x', s"a)=sa+(e',0), 0-+ E'-*F'-*M- 0 -aE-)F 0-) E-aF->M--)0 F = F = (s'a+e',s'bd')F'(x',Y')F1. = (a,b)M+(e,d')El(s'a,s'b)F'(e';E, =((0,x"),sa)F+((0,x"),(e',0))E =0 7r + ((e',0),(d',0))E (sa, sb) Ko(o) --fK°(n) : FO(45) F + (y', s"b)=sb+(d',0) (sa, (d',0)) KO (5) M --*0 F + ((e',0),sb) ) 0 F 237 238 Chapter III. Riemann-Roch Theory If now0--* E'-> F'-+M__> 0,and0, E"-* F"-* M-+0are two arbitrary metrized projective resolutions of M, then by (5.3) we find a third one, 0 E --> F -+ M -> 0, dominating both, such that [F'] - [E'] = [F] - [E] = [F"] - [E"]. This shows that the map,-r : F°(o) -+ Ko(o) is well-defined. We now show that it factorizes via KO (0) = F°(o)/R°(o). Let 0 -* M' -* M--*M" -* 0 be a short exact sequence of metrized coherent o-modules. By (5.2), we can pick a metrized projective resolution 0 -+ E -+ Ff)M -> 0. Then clearly 0 -+ E" -* F M" -+ 0 is a short exact sequence of metrized o-modules as well, where we write f" = a o f and E" = ker(f"). We thus get the commutative diagram 0. E , F f) M ' 0 lid I .....LLLLL n 1- 0> E" > Ff M") 0 and the snake lemma gives the exact sequence of o-modules 0) E) E"-f-> M') 0. It is actually a short exact sequence of metrized o-modules, for E is mapped isometrically by f onto M, so that E"L C E' is mapped isometrically by f onto Mc". We therefore obtain in K0(5) the identity rr(M') - ir(M) +n{M"} = [E"] - [E] - ([F] - [E]) + [F] - [E"] = 0. It shows that r : F° (5) --* K° (o) does indeed factorize via a homomorphism K°(o) -+ Ko(5). It is the inverse of the Poincare homomorphism because the composed maps Ko(o) -) K°(o) -) Ko(o)andK°(o) -) KO (5) -> K°(o) are the identity homomorphisms. Indeed, if 0 -> E --* F -* M -3 0 is a projective resolution of M, and M is projective, resp. coherent, then in K°((5), resp. K°(5), one has the identity [M] = [F] - [E]. 0 The preceding theorem shows that the Grothendieck group K0(5) does not just accommodate all projective metrized o-modules, but in fact all coherent metrized o-modules. This fact has fundamental significance. For when o-module L(a)corresponding totherepleteidealaofK,sothatweget transferred viatheisomorphism N,-CL=afo.Thusofbecomesthemetrized we restrictthemetricfrom MtoNandchooseonofthemetricwhichis of o-modules.This.becomes anexactsequenceofmetrizedo-modulesonce sequence o-module admitsasquotientafractionalidealaf,i.e., wehaveanexact Proof: LetMbeaprojectivemetrizedo-module.By(4.3), theunderlying ments [a]. (5.6) Proposition.K°(o)isgeneratedasanadditive groupbytheele- metric. corresponds theclass1=[o]ofo-moduleoequipped withthetrivial module L(a)inK°(o)by[a].Inparticular,totherepleteidealo=flp° representative, thenwehavea=b[a],forsomerepleteprincipalideal[a], [L(a)] _[L(b)].Thecorrespondenceisamultiplicativehomomorphismas and themetrizedo-modulesL(a)L(b)areisometricby(4.5),(i),sothat a repleteidealinsidetheclass[a]EPic(o).Indeed,ifbisanother Proof :Thecorrespondence[a]r-[L(a)]isindependentofthechoice into theunitgroupofringKo(o). module L(a)yieldsahomomorphism (5.5) Proposition.AssociatingtoarepleteidealofKthemetrizedo- and therepletePicardgroupPic(o),whichwasintroducedin§1. advanced theory. ring K0(5),becauseonlythiscanbeimmediatelysubjectedtoamore classes inK°(5),however,canactouttheirimportantrolesonlyinsidethe dealing withprojectivemodules,oneisledveryquicklytonon-projective modules, forinstance,totheresidueclassringso/a.Thecorresponding § 5.GrothendieckGroups In thesequel,wesimplydenoteclassofametrizedinvertibleo- The followingrelationshipholdsbetweentheGrothendieckringK°(o) ,a..- 9.2 ... [L(ab)] =[L(a)®oL(b)][L(a)][L(b)]. Pic(a) --)Ko(d)*, 0-) N-)M ,-: O0) ,Q, [a] E--)[L(a)], ) af+0 239 d , for Y slap (cl57 Chapter III. Riemann-Roch Theory III. Riemann-Roch Chapter A', f of such functions, we consider on the Q A / = ®a(cr)x,Ya [a] + [6] _ [a6] + 1. axy 8 Y ([a] - 1)([b] - 1) = 0. ([a] - 1)([b] - 1) = [Ml = [at ] + ... + [a,.] . [Ml = [at \ ,fl 0--) af-+ afED 6f-) 6f- 0. we write ' )bY(afbfA).GivenanYtwomatricesA= (x ®Y,x'(D y')A =axx'+yxy'+Syx'+,BYY. Y' j61 1 a' Y Let a and b be fractional ideals of K. We have to prove the formula Let a and b be fractional ideals of K. The elements [a] in K0(o) satisfy the following remarkable relation. the following [a] in K0(o) satisfy The elements another, because if necessary we may pass to replete ideals a' = a[a], another, because if necessary we may = afa, bf = bfb without changing the 6' = b[b] with corresponding ideals a' the o-module af, when metrized classes [a], [6], [a6] in Ko(o). We denote of ®6f, metrized by (, )A)an by ax/y`, by (af,a), and the o-module A = i y Y A'= if [(af (D 6f), A] = [(af ® bf), A] in Ko(o). We now consider the canonical exact sequence (5.7) Proposition. For any two replete ideals a and 6 of K we have in For any two replete ideals a and 6 (5.7) Proposition. K0(5) the equation us consider on the For every function a : X (C) -+ C let Proof (TAMME): ®,,X(C) C the form Kc-module Kc = For every matrix A = KC -module Kc ® Ko the form on Kc, resp. on Kc ® Kc, if ax, resp. ()A, is an F -invariant metric = a(5)) and a(a) E R+, resp. if all and only if a is F-invariant (i.e., a(a) a(o), 8(a) E R+ and S = y, and if the functions a,y,S are F-invariant, assume this in what follows. moreover det A = a,8 - y j7 > 0. We now ideals relatively prime to one We may assume that of and b f are integral 240 for shows that rank on the Induction in Ko(o). [N] + [a] [M] = identity a decomposition M, there is metrized o-module every projective a +bEKr_®KCsuchthat Indeed, intheexactsequence we Once weequipof®bfwiththemetric(,)AwhichisgivenbyA=(Y- §5. GrothendieckGroups This showsthat(*)isashortexactsequenceofmetrizedo-modules,i.e., on Vintothemetric8x,where3isdeterminedbyrule for allxEK(c,sothat and theorthogonalcomplementVofKc®{0}consistsallelements the restrictionof (*) Applying thesameproceduretoexactsequence0 Choosing a f-*0andthemetric(Y 0 S ={7r-t(1)n-1(1))A((-y/a)1®1,(-y/a)1 the followingexactsequenceofmetrizedo-modules: P =aaZ -Y--ya+fl_fl- by ,B+Y7,weget YY (af, a) 0 -Kc V -* KC,(-y/a)b®bF--)b,transfersthemetric(,) (xED 0,aEDb)=axa+yxb=0, /3'=P+YY, a'= CY )A toK(C®{0}yieldsthemetricaxyonKc, CY Y (af ®bf,A)-(bfl fj+Ya I CaYfit ,) Y Kc (D-±KC-f0, 01 on of®bf,weobtain ;')-(O C0 YY II? a a,8 .CD Ya ). ICI YY) bf -+of®- (P 1)A , 0. a Yobtain 241 A 242 Chapter III. Riemann-Roch Theory makes the matrices on the left equal, and yields a8 (a 0) (fi+vv 0 OCL Ya or, if we put 8 = ,B + , (0 S) ti (0 8 which is valid for any F-invariant function 8 : X (C) -+ R such that 8 > This implies furthermore ,0) 0) for any two F-invariant functions 8, s : X (C) -* I1 .For if K : X (C) -* R is an F-invariant function such that K > 8, x > E, then (**) gives (0 C (0 8)x(0K) - Now putting 8 = and s = 1 in (* * *), we find [(af, a)] + [(bf, fi)] = [(af, ak)] + [bf]. For the replete ideals a = Hp PIP, b = rjp PIP, this means (1) [a] + [b] _ [abj + [bf], for if we put a(a) = e2DPa,P(or) = e2uPQ 'then we have (af,a) =L(a), L(b),(af,ap) = L(ab.) On the other hand, we obtain the formula (2) [a] + [bf] = [abf] + 1 in the following manner. We have two exact sequences of coherent metrized o-modules: 0 -* (afbf,a) (af,a) af/afbf -+ 0, 0 -± (bf, 1)-* (o, l) -* o/bf * 0. As a f and bf are relatively prime, i.e., a f + bf = o, it follows that of/afbf -) o/bf Proof: By(5.6),everyelement replete idealsofK. by theelements[a]-1,resp.([a]1)([b]1),where a,bvaryoverthe (6.1) Proposition.TheidealI,resp.I2,isgeneratedas anadditivegroup the augmentationideal. It iscalledtheaugmentationofK°(o)anditskernelI =ker(rk)iscalled rk(M) =rk(M')+rk(M"),andsork((M'}-(M){M"}) =0.Thusrkis bII zero ontheidealR°(o)andinducesthereforeahomomorphism KO(5)-+Z. sequence 0 extends bylinearitytoaringhomomorphismF°(o)-*Z.Forshortexact metrized o-modulestherank Indeed, therulewhichassociatestoeveryisometryclass(M}ofprojective homomorphism in K°(o). In viewoftheisomorphismK°(o)-K°(5),thisisindeedanidentity From (1)and(2)itnowfollowsthat and so [af/afbf] _[o/bf],andtherefore is anisomorphism,sothatinthegroupK°(o)onehasidentity § 6.TheChemCharacter C°. The GrothendieckringK°(o)isequippedwithacanonicalsurjective 5'5 [a] +[b]=[ab.][bf][abocbf]1[ab]1. [(af, a)]-[(afbf,=[(0,1)][(bf,1)], M' --?M-*M"-+0ofmetrizedo-modulesonehas § 6.TheChernCharacter 040 rk(M} =dimK(M(9oK) [a]-[abf]=1-[bf]. rk:K°(5)-f Z. E K°(o)isoftheform i=1 ''C1 243 244 Chapter III. Riemann-Roch Theory IfE I, then ni = 0, and thus ni[ai]-ni =>ni([ai]-1). The ideal 12 is therefore generated by the elements ([a] - 1)([b] - 1). As [c]([a] - 1)([b] - 1) = (([ca] - 1) - ([c] - 1)) ([b] - 1), these elements already form a system of generators of the abelian group 12. By (5.7), this gives us the (6.2) Corollary. 12 = 0. We now define grKo(n)=Z®1 and turn this additive group into a ring by putting xy = 0 for x, y E I. (6.3) Definition. The additive homomorphism c1 : Ko(o) -) I,c1(.) = - rk(4) is called the first Chern class. The mapping ch : Ko(o) -> grKo(o),ch(i) = is called the Chern character of Ko(o). (6.4) Proposition. The Chern character ch : KO (5) -k grKo(n) is an isomorphism of rings. Proof: The mappings rk and c1 are homomorphisms of additive groups, and both are also multiplicative. For rk this is clear, and for cl it is enough to check it on the generators x = [a], y = [b]. This works because c1(xY)=xy-1 =(x-1)+(y-1)+(x-1)(Y-1)=c1(x)+c1(Y), because (x - 1) (y - 1) = 0 by (5.7). Therefore ch is a ring homomorphism. The mapping Zee -+ Ko(o), n® I--> E+ n, is obviously an inverse mapping, so that ch is even an isomorphism. § 6. The Chern Character 245 We obtain a complete and explicit description of the Chem character by taking into account another homomorphism, as well as the homomorphism rk : K0(5) -+ 7L, namely det : Ko(o) -) Pic(as) which is induced by taking determinants det M of projective o-modules M as follows (see § 4). det M is an invertible metrized o-module, and therefore of the form L (a) for some replete ideal a, which is well determined up to isomorphism. Denoting by [det M] the class of a in Pic(o), the linear extension of the map (M) ra [detM] gives a homomorphism det : FO (5) ---) Pic(o). It maps the subgroup Ro(o) to 1, because it is generated by the elements {M') - (M) + {M"} which arise from short exact sequences 0-) M'---M-aM"-±0 of projective metrized o-modules and which, by (4.7), satisfy det(M) = [detM] = [detM' ® detM"] = [det M'][det M"] = det{M'} det{M"} . Thus we get an induced homomorphism det : Ko(o) -+ Pic(a). It satisfies the following proposition. (6.5) Proposition. (i) The canonical homomorphism Pic(a) -) Ko(o)* is injective. (ii)The restriction of det to I, det : I - Pic(o), is an isomorphism. Proof: (i) The composite of both mappings Pic(o) -) Ko(o)* d Pic(o) is the identity, since for an invertible metrized o-module M, one clearly has det M = M. This gives (i). (ii) Next, viewing the elements of Pic(o) as elements of Ko(o), 8: Pic(o) -) I,8(x) = x - 1, gives us an inverse mapping to det :I -+ Pic(as). In fact, one has det o 8 = id since det([a] - 1) = det[a] = [a], and 8 o det = id since 8(det([a] - 1)) = 8(det[a]) = 6([a]) = [a] - 1 and because of the fact that I is generated by elements of the form [a] - 1 (see (6.1)). .ti 3-d 0 and the : o 7L ®Pic(o) Chapter M. Riemann-Roch Theory, M. Riemann-Roch Chapter : K0(O) -> Ko(o), ch(i) = ) M"-) 0 this yields the explicit description this yields the explicit i* `d®a' I -- f Pic(a), we now obtain an obtain we now f Pic(a), I -- and : 80. 1) = gr Ko(o) --> Z ®Pic(o) gr Ko(o) ,.-> 0-) M' -aM - K0(o) `+ grK0(o) K0(o) `+ § 7. Grothendieck-Riemann-Roch (M®o0)C=M®o0®zC=MC®KcLC, det( - : K0(5) -) Ko(O) i* ch : Ko(o) -* Z ® Pic(o), ch : Ko(o) -* Z ® t'. From the isomorphism det the isomorphism From realization map note that this homomorphism is a The expert should LIK of algebraic number fields We now consider a finite extension If M is a projective metrized o-module, then M ®o 0 is a projective If M is a projective metrized o-module, 246 isomorphism and the composite Observing that the Chem character of Ko(o). will again be called group Ko (5): of the Grothendieck Chem character gives an isomorphism (6.6) Theorem. The Identifying Pic(o) with the divisor from K-theory into Chow-theory. Z ® Pic(o) as the "replete" Chow class group CH1(o), we have to view ring CH(o). Grothendieck groups of L and K. and study the relations between the Q. The inclusion i X(C) = Hom(K, C), Y(C) = Hom(L, give two canonical homomorphisms surjection Y (C) -> X (C), Or t-> (T I K, Let o, resp. 0, be the ring of integers of K, resp. L, and write Let o, resp. 0, be the ring of integers defined as follows. 0-module. As the hermitian metric on the Kc-module Mr extends canonically to an F-invariant metric of the Lc-module (M ®o O)n. Therefore M (9o 0 is automatically a metrized 0-module, which we denote by i*M. If gives usawell-defined(additive) homomorphism this iswhythecorrespondence is clearlyanexactsequenceofprojectivemetrizedo'-modules. Asbefore, metrized 0-modules,then denote themetrizedo-moduleMthusconstructedbyi*M. is clearlyguaranteedbytheF-invarianceoforiginal metric(,)M.We This givesahermitianmetricontheKc-moduleM(C,whose F-invariance metric (,M.ontheC-vectorspaceMQtobeorthogonalsum The C-vectorspacesMtcarryhermitianmetrics(,)Mi where M==M®o, Mr =M®zCwehavethedecomposition is The readermayverifyforhimselfthatthisisinfactaringhomomorphism. gives awell-definedhomomorphism Ko(o) =Fo(o)lRo(o)), This iswhymapping,intheusualway(i.e., simply extendLc-sesquilinearlytometricsinthesequenceofLc-modules o-module andthemetricsinsequence is ashortexactsequenceofmetrized0-modules,because0projective is ashortexactsequenceofprojectivemetrizedo-modules,then §7. Grothendieck-Riemann-Roch If 0-+M`-*M On theotherhand,ifMisaprojectivemetrized0-module,then automatically alsoaprojectiveo-module.Forthecomplexification °"C°. 0) M,C®KcLc->Me®KcLc+MC"®KcLc-*0. 0) M'®oO)M®oO Mc _®Mt=M, 0-) 0--fMC) M'C t C and VEY(C) Ma=M®o,QC=®Mt. M (x,Y)MQ =E(xt,Y)MT. i* M" -0isashortexactsequenceofprojective i*:K0(o)-->K0(O). : K0(O)--*Ko(o). ."J M i[i*M] 'T1 [i *Ml=[M®oOl oEX(C) rlQ tJo tJo ) M"(ga0 QEX(C) via therepresentation , andwedefinethe , ) 0 +0i 247 w^7 ,.d ti] . Chapter III. Riemann-Roch Theory III. Riemann-Roch Chapter ria / Ko(n) > Ko(5) Ko(n) KO (0) -= KO (0) KO (0) x x i*(M ®o i*N) - i*M ®o N i*(M ®o i*N) - i*M NLIK(detM) := L(NLIK(2I)) (,)Mr(, )NQ =(E(, )Mr)(')Na t*1 Ko(o) Ko(O) tic MC ®Lc (NN ®Kc Lc) = MC OK, Nc. rk(i*M) = rk(M) rk(O), det(i*M) = NLIK(detM) (9o (deti*0)x(M) an isometry of metrized K(C -modules results from the an isometry of metrized K(C -modules is M®o(N0,0) - M®,N, a®(b(9 c)Hca®b. M®o(N0,0) - The Riemann-Roch problem in Grothendieck's perspective is the task The Riemann-Roch problem in Grothendieck's Here we have rk(O) _ [L : K]. of computing the Chern character ch(i*M) for a projective metrized 0- of computing the Chern character ch(i*M) this amounts to computing det(i*M) module M in terms of ch(M). By (6.6), metrized 0-module and is in terms of detM. But detM is an invertible 0-module L(%) of a replete therefore isometric by (4.5) to the metrized ideal of K, and we put ideal 2L of L. NLI K(2L) is then a replete which is well determined by M up This is an invertible metrized o-module establish the following theorem. to isometry. With this notation we first 0-module M one has: (7.2) Theorem. For any projective metrized Proof: If M, resp. N, is a projective metrized O-module, resp. o-module, N, is a projective metrized O-module, Proof: If M, resp. there is an isometry of the o-modules. Indeed, we have an isomorphism of projective metrized underlying o-modules induces an isomorphism Tensoring with C, it That this distributivity by applying mathematical grammar. 248 diagram The Formula). (Projection Proposition (7.1) the horizontal arrows are multiplication. is commutative, where In ordertoprovethesecondequation,wefirstreduceaspecialcase.Let therefore Proof: OnehasMK:=M®,K=M®o0®oK=M®oL=:MLand § 7.Grothendieck-Riemann-Roch If 0andowereprincipalidealdomains,itwouldbeobvious. Infact,inthat For theunderlyingo-modulesthisamountstoidentity we havetoestablishtheisomorphism reduce byinductiononrk(M)tothecaseM=L(2t).Here1,so module oftheformL(21)forsomerepleteideal21.Thus(*)allowsusto the proofof(5.6),wenowmakeusefactthateveryprojectivemetrized the multiplicativityofnormNLIKandadditivityrankrk.Asin i*M -+i*M"-*0by(4.7),andtheoneonrightfrom(4.7)also, The isomorphismontheleftfollowsfromexactsequence0--*i*M'- (*) metrized 0-modules,onehas If 0-+M'-kM-M"-*isashortexactsequenceofprojective produce aprincipalidealdomainasdesiredbypassing fromO1otothe the left-handside,resp.right-handof(**).But wemayalways the left-handside,resp.right-handofwhichwould, by(1.6),generate transformation Ta:L-3L,xHax,wewouldgettheequation of 0overo.SinceNLIK(a)isbydefinitionthedeterminant det(Ta)ofthe case wecouldchooseageneratorof2tfandanintegral basiswl, which hastobeviewedasinsidedetKLandisprovedfollows. (**) O-module Mprojectsviaanadmissibleepimorphismontoasuitable0- rk(i*M) =dirK(MK)dimK(ML)dimL[L:K]rk(M)rk(O). equality (**). and sincethisidentityisvalid forallprimeidealspofo,wededucethe exercise 4).Theprecedingargument thenshowsthat localization Op(opforeveryprimeidealpofo(seechap. I,§11and3, (deto 2tf)p=detop2fpNLI K(21fp)detonOp=(NLJK(2tf)detn(9)p, X(M) =det(i*M) L+, A(M) =?(M')(9o),(M") aWl A...AaWn=NLIK(a)(WiAWn), det(i*L(2t)) =L(NL!K(2t))®odet,O. deto Qf=NLIK(2if)det,0, and p(M)=NL1K(detM)®oi*O)rk(M) and p(M) =®o p(M"). ... ,wn . 249 yields thefollowingcorollary. by linearity,andpassingto thequotientgroupKo(O)=Fo(O)/Ro(O) This provesourtheorem. we doindeedget Then, observingthat rt, ...,r0,putVk=VTTkandformthematrices Let with vp=E13IpfT,pvq.Then For this,let2(.=rj where aEHom(K,QandrHom(L,C)issuchthatIor.Wehaveto metrics onthecomponents 250 show that,for,itEdetcMQanda,bC,onehastheidentity o-modules. OnehasMr=NCLCandacKc,weconsiderthe M =L(2t),NL(O),aNLIK(2t)andweviewM,N,asmetrized Extending theformulasof(7.2)tofreeabeliangroup In ordertoprovethatthemetricsagreeonbothsidesof(**),weput = x1A...AX,,17yly,.WenumbertheembeddingsrIcr, (a (x,Y)N, =ExrYr, (a, b)aQ=e2v,,ab , brj)detM,= A =(xirk), det(D) =fleaTTofpipQ"Te"- MQ=®C> NQ=®C,aQ=cC, au= via (a = = abdet((AD)(BD)t)(detD)2det(ABt) l e2uyo ab ab ( , b17)detMo=(a,b)M a00 =NLIK(.)rj rla B =(Yirk), 93°T, sothatonegets I Fo(O) _®7G(M} , 17)detM, (x,Y)M0 = l7)det `Ipo {M} vpQ =EfTlpoVT via , r=(a, D = Chapter III.Riemann-RochTheory, 'I3lpo r1a PI- Ee2vtxryv, e°t rj)detN,, b ) 0 pVP ( , rj)detNo eV7 0 via . ,`°^, V r. 0 § 7. Grothendieck-Riemann-Roch 251 (7.3) Corollary. For every classE K0(O), one has the formulas [LK]rk(l;), det(i*I;') = [deti*0]7k( The square of the metrized o-module det i*O appearing in the second formula can be computed to be the discriminant DL1K of the extension L I K, which we view as a metrized o-module with the trivial metric. (7.4) Proposition. There is a canonical isomorphism (deti*O)®2 = DLlK of metrized o-modules. Proof : Consider on 0 the bilinear trace map T : O x O --) o,(x, y) --a TrLIK (xY) It induces an o-module homomorphism T : det O ® det O - c, given by T ((al A ... A an) ®(PI n ... A ,en)) = det(TrLIK The image of T is the discriminant ideal -OLIK, which, by definition, is generated by the discriminants d(wt, ..,w,)=det(TrLIK(wic)i)) of all bases of L I K which are contained in O. This is clear if 0 admits an integral basis over o, since the ai and fii can be written in terms of such a basis with coefficients in o. If there is no such integral basis, it will exist after localizing Op1op at every prime ideal p (see chap. I, (2.10)). The image of Tp : (det Op) (9 (det Op) -) op is therefore the discriminant ideal of Op l op and at the same time the localization of the image of T. Since two ideals are equal when their localizations are, we find image(T) = c1LIK. Furthermore, T has to be injective since (det O)®2 is an invertible o-module. Therefore T is an o-module isomorphism. 252 Chapter III. Riemann-Roch Theory We now check that Tc : (det(O)®2)C )(cDLIK)c is indeed an isometry. For O(c = 0 ®z C, we obtain the Kc -module decomposition where or varies over the set Hom(K, Q, and the direct sum Oa- ®(O®o,,Q= (DC is taken over all r E Hom(L, C) such that r I K = or. The mapping Oc -+ Kc induced by TrL K : O -* o is given, for x = ®a xa , xa E Oa , by TrLIK(x) _ Tro(xa), where Tra (x,) xa, r , the xa, r E C being the components of xa . The metric on (i*O)c = Oc is the orthogonal sum of the standard metrics (x, Y), = ExrYr =Tra(xY) r1a on the C-vector spaces (i*O)a = Oa = ®,I, C. Now let xi, yir= Oa, i = 1, ...,n,andwrite x=x1A...Ax,,,y=ylA...Ay, Edet(Oa).The map TT splits into the direct sum TT = ®a Ta of the maps Ta : det(Or) (&c det(Or) -3 (Z)LIK)a = C which are given by Tr(x 0 y) = det(Tra(xiyj)). For any two n-tuples x', y! E Oa we form the matrices A = (Tra(xtyi)), A' = (Tra(xiY1)), B = (Tra(xix')), B' = (Tra(YiYj)). Then one has AA' = BB', and we obtain (Ta(x (0 Y),Tc(x' ®Y'))("')Q =Ta(x 0Y)Ta(x' ®Y') = det(Tra (xi yj)) det(Trr (x yj')) = det(AA') = det(BB') = det(Tra(xixf)) det(Trr(yiyf)) = det((xi,xj)r) det((yi,y1)a) _ (x,x')det0,(Y,Y')detOa = (x (9 Y,X' ® Y')(detoa)®2. This shows that TT is an isometry. between thekernelsofbothrankhomomorphisms,sothatthereisa Therefore i*inducesahomomorphism because [L:K]timestherankofan0-moduleMisitsaso-module. there isthecommutativediagram language ofGROTHENOIECK'sgeneralformalism.Forthehomomorphismi* §7. Grothendieck-Riemann-Roch is commutative. (7.5) Corollary.Thediagram description ofit. It iscalledtheGysinmap.(7.3)immediatelygivesfollowingexplicit homomorphism by theToddclass,whichisdefined asfollows. which willbeprovidedviathe moduleofdifferentials(withtrivialmetric), is notcommutative.Inorder tomakeitcommute,weneedacorrection, problem. Thedifficultythatconfronts ushereliesinthefactthatdiagram the determinationofcomposite choi*ispreciselytheRiemann-Roch where theGysinmapi*onrightisexplicitlygiven by(7.5),whereas We nowsetouttorewritetheresultsobtainedin(7.2)and(7.4) We nowconsiderthefollowingdiagram coo grKo(O) grKo(5) i* :gr'Ko(O) i* Ko(O) chigrKo(6) Ko(d) ch) i. J- i* Ko(O) `k Ko (o`) i.1 : 1(O)-1(0) id ®det id®det _ rk ) gr Ko(5) ZED Pic(O) Z EDPic(n) 1 gr'Ko(a) .-r Z' z I [L:K] t [L:K]®NLIK i. 253 . E gr'Ko(O) ®Z[2] li* S21010 - 0 Chapter III. Riemann-Roch Theory III. Riemann-Roch Chapter OVA grKo(0) grKo(5) K°(O), ch )0 = 1 - i*(Td(OIo) ch(i)) Td(o1o)ch) the Todd class does not belong to the ring the Todd class does Ko(O) J- ) J L I K i* Ko(0) Ko(n) of differentials is only a coherent, and not a projective and not coherent, is only a of differentials 0 E Ko (O) , we have to show the identity Td(Olo)=1 +2([JLIK]-1)= 2(1+[OLIK]). Td(Olo) = 1 - 2ci([52o0]) The module S21 The module 2 , Because of the factor 254 via the element of K0(O) is viewed as an But its class [S2aio] 0-module. lies in I (O). rko(S2o,o) = 0, it and since the element Todd class of 0 l o is defined to be (7.6) Definition. The [ 2 ]. The module is only an element of gr Ko (O) ®Z gr Ko (O) itself, but of the extension- is connected with the different 2LIK of differentials S2o1o describe the Todd class also by the [Slo1o] = 1 - [TLIK]. We may therefore different: Poincare isomorphism L l K by the exact sequence §2, exercise 3). This implies that of 0-modules (with trivial metrics) (see using the Todd class. The main result now follows from (7.3) The diagram (7.7) Theorem (Grothendieck-Riemann-Roch). is commutative. Proof : For case ofanextensionKIQ,amounts torevisitingtheRiemann-Rochtheory realizes inthepresentcontextoldideaofviewingnumber theoryaspart metrized o-modulesasvectorbundles,theinvertible o-modulesas Thus theringoisinterpretedasschemeX=Spec(o), theprojective of geometry. of schemes,theclassSlobascotangentelement,etc. Inthiswayone line bundles,theinclusioni What isneededtheuseofgeometriclanguagefor objectsconsidered. completely intoageneralprogrammeofalgebraicgeometry asaspecialcase. But (e)isthesecondidentityof(7.3),and(f)followsfrom(7.4)(2.9). (e) reduced bythecommutativediagram(7.5)toequations rk(i,O) _[L:K].Toshow(b)and(d),weapplydettobothsidesare in grK0(o).Theequations(a)and(c)areclearbecauseofrk(i*[O])_ and it sufficestochecktheequations observing that Decomposing (f) IIo (d) (c) (b) (a) § 8.TheEuler-MinkowskiCharacteristic Considering thetheoremof Grothendieck-Riemann-Rochinthespecial With thisfinaltheorem,thetheoryofalgebraicintegerscanbeintegrated Td(O Io)ch(i)=(1+z ova § 8.TheEuler-Minkowski Characteristic (deti*O)®2 =NLIK(det`DLIK). 2c1(i*[O]) rk(i*[O]) =i.(1), det(i.) =NLIK rk(i )= = : o-+0asmorphismfY=Spec(O)X = ix([DLIK]-1) i*(c1()) +rk()ct(i*[O]) [c1() +i rk(i*[O]) and ch(i)= [det i*Oyk(g) 1)) (rk()+c1()) , cad C]. ^J. 1)], r-. and 255 Euler-Minkowski characteristic. This pointofviewleadsusnecessarilytothefollowingdefinition of rank1,itshouldbeviewedasametrized7G-module[K:Q]. higher rank.Moreprecisely,insteadofconsideringaasmetrizedo-module fundamental meshofthelatticeinMinkowskispaceKR=a®zJRdefinedby of repleteidealsaK.Here,vol(a)wasthecanonicalmeasure Euler-Minkowski characteristic of §3fromournewpointview.Atthecenterthattheorywas 256 is givenbythecompositeofhomomorphisms homomorphism Ko(a) elements [a]EPic(o),themapdegKonPic(o)determinesaunique Proof: Since,by(5.6),K0(5)isgeneratedasanadditivegroupthe and calledtheEuler-MinkowskicharacteristicoverK. extends uniquelytoahomomorphism (8.1) Proposition.Thedegreemap a. Thisdefinitionisproperlyexplainedinthetheoryofmetrizedmodules (see (1.6),(iii))givesthe formula (7.2)andusing inclusion ofthemaximalorders ofK,resp.L,thenapplyingdegKtothe If LIKisanextensionofalgebraicnumberfieldsand i We defineinwhatfollowsXK(M)=([M])forametrized o-moduleM. modules. Thenproposition(8.1)isequallyvalidforK°(O) asforK0(5). det andXKtotheGrothendieckgroupK°(o)ofcoherent metrizedo- as thecompositePic(o)vK0(5) on K0(5),andtherebyK°(5).Itisgivenby Via thePoincareisomorphismKo(o) deft NA)=-log'7t(?t)_-logOt(NLIK(2L)) degK :Pic(o)-*1R, Ko(o) IIR whichextendsdegK.Butsuchahomomorphism X (a)=-logvol(a) XK:Ko(o)-SIR I2. XK =degodet d- Pic(o) --e-g,R, a- degK ([a])_-logN(a), Pic(a) istheidentity. Chapter 1111.Riemann-RochTheory K°(5), wetransferthemaps = cad degK(NLIK(2)) : o-±0the NBA metric onMissimplygivenbyahermitianscalarproducttheC-vector Indeed, sinceQhasonlyasingleembeddingintoC,i.e.,Qc= abelian groupMtogetherwithaeuclideanmetricontherealvectorspace consider metrized7L-modules.Suchamoduleissimplyfinitelygenerated is valid,andinparticular,foraninvertiblemetrizedO-moduleM,wehave formula (8.2) Theorem.Foreverycoherent0-moduleM,theRiemann-Roch § 8.TheEuler-MinkowskiCharacteristic is givenby on I[8".Withrespecttothismeasure,thevolumeoffundamental mesh e1, ..., defines aHaarmeasureonMR.Oncewechooseanorthonormalbasis is afundamentalmeshofthelatticeM.Theeuclideanmetric(,),yt Z-basis ofM,thentheset a 1-±(91,identifiesMwithcompletelatticeinMR.If1,...a,tis is afinitelygeneratedfreeabeliangroup.ThecanonicalmapM sesquilinear extensionofwhichreproducestheoriginalmetric. space MC=MR®C.Restrictingthistogivesaeuclideanmetricthe a generator(forinstance,x =alA...a"),then metric ( metrized 7L-moduledetM. MRisaone-dimensionalR-vectorspacewith coefficients, hencehasdeterminantofabsolutevalue1. by amatrixwithintegercoefficientswhichalsohasaninverse withinteger 7L-basis al, It willbedenotedbyvol(M)forshort.doesnotdepend onthechoiceof MR -IR",x1e1+ We nowspecializetothecaseofbasefieldK=Q,thatis,we If MisaprojectivemetrizedZ-module,thentheunderlying7L-module A moreelegantdefinitionofvol(M)canbegiveninterms oftheinvertible e,t ofMR,thisHaarmeasurecanbeexpressed,viatheisomorphism , y'° ) detM,andwiththelattice MisomorphictoZ.IfxEdetis ... ,a, XK(i*M) =degL(detM)+rk(M)XK(i*O) because adifferentchoiceislinkedtotheoriginalone XK(i*M) =degt(M)+XK(i*O). Vol(M) =IIXIIdetM vol((P) =det((ai,aj))I1/2 'i7 +x,te" I 0={xlal+...+xa,I MR =M®zR. (x1, x, EI[8,0 In the present case, where the base field is Q, the degree map deg : Pic(Z) I8 is an isomorphism (see § 1, exercise 3), and we call the unique homomorphism arising from this, x=degodet:K°(76) R, the Euler-Minkowski characteristic. It is computed explicitly as follows. (8.3) Proposition. For a coherent metrized Z -module M, one has X (M) = log #Mtor - log In this formula Mtor denotes the torsion subgroup of M and M/Mtor the projective metrized Z-module which receives its metric from M via M ®R M/Mtor ®R. Proof of (8.3): If M is a finite Z-module, then the determinant of the class [M] E K°(Z) is computed from a free resolution 0-*E-->F-M-*0, where F = Z and E = ker(a) Z'. If we equip F ® JR = E ® J = IE8" with the standard metric, the sequence becomes a short exact sequence of metrized Z -modules, because M ® JR = 0. We therefore have in K ° (2): [M] = [F] - [E]. Let A be the matrix corresponding to the change of basis from the standard basis e1, ..., e of F to a Z-basis e1..... e;, of E. Then x = el A ... A en, resp. x' = e' A ... A e;,, is a generator of det F, resp. det E, and The metric IIII on det E is the same as that on det F, so that x(E) = deg(det E)-log IIx'II = -log(#MIIx II) _log #M + x (F), and then x (M) = x ([F] - [E]) = X (F) - x (E) = log #M. For an arbitrary coherent metrized Z-module M we have the direct sum decomposition M = Mtor ® M/Mtor to KR=1plooK.istherefore givenby(xp)r-(e°pxp).ThelatticeTofis extend rtoKpt.Therestriction ofthetransformationT:(xi)-*(e'p=x,) maps anelement(xp)pjooto theelement(xt)TEX(C)withxt=rxpt.Herewe In therepresentationKR=fps.Kp,canonicalembedding canonical measuredefinedby( the volumeofafundamentalmeshlatticeTf with respecttothe respect totheHaarmeasuredefinedbyeuclideanmetric onKRisthen The volumevol(i,L(a))ofafundamentalmeshthelattice ofinKRwith Equivalently, transformation where pt Proof: Leta=afaooafFIpI,,c,p'p.Themetric(,)i,L(a)ontheC-vector (8.4) Proposition.X(a)=(i,L(a)). the metrizedZ-modulei,L(a)ofrank[K:Q],weget us. Indeed,viewingthemetrizedo-moduleL(a)ofrank1associatedtoaas metrized Z-modulestowhichthedetaileddevelopmentoftheoryhasled r space Kc=fltEx(C)Cisthengivenby appears asasimplespecialcaseoftheEuler-Minkowskicharacteristicfor which wedefinedadhocin§3viatheMinkowskimeasurevol(a)'now conclude that = deg(detM/Mtor)-109114-logvol(M/Mtor).Wetherefore then x=atA...a,, into metrizedZ-modules.Ifat,...,a,,isabasisofthelatticeM/Mtor, § 8.TheEuler-MinkowskiCharacteristic : K The Euler-Minkowskicharacteristicofarepleteideala, X (M)=(Mtor)+(M/Mtor)log#Mtor-vol(M/Mtor). C. Itresultsfromthestandardmetric(,)viaF-invariant is theinfiniteplaceofKcorrespondingtoembedding T :KC-)KC, KR=K®QR-*KC=K®QC (X,Y)i*L(a) =Fe2vp`xryT, vol(i,L(a)) =vol(Taf). (x,Y)i*L(a) =(Tx,Ty). is ageneratorofdetM/Mtor;thenX(M/Mtor) X (a)=-logvol(a), , (XT)TEX(C) F±(euPTxr)TEX(C) ). Thus r coo 'l7 259 O 260 Chapter III. Riemann-Roch Theory then the same lattice which was denoted a in § 3. So we obtain vol(i*L(a)) = vol(a), i.e.,X (i*L (a)) = X (a). Given this identification, the Riemann-Roch theorem (3.4) proven in § 3 for replete ideals a, X (a) = deg(a) + X (o), now appears as a special case of theorem (8.2), which says that X(i.L(a)) = deg(L(a)) + X(i.o). The subgroup(cp)={cp"InEZ)hasthesamefixedfieldIFpaswhole Galois theorydoesnotremaintrueforinfiniteextensions.Letusexplainthis theory. Butsuchanattemptfacesthedifficultythatmaintheoremof of k.ThisiswhyitreasonabletotrygiveaprominentplaceinGalois does, however,havetheadvantageofcollectingallfiniteGaloisextensions Galois groupofk.Asarule,thisextensionwillhaveinfinitedegree.It separable closurekIk.ItsGaloisgroupGk=G(Elk)iscalledtheabsolute whenever mIn,butsuchthatthereisnointegerasatisfying a"-amodn satisfying which doesnotbelongto(pp).Wechooseasequence{a"}FENofintegers find ((p)0GFP.Inordertocheckthis,letusconstructanelementi/rEGrP of GrP.ButcontrarytowhatweareusedinfiniteGaloistheory, p elementscontainstheFrobeniusautomorphismcpwhichisgivenby Example :TheabsoluteGaloisgroupGrP(G(FpIIFp)ofthefieldFpwith in thefollowing for alln,whichiswhatweruled outbyconstruction. a EZ,wouldimply*Ir'Pn = Pan Observe thatoIIFP,,hasorderm.Thereforethei/rndefine anautomorphism If Fpmc]Fpn,then we writen=n'pVPi'l,(n',p)1,and1n'x,,+pP(")y,,. Nowput for allneN.Anexampleofsuchasequenceisgivenby a"=n'x,,,where Every fieldkisequippedwithadistinguishedGaloisextension:the of FP=Un_1Fpn.Now,lr cannot belongto((p)because1/r=qpa,for 4'. Abstract ClassFieldTheory 'm /n IFPm=Vale(PamI.PmY'm § 1.InfiniteGaloisTheory In, sothata"=ammodm,andtherefore ... *n =spanIF an -ammodm Chapter IV If, ,, Pn = E G(FpnIIFp). ,a IF x E]FP. P . and hencea"-amodn .C: 111 as abasisofneighbourhoodsor,withKIkrangingoverfiniteGalois take thecosets called theKrulltopologyandisobtainedasfollows.ForeveryorEGwe of anyGaloisextensionS2Ikcarriesacanonicaltopology.Thistopologyis have toamenditusingtheobservationthatGaloisgroupG=G(S21k) theorem ofGaloistheoryaltogetherinthecaseinfiniteextensions.Wejust 262 open setsoftheproductUKG(KIk),whereKokvaries overthefinite to The homomorphismhisinjective,becauseCrIK=1for allKisequivalent therefore acompacttopologicalspace,byTykhonov'stheorem (see[98]). groups G(K1k)asdiscretecompacttopologicalgroups. Theirproductis where KIkvariesoverthefiniteGaloissubextensions. Weviewthefinite to provecompactness,considerthemapping thus aG(S2IK)ntG(S2I=0.ThisshowsthatGisHausdorff.Inorder Klk ofSQIksuchthataIK0xIK,soaG(S21K):rG(S21K)and Proof: Ifa,zEGanda#r,thenthereexistsfiniteGaloissubextension (1.1) Proposition.Forevery(finiteorinfinite)GaloisextensionSZIkthe which satisfiesthefollowing aG(S2IK) xrG(S2IK),resp.aG(S21K).ThusGisatopologicalgroup bourhood arG(S2IK),resp.or are continuousmaps,sincethepreimageofafundamentalopenneigh- subextensions ofS21k.Themultiplicationandtheinversemap finite Galoissubextensionsof 91k,theset compact set homeomorphism. Ittherefore sufficestoshowthath(G)isclosedinthe h(cr G(S2IKo))=h(G)nu,soh of v,thenh-'(U)=aG(QIK0).Thush subextensions ofQ1kandiiEG(Kok). topology. Galois groupG=G(S2Ik)iscompactHausdorffwithrespecttotheKrull G^. The exampledoesnotmean,however,thatwehavetochuckthemain or =1.ThesetsUHK#KoG(KIk)x(F)form asubbasisof >.. .7' GxG -)G,(a,r)l i tK ML'IL= F1UKETTG(Klk)I aL'IL=aL}. G(K Ik).Toseethisweconsider, foreachpairL'QLof h:G--) fG(Klk), K K 'C7 >as, 1 G(S21K),containstheopenneighbourhood aG(S2I K) K : Gr-ah(G)isopen,andthusa Chapter IV.AbstractClassFieldTheory and If is K or continuous. Moreover E Gisapreimage ar-*o'-1, i.e., ML'ILisindeedclosed. extensions oftoL',then is closed.ButifG(L1k)=(o,...,Qn},andSic_(L'thesetof One clearlyhash(G)= is closed. of KIk.Ifkisanarbitrarysubextension,then neighbourhood cmG(S2IN)CG(S21K),whereN1kisthenormalclosure then G(171K)isopen,becauseeachvEG(171K)admitstheopen complement oftheunionitsopencosets.IfKIkisafinitesubextension, Proof :EveryopensubgroupofG(S21k)isalsoclosed,becauseitthe precisely tothefinitesubextensionsofS21k. closed subgroupsofG(121k).TheopenG(S21k)correspond is a1-1-correspondencebetweenthesubextensionsK1kof01kand assignment (1.2) Theorem.Let121kbea(finiteorinfinite)Galoisextension.Thenthe formulated asfollows. § 1.InfiniteGaloisTheory theory forfiniteextensions. ThuswemaychoosearEHsuchthat field Kandisthereforeequal toG(LIK),bythemaintheoremofGalois map H-±G(LIK)iscertainly surjective,becausetheimageHhasfixed then aG(S2IL)isafundamental openneighbourhoodof6inG(S2IK).The Conversely, letorEG(S2IK). IfLIKisafiniteGaloissubextensionofS2K, where KisthefixedfieldofH.TheinclusionHCG(S21K) istrivial. closed subgroupHofG(S2Ik),wealwayshave G(S2 IK).Toprovesurjectivity,wehavetoshowthat, givenanarbitrary where KiIkvariesoverthefinitesubextensionsofKIk.Therefore G(S21K) The maintheoremofGaloistheoryforinfiniteextensionscannowbe The assignmentKFaG(S2IK)isinjective,sincethe fixedfieldof "'v fit ML'IL =U(IIG(KIk)xSifail) -fl G (S2IK)=nG(S21Ki), 't7 I i=1 KOL,L' IL'2L ML'IL K HG(S21K) H=G(S21K), i So itsufficestoshowthatML'IL 000++ CAD 263 j 0i.IfK;flK=kforalli,then onehasatopologicalisomorphism composite ofallK11k,andK,'Ik thecompositeofextensionsKjlksuchthat Exercise 2.Givenafamilyof GaloisextensionsK,lkinS2Ik,letKIkbethe homeomorphism oftopologicalspaces. ..C is awl contained inacommonextensionS2Ik.IfLflK=k,thenthe mapping Exercise 1.LetLIkbeaGaloisextensionandKanarbitrary extension,both occurs invariousplacesnumbertheoryandwhichwe will introducenext. description ofthisweneedthenotionprojectivelimit, whichnaturally be reconstitutedrathereasilyfromtheirfinitequotients. Fortheprecise shows immediatelythatHisopenifandonlytheindex(G:H)finite. profinite group.Thedisjointcosetdecomposition connected component.EveryclosedsubgroupHofGisobviouslyagaina disconnected, i.e.,totheconditionthateachelementofGisequalitsown Hausdorff andcompact,whichadmitsabasisofneighbourhoods (1.3) Definition.AprofinitegroupisatopologicalGwhich abstract, purelygroup-theoreticalnotionofaprofinitegroup. that thereisafundamentalsystemofneighbourhoodstheneutralelement has finiteindexinG(S2Ik),andthisimpliesthatKIkdegree. cover thegroup.Thusthereisonlyafinitenumberofthem;H=G(S?IK) cosets ofH.SinceG(S2Ik)iscompact,afinitenumbersufficesto of theformH=G(S2IK).ButG(S21k)isdisjointunionopen closure ofHinG(S2IK),andthustoitself,sothat=G(S2IK). 264 1 EGconsistingofnormalsubgroups. 1 EGwhichconsistsofnormalsubgroups.Thispropertyleadsustothe rIL =OIL,i.e.,rEHflcrG(S2IL).Thisshowsthatorbelongstothe Profinite groupsarefairlycloserelativesoffinitegroups. Theycan It canbeshownthatthelastconditionistantamounttoGbeingtotally The topologicalGaloisgroupsG=(S2Ik)havethespecialproperty If HisanopensubgroupofG(S2Ik),thenitalsoclosed,andtherefore a topological isomorphism, G(LKIK) -+G(LIk), fir' G(KJk) =HG(K;1k). °a) that '+- G=UajH 112 211 is, i Chapter IV.AbstractClassFieldTheory an isomorphism yr-- QIL, III lad of groups f3" 3'G and a ... resp. thedisjointunionU1 X. resp. inductivesystem(Xi,fij },wemakeuseofthedirectproductIIiEIXi, L1. when i fik =fijofjk, Min X j)byfijandobtainasystem(Xi,)ofsets iEI Xi, Dew ``1 (resp. coo resp- resp. i, jEI,i Xi =UXi). ,[°', iEI 265 11i,21 Xi, by Xk -k Xi are i < j} 1 tkE1 topology. If the Xi are topology. If the Xi : are nonempty. In fact, I g(x) = f(x)}. But in I g(x) = f(x)}. But Xi of nonempty compact for i ffk(Xi). fl k,1 Xk -- Xi for the i -th fl k,1 Xk -- Xi for : l lkXk Xi the product Chapter IV. Abstract Class Field Theory Field Class IV. Abstract Chapter iEl iEI fij(xj) =xi i In the applications, the projective and inductive systems {Xi, fi j) that occur will not just be systems of topological spaces and continuous maps, but the Xi will usually be topological groups, rings or modules, etc., and the fi j will be continuous homomorphisms. In what follows, we will deal explicitly only with projective and inductive systems {Gi , gij } of topological groups. But since everything works exactly the same way for systems of rings or modules, these cases may be thought of tacitly as being treated as well. Let {Gi, gij} be a projective, resp. inductive system of topological groups. Then the projective, resp. inductive limit G = l'm Gi,resp.G = ltw Gi idd iEI isa topological group as well. The multiplication in the projective limitis induced by the componentwise multiplication in the product fl Gi. In the case of the inductive limit, given two equivalence classes x, y E G = liter tGi, one has to choose representatives xk and yk in the iEI same Gk in order to define xy = equivalence class of xkYk We leave it to the reader to check that this definition is independent of the choice of representatives, and that the operation thus defined makes G into a group. The projections pi: HiEI Gi Gi, resp. the inclusions ti: Gi - 11i EIGi, induce a family of continuous homomorphisms gi : G --j Gi,resp.gi: Gi --) G such that gi = gij o g j , resp. gi = g j o gij, for i < j. This family has the following universal property. (2.5) Proposition. If H is a topological group and hi:H -) Gi,resp.hi:Gi is a family of continuous homomorphisms such that hi = gi j o hj,resp.hi = hj o 9i j for i < j, then there exists a unique continuous homomorphism m Gi,resp. h:G= liter Gi-+H i satisfying hi = gi o h, resp. hi = h o gi, for all i E I. commutative diagram Proof: LetG'=lira is alsoexact. is exactforeveryiEI.Thentheinducedsequence such thatthesequence (Gt (2.6) Proposition.Leta limit litt,exactnessholdswithoutrestrictions.Inotherwords,onehasthe property ofthisfunctoris In thiswayl,im which inducesahomomorphism commute fori ... , g",} bemorphismsbetweeninductivesystemsoftopologicalgroups f :1mGi-)1tnG;, iEI gji f :flGi-*rjG; G! Gi iEI f =9T , resp. iEI G' G'am` fg iEI G G' LT ,becomesafunctor.Aparticularlyimportant iEl : g,,, ; G =lin-t G; {G; its so-called"exactness".Fortheinductive , , lim Gi g;j } iEl resp. resp. resp. G G t gi Gi 14 Gi, G"=literr -a (Gi,gif}and0 Chapter IV.AbstractClassFieldTheory f :jIGi-->LIG'i : Gi->G',iE1,suchthatthe f : Gi' iGI giI iEI 1 G". G' lim Gz iEI Gj Gi of topologicalgroupsisa gill Gi -liraG. G". Weconsiderthe f' fG'. iE1 G; : lsi iEI {Gi, IC. , gii }-3 groups withrespecttothediscrete topology). projective limitsoffinitegroups (whichweviewascompacttopological of normalsubgroups.Thenext propositionshowsthattheyarepreciselythe i.e., theyadmitabasisofneighbourhoods oftheneutralelementconsisting topological groupswhichare Hausdorff,compactandtotallydisconnected, return toourstartingpoint,theprofinitegroups.Recall thatthesearethe a mapseveryelementyEYtoX. means thattheprojectivelimitY=mYic all iEI.ThepreimagesYi P-4 of nonemptyclosed,andhencecompactsubsetsthe Proof: Letx=(xi)iitEmGiandt4(x)1, is againanexactsequenceofcompacttopologicalgroups. is exactforeveryiEI.Then groups suchthatthesequence (2.7) Proposition.Leta compact groups,sothatwehavethe ai (yi)=xi.Puttingyg'(yi),wehavea(y)x. may thereforeassumethatPi(xi)=1,sothereexistsyiEGisuch there existsj?isuchthatPi(xi)equals-'linG.Changingnotation,we that gi(xi-)=x.As Let XEGbesuchthat,b(x)=1.ThenthereexistsaniandxiGi § 2.ProjectiveandInductiveLimits Now thatwehaveatourdisposalthenotionofprojective limit,we "'1 The projectivelimitisnotexactincompletegenerality,butonlyfor be morphismsbetweenprojectivesystemsofcompacttopological g' fl(xi)=,3gi(xi)=P(x)1, Gl cad : G' -*Gi-L'>.Gt i a-' «) 'bO (G' , g'i }-*(Gi,gii (xi) c_Githenformaprojectivesystem ` Gi -!+lrnG,' .L' CAD ` and P so thatPi(xi)=1for G isnonempty,and : By (2.3),this (Gi,gii }- 269 groups, then If conversely{G1,gij}isaprojectivesystemoffinite(orevenprofinite) normal subgroupsofG,thenonehas,algebraicallyaswelltopologically, (2.8) Proposition.IfGisaprofinitegroup,andifNvariesovertheopen 270 is theintersectionf1iEiNi,whichequals{1}becauseGHausdorffand is anisomorphismandahomeomorphism.finjectivebecauseitskernel show thatthehomomorphism projective system{Gi,gij}offinite,andhencediscrete,compactgroups.We compact. Fori Conversely, let{G1,gij}be aprojectivesystemofprofinitegroups.As fem. m G1.Ontheotherhanditisalsodense.Forifx=(xi)i EJ f :G-) Gi) isafundamentalneighbourhoodofx,thenwemay °°o iel Us =flGix1111Gi}, Gi, G -1toGIN. G=1}m Gi igs a i-fJai, Gi, andsof(G)= N Chapter IV.AbstractClassFieldTheory iEl C1. ran iES Gi isanisomorphismanda i EI}bethefamilyofits CT' ai =ormodNi, : Gj--*Giandobtaina coo Gi) =liEsNi,wesee Gi. SinceG G/Nk, '°h p.. G, v'' 271 Z/pmZ, °f1 . fl (a mod pn) nEN iEs U/U(n) Z/p'2Z a n , igs 211 U = 7Lp = >;o o/pn Us=r[GixfJN, G(S2Ik) = lfirn G(KJk) n .tip o = Let us now illustrate the notions of profinite group and projective limit by the notions of profinite group and projective Let us now illustrate The group of units U = o* is closed in o, hence Hausdorff and compact, The group of units U = o* is closed and the subgroups U(n) = 1 + pn form a basis of neighbourhoods of 1 E U. Thus is also a profinite group. In fact, we have seen all this already in chap. II, § 4. of (2.8), we therefore obtain the Galois group G(S2 Ik) as the projective limit of (2.8), we therefore obtain the Galois the projections Z/p'2Z a projective system with respect to for n > m. The projective limit II, § 1). is the ring of p-adic integers (see chap. in a p-adic number field K and p its Example 3: Let a be the valuation ring make up a basis of neighbourhoods of maximal ideal. The ideals pn, n E N, and compact, and so is a profinite ring. the zero element 0 in o. o is Hausdorff have a topological isomorphism The rings o/pn, n E N, are finite and we a few concrete examples. extension S2Ik group G = G(S2Ik) of a Galois Example 1: The Galois This was already with respect to the Krull topology. is a profinite group of S2 I k, then, k varies over the finite Galois subextensions stated in § 1. If K I the open normal Krull topology, G(S2IK) varies over by definition of the G(KJk) = G(QIk)/G(S2IK) and subgroups of G. In view of the identity of the finite Galois groups G (K 1 k). then the rings Z/p'2Z, n e N, form Example 2: If p is a prime number, § 2. Projective and Inductive Limits Inductive and Projective § 2. element of the neutral of neighbourhoods a basis varies over If Ni by (2.3). the groups subgroups, then consists of normal in Gi which up a basis of neighbourhoods subsets of I, make over the finite with S varying normal subgroups. The Gt consisting of normal element in III El of the neutral of the Gi therefore form a basis of neighbourhoods subgroups us n 1tn 1 to Gi ; thus 1tn G, is a profinite group. neutral element in c0. which doesnotbelongtoZ. what wedidamountedtowriting downtheelement not belongto(rp).Infact,lookingatitviatheisomorphism G(lFqIlFq)=7G, chapter, howwewereabletoconstructanelementi/rEG(lFq IlFq)whichdid of Z.Giventhis,itisnowclear,intheexampleat beginningofthis subgroup (rp)_{rpnInE7L}ontothedense(butnot closed) subgroupZ which sendstheFrobeniusautomorphismrpEG(FqIIFq) to1EZ,andthe Taking, foreachnaturalnumbern,theprimefactorizationn=flPpuP, Example 4:Therings7L/n7L,nEN,formaprojectivesystemwithrespect 1 modnZ.Passingtotheprojectivelimitgivesanisomorphism one foreverynEN,bymappingtheFrobeniusautomorphism rpn Example 5:ForthefieldlFqwithqelements,wegetisomorphisms Z -+fp7LP,aH(a,a,...). This takesthenaturalembeddingofZintotodiagonal and passingtotheprojectivelimit, Chinese remaindertheoremimpliesthedecomposition precisely theopensubgroupsof7L,anditiseasytoverifythat in whatfollows.ItcontainsZasadensesubring.ThegroupsnZ,nEN,are (or "zee-hat").Thisringisgoingtooccupyquiteanimportantposition customary torefer was originallycalledthePriiferring,whereasnowadaysithasbecome given bydivisibility,nIm.Theprojectivelimit to theprojectionsZ/nZ-*Z/mZ,nI'm,whereorderingonNisnow 272 (...,0,0,1p,0,0, ...)Er[Zt =2, it bythesomewhatcurtabbreviation"zed-hat" G(FgnllFq) =7L/n7L, 7L/n7L =flZ/pPZ, 7L/n7L =7L/n7L. G(FgllFq) Z= l4mZ/n7L 7L =r[zp. n P P II? Chapter N.AbstractClassFieldTheory = 7L e POD to with finitegroupsX(A1).If for instance, where A,variesoverthefinitesubgroupsofandthus is aprofinitegroup.Foronehas Example 8:LetAbeanabeliantorsiongroup.Thenthe Pontryagindual coy surjection Z-*G. and inviewof(2.7),passingtotheprojectivelimityieldsacontinuous then wehaveforeverynthesurjectivehomomorphism then G"CHCGandn=(G:H)<(G:G") 76/nZ-+ G/G", bhp G(Q(l-tn)IQ) _(Z/nZ)*, A =Q/7U-Z/Z, X (A)=Hom(A,Q/Z) X(A) and Zare(additive)specialcasesoftheclass A.. G(QIQ) =Z A =UAi, i.e., Gistheclosure(a)ofsubgroup i nEN X(Ai) modG", 273 finite index,thentheprofinitegroup Example 9:IfGisanygroupandNvariesoverallnormalsubgroupsof then X(n7L/7L)=7L/n7L,sothat 274 Exercise 3.Apro-p-groupisaprofinitegroupGwhosequotientsGIN,moduloall Exercise 2.IfaEGand=lima,ZwithZ,thenasa'iisin'G. Exercise 1.Showthat,foraprofinitegroupG,thepowermapGx7L-* example, isthegroupZ=m7L/n7L. is calledtheprofinitecompletionofG.TheZ,for then Gab= Exercise 6.If[G;)isaprojectivesystemofprofinitegroups andG= Exercise 5.Whatisthep-SylowsubgroupofZand7La? (iii) Everytwop-SylowsubgroupsofGareconjugate. (ii) Everypro-p-subgroupofGiscontainedinap-Sylowsubgroup. (i) Foreveryprimenumberp,thereexistsap-Sylowsubgroup ofG. p-Sylow subgroupofGIN.Show: subgroup ofGif,foreveryopennormalNG,thegroupHN/Nisa Exercise 4.AclosedsubgroupHofaprofinitegroupGiscalledp-Sylow the powersa",forallaEGandZ open normalsubgroupsN,arefinitep-groups.Imitatingexercise1,makesenseof and thatonehas(a°)b=a°ba°}baaabifGisabelian. (a, n)i-- a", extendstoacontinuousmap G°b (see§1,exercise5). X (Q/7L)-1rn7L/n7L=7L. GxZ-aG, (a,a)Ha", 1 n 00 G = P1 N n Chapter IV.AbstractClassFieldTheory GIN F7' i- G; , the globalcaseitisamodificationofidealclassgroup.Atheart the basefield.Inlocalcase,AKismultiplicativegroupK*andin classified bycertainsubgroupsNL=NLIKALofagroupAKattachedto extensions ofalgebraicandp-adicnumberfields.TheseLIKare The endeavourstogeneralizethislawfinallyproducedatheoryoftheabelian number theorythebeginningofwhichwasGauss'sreciprocitylaw § 3.AbstractGaloisTheory extension kIk;thiswillallowthereadertorelyonhisstandard knowledge profinite groupisindeedtheGaloisG=(k1 k) ofaGaloisfield relations. WewillthereforeformallyinterprettheprofinitegroupGasa language offieldtheoryallowsimmediateinsightsintothegrouptheoretical However, theonlyapplicationswehaveinmindarefieldtheoretical,and The theoryweareabouttodevelopispurelygrouptheoreticalinnature. foundation, towhichwewillnowturn. basis. Inthisway,classfieldtheorycanbegivenanabstract,butelementary from thefield-theoreticsituationandtreatedonapurelygrouptheoretical law initsmostgeneralform.Now,thismapcanbeabstractedcompletely which -ifweviewthingsintherightwayencapsulatesreciprocity this theorythereisamysteriouscanonicalisomorphism of finiteindexinGK,andthis index L IKafieldextension. iscalledafiniteextension,ifGLopen,i.e., generated byanelemento,EGissimplycalledthefixed fieldofor. field belongingtotheclosure(a)ofcyclicgroup (a)={akkcZ) is calledthebasefield,andkdenotesfieldsatisfying Gk=(1}.The "fields"; KwillbecalledthefixedfieldofGK.The ksuchthatGk=G alone wouldseemodd.) of Galoistheorywhenevertheformaldevelopmentinterms ofgrouptheory Galois groupinthefollowingway.(Letusremark passing thatevery Class fieldtheoryisthefinaloutcomeofalongdevelopmentalgebraic We beginourconsiderationsbygivingourselvesaprofinitegroupG. We writeformallyKCLor LIKifGLcGK,andwecallthepair We denotetheclosedsubgroupsofGbyGK,andcall theseindicesK § 3.AbstractGaloisTheory G(LIK) =AKINLIKAL, b)(a [L :K] vin COD - (GK :GL) (-1)n21 11y 2 275 276 Chapter IV. Abstract Class Field Theory' will be called the degree of L I K. L I K is said to be normal or Galois if GL is a normal subgroup of GK. If this is the case, we define the Galois group of L I K by G(LIK) = GK/GL If N D L D K are Galois extensions of K, we define the restriction of an element a E G (N I K) to L by aIL = a mod G(NIL) E G(LIK). This gives a homomorphism G(NIL). The extension L I K is called cyclic, abelian, solvable, etc., if the Galois group G (L I K) has these properties. We put K = n Ki ("intersection") i if GK is topologically generated by the subgroups GK,, and K = fl Ki("composite") if GK =niGK;.IfGK'=a-1GKCrfor or EG,wewrite K'=K°. Now let A be a continuous multiplicative G-module. By this we mean a multiplicative abelian group A on which the elements Cr E G operate as automorphisms on the right, or: A -* A, a a°. This action must satisfy (i)al = a, (ii)(ab)° = a°b°, (iii) a°Z = (a°)T (iv) A= U[K:k] = `a' ilt=o a``. L 'L1 Aa-1 =jaa-1 as-1 := AL L = AK. J aEAL} NLIK(a) =flan, , aaa-1 the operationofGonAappears . + ork-1)(a-1)implies a 277 map ar>a',andµ,,=µn( with finitecyclickernelp G-homomorphism .C' extensions LIK. (3.1) Axiom.OnehasH-'(G(LIK),AL)=1 278 on thechoiceofa.Thus,foraKummerextension L=K(p-1(A))_ K (p and itsGaloisgroupisabelianofexponentn.Indeed,foranextension where dc_AK.AKummerextensionK(p-'(L))IKisalwaysGalois, extension (withrespecttop)isbydefinitionanoftheform of GK let K(B)denotethefixedfieldofclosedsubgroup of unity"inA. the operatorp.Thecaseofprimeinteresttousiswhenpn-thpower If inparticular,LIKiscyclic, thenwefindL=K(a)withar'aEAK. (3.2) Proposition.IfLIKis anabelianextensionofexponentn,then L IKofexponentnisaKummerextension. is aninjectivehomomorphism. i taed where aEp-'(a).SinceIt,,cAK,thisdefinitiondoesnotdepend The theoryweareabouttodevelopmakesreferenceasurjective We nowfixafieldKsuchthatt,,GAK.ForeverysubsetBCA, The followingpropositionsaysthatconversely,anyabelian extension K(p-(a)), thecompositemap . (a)) IK,wehavetheinjectivehomomorphism If BisGK-invariant,thenK(B)IobviouslyGalois.AKummer G(LIK) --flG(K(bo-1(a))IK)-A G(K(do-1(a))IK) L =K(p-1(A)) H={a EGKIb°=bforallbEB} p:A-3A, aHap, . The aE4 K(p-1(a)) IK, order n=#µ,s,iscalledtheexponentof E AI with (AD Chapter IV.AbstractClassFieldTheory F,p, A =AinAK. 1) isthegroupof"n-throots a j for as-1 all finite cyclic Let nowMIKbeacyclicsubextensionofLK.Itsufficestoshowthat groups, whichmaybeinterpretedasGaloisgroupsofcyclicsubextensions. and theGaloisgroupofafinitesubextensionisproductcyclic its cyclicsubextensions.Foritisthecompositeoffinitesubextensions, K(r-1(d)) CL.Ontheotherhand,extensionLIKiscompositeof a EAL,thenx=iALforsome Proof: Wehavep-1(A)CAL,forifxeAandxP=aPaEAK, § 3.AbstractGaloisTheory is a1-1-correspondencebetweenthegroupsAsuchthatAKCand (3.3) Theorem.Thecorrespondence a Er-1(A),andthereforeMCK(g-1(d)). K (a)=M.But But a"='a.Thusa°'aisequivalenttoi (3.1) showsthat of u,,.Letd=[M:K],d'=n/d M C_K(p-1(A)).LetorbeageneratorofG(MIK)and where X,(or)=a°-1,aEP-1(a).Since find L=K(Y-1(A))withAALflAK.Weconsiderthe homomorphism Proof :LetLIKbeanabelianextensionofexponentn. By(3.2),wethen a EP-1(a). canonical isomorphism the abelianextensionsLIKofexponentn. G(LIK) -*G(MIK)-Ip.. Letorbea'generatorofG(MIK).Since X definesacyclicextension MIKandisthecompositeofhomomorphisms it hasthekernelAKp.Toprove thesurjectivity,weletXEHom(G(LIK),Ap). where thecharacterXa:G(LIK) If AandLcorrespondtoeachother,thenAPflAK=a,wehavea As themainresultofgeneralKummertheory,wenowobtainfollowing A/AK -Hom(G(LIK),Ap) Xa A --Horn = a"forsomeaEAm.ThusKCK(a)C_M. A 1 = aEAK as-1=1 forallorEG(LIK) (ac- I?=V1,sothatacry'EAK;then L =K(p-1(a)) A,, isgivenbyXa(a)=a°-1,for , a=up EAKI a modAKHXa, 'p. a HXa, E µg, 0 modd,sothat I+, AK. Therefore a generator 279 Then axiom(3.1)isindeedsatisfied, forwehave,incompletegenerality: the operator closure ofk,thenAmaybe chosentobetheadditivegroupkandp 3) Ifkisanactualfieldofpositivecharacteristicpand kistheseparable we haveby(3.3)that exponent n.ItisgivenbyK= of thesametype,andallthemlieinmaximalabelian extensionof 2) ThecompositeoftwoabelianextensionsKexponent nisagain G(LIK). X :G(LIK)-+A,,,i.e.,asthecharactergroupoftopological has tobeinterpretedasthegroupofallcontinuoushomomorphisms Remarks andExamples:1)IfLIKisinfinite,thenHom(G(LJK),µ,,) a 1-1-correspondence,asclaimed.Thisfinishestheproofoftheorem. Hom(G(LJK)/H,A&,,) =Hom(G(LIK),µr,).Itfollowsthat4/AK of P`(A),andasK('(4))=L,wefindthatH1,so As a°-I=Xa(o-) Hom(G(LIK)/H, Ap),where The subgroup4/AKcorrespondsunderPontryagindualitytothe has 4'/AK, i.e.,4=A. 4 =AKnAK.Infact,putting4'ALnAK,wehavejustseenthatone If 4isanygroupbetweenAKandifL=K(p-I(4)),then so thatX=Xa.Thisprovesthesurjectivity,andweobtainanisomorphism AP' nAK=A.ForrEG(LIK),onehasX(r)X(rIM)az-tXa(r), some aEAM.Now, NMIK (X(6))=X(0.)1M:KI1,wededucefrom(3.1)thatX(a)a°-Ifor 280 It isthereforeclearthatthecorrespondence4HL=K G(KIK)'*=Hom(G(K[K),Q/Z) =Hom(G(KJK),,u,) H={QEG(LIK)IXa(o')=1 forallaE4}. m^I P:k-*k, ar-),a=ap-a. 4'/AK =Hom(G(LIK),A,) d/AK =Hom(G(LjK),p.6,). for aEp-1(a),Hleaves G(KIK)* =AK/AK. (aff-t = K(p-t (AK)),andforthePontryagindual Chapter IV.AbstractClassFieldTheory' = X(o)P1,sothataa'E . cam fixed theelements (p-I (4))is § 3- Abstract Galois Theory 281 (3.4) Proposition. For every cyclic finite field extension L IK,one has H-'(G(LIK),L) = 1. Proof: The extension L I K always admits a normal basis {a c I or E G (L I K) }, so that L = ®Q Kac. This means that L is a G(LIK)-induced module in the sense of §7, and then H-' (G (L I K), L) = 1, by (7.4). The Kummer theory with respect to the operator pa = a P - a is usually called Artin-Schreier theory. 4) The chief application of the theory developed above is to the case where G is the absolute Galois group G (k 1 k) of an actual field k, A is the multiplicative group k* of the algebraic closure, and p is the n-th power map a H a", for some natural number n which is relatively prime to the characteristic of k (in particular, n is arbitrary if char(k) = 0). Axiom (3.1) is always satisfied in this case and is called Hilbert 90 because this statement occurs as Satz number 90 among the 169 theorems in Hilbert's famous "Zahlbericht" [72]. Thus we have the (3.5) Theorem (Hilbert 90). For a cyclic field extension L I K one always has H-'(G(LIK),L*) = 1. In other words: An element a E L* of norm NLIK (a) = 1 is of the form a where P E L* and or is a generator of G (L I K). Proof: Let n = [L : K]. By virtue of the linear independence of the automor- phisms 1, a, ... , a"-' (see [15], chap. 5, § 7, no. 5), there exists an element y E L* such that 2Yall-1 P = Y + aya +aI+a yoZ+ ... + 0. As NLIK (a) = 1, one gets a =,61-Q If now the field K contains the group µ1z of n-th roots of unity, the operator p (a) = a' has exponent n, and we obtain the following corollary, which is the most important special case of theorem (3.3). i=0 k-I f(a)_...= l f(a)a(, = fa(a)T fa(t) (a)a as-l Chapter IV. Abstract Class Field Theory' Field Class IV. Abstract Chapter i (aa-l)zar-1 ((rk-2)a2f fa(a) = f (a)0, = f ((Y") = f (1) = 1, f (at) = f (a),f (r) f1 i=o n- (a) = f G(LIK) = Hom(A/K*", µ"). G(LIK) = Hom(A/K*", H1(G, A) = Z1(G, A)/B 1(G, A) NGf (a) _ (ak-1)af fa(ct) = a"-' = Then the abelian extensions L I K of exponent n correspond 1-1 to the n correspond 1-1 L I K of exponent abelian extensions Then the admits the which is the main basis of this corollary, Hilbert's theorem 90, f (ak) = f and obtain as a first result about this group the and obtain as a first result about this group H 1(G, A) = H-1(G, A). (3.7) Proposition. If G is cyclic, then then fork > 1 Proof: Let G = (a). If f E Z'(G, A), and f (1) = 1 because f (1) = f (1) f (1). If n = #G, then and we have characteristic of the field K, and assume that µ" c K. and assume that of the field K, characteristic K, which goes back to arbitrary Galois extensions L I following generalization G be a finite group EMMY NOETHER (1882-1935). Let to the mathematician homomorphism, G-module. A 1-cocycle, or crossed and A a multiplicative A is a function f : G -a A satisfying of G with values in an abelian group Z1(G, A). For every for all a, r E G. The 1-cocycles form a E A, the function is a 1-cocycle, for one has and form a subgroup B 1(G, A) The functions fa are called 1-coboundaries of Z1(G, A). We define 282 the prime to relatively which is number a natural Let n be Corollary. (3.6) rule K*", via the A c K* which contain subgroups `ti onto IGA,becausefEB1(G,A) isomorphism betweenZ1(G,A)andNGA.ThismapsB The readerisinvitedtocheckthis.mapf and- obtain, foreveryaEAsuchthatNGa=1,1-cocyclebyputtingf(Q) so thatf(v)ENGA={aAINGafoaa=1}.Converselywe § 3.AbstractGaloisTheory i.e., f(r)= no. 5),wecanchoosecEL*suchthata¢0.ForeG(LIK),obtain Since theautomorphismsaarelinearlyindependent(see[15],chap.5,§7, Proof: Letf:G (3.8) Proposition.ForafiniteGaloisfieldextensionLIK,onehasthat f (a)=as-1 Kummer theoryfortheabelian extensionsofexponentp". is ahomomorphismwithcyclic kernelj Exercise 3.Showthattheoperator axiom (3.1)holdsfortheG(kI k) -moduleA= additive group n >1,considerintheringofWittvectorsW(k)(seechap.II, § 4,exercise2-6)the Al Exercise 2.Letkbeafieldofcharacteristicpanditsseparable closure.Forfixed Hint: Usethenormalbasistheorem. group LofaGaloisextensionLIK. Exercise 1.ShowthatHilbert90inNoether'sformulationalso holds fortheadditive (2.5)). This propositionwillonlybeappliedonceinthisbook (seechap.VI, Noether's generalizationofHilbert'stheorem90nowreads: v0) at =Ef ""b z-1 with a f (a)EIGA. da:WW(k)-.W.(k), pa=Fa-a, of truncatedWittvectorsa=(a0,a1.Showthat L* bea1-cocycle.ForcEL*,weput P =a-1. H1(G(LIK),L*) =1. f (ak)=flau'. aEG(LIK) E f a E f(U)C' k-1 f (,.k)= t=o (z)-1f of orderp°.Discussthecorresponding (ar)cat =f(t)-1a, a°) ° aak-1 for somefixeda4=* f (Q)thereforeisan f'1 283 by chap.II,§9,p.173andexample 5in§2,wehavecanonicalisomorphisms maximal unramifiedextension kI:if]Fqistheresidueclassfieldofk,then, number fieldk,suchasurjective homomorphismd:G--Zarisesviathe in thecasewhereGisabsoluteGaloisgroupGk= G(klk)ofap-adic abstract reflectionoftheramificationtheoryp-adicnumber fields.Indeed, from theprofinitegroupGontoprocyclic Z =1tnZ/nZ(see Exercise 8.IfnisnotdivisiblebythecharacteristicoffieldKandifpt,,denotes Remark :ThegroupH'(G,A)isonlythefirsttermofawholeseriesgroups Exercise 5.ShowthatH1(G,A)=lir § 2,example4).Thishomomorphismwillproducea theory whichisan surjective continuoushomomorphism the groupofn-throotsunityinseparableclosurek,then H'(G(LIK), L*)=1. Exercise 7.EvenforinfiniteGaloisextensionsLIK,onehasHilbert'stheorem90: Class fieldtheorycanalsobebuiltuponthis(see[10],[108]). H` (G,A),i=1,2,3,...,whicharetheobjectsofgroupcohomology(see[145]). G-modules, thenonehasanexactsequence Exercise 6.If1 the opennormalsubgroupsofG. subgroup ofG,thenonehasanexactsequence all functionsoftheformfa(a)=as-1,aEA.Showthatifgisclosednormal discrete topologyonA)suchthatf(ar)=(a)z(-r),andB1(G,consistsof where Z'(G,A)consistsofallcontinuousmapsf:G-->A(withrespecttothe Exercise 4.LetGbeaprofinitegroupandAcontinuousG-module.Put 284 The furtherdevelopmentwillnowbebasedonafixed choiceofa 1-->AG-* BG-->CG+H'(G,A),H1(G,B)->H'(G,C). '-' --> A--*B->C-*1 1 ->H1(G/g,As)-H'(G,A)1(g,A). § 4.AbstractValuationTheory I., H' (G G(klk) =G(FgIlFq)-Z H'(GK,A,,) f---VIK". , A)= Z' (G,A)/B1(G,A), H1(G/N,A^'), whereNvariesoverall Chapter IV.AbstractClassFieldTheory' is anexactsequenceofcontinuous 0.O is 285 cad a E o, eK=(1:IK) eMIK = eLIK eMIL . for and eLIK=(IK:IL). E Z the Frobenius automorphism the Frobenius E Z 1 ,.y and K=Kk. , .a) dK:G(KIK)-+ 2. dK = I d: GK a' = a4 mod p a' = a4 mod the element 'K =GK flI = GK flG= GK, 'K =GK flI = GK fK = ( 2 : d(GK))., to fMIK = fLIK fMIL fLIK=(d(GK):d(GL)) In the abstract situation, the initial choice of a surjective homomorphism the initial choice of a surjective In the abstract situation, More generally, for any field K we denote by IK the kernel of the any field K we denote by IK the More generally, for For a field extension L K we define the inertia degree fL IK and the For a field extension L K we define § 4. Abstract Valuation Theory Valuation Abstract § 4. associate which It is defined by cp E G(kIk). as the composite d : G --> E mimics the p-adic case, but the applications of the theory are by the p-adic case, but the applications of d : G --> E mimics I of d has a certain to p-adic number fields. The kernel no means confined = E. d induces an isomorphism G (k I k) fixed field k I k, and K. Since -* Z, and call it the inertia group over restriction d : GK where o, resp. P, denote the valuation ring of k, resp. its maximal ideal. The resp. its maximal valuation ring of k, P, denote the where o, resp. case, given, in this concrete in question is then d : G -+ 7G homomorphism the fixed field k of IK is the composite extension of K. We put We call K IK the maximal unramified homomorphism and obtain, when fK is finite, a surjective with kernel IK, and an isomorphism G (K 1 K) such that dK (coK) = 1 (4.1) Definition. The element cOK E called the Frobenius over K. ramification index eL I K by For a tower of fields K C L C M this definition obviously implies that ti 1 = .fLlfK, and we I . d(GL) 7G 2. d(GK)) 1 ? d(GK)/d(GL)- I. ? d(GK)/d(GL)- Chapter IV. Abstract Class Field Theory Field Class IV. Abstract Chapter I GL f =LK, >GK KLIK GL dL GK .fLIK=[LnK:K]. G(KIK) -->. G(LIK) [L : K] = fLIK eLIK [L : K] = I 1>IK 1 -+ IL 1aIK/IL-> G(LIK) .ti LIK is called unramified if eLIK = 1, i.e., if L C_ K. LIK is called LIK is called unramified if eLIK = For an arbitrary extension L I K one has (4.3) Proposition. If fK and fL are finite, then fLIK (4.3) Proposition. If fK and fL are In particular, one has L Ig = If L I K is not Galois, we pass to a Galois extension M I K containing L, and we pass to a Galois extension M I K If L I K is not Galois, f. the above transitivity rules for e and get the result from n k = K. In the unramified case, we totally ramified if fLJK = 1, i.e., if L have the surjective homomorphism of cpK the Frobenius automorphism and, if fK < oo, we call the image cPLIK of LIK. n KKK is the maximal unramified since LK = LKk = Lk = L, and L subextension of L I K. It clearly has degree Equally obvious is the have the commutative diagram 286 "fundamental have the I K we L every extension For Proposition. (4.2) identity" diagram exact commutative Proof : The if L I K is Galois, the exact sequence immediately yields, § 4. Abstract Valuation Theory 287 The Frobenius automorphism governs the entire class field theory like a king. It is therefore most remarkable that in the case of a finite Galois extension L IK, every or E G(L I K) becomes a Frobenius automorphism once it is manceuvered into the right position. This is achieved in the following manner. For what follows, let us assume systematically that fK < oo,. We pass from the Galois extension L I K to the extension L I K and consider in the Galois group G (E I K) the semigroup Frob(LIK) _Cr {E G(LIK) I dK(a) E N} . Observe here that dK : GK --) Z factorizes through G(LIK) because GL = IL c IK ; recall also that 0 0 N. Firstly, we have the (4.4) Proposition. For a finite Galois extension L I K the mapping Frob(LIK) ---) G(LIK),a 'a IL, is surjective. Proof: Let Q E G (L I K) and let (p E G (L I K) be an element such that dK (cp) = 1. Then tp Ik = 6°Kand w ILnk = cLnKIK.Restricting u to the maximal unramified subextension L n K 1 K, it becomes a power of the Frobenius automorphism, or I Lnk = OLfKJK so we may choose n in N. As L = LK, we have G(LIK) - G(LIL n k). If now r E G (L I K) is mapped to Qrp-" I L under this isomorphism, then v = crp" is an element satisfying & I L = rip" I L = IL = v and °IK = oK. Hence dK(ff) = n, and so Q E Frob(LIK). Thus every elementorE G (L I K) may be liftedto an element U E Frob(L IK). The following proposition shows that this lifting, considered over its fixed field, is actually the Frobenius automorphism. (4.5) Proposition. Let CT E Frob(LIK), and let E be the fixed field of Q. Then we have: (i)ffjK = dK(5), (ii)[E : K] < oo, (iii)E = L, (iv)jr = wz- extension Kofk. valuation v:Ak->Zhasthe propertyofreproducingitselfovereveryfinite (ii) (i) v(Ak)=ZDZandZ/nZ-7L/n7LforallnEN, satisfying thefollowingproperties: a homomorphism (4.6) Definition.AhenselianvaluationofAkwithrespect tod:G->Zis a henselianvaluation. module A,whichweequipwithahomomorphismthatis toplaytheroleof homomorphism d:G-)-Z.Wenowaddtothedataa multiplicativeG- All theprecedingdiscussionsaroseentirelyfrominitialdatumof which oneshouldkeepinmindforthesequel. (iv) fLIKdE(d)=dK(d)fEIK;thusdE(d)1,andsodroz. and soisF-+Z.ButG(LIE')=G(EIE)impliesthatLE. n EN(see§2,p.273).ThustheinducedmapsF/1"'-7G/n7Larebijective bijective. Forsincer_(8)isprocyclic,onefinds(T:I") AL UL) 7L ^C3 1 AK fK aaa-lra) NLIK UK= {uEAK I VK(u)=O}. VK = = vK(NLIK(a)) . (a). = VK fLIKVL(a) = fLIK Tv(NLIk(a)) = vKa(a') = f1 V(fj that For an unramified extension L I K, § 4. Abstract Valuation Theory Valuation Abstract § 4. Gk/GKa. Hence we have, for a E AK, (ii) For a E AL one has: (4.7) Proposition. For every field K which is finite over k, the formula over k, the is finite K which field For every Proposition. (4.7) following properties: satisfying the surjective homomorphism defines a for all rcG. (i) VK=VKooor diagram extension LIK, one has the commutative (ii) For every finite Proof: (i) If c runs through a system of representatives of Gk/GK, then through a system of representatives Proof: (i) If c runs representatives of Gk/c-1GKa = o -lro sweeps across a system of AK is an element nK E AK such that (4.8) Definition. A prime element of VK (IrK) = 1. We Put a fLIK = [L : K], we have from (4.7), (ii) that VLIAK = vK. In particular, prime element of AK is itself also a prime element of AL. If on the other hand, LIK is totally ramified, i.e., fLIK = 1, and if JrL is a prime element of AL, then 7rK = NLJK(7rL) is a prime element of AK. Exercise 2.ForafiniteextensionLIKandmicroprimeTIpofL,let set 7r-I(p)isfinite.Wewriteg3Iptomeanq3E7r-1(p). Above anymicroprimepthereareonlyfinitelymanymicroprimes93ofL,i.e.,the canonical mapping set ofallmicroprimesK.ShowthatifLIKisafiniteextension,thenthere o'", n>1,ofsomeotherFrobeniuselementQ'EFrob(kjK).Letspec(K)bethe bib be afiniteextension.AmicroprimepofKisbydefinitionconjugacyclass Exercise 1.AssumethateveryclosedabeliansubgroupofGisprocyclic.LetKIk 290 whenever itoccurswithoutembellishmentsorcommentary tothecontrary, respect tod.Inthefollowingweintroduceconvention thattheletterK, such thatdiscontinuousandsurjectivevahenselian valuationwith profinite groupG,acontinuousG-moduleA,andpair ofhomomorphisms L IZcpisunramified. is cyclic,andifZT=LGP(LIK)the"decompositionfield"ofq3Espec(L),then transitively onspec(L).The"decompositiongroup" Exercise 4.ShowthatifLIKisGalois,thentheGaloisgroupG(LK)operates of k? where L.IKvariesoverthefinitesubextensionsofLK.Whataremicroprimes Exercise 3.ForaninfiniteextensionLIK,let fqit p=d(T)/d(p).Showthat (v) cGKofsomeFrobeniuselementaEFrob(kIK)whichisnotproperpower (5.1) Axiom.Foreveryunramified finiteextensionLIKonehas sequel. following axiomaticcondition,whichwillbesystematically assumedinthe will alwaysdenoteafieldoffinitedegreeoverk.Wefurthermore imposethe Continuing withthenotationofprevioussection,we consider againa rid H` (G(LIK),UL)=1 Ge(LIK)=[aEG(LIK)IqF _P] d:G-) Z, § 5.TheReciprocityMap spec(L) = jr : spec(L)->spec(K). Ef°Ip=[L:K). Ip a v:Ak Chapter IV.AbstractClassFieldTheory spec(L.), for try i =0,-1. >Z, Frobenius automorphismrpyoverE.ThedefinitionofrLIK(a)doesnot (5.2) Definition.Thereciprocitymap extension LIKanddefinefirstamappingonthesemigroup for everyfiniteGaloisextensionLIK.Tothisend,wepassfromKtothe with MIKvaryingoverthefinitesubextensionsofLK. § 5.TheReciprocityMap In formalnotation,thisgivesan= this, weconsiderforeveryaEG(L1K)andnN the endomorphisms Next wewanttoshowthatthereciprocitymaprLIKismultiplicative. Todo the unramifiedextensionMIE,onefindsu=NME(e),sUM,andthus of LIK.Toseethis,wemayclearlyassumethatEC_M.Applying(5.1)to so thatNEIK(u)ENMIKAMforeveryfiniteGaloissubextensionMIK only byanelementuEUE,andforthiswehaveNEIK(u)NLIKAL, depend onthechoiceofelementJrE.Foranotheronediffersfrom7rE where Eisthefixedfieldofaand7rEAEprimeelement. is definedby and provetwolemmasforit. Now weintroducethehomomorphism Our goalistodefineacanonicalhomomorphism For aninfiniteextensionLIKweset Observe thatEisoffinitedegreeoverKby(4.5),andorbecomesthe NEIK(u) =NEIK(NMIE(E))NMIK(e)ENMIKAM. `', a-1:AL-) AL, 'C7 tin Frob(LIK) ={aEG(LIK)IdK(o)NI. an:AL (a-1)oan =ano(a-1)=an-1. rLIK(a) =NEIK(JTE)modNLIKAL, rLIK :Frob(LIK)-AK/NLIKAL rLIK :G(LIK) .-, ate'. NLIKAL =nNMIKAM, N=NLIK:AL-+AK . AL, v" a-1' M - 1 aFafla° AKINLIKAL and wefindthat i=0 n-! . Sri 291 Q" =co'.By(5.1),wecanthen findelementsu,uiEUEsuchthat let EnIEbetheunramified extensionofdegreen,i.e.,thefixedfield n =[M:K],aco'and let EDMbethefixedfieldofa.Further, N(u) ENMIKUM,wemayassumethatu,uicUM andLCM.Let Let MIKbeafiniteGaloissubextensionofLK.In order toprovethat Proof: Letx=umodIG(LIK)UL,withx(P-11,sothat (*) such thatdK(cp)=1,i.e.,x'x,then (5.4) Lemma.IfxEHO(G(LIK),UL)isfixedbyanelementqG(LIK) have thefollowinglemma. on thequotientgroupHo(G(LIK),UL)=UL/IG(LIThUL.Forthisgroup,we N =NLIK:UL form uT-1,uEUL,rG(LIK),ismappedto1bythehomomorphism The lastequationfollowsfromcpG(LIK)=G(Lk)v. NEIEO =NIA,.ForaEAEwethusget the otherhand,onehasEK=LandEflkE0,therefore cPEOIK =4PKIEocPKEO=VEo.Consequently,NEo(PnA,.o.On and itsGaloisgroupG(E°IK)isgeneratedbytheFrobeniusautomorphism Proof: ThemaximalunramifiedsubextensionEO=EflkIKisofdegreen, (5.3) Lemma.Letcp,orcFrob(LIK)withdK(co)=1,(Q)n.IfEis 292 the fixedfieldoforandaEAE,then The subgroupIG(LIk)UL,whichisgeneratedbyallelementsofthe NEIK(a) =NEoIK(NEIEo(a))N(a)P" u = ac W-1 = NEIK (a)=(NaV.)(co,oN) UK. Wethereforeobtainaninducedhomomorphism i=1 r N :Ho(G(LIK),UL)---UK ui III tt-1 N(x) ENEIKUL. u°", ui EUL, ui = Chapter IV.AbstractClassFieldTheory C"2 Tj EG(LIK). u!°' § 5. The Reciprocity Map 293 By (*), the elements u`p-1 and fli uj'-1 only differ by an element 2 E Uz such that NE I E (x") = 1. Hence - again by (5.1) - they differ by an element of the form ya-1, with y E U_ r,,. We may thus write (VPf)'P-1 ut;-1 U0-1 =yp -111ult`-1 = F1 i Applying N gives N(u)P-1 = so that N(u) = N(ywn)z, for some z E UK such that zP-1 = 1; therefore z`1 = z, and z E UK. Finally, applying an and putting y = ya = E U_r, we obtain, observing n = [M : K] and using (5.3), that N(u) = N(u)an = N(y°n.)z' = NEIK(y)NMIK(z) E NMIKUM. (5.5) Proposition. The reciprocity map rLIK : Frob(LIK) --) AK/NEIKAL is multiplicative. Proof: Let ata2 = a3 be an equation in Frob(LIK), ni = dK(ai), Ei the fixed field of ai and 7ri E A_rt a prime element, for i = 1, 2, 3. We have to show that NE,IK(7r1)N_r2IK(7r2) = NE3jK(ir3) mod NLIKAL. Choose a fixed q E G (L I K) such that dK (rp) = 1 and put ti =ai- Ian' EG(LIK). From a, a2 = a3 and n1 + n2 = n3, we then deduce that t3 = a2 lai 1(pn2+n, =or 2 _r4 112 "2 Putting as = P-n2 a1 (Pn2, n4 = dK(a4) = n i , = 1 74=71 E A _r4 and r4=a411pn4,wefind r3=t2r4and ' NE4I K (n4) = NE'1 I K (nt) We may therefore pass to the congruence NE3IK(Jr3) = Nz2lK(Jr2)NE'4IK(Jr4) mod NLIKAL this, however,lemma(5.4)givesusallthatweneed. then thecongruenceamountssimplytorelationN(u)ENLIKAL.For NE;IK(Jri) =N(7rV"').Thus,ifweput the proofofwhichusesidentityr3=r2r4From(5.3),wehave 294 For theelementx=umodIG(LIK}ULEHo(G(L1K),UL),thismeansthat we obtain From theidentityr3=r2r4,itfollowsthat(r3-1)+(1r2)r4)_ 7rz EAzisaprimeelement. Itiscalledthereciprocityhomomorphism given by (5.6) Proposition.ForeveryfiniteGaloisextension LIK,there and thefactthatNLIKALcNLI (1 -r2)(1r4).Puttingnow of LIK. where Eisthefixedfieldofa preimagevEFrob(LIK)ofQGand canonical homomorphism Since rp,,,o(gyp-1)=o'ii1andjr Can From thesurjectivityofmapping 1, andsox1"=x;thenby(5.4),wedogetN(u)N(x)ENLIKAL. 73 =u374, rLIK(o,) =NEIK(Jr)modNLIKAL, rLIK :G(LIK) 7r2 =u21n4, Frob(LI K)->G(LIK) U - u'P-1 = 3 3 r3-11-r4 , wenowhavethe 4 i=2 F1 a N"' -7r4 Chapter IV.AbstractClassFieldTheory "i-1 ur-1 JC42 =u4n4, AKINLIKAL 2 = ni' N.. 2 1-r2 v-1W"'-1 U, EUL, = 7r' 7 1 , we have is a Q', 295 G (L I K) corresponds tats AKINLIKAL = rLI K (Q) rLIK (r) = rLIK (Q) . = rLI K (Q) rLIK (r) rLIK(cOLIK) = nK mod NLIKAL, rLIK : G(LIK) :a^ by the valuation : K], where the second map is induced G(LIK) - AK/NLIKAL -+ Z/nZ = Z/nZ, G(LIK) - AK/NLIKAL -+ Z/nZ : AK -± Z because VK(NLIKAL) c nZ. It is an isomorphism, for : AK -± Z because VK(NLIKAL) The reciprocity homomorphism rLIK exhibits the following functorial The fact that the mapping is a homomorphism now follows directly is a homomorphism now follows The fact that the mapping principle reciprocity map expresses the fundamental The definition of the VK - 1 mod NLIKAL . On the side of E E UL , by (5.1), we find a = NL I K (ETrK) IrK mod NLIKAL, and 1 mod nZ the homomorphisms, the generators cOLIK, correspond to each other, and everything is proved. behaviour. is given by and is an isomorphism. cDK E G (K I K) is a preimage of Proof: In this case one has f = K and = IrK mod NLIKAL. The fact that q'LIK with fixed field K, i.e., rLIK((PLIK) the composite we have an isomorphism is seen from with n = [L and since u = NLIK(E) for some if vK(a) =- 0 mod nZ, then a = uirK, to prime elements: the element a = APE E G (LIE) is map ed to nE E AE ; the element a = APE E G (LIE) is map to prime elements: the definition of rLIK(Q) is already to the norm map NEIK : AE -3 AK. So This principle appears at its purest in forced upon us by these requirements. the extension, then the reciprocity (5.7) Proposition. If L I K is an unramified map and there is nothing to show. However, if we have, say, dK (Q) < dK (Q'), have, say, dK (Q) However, if we is nothing to show. and there field E" ?1L = 1. The fixed (L I K) , and d f for some f E Frob then v' = It follows rLIK(f) - NE"IK(7rE'") ° 1 mod NLIKAL. of f contains L, so (c) therefore that rLIK E G (L I K), then E Frob(L I K) are preimages of al, 0-2 from (5.5) : if &1, Q2 of o-3 = Ql v2. Q3 = &I d'2 is a preimage correspond to the effect that Frobenius automorphisms of class field theory the inclusion G (L I Z) for reasons of functoriality, § 5. The Reciprocity Map Reciprocity The § 5. the of is independent rLIK(Q) of the definition show that We first Proof: For this, let 8' E Frob(LIK) Frob(LIK) of v. the preimage v r= choice of element. E AE, a prime fixed field and 7rE' preimage, E' its be another d = I L = aI L, so that I g = 6' I is and a = dK (8'), then & If dK (8) cad 4-i follows (see[75],chap.IV,§ 1). which iscalledtransferfrom GtoH.Thishomomorphismisdefinedas -3a finite index,wehaveacanonicalhomomorphism for themaximalabelianquotientgroup.IfthenHC G isasubgroupof G, letG'denotethecommutatorsubgroupandwrite obtained viathetransfer(VerlagerunginGerman).For anarbitrarygroup therefore followsfromtheequalityofnorms is aprimeelementofE.Thecommutativitythediagramonright field ofor-1faILo with fixedfieldE,andfEGaliftingoftok.ThenEQisthe On theotherhand,letrEG(LIK),andfbeapreimageinFrob(LK) follows fromtheequalityofnorms prime elementofE.Thecdmmutativitythediagramonlefttherefore JE' EAE,isaprimeelementofE',thennE=NE'IE(7r that dK(&) is apreimageofa',thenjr=Q'ILEFrob(LK)such Proof: Leta'EG(L'IK')anda=a'ILG(LIK).If5'Frob(L'IK') G(L'IK') E =E'nListhefixedfieldofCrandfE'IE1.Ifnow conjugation tr->o-Iicr. K CK'andLL',leto-EG.Thenwehavethecommutativediagrams (5.8) Proposition.LetLIKandL'K'befiniteGaloisextensions,sothat 296 NEIK(nE) =NEIK(NE'IE(Jr'))NE'IK(ir')NK'IK(NEr1K'(JrE')). where theverticalarrowsonleftaregivenbya'F-+IL G(LIK) Another veryinterestingfunctorialpropertyofthereciprocity mapis 1111V `IK> AKINLIKAL fK'IKdK'(v') ) AK,/NLIK,AL CDR , and -ay if 7tEAisaprimeelementofZ,then7r°Eo NEoIK°(Yr) =NEIK(7r)'. I NK'IK Ver :Gab Gab =GIG' N. LetE'bethefixedfieldofv'.Then G(L0 iK) G(LIK) Chapter IV.AbstractClassFieldTheory ) Hab I u* ,-L rLuc-T AKQ/NLuIKQALQ AKINLIKAL , resp.by ) EAEisa 1 or the § 5. The Reciprocity Map 297 Let R be a system of representatives for the left cosets of H in G, G=RH,IER.If or EGwe write, for every pER, aP = Pap , Up E H, P' E R, and we define Ver(a mod G') = r[ ap mod H'_ PER Another description of the transfer results from the double coset decomposi- tion G = U (a)rH r of G in terms of the subgroups (a) and H. Letting f (r) denote the smallest natural number such that a, = r_laf(T')r E H, one hasHfl(r-iar) = (a1), and we find that Ver(a mod G') _ fl at mod H' This formula is obtained from the one above by choosing for R the set {a` r Ii = 1, ... , f (r)}. Applying this to the reciprocity homomorphism rLIK :G(LIK)ab ) AKINLIKAL we get the (5.9) Proposition. Let L I K be a finiteGalois extension and K' an intermediate field. Then we have the commutative diagram G(LIK')abr`I 'AK'INLIK'AL T G(LIK)ab rLIK AKINLIKAL, where the arrow on the right is induced by inclusion. Proof: Let us write temporarily G = G (L I K) and H = G (L I K'). Let a E G (L I K), and let v_be a preimage in Frob(L I K) with fixed field E and S = G(L I E) _ (a). We consider the double coset decomposition G = USrH and put Sr = r-'Sr n H and Qr = r-af(t)r as above. Let G=G(LIK), H=G(LIK'), S=(a), r=TIL and aT=51IL Then we obviously have G=USYH, z prime elementofET,andthusalsoE.InViewtheabovedoublecoset Let EtbethefixedfieldofCr,,i.e.,ST.Eris Then onehas For everyr,letco,varyoverasystemofrightcosetrepresentativesHIS,. and therefore 298 and sinceQt decomposition, wethereforefind in LIEr.Ifnow7rEA_risaprimeelementofE,then7rrAET field ofr-tdrsothatETIEtistheunramifiedsubextensiondegreef(r) primes inP. whose quotientsGINbyopennormalsubgroupsNhaveorder divisibleonlyby a setofprimenumbersandletGbepro-P-group,i.e.,profinite groupallof Exercise 2.Generalizethetheorydevelopedsofarinfollowing way.LetPbe with yEUL,thenrLIK(c)=N(y)modNLIKAL,whereNNLIK(B.DWORK,see of AL.IfthenoEG(LIK)and Exercise 1.LetLIKbeabelianandtotallyramified,let7raprimeelement extensions LIK,provetheexistence ofacanonicalreciprocityhomomorphismrLIK (ii) u(NKIkAK)=fKZforallfinite extensionsKIk,wherefK=(d(G):d(GK)). divisible onlybyprimesinP, (i) v(AK)=ZDandZ/nZ-Z/n7Lforallnaturalnumbers nwhichare which satisfiesthefollowingproperties: a homomorphism and letAbeaG-module.henselianP-valuationwithrespect todisbydefinition [122], chap.XIIl,§5). G (LIK)°h-AK/NLIKALfor every finiteGaloisextensionLIK. rLIK(a) =flrLIK'(cr)mrLIK'(lrcrT)rLIK(Ver(crmodG(LIK)')). Under thehypothesisthatH`(G (LIK),UL)=fori0,-1,allunramified Let d:G ... :tea NEIK(7r) =r[nt(DTfl(fl(nt)°'T)=FINZTIK'(7r'), T Ver(cr modG(LIK)')=flotG(LIK')'. Frob(L IK')isapreimageoforEG(LK),itfollowsthat Z Pbeasurjectivehomomorphismontothegroupp=H H =USTmt r, OJT ay' CUT v: Ak+Zp yV-1 = and G=USrmr. `2" T 112 r (UT Chapter IV.AbstractClassFieldTheory 7r T a-1 ICI r, WT r . PEP Z P' 0 and showthat,forafiniteextension,thediagram (see §4,exercise1-5).Defineacanonicalmapping satisfies axiom(5.1),andletv:Ak-+7Lbeahenselianvaluationwithrespecttod. Exercise 3.Letd:G-->7Lbeasurjectivehomomorphism,AG-modulewhich § 6.TheGeneralReciprocityLaw surjective homomorphismvE:AE-aZandaNECK-)-AK. limit istakenwithrespecttothenormmapsNK,IK,:AK',-*AK..Thentherea where K.IKvariesoverthefinitesubextensionsofEK,andprojective Q and Hint: LetVEGKbeanelementsuchthatdK(V)N.E'thefixedfieldof the reciprocityisomorphism commutes. Showfurthermorethat,foreveryfiniteGaloisextensionLIK,rKinduces (6.1) ClassFieldAxiom.ForeverycyclicextensionLI K onehas (6.2) Proposition.Forafinite unramifiedextensionLIK,onehas that onehas For themtheaboveconditionamountspreciselytorequiring axiom(5.1),so Let KIkbeafiniteextensionandletspec(K)thesetofmicroprimes Among thecyclicextensionsthereareinparticularunramified ones. We nowimposeonthecontinuousG-moduleAfollowing condition. #H`(G(LIK),AL) = § 6.TheGeneralReciprocityLaw H`(G(LIK),UL) =1for i=0,-1. rLIK :G(LIK)-AKINLIKAL. spec(K) rK>AK/NkjKAT rK :spec(K)-+AKINIJKAk, spec(L) `.' a .77 I BCD AE =.AKa, r` a j[L:K] ) ALIN-1.Ak I Ntlx `D- for i=-1. for i=0, coo 299 commutative exactdiagram Proof: IfMIKisaGaloissubextensionofL is anisomorphism. general reciprocitylaw. represents themaintheoremofclassfieldtheory,andwhichwewillcall But theclassfieldaxiomyieldsmoreoverfollowingtheorem,which homomorphism where AisaG-modulewhichsatisfiesaxiom(6.1),dsurjective prime elementofAL.AsH-1(G(LIK),AL)=1,everyuEUL (6.3) Theorem.ForeveryfiniteGaloisextensionLIK, the homomorphism (6.2) and§5,weobtainforeveryfiniteGaloisextensionLIK,thereciprocity continuous homomorphism,andvisahenselianvaluation.Fromproposition so wegetinfactaEUL.ThisprovesthatH°(G(LIK),UL)=1. with aEAL,sincevK(u)=0.But0VK(U)vK(NLIK(a))nvL(a), bijective as#AK/NLIKAL=n.IfnowueUK,thenwehaveNLIK(a) homomorphism issurjectiveasvK(7rKmodNLIKAL)=1nZ,andit where n=[L homomorphism H-1(G(LIK),UL) =1. So writinga=EnK,EUL,weobtainus°.Thisshowsthat such thatNLIK(u)=1isoftheformuas-',withaEAL,QVLIK Proof :SinceLIKisunramified,aprimeelementnKofAKalso 300 1-* G(LIM) We usethisdiagramtomake threereductions. By definition,aclassfieldtheoryispairofhomomorphisms On theotherhand,homomorphismvK:AK-Zgivesrisetoa AMINLIMAL .:, rLIM : K]=fLIK,becauseVK(NL(KAL)_!LIKZnZ.This VK AKINLIKAL-->Z/nZ=Z/n7L, rLIK : r.LIK : NM BCD (d :G-7,v:A--->7L), G(LIK)ab G(LIK)ab AKINLIKAL G(LIK) ----G(MIK)-=r1 1 rLIK Chapter IV.AbstractClassfieldTheory ) AKINLIKAL ) AKINLIKAL ) AKINMIKAM--1 , weget rMIK mar. from (5.8)the , ,-:,:, . 1 301 SP is I K . vim abelian. For if the For abelian. therefore of rL I K = 1, and so E fl K = K. K], p) = 1, SP I K, and find G (L I K )ab = G (M I K), and the I K )ab = G (M I K), find G (L I K of & that f r , we : AKINLIKAL -) AMINLIMAL, : AKINLIKAL -) i ,flw Now let LIK be cyclic and totally ramified, i.e., fLIK = 1. Let a be a Now let LIK be cyclic and totally ramified, § 6. The General Reciprocity Law Reciprocity General The § 6. First reduction. We may assume that G(LIK) is G(LIK) that assume We may reduction. First maximal M = Lab the case, then, putting is proved in this theorem K of L I abelian subextension of rL I K, precisely the kernel M) of G (L I K) is subgroup G (L I commutator by The surjectivity follows is injective. -+ AKINLIKAL i.e., G(LIK)ab solvable, where G(LIK) is in the case on the degree. Indeed, induction and rLIM are = L or [L : M] < [L : K], and if rMIK one has either M the fixed field of a is rLIK. In the general case, let M be surjective, then so use the left part M I K need not be Galois, but we may p-Sylow subgroup. to show that the rLIM is surjective. It then suffices of the diagram, where That this holds p-Sylow subgroup SP of AK/NLIKAL. image of NMIK is the Now the inclusion to saying that rLIK is surjective. true for all p amounts a homomorphism AK C AM induces o i = [M : K]. As ([M : such that NMIK and hence the same is true of rL I K G (L I K) is abelian, this map is injective surjective, so SP lies in the image of NM L I K is cyclic. For if M I K varies Second reduction. We may assume that then the diagram shows that the kernel over all cyclic subextensions of L I K, G(L I K) - JM G(M I K). Since of rLIK lies in the kernel of the map M I K of L I K, surjectivity follows by Choosing a proper cyclic subextension reduction for solvable extensions. induction on the degree as in the first We may assume that fLIK = 1. To Third reduction. Let L I K be cyclic. unramified subextension of L I K. see this, let M = L fl k be the maximal by (5.7). In the bottom Then fLIM = 1 and rMIK is an isomorphism is injective because the groups sequence of our diagram, the map NMIK orders [L : M], [L : K], [M : K] by in this sequence have the respective if rLIM is. axiom (6.1). Therefore rLIK is an isomorphism the isomorphism G (L I K) - G (L I k) generator of G (L I K). We view o via as an element of G(L I K), and obtain the element v = o coL E Frob(L IK), as an element of G(L I K), and obtain that dK (Q) = dK (cPL) = which is a preimage of or E G (L I K) such We thus find for the fixed field Let M I K be a finite Galois subextension of L I K containing E and L, let M° = M fl k be the maximal unramified subextension of M I K, and put N=NMIMo.AS fEIK =.fLIK = 1, one finds NIAE =NEIK,NIAL =NLIK (see the proof of (5.3)). a EAm,andfindinAMtheequation that N(u-'v)=1.Fromaxiom(6.1),wemaywriteu-IVas-'forsome From rLIK(6k)=1,itthusfollowsthatN(at)N(v)forsomevEUL,so U EUM,weobtain 0 V)°-1 Chapter IV.AbstractClassFieldTheory a G(LIK)ab AKa AK t = III f Ver (aa-l)a-I ( III ( LalKa) = . G(LaIKa)ab (aa k) = 1,where t a* RD' § 6. The General Reciprocity Law 303 The definition of the norm residue symbol automatically extends to infinite Galois extensions L I K. For if L t 1 K varies over the finite Galois subextensions, then G(L IK)ab = G(L; I K)ab (see §2, exercise 6). As (a, Lj/ I K) I Lab = (a, L; I K) for Li, D L; ,the individual norm residue symbols (a, L; I K), a E AK, determine an element (a,LIK) E G(LIK)ab. In the special case of the extension KIK , we find the following intimate connection between the maps dK, VK, and (,KIK). (6.5) Proposition. One has (a) (a, KIK) = cpK , and thusdKo( ,KIK)=VK- Proof : Let L I K be the subextension of KIK of degree f. As Z/ f Z = 7Z/fZ, we have vK(a) = n + fz, with n E Z, z E Z; that is, a = u7rKbf, with u E UK, b E AK. From (5.7), we obtain (a,KIK)IL= (a,LIK) = (7rK,LIK)n(b,LIK)f =c0K(a)IL. (a) Thus we must have (a, k I K) = O K El The main goal of field theory is to classify all algebraic extensions of a given field K. The law governing the constitution of extensions of K is hidden in the inner structure of the base field K itself, and should therefore be expressed in terms of entities directly associated with it. Class field theory solves this problem as far as the abelian extensions of K are concerned. It establishes a 1-1-correspondence between these extensions and certain subgroups of AK. More precisely, this is done as follows. For every field K, we equip the group AK with a topology by declaring the cosets aNLIK AL to be a basis of neighbourhoods of a E AK, where LIK varies over all finite Galois extensions of K. We call this topology the norm topology of AK. (6.6) Proposition. (i) The open subgroups of AK are precisely the closed subgroups of finite index. (ii)The valuation VK : AK + Z is continuous. ltd follows , then it NL,nL2 =NL,NL2. Al is closed, and so Al Al is closed, and so Chapter IV. Abstract Class Field Theory Field Class IV. Abstract Chapter WOAr L AK = (1 NLIKAL Ar =AK-1 UaN. G(LIK) = AK/N. LHNL=NLIKAL cam NL2, NL,L2 =NL, n NL2, VK(NLIKAL) _ fvL(AL) C .f7L, y., NMIKAM, NLIK(NMLILAML) = NMLIKAML C If L I K is a finite extension, then NL I K : AL -+ A K is continuous. -+ A K I K : AL then NL extension, K is a finite If L I The finite abelian extensions L I K are now classified as follows. The finite abelian extensions L I K are The field L corresponding to the subgroup Al of AK is called the class 304 (iii) only if the group Hausdorff if and (iv) AK is norms is trivial. of universal subgroup of AK, then Proof: (i) If Ar is a of neighbourhoods neighbourhoods NLIK AL of the basis to contain one of the and of finite index, index. If, conversely, Al is closed of 1, Al is also of finite the finitely many cosets aN then the union of extension of degree f (see § 2), and if L I K is the unramified L1 c L2 Now, if N is open, so are all cosets aN, so that Al is closed, and since Al has so are all cosets aN, so that Al is closed, Now, if N is open, is open. of 0 E Z Z, f E N, form a basis of neighbourhoods (ii) The groups f from (4.7) that which shows the continuity of UK. of 1 E A K. Then (iii) Let NM I K AM be an open neighbourhood which shows the continuity of NL I K- (iv) is self-evident. (6.7) Theorem. Associating the finite abelian extensions L I K and sets up a 1-1-correspondence between one has the open subgroups Al of AK. Furthermore, field associated with Al. By (6.3), it satisfies Then (d',v')isalsoap-classfield theory.Thenormresiduesymbolsassociatedto .°C field theories(d:G--;7L,v:Ak--->ZM (d, v)and(d',v')coincide.(No analogousstatementholdsinthecaseof7L-class composite of homomorphism, and field theoryinthesenseof§5,exercise2.Letd':G-->Z,be anothersurjective Exercise 3.Letpbeaprimenumberand(d:G-Z,,,v Ak->Z)ap-class bilinear mapping(theabstract"Hilbertsymbol") yield, viaPontryagindualityG(LIK)xHorn(G(LIK),A.")--> p.,,,anondegenerate Exercise 2.Underthehypothesesofexercise1,Kummertheory andclassfieldtheory L IKbethemaximalabelianextensionofexponentn.If isfinite,thenonehas Exercise 1.Letnbeanaturalnumber,andassumethegroupµ This showstheinjectivityofcorrespondenceLHNL. NLIKAL =A'K'' is cyclicofordern,andACA".LetKbeafieldsuchthat abelian extensionLIK.ButNL,S;NLimpliesc1nL2,sothat As NL1NL2isopen,wehavejustshownthatNLINL2=NLforsomefinite L1 nL2S;LiimpliesthatNL1nLZ2NL,, correspondence LHNLissurjective. kernel of( G(LIL') underthemap(,LIK):AK-#G(LIK),andthusitisfull field L'ofLIK.SinceN2NL assume LIKtobeabelian.But(N,K)=G(LL')forsomeintermediate of someGaloisextensionLIK.(6.3)impliesthatNL=NLab therefore haveNL,L2=NL,nNL2,andso is, (a,LiIK)=1fori1,2.Thus(a,L1L2jK)1,i.e.,aENLL2.We the element(a,LIL2IK)EG(L1L2IprojectstriviallyontoG(L1IK),that of thenormimpliesNL,LZCNL,nNL2.If,conversely,aEnNLZ,then Proof of(6.7):IfL1andL2areabelianextensionsK,thenthetransitivity § 6.TheGeneralReciprocityLaw Eli 0°) Finally, If Nisanyopensubgroup,thenitcontainsthenormgroupML=NLIKAL NL1 2XL2{#==>ML1nNL,=NL1L2=NL2 , the L'IK) :AK-G(L'IK).ThusN=NL,.Thisshowsthatthe equality NL1nL2=NL,NL, K the7LP-extensiondefinedbyd'.Letv':Ak->7Gpbe ( Ak NL1.NL2 =NL2NL1nL2. ):AK/A'KxAK/A'. c NL,nL2 2NL1NL2 1C'IKT G(K 1K) , =[L2:K]4L1CL2. the groupNisfullpreimageof i =1,2,andthus . zP is obtained as [L1L2:K] c AK,andlet E AI"=1} , sowemay CAD follows, 305 taken withrespecttothenormmapsNK,JK,,:AKA-+AKa.Show: where K.IkvariesoverthefinitesubextensionsofKIkandprojectivelimitis every finiteextensionK1k.Foraninfinite1k,put be aclassfieldtheory.WeassumethatthekernelUkofvk:Ak-.7Giscompactfor Exercise 4.(Generalizationtoinfiniteextensions.)Let(d:G 306 canonical way,andthetransferfromGKtoGLisahomomorphism Exercise 5.IfLIKisafiniteGaloisextension,thenGjbG(LK)-modulein or infinite)GaloisextensionswithLCL',thenonehasacommutativediagram 3) IfKCK'areextensionsofkwithfK, Show thatforeverycyclicextensionLIKonehas N.. Abstract classfieldtheoryacquiresamuchbroaderrangeofapplications ifitis AK viathetransfer.ThenisidentifiedwithAGK. ( #H' (G(LIK),AL)= o.° c]' .1Q AK AK Ver :GK 112 Ver NLIK :AL-AK, AK, gab : GK vas ( (LIK) >G(LIK)an L'IK') = 1 ([L :K] (Gib)G(LIK) (GL)G(LIK) .-. Chapter IV.AbstractClassFieldTheory G(L,IK )ab AK, , : AK-+AL.IffurthermoreLIKis I for i=-1, for i=0, : G-+7L,theinducedmap Paz 7L, v:Ak-Z) ".. of 7r1(Y,y).Inthiscase,onehasacanonicalisomorphism y =f(X)EY.ShowthatisGaloisifandonly7r1(X,x)anormalsubgroup Exercise 8.Letf:X-aYbeamapofconnectedfiniteG-sets,andletxEX, the fibresf-'(y). f iscalledGaloisifXandYareconnectedG(XIY)operatestransitivelyon B(G), weput is calledthefundamentalgroupofXwithbasepointx.Foramapf:-*Yin § 6.TheGeneralReciprocityLaw h (X):A(X)->BrepresentingnaturaltransformationsA*-+*andA.-aB*. A homomorphismh:-*Bofdoublefunctorsisafamilyhomomorphisms If f:X for allXEB(G).Wedefine the category(ab)ofabeliangroupsiscalledadoublefunctorif consisting ofacontravariantfunctorA*andcovariantfromB(G)to Exercise 9.G-modulationsform anabeliancategory. the nameofMackeyfunctors(see[32]). Remark: G-modulationswereintroducedinageneralcontext byA.DRESSunder commutative. in B(G),resp.(ab),theoneonleftiscartesian,then oneontherightis (ii) Ifamongthetwodiagrams (i) A(XUY)=A(X)xA(Y). norm NLIK. resp. f*:AL-+AK,induced byf:G/GL-*G/GK,istheinclusion,resp. G-modulation Ainsuchaway that,foranextensionLIK,themapf*:AK Exercise 10.IfAisaG-module, thenthefunctionA(G/GK)=AGKextendstoa If XisafiniteG-setandxEX,then A G-modulationisdefinedtobeadoublefunctorsuchthat v,' A pairoffunctors Y isamorphisminB(G),thenweput fI 0'Q X ( '^C Y < s 8 nr,(X,x)=G_T={cEGlax=x} A4(f)=f' a5' G(XIY) =jr1(Y,y)1ni(X,x). A =(A*,A*):B(G)-+(ab), Y' X' If' A*(X) =_:A(X) G(XIY) =Auty(X). v,, AK =A(GIGK) 4"' and and A5(f)=f4. f* A(X) A(Y) 8-A(Y') I s, ) A(X') I f: 307 AL, where theprojectivelimitistaken overallGaloismapsf:Y-->X. which, forconnectedX,isgiven by Exercise 14.IfAisaG-modulation, thenthecompletionAisagainaG-modulation norms. function NA(X)definesaG-submodulationNAofA,themodulation ofuniversal where theintersectionistakenoverallGaloismapsf:Y -* X.Showthatthe Exercise 13.LetAbeaG-modulation.Foreveryconnectedfinite G-setX,let by thetransfer,resp.inclusionGL-3GK. +~G into thecategoryofpro-abeliangroups.Thus,foranextension LIK,themaps Exercise 12.ThefunctionJrab(G/GK)=GKextendstoaG-modulation f of allG-equivariantsectionsX group is againaG-set,becausec(a),A(G/Gc)=A(G/G°,t).DefineA(X)tobethe Hint: IfXisanarbitraryfiniteG-set,thenthedisjointunion then AextendsuniquelytoaG-modulation:B(G)-->(ab). If foranythreeopensubgroupsU,VCWofG,onehastheinductionformula (U CV),resp.c(a)G/U-*G/aUa-'(aEG),define TU H rr over theopensubgroupsofG,andwhosemorphismsarejustprojections B0(G) bethecategorywhoseobjectsareG-setsG/U,whereUvaries Exercise 11.G-modulationsareexplicitlygivenbythefollowingdata.Let for everyGaloismappingf:X-+Y,isanisomorphism. such that category ofG-modulationswith"Galoisdescent",i.e.,thoseA 308 : G/U-+G/VforUCV,aswellthemapsc(a)-->G/aUa't, : GK-)-GL,resp. Let A=(A*,A,):B0(G)-+(ab)beadoublefunctorandfor7rG/U-->G/V The ruleAr arc TUa-1 Rest oIndW= = Ta-'(aUa-'),foraEG. c(a), =A,(c(a)):A(G/U) A isanequivalencebetweenthecategoryofG-modulesand Rest =A*(Jr):A(G/V)--A(G/U), Indu =A,(7r):A(G/U)-->A(G/V), A(X) _mA(X)/f,A(Y), .nom ULe f* :A(Y)-A(X)G(X'Y), Trab :B(G)-4(pro-ab) U\W/V A(X) =Homx(X,Ax) NA(X) =n.f,,A(Y), Ax =UA(G/G.,) Ax oftheprojection-*X. GK xEX , inducedbyf:G/GL-*G/GKare Chapter IV.AbstractClassFieldTheory oc(a), oResvr, 0C, ju° , ".y A(G/aUa-1). given for everyGaloismapping_f-.4 ,Y. so is is anisomorphismforeveryGalois mappingf:X where [X:Y]=#f-1(y),withy EY,anduisageneratorofG(XIY).Thenifrxly Galois mappingf:X--aYwithcyclicgroupG(XIY), onehas Exercise 19.Assume,beyondtheconditionrequiredinexercise 18,thatforevery homomorphism" is exact,andthatA(Y)Ux;Y)cf*A(X)foreveryGaloismapping f:X--Y(the the pair(d,v)gives,foreveryGaloismappingf:X-*Y,acanonical "reciprocity latter isaconsequenceoftheconditionwhichwillbeimposed in exercise19).Then G-sets, thesequence Exercise 18.Assumethatforeveryunramifiedmapf:X-->Yofconnectedfinite of aG-moduleA. Exercise 17.Comparethisdefinitionwiththe(4.6)ofahenselianvaluation contains ZandsatisfiesZ/nZ=forallnEN.LetUdenotethekernelofA. such thatthesubmodulationv(A)ofZcomesfromasubgroupCwhich homomorphism Let cpxjyEG(XIY)betheelementwhichismappedto1modfx,yZ. Exercise 16.AnunramifiedmapfX-+YofconnectedfiniteG-setsisGalois, This givesahomomorphismofG-modulations and dinducesanisomorphism to amappingf:X--*YofconnectedG-sets,aregivenby Exercise 15.ddefinesaG-modulationZsuchthatthemapsf*,f*,corresponding unramified ifexly=1. where Iy,resp.I,,isthekernelofd:Gy-+Z,G,Z.fcalled inertia degree,resp.theramificationindex,offisdefinedby f :X-*YbeamapofconnectedfiniteG-setsandEX,y=(x)Y.The § 6.TheGeneralReciprocityLaw Let AbeaG-modulation.Wedefinehenselianvaluationofto C17 For thefollowing,letd:G--Zbeafixedsurjectivehomomorphism.Let (A(Y) :f*A(X))=[XY] 0 fxlr =(d(Gy):d(G.,)), U(Y) rxly :G(X rx1y :G(XIY)->A(Y)lf*A(X). Z(Y) =ZE G(XIY) U(X) PXY dn'ab>Z. ly)ab v:A -*Z fXIY X,y and ZlfxlyZ. resp. A(Y)/f*A(X), U(X) -t`-*U(Y)-*0 Z =Z(X). ker f*=im((Y*-1), exir =(ly Y ofprimedegree[X:Y], C,, : Ix), '.a 309 Qtr °C7 induces anisomorphism whose kernelistheG-modulationNAofuniversalnorms(seeexercise13).It homomorphism ofG-modulations Exercise 20.Underthehypothesesofexercise18and19oneobtainsacanonical 310 The correspondenceKr-*K2(K)maybeextendedtoaG-modulationK2.Itdoes as aspecialcaseofthetheorysketchedabove.Thebasicideaisfollowing. Russian mathematicianI.FesENxo(see[36],[37],[38]).Hisproofmaybeviewed Kato's proofisintricateandneedsheavymachinery.Itwassimplifiedbythe mathematician KAZUYAKATO(see[83])hasestablishedacanonicalisomorphism with residueclassfieldalocal(e.g.,Q((x)),FP((x))((y)))-theJapanese extensions LIKof"2-localfields"-thesearediscretelyvaluedcompletefields the fieldK.ThegroupK2(K)isdefinedtobequotient The multiplicativegroupK*maybeinterpretedinK-theoryastheK,(K)of isomorphism we willshowthat,foraGaloisextensionLIKoflocalfields,thereisreciprocity interesting applicationtohigherdimensionalclassfieldtheory.Inchap.V,(1.3), Remark :Thetheorysketchedaboveandcontainedintheexerciseshasavery of thecompletionA(seeexercise14). makes KATO'stheoremintoaspecialcaseofthetheorydeveloped above. the hypothesisofexercise18and19whenviewingKZPasa G-modulation.This for everyGaloisextensionLIK,andthatK2"'(K)satisfiesproperties whichimply shows that where A2(K)istheintersectionofallopenneighbourhoods 1inK2(K),andhe one hasx,,+y-a0,-x--0whenever canonical map( theory directlytoK2.ButFESENxoconsidersonK2thefinesttopologyforwhich not satisfythehypothesisofexercise15,sothatonemayapplyabstract where Risgeneratedbyallelementsoftheformx®(1-x).TreatingGalois be abletoapplyittheconcrete situationsencounteredinnumbertheory, The precedingsectionconcluded abstractclassfieldtheory.Inorderto ... oc0 , ) Ki'(K)INLIxKzP(L) =K2(K)INLIKK2(L) (CD : K*x-+K2(K)issequentiallycontinuous,andforwhich § 7.TheHerbrandQuotient G(LIK)ab = K2a'(K) =K2(K)/A2(K), G(LIK)ab =K*/NLIKL* K2(K) =(K*(9K*)/R, A-+rab A KZ(K)INLIKK2(L) ) 7rab 112 Chapter IV.AbstractClassFieldTheory 0, y->0.Heputs .ti G-module. Asbefore,weformthetwogroups formalism, whichwedevelophereforfutureuse. An excellenttoolforthisistheHerbrandquotient.Itagroup-theoretic it isallimportanttoverifytheclassfieldaxiom(6.1)inthesecontexts. § 7.TheHerbrandQuotient where The exactnessatallotherplaces isseenevenmoreeasily. that NGbEA.Themapf6isnowgivenbycmodIGA H NGbmodNGA. let bEBbeanelementsuchthatj(b)=c.Thenj(NGb) =NGC1,so by cmodNGCHba-1IGA.Inordertodefinef6, letceNGC,and NG(ba-1) =NG(ba)/NG(b)1,sothatba-1EN0A. f3isthusdefined an elementsuchthatj(b)=c.Thenwehave inclusion. Thenf3isdefinedasfollows.LetcECG andletbEBbe and B-]+C.WeidentifyAwithitsimageinsothat ibecomesthe Proof: Thehomomorphismsfl,f4andf2,f5areinduced byA-' G -modules,thenweobtainanexacthexagon (7.1) Proposition.If1 with bEB.Writingc=j(b), wefindf3(cmodNGC)=aIGA. is deducedasfollows:leta ENGAandf4(amodIGA)=1,i.e.,ab", c =j(b),wefindf6(cmod IGC)=amodNGA.ExactnessatH(G,A) fl (amodNGA)=1;inother words,a=NGbforsomebEB.Writing Let Gbeafinitecyclicgroupofordern,letcrgenerator,andA We nowproveexactnessattheplaceH°(G,A).Leta E AGsuchthat NGA={aEAINGa=1}, AG={aEAIa'=a}, H-1 (G,C) H°(G,A) =AG/NGA fs H-1(G, B), H°(G,A) 4\'. --> A--fB--+C-*1isanexactsequenceof and NGA={NGa= Fjaa`IaEA}, IGA={aa-'I aEA}. f4 f H-1 (G,A) H°(G,B H-1(G, A) = NcA/1GA, (ba-1) =ca-11and i=o n-1 a H° (G,C) .Y. 311 B may associatetheso-called induced G-module show that#A=#AG#IGA#NGA#NGA,andh(G, A)=1. exact sequences h(G,B) =h(G,A)h(G,C).Finally,ifAisafiniteG-module, thenthe defined, thensoisthethird.Andfromlastequation, weobtain At thesametime,weseethatifanytwoofquotients arewell- and thus image offi,wefind provided thatbothordersarefinite. 1 ---)A`'-A-a Proof: Weconsidertheexacthexagon(7.1).Callingn;orderof and theidentityholds. in thesensethat,whenevertwoofthesequotientsaredefined,soisthird G -modules,thenonehas (7.3) Proposition.If1--A-*BC to be (7.2) Definition.TheHerbrandquotientoftheG-moduleAisdefined 312 #H-1(G, A) #H°(G, A)=n6n1, If Gisanarbitrarygroupand gasubgroup,thentoanyg-moduleB,we CAD For afiniteG-moduleA,onehash(G,A)=1. The salientpropertyoftheHerbrandquotientisitsmultiplicativity. #H°(G, A)C)#H-1(G,B) = n3n4, = #H°(G,B)-#H-1(G,A)C). h(G, B)=A)C) IGA -)1, h(G A)= #H-1(G, B)=n4n5, #H°(G, B)=n1n2, A =Indg(B). #H- (G,A) #H°1 G,A) 1 --)N0A-N-G+NG-->1, Chapter IV.AbstractClassFieldTheory' 1 isanexactsequenceof #H-1(G, C) #H°(G, C)=n2n3, = n5n6, , 313 fQ pER f F- fl f(p) i =0, - 1. 21' for or E g, for a 'g, ) B, 1 for 1 ba Ba, f t--r .f(1), Ba, aEG/g v:A aEG/g ) fb(a)={ (x) = f ((Tx). fa A = II 211 bi rroNG=Ngov, IndG(B) = fl : IndG(B) -- B, f H f(1); Jr B'= (fEIndG(B)I f(x)=1 forx0g} B'= (fEIndG(B)I ) B, H`(G, IndG(B)) - H'(g, B) s:B-+A, rr:A If conversely A is a G -module with a g-submodule B such that A is the If conversely A is a G -module with a Indeed, for any f c- IndG (B) we have a unique factorization f = Indeed, for any f c- IndG (B) we have § 7. The Herbrand Quotient Herbrand The § 7. all f (x)t for f (xr) = such that G ->. B f : all functions of It consists G is given by operation of Or E r E g. The canonical We have a instead of IndG(B). we write IndG(B) If g = (1}, g -homomorphism which maps the g-submodule index, we find B. We identify B' with B. If g is of finite isomorphically onto a system of left or E Gig signifies that or varies over where the notation is determined by fa (1)-= f (a-1). where fa denotes the function in B' which then A - IndG (B) via B - B'. cyclic group, g a subgroup and B a (7.4) Proposition. Let G be a finite g -module. Then we have canonically coset representatives of G/g. coset representatives direct product system of right coset representatives Proof: Let A = IndG(B) and let R be a for G/g with 1 E R. We consider the g-homomorphisms Both admit the g-homomorphism as a section, i.e., rr o s = v o s = id, and we have If fEAG,then(v)=(1)forallorG,and(r)(1)T because onefindsthat,forfEA, 314 It issurjectivesincevos=id.WeshowthatIGAthepreimageofIgB. g-homomorphism Therefore H°(G,A)=H°(g,B). one hand,andontheother,Ng(B)=Ng(vsB)_Jr(NG(sB))S;Tr(NGA). It sendsNGAontoNgB all rEg.Themapitthereforeinducesanisomorphism Let fENGAbesuchthatv(f)_Fl!ogyp`=1.Definethefunction show ker(v)cIGA.LetG=(gyp),n(G: is apreimageasv(ft-1)= On theotherhand,for aP Eg,weobtain has lgB=(bT-'IbEB,rEg).Writingnowup=p'rr,withp,p'ER, a, =o'p,then IGA consistsofallelementsf'-',fEA,ClG.ForifG=(oo)and provided allquotientsaredefined. Exercise 2.Letf,gbetwocommuting endomorphismsofanabeliangroupA. is multiplicative. f og=0.Makesenseofthefollowingstatement.The quotient Exercise f =hl- h EAbyh(1)=1,h(rPk)nk-tf((p').Then(wk) Show that (NGf)(1)= n11fPT(1)=flflf(pr)=II(flf(P))T=Ng(v(f)). r°, As Ngov=rrNG,theg-homomorphismv:A--).Binducesa 1. Let f,gbeendomorphismsofanabeliangroupAsuch that E IGA.ThuswefinallygetH(G,A)=H-1(g,B). `ti TEg PER for 0 E 1gB,thefunctionfr-1,withf=sb, r = F1 P has 7r(NGA)=Ng(vA)cNgBonthe P' (ker gimf) (ker fimg) Vs(b)T-1 =br-t.Afterthisitremainsto f(P')iP Chapter IV.AbstractClassFieldTheory f (p) ) Bg. E IGA.Inthesameway,one r _ flby-1EIgB. g), R={l,(p, `ti P P i=O f (cpt)=1.Hence h(cpk-1) _ G (LIK)-module. compute theHerbrandquotientofgroupunitsoLL,viewedasa one has is isomorphicto7L[fl,for1;aprimitivep-throotofunity,andthatinthisring Let N=1-I-o-+ Hint: Usetheexactsequence such thatq0(A)isdefined.Show Exercise 3.LetGbeacyclicgroupofprimeorderp,andletAG-module § 7.TheHerbrandQuotient H' (G,Ind'(A))25(g,A). Exercise 5.IfGisagroup,gnormalsubgroupandAg-module,then Exercise 4.LetLIKbeacyclicextensionofprimedegree.Usingexercise3, where aisunitin7L[G]/7LN. ti' + vp-"inthegroupringZ[G].Showthat7L[G]/7LN h(G, A)p-' 0->AGA*A°-'-'0. p =(d- 1)p-te, =go,,(AG)plgo,p(A). 315 fori =0, fori =-1. [L : K] 11 Chapter V Chapter A § 1. The Local Reciprocity Law Local Reciprocity § 1. The .r~ Local Class Field Theory Field Class Local #H`(G(LIK),L*) __ = 1 + pK the group of n-th higher units, n = 1, 2, ... , = 1 + pK the group of n-th higher units, (K) = An n K*. the group of n-th roots of unity, and An K = oK/PK the residue class field, q =qK=#K, An of K, i.e., PK = IOK IrK, or simply ir, denotes a prime element pK = (a E K VK (a) > 0} the maximal ideal, pK = (a E K VK (a) vK be the discrete valuation normalized by VK (K*) = 7G, vK be the discrete valuation OK = { a E K VK (a) ? 0} the valuation ring, OK = { a E K VK UK = { a E K* VK (a) = 0} the unit group, UK = { a E K* VK (a) = 0} the unit Jalp = q-"K(a) the normalized p-adic absolute value, Jalp = q-"K(a) the normalized p-adic The abstract class field theory that we have developed in the last chapter theory that we have developed in The abstract class field In local class field theory, the role of the profinite group G of abstract In local class field theory, the role of UK) is now going to be applied to the case of a local field, i.e., to a field which is applied to the case of a local field, i.e., is now going to be has a finite residue to a discrete valuation, and which complete with respect extensions K of II, (5.2), these are precisely the finite class field. By chap. Let We will use the following notation. the fields Qp or lFp((t)). class field theory is taken by the absolute Galois group G (k 1k) of a fixed class field theory is taken by the absolute A by the multiplicative group k* local field k, and that of the G-module a finite extension K1k we thus have of the separable closure k of k. For verify for the multiplicative group of AK = K*, and the crucial point is to a local field the class field axiom: L I K of local fields, one has (1.1) Theorem. For a cyclic extension .-, Vn+1, ono OKa°. Then fir ai+l E (Vn+i+l)G Chapter V. Local Class Field Theory Field Class V. Local Chapter n=1,2,..., aEG : K], where we have put G = G(L !K). we have put G : K], where bi E Vn+i ) UL>L*-!-L+ Z)0, ) UL>L*-!-L+ 1 Vn=1+7r"M, sac h(G, UL) = h(G, UL/V") h(G, V") = 1 h(G,L*) = h(G,Z) h(G, UL) = [L : K]h(G,UL). h(G,L*) = h(G,Z) ai = (NGbi)ai+1, (7rgM)(7rj{M) =7r2 MM C7r2 OL c ,.2 -NM C 7rKM. (7rgM)(7rj{M) =7r2 MM C7r2 OL V'/ V'+1 = MI7rKM = ® (OKIpK)a° = IndG(OK/pK) V'/ V'+1 = MI7rKM = ® (OKIpK)a° because UL/V" is finite. (of finite index) of UL, because we have ,'7,KVia the correspondence does (1 - 7rKµ)- = 1 + 7r" (E°_1 G-isomorphisms as in II, (3.10), 1 + 7rKa H a mod 7rK M, we obtain (G, Vn/Vn+1) = 1 for i = 0, - 1 So by chap. IV, (7.4), we have H` that H' (G, V") = 1 for i = 0, - 1 and n > N. This in turn implies 0 and a E (V")G, then a = (NGbO)al, and n > N. Indeed, if for instance i = a1 = (NGb1)a2i for some b1 E with b° E V", a1 E (Vn+1)G and thus a2 E (V"+2)G, etc.; in general, product b = TlO°° bi E V", This yields a = NGb, with the convergent way we have for a E V" such that so that H°(G, V") = 1. In the same E V", where or is a generator of G. NGa = 1, that a = b°-1, for some b Thus H-1(G, V") = 1. We now obtain in which Z has to be viewed as the trivial G -module, yields, by chap. IV, viewed as the trivial G -module, yields, in which Z has to be (7.3), choose a normal show that h(G, UL) = 1. For this we Hence we have to 20), a E OL, and of LI K (see [93], chap. VIII, § 12, th. basis {a° I O E G} open (and closed) G-module M = consider in OL the the open sets is open, we have neighbourhoods of 1 in UL. Since M form a basis of open n > N the V" are even subgroups 7rK OL C M for suitable N, and for - 7rKµ, for µ E M, lies in V", so Hence V"V" C V", and since 1 318 90") (3.5) ("Hilbert of proposition is the claim -1 this For i = Proof: is the Herbrand quotient to show is that IV. So all we have in chap. [L = #HO(G,L*) = h(G,L*) sequence The exact con 319 for n=1,2,... = 1 4., = 1, (n) UL - - a - 1, >xq'=0} b019 Then K*':CN,anditisenough toshowthatK*"containsagroupof The casechar(K)fin.LetN beasubgroupoffiniteindexn=(K*:)V). first in the compactgroupUL,and has finiteindexinUK.Weprovetheconverse subgroup NLIKULwhichitselfisopen,foritclosed, beingtheimageof By (1.3),thishasfiniteindexinK*.Itisalsoopenbecause itcontainsthe open inthenormtopologycontainsbydefinitionagroupof normsNLIKL*. finite indexwhichareopeninthevaluationtopology.Asubgroup Nwhichis of K*whichareopeninthenormtopologyprecisely thesubgroupsof Proof: Bychap.IV,(6.7),allwehavetoshowisthat thesubgroupsN cry field KandtheopensubgroupsNoffiniteindexinK*.Furthermore, gives a1-1-correspondencebetweenthefiniteabelianextensions ofalocal (1.4) Theorem.Therule extensions ofalocalfieldK.Itisformulatedinthefollowing is againthegroupofnormsNCIRC*={zzIzE(C*}. The kernelof( by residue symbol law: fortheuniquenon-trivialGaloisextensionGIR,wedefinenorm R =Qalongwiththep-adicnumberfields It issurjectiveandhaskernelNLIKL*. and ir-rEisaprimeelement,then or, inotherwords,&IK=AK,forsomenEN.IfisthefixedfieldofQ § 1.TheLocalReciprocityLaw L1 CL2NL1RNL2,NL,L2=NL,fNL2,NL,nL2 The reciprocitylawgivesusaverysimpleclassificationoftheabelian In globalclassfieldtheorywewillhavetotakeintoaccountthe Inverting rLIKgivesusthelocalnormresiduesymbol , C IR)isthegroupR+ofallpositiverealnumbers,which can 1rLIK(a) =NEIK(t)modNLIKL*. ( (a, CIR) ( LIK) :K* LH.NL=NLIKL* cCIR) :R*_)G(CIR) G(LIK)ab sgn(a) F. It also admitsareciprocity r.. 321 =NL,NL2. a GaloisextensioncontainingL1,then put K,=K(p").IfKincontainsagroupofnormsNL,ILi,andis norms. ForthisweuseKummertheory(seechap.IV,§3).Wemayassume 322 elaborate. Sincetheresulthasnofurtheruseinremainderofthisbook, theory, attheexpenseofadhocargumentswhichturnouttobesomewhat theory whichwewilldevelopin§4.Butitisalsopossibletodowithoutthis The casechar(K)=pin.InthistheproofwillfollowfromLubin-Tate and (*)yields is isomorphictoG(LIK)andhasexponentn,wehavethatK*nc_NLIKL*, By chap.II,(5.8),K*/K*"isfinite,andthensoG(LIK).SinceK*/NLIKL* exponent n.Thenbychap.IV,§3,wehave So letµ"CK,andL=K(n*)bethemaximalabelianextensionof that K*containsthegrouppnofn-throotsunity.Forifitdoesnot,we neighbourhoods of1inK*. We putUK)=UKanddefine: exists anabelianextension LIKsuchthatNLIKL*=Al.Thisisthe istic ofKdoesnotdividen,thentheextensionL=K(n K*) IKisfinite,and (1.5) Proposition.IfKcontainsthen-throotsofunity,and ifthecharacter- and chap.XIV,§6. we simplyreferthereadertobeautifultreatmentin[122],chap.XI,§5, and thereforeK*"=NLIKL*. (*) of K*containssomehigher unitgroupU,astheseformabasisof finite indexisautomatically open -seechap.II,(5.7).)Everysubgroup "class field"ofAl.(Incidentally, whenchar(K)=0,everysubgroupof statement isthat,foreveryopensubgroupNoffinite index inK*,there one has Theorem (1.4)iscalledtheexistencetheorem,because itsessential The proofalsoshowsthefollowing NLIKL* =NKIIK(NLIKLL*)CNKIIK(NLtIK1Ll) NLIKL* =K*n #K*/K*n =#G(LJK)#K*/NLIKL* Hom(G(LjK),µn) =K*/K*" C NKIIK(Kln)K*n. and L'1 G(LIK) =K*/K*" Chapter V.LocalClassFieldTheory 112 .'h , tom-. bob is calledtheconductorofLIK. number >0suchthat unramified. then NMIKM*=(irk-)xUKcNLIKL*,andMDL,i.e.,LIKis for n=(K*:NLIKL*).IfMIKistheunramifiedextensionofdegreen, f =1.Ifconversely1 Proof :IfUKisunramified,then=NLKULby(1.2),sothat if itsconductorisf=1. (1.7) Proposition.AfiniteabelianextensionLIKisunramifiedifandonly (1.6) Definition.LetL1Kbeafiniteabelianextension,andnthesmallest § 1.TheLocalReciprocityLaw group (p)xUW the fieldQp(µP")ofp"-throotsunity: ground fieldK=Qp,theclassofgroup(p)x immediate analoguesofthecyclotomicfieldsoverQp.Incase are particularlyimportant.Wewillcharacterizethemexplicitlyin§5,as group (2rf)xUK).Thereforetheclassfieldsforgroups(nrf) finite abelianextensionLIKiscontainedintheclassfieldofsucha form (Trf)XUKU.Thisisagainopenandoffiniteindex.Henceevery for v>1,providedp isomorphism We nowconsidertheexponentialmapofQp.Bychap. II, (5.5),itgivesan root ofunity,then1- L IKistotallyramifiedofdegreepn-1(p-1),andif1" is aprimitivep"-th Proof :LetK=QpandLQp(ip").Bychap.II,(7.13), theextension (1.8) Proposition.Thegroupofnormstheextension Q (UK ))Pa-1(P-1)=UK)if into theisomorphismUK) transforms theisomorphism pK-,s-1givenbyaHps-1(p-1)a, Every opensubgroupNoffiniteindexinK*containsagroupthe try QP. UK) is aprimeelementofLnormNLIK(1-)=p. C NLIKL*.Thentheideal p 542,and(UK exp :pK-UK f then UKcNL1KULandnKEKL UK +s-t)givenbyxHxP" (P-1),sothat 2, andforv>even ifp=2. pK ) ))2"-2 = UK p (µP")I UK) if p=2,n>1 is precisely Q pisthe UK) 4-, 323 It 324 Chapter V. Local Class Field Theory (the case p = 2, n = 1 is trivial). Consequently, we have UK ) c_ NL K L * if p 0 2. For p = 2 we note that UKl=UK>U 5UK31 = (UK j)2 U 5(UKl)2, because a number that is congruent to 1 mod 4 is congruent to 1 or 5 mod 8. Hence U( n)-(U(2) 2" U J2rs-2 (UK)) 2n-1 It is easy to show that 52 -2 = NLIK(2+i), so UK) C NLJKL* holds also in case p = 2. Since p = NLIK(1 - 0, we have (p) x U(n) C NLIKL*, and since both groups have index pn-1(p - 1) in K*, we do find that NLJKL* = (p) x UK) as claimed. As an immediate consequence of this last proposition, we obtain a local version of the famous theorem of Kronecker-Weber, to the effect that every finite abelian extension of Q is contained in a cyclotomic field. (1.9) Corollary. Every finite abelian extension of L I Q p is contained in a field Q () . whereis a root of unity. In other words: The maximal abelian extension QpbI Q p is generated by adjoining all roots of unity. Proof: For suitable f and n, we have (pf) x U(n) C NLIKL*. Therefore L is contained in the class field M of the group (p1) x ((pf) x UQP) n ((p) x By (1.4), M is the composite of the class field for (pf) x UQP - this being the unramified extension of degree f - and the class field for (p) x UUP. M is therefore generated by the (pf - 1)pn-th roots of unity. From the local Kronecker-Weber theorem, one may readily deduce the global, classical Theorem of Kronecker-Weber. (1.10) Theorem. Every finite abelian extension L IQ is contained in a field Q generated by a root of unity . is thegroup If nisnotdivisiblebytheresidue characteristicp,thentheorthogonalcomplementof of finitegroupsgivenby erate pairing If Kisap-adicnumberfieldandnnaturalnumber,thenthere existsanondegen- Exercise 2.ForanarbitraryfieldKandaGK-moduleA,put the latterprovidedthatnisnotdivisiblebyresiduecharacteristic. Exercise 1.FortheGaloisgroupF=G(KIK),onehascanonically L CQ(L1). and therefore[M:Q}=[Q(An)Q},sothatMQ(Itn).Thisshows hand wehave theorem fromchap.III,(2.18),itequalsQ,i.e.,I=G(MIQ).Ontheother p ES.ThefixedfieldofIisthenunramified,andhencebyMinkowski's is Euler'sfunction.LetIbethesubgroupofG(MIQ)generatedbyall1p, the groupG(Qp(/.tpep)IQ),andconsequentlyhasorderqp(pep) Q (ILpepn')IQp.TheinertiagroupIpofM.IQ,isthereforeisomorphicto with (n',p)=1.Q gives thecompletionLp,then completion withrespecttoaprimeofMabovepwhoserestrictionL We willshowthatLCQ(µn).ForthisletM=L(I0).ThenMIQis Let pepbetheprecisepowerofpdividingnp,andlet Then LpIQisabelian,andtherefore abelian, andifpisramifiedinMIQ,thenmustlieS.IfMpthe let L Proof :LetSbethesetofallprimenumberspthatareramifiedinL,and § 1.TheLocalReciprocityLaw The followingexercises1-3presuppose4-8ofchap.N,§3. ICS p be thecompletionofLwithrespecttosomeprimelyingabovep. H'(I',Z/nZ) #I < Mp =Lp(ILn)Q H,;,.(K,Z/n) :=H'(G(KIK),Z/nZ) CH'(K,Z/nZ) H,,,(K,!-tn) pES fl H'(K,Z/nZ) xH'(K,Ft,,)--}Z/nZ p (µn')IQisthemaximalunramifiedsubextensionof #Ip = H'(K,A) =H'(GK,A). 211 Z/nZ (X, a)HX((a,KIK)) CD' pES flI H'(G(KIK)>Fr,,) 9H'(K,An) p (IJpePn')=Q(ApeP)(An'), n =I-P o(peP) =So(n)[Q(pn):Q] and pES H'(P, p.")=UKK`"/K'" p _CQ(µ17P),forasuitablenp. , where 325 O .00 Exercise 5.LetLIKbethecompositeofall7Lp-extensionsap-adicnumberfield Hint: Useexercises4-8ofchap.IV,§3. Hom(A, K*)insteadofN.,,. 3 generalizetoanarbitraryfiniteGK-moduleAinsteadofZ/nZ,andA' Exercise 4(LocalTateDuality).Showthatthestatementsofexercises2and commutative diagram Exercise 3.IfLIKisafiniteextensionofp-adicnumberfields,thenonehas 326 principal unitsUK'sontotheramification groupR(KabIK). yields anisomorphism mapping dK:GK--)-Z.Show: Exercise 11.TheWellgroupofalocalfieldKisthepreimage WKofZunderthe p-adic numberfieldK. Exercise 10.Determineallclassfieldtheories(d:GK a p-adicnumberfieldK. Exercise 9.Determineallp-classfieldtheories(d:GK-a77p, v:K*-+74p)over theory (seechap.IV,§5,exercise2). defines ahenselianvaluationwithrespecttod,inthesenseof abstractp-classfield the absoluteGaloisgroup.Show: be achosenisomorphism,andletd:GQ-+Zptheinducedhomomorphismof Exercise S.LetQpIbethecyclotomicZ G (LIK)isaZ by allrootsofunityp-powerorder.Thefixedfieldthetorsionsubgroup Exercise 7.LetpbetheresiduecharacteristicofK,andletLfieldgenerated unity. Exercise 6.Thereisonlyoneunramified7Lp-extensionofK.Generateitbyroots Hint: Usechap.II,(5.7). G (LIK)isafree,finitelygeneratedZ,-moduleanddetermineitsrank. K (i.e.,extensionswithGaloisgroupisomorphicto77p).Showthatthe which mapstheunitgroupUK ontotheinertiagroupI(KabK),andof The normresiduesymbol(,K" IK)ofthemaximalabelianextensionKabK For asuitabletopologicalgeneratoruofthegroupprincipalunitsQp, 9-0 4., [rte ... P -extension. ItiscalledthecyclotomicZ a.. H' H'(L, Z/nZ) bpi '±' ( D,'^' , KabjK) :K*-!-*WKb, x H' (K,p.) p -extensionofQ,,letG Chapter V.LocalClassFieldTheory log it log a -* P -extension. 7L/n7L. v :K*-37L)overa (&, IQ,)=Z, C]. _ use ofthecompletionKmaximalunramifiedextensionalocal known todate.Wehaveinvokeatranscendentalmethodwhichmakes chap. VI,§5).Unfortunately,thereisnodirectalgebraicproofofthisfact r -u-1modp°.Thisresultisimportant,notonlyinthelocalsituation,but where a=up"P(a),and"-1isthepower-rwithanyrationalinteger symbol fortheextensionQ But if is unramified,andthenormresiduesymbolobviouslygivenby § 2.TheNormResidueSymboloverQ has asolution,00inK=oK/pK,i.e., of K,andwehavethegyp-invariantisomorphisms Proof: Let7rbeaprimeelementofK.Then.7risalso aprimeelement admits asolutioninoK,resp.UK.Ifx9'=xforxEok,thenOK. (2.1) Lemma.ForeverycEoK,resp.Uk,theequation prove the field K.WeextendtheFrobeniusVEG(KIK)tokbycontinuity.Firstwe it willplayanessentialrolewhenwedevelopglobalclassfieldtheory(see have tosolvethecongruence y2-Y2bl=0mod7r,orequivalentlythe x2 =1+yen,gives a2 E For similarreasons,wefind thata1=x2-1a2,forsomeEUK)and field Kofkisalgebraicallyclosed,theequation.o=y q=x-c(qqK) (see chap.II,(3.10)).LetCEUKandc=modpK.Since theresidueclass If UKl, sothatc is aprimitivem-throotofunity,with(m,p)=1,thenQp(0I is aprimitivep'1-throotofunity,thenweobtainthenormresidue coy § 2.TheNormResidueSymboloverQp coy UK/Uk =K*, c=x° Ix! x`° -x=c, a1x2- = (x1x2)`O-1a2.Indeed,putting a1=1_+bin, at, Ill xl EUK, 1 -(y2y2b1)7rmod7r2, i.e.,we i.+ resp. explicitly inthesimpleform UK )/UKFl)- III acs xv-1 =C'. rpup(a) a1 EUK'). 211 per' C7" K G`' ion 327 form withcoefficientsaiEoK suchthat e(X) E£nbeLubin-Tateseries. LetL(X1,...,Xn)_F_n1alXibealinear (2.2) Proposition.LetJrand nbeprimeelementsofK,andlete(X)E£, polynomial fortheprimeelementp. In thecaseK=Qforexample,e(X)(1+X)P- 1 isaLubin-Tate polynomials. Thesimplestoneamongthemisthepolynomial Xq+JrX. where u,aiEOKandu=1modJt.Thesearecalled theLubin-Tate £n thereareinparticularthepolynomials class fieldofK.ThetotalityallLubin-Tateseriesis denotedbyE.In where q=qKdenotes,asalways,thenumberofelementsinresidue definition apowerseriese(X)EOK[[X]]withtheproperties power seriesinOK[[X1,...,Xn]]whicharisesfromFbyapplyingqptothe coefficients ofF.ALubin-TateseriesforaprimeelementJrKisby in theequation(*)givesx=limn-).oaxnEOK,becauseKiscomplete. xn+1 with cnEOK,dooK,andthereforex=(xn+cnrrn)Jrn+l implies furthermorethaty,'=y,,,sowegetasaboveyncn+Jrdn, therefore x=x1+7r(b+c)x1+7ry1,y1Eok.TheequationXn+Jrnyn implies a`'=amodJr.Hencex1+Jrc,withEOK,Cok,and Indeed, forn=1wehavexa+Jrb,withEoK,bandx° isomorphisms pK/pK1=K. The solvabilityoftheequationx°-x=cfollowsanalogously,using and passingtothelimitfinallygivesc= closed. Continuinginthisway,weget equation y2-Y2b1=0inK.Thisispossiblebecausexalgebraically 328 (*) For apowerseriesF(X1, Now letxEoKandxw=x.Then,foreveryn>1,onehas +rn+lyn+1, for c =(x1x2...xn)'P1an e(X) =_JrXmoddeg2 e(X) =uXq+Jr(aq_1Xq-1+--+a2X2)+JrX, Sao X =Xn+Jrnyn 7rL(X1, ...X,) some xn+iEOK,Yn+1oK.Nowpassingtothelimit III 112 ..., X,)EOK[[Xi,...,X,,]],letFvbethe With xnEOKandY,,oK. , and +'A x,1 E Chapter V.LocalClassFieldTheory e(X) -XqmodJr, U(n-1) .--,Xn)- a,: EUK x =Hn°1x,TEU. (n) (3) mod deg(r+1)and with Gr+t(X)=e(Fr(X))- F,°(e(X))E0[[X]].WehaveGr+t(X) (2) show thatE,.+1(X)hastosatisfythecongruence congruences Er+I (X)Eo[X]ofdegreer+1whichhasyettobe determined.The We thenputFr+1(X)=Fr(X)+Er+twithahomogeneous polynomial already beenfound,andthattheyareuniquelydetermined bycondition(1). for r=1byhypothesis.AssumethattheE,(X),v1,...,r,have v =1weareforcedtotakeE1(X)L(X).Condition(1)isthensatisfied for everyr>1.WedeterminethepolynomialsE(X)inductively.For L (X)and Clearly, F(X)isasolutionoftheaboveproblemifandonlyF1 be apowerseries,E(X)itshomogeneouspartofdegreev,andlet Proof: Letobeacompletesubringofogsuchthato`°=o,which If thecoefficientsofe,LlieinacompletesubringOOksuchthatoP=o, (1) (e (Xt),... contains thecoefficientsofe,L.WeputX=(X1, then Fhascoefficientsinoaswell. E Og[[XI, T hen § 2.TheNormResidueSymboloverQ there Gr+I (X) F+1(e(X)) e(F,-+1(X)) .l+ , e(X.)). e(F(Xi, ...,X,))=F°(e(X1),...,e(X,)). . is , ice-. Xn]] satisfying Gr+t(X) =Fr(X)4-F"(XQ) -0mod7r F(X1, ...,X,:)=L(XI,...,Xn)moddeg2, e(Fr(X)) -F,P(e(X))moddeg(r+1) a III III Jr Er+t(X)-n ion Let uniquely determinedpowerseriesF(XI III F,?(e(X)) +n''+1E,+1(X)moddeg(r+2) e(Fr(X)) +7rEr+I(X)moddeg(r+2), ICI F(X) =F Fr(X) v=1 00 r+t E++1(X) V=1 > E,M. r E o[[X]] _= 0moddeg(r+2) coo ... , .U+ 'LS and e(X) , ... ;X) 329 0 330 Chapter V. Local Class Field Theory because e (X) = jq) X q modrr and av = a' mod jr for a E 0. Now let X' = XI' Xn' be a monomial of degree r + 1 in o[X ]. By (3), the coefficient of X' in Gr+i is of the form -7r,B, with S E o. Let a be the coefficient of the same monomial X` in Er+1 Then rra - ira is the coefficient of X' in 7r Er+i - irE +I.Since Gr+1(X) = 0 mod deg(r + 1), (2) holds if and only if the coefficient a of X' in E,.+1 satisfies the equation (4) -7rf + Ira - 0 for every monomial X ` of degree r + 1. This equation has a unique solution a in oK, which actually belongs to o. For if we put y = n-'5-rr+1 we obtain the equation a-yca`°=/3, which is clearly solved by the series a=,B+y/w+YI+`pp,2 +...E o (the series converges because vR(y) >1).If a' is another solution, then a - a' = y (a`' - hence vK (a - a') = vK (Y) + vk ((a - a')9) = vK (y) + vK (a - a'), i.e., vK (a - a') = co because vK (y) ? 1, and therefore a = a'. As a consequence, for every monomial X' of degree r + 1, equation (4) has a unique solution a in o, i.e.,there exists a unique Er+1(X) E o[X] satisfying (2). This finishes the proof. (2.3) Corollary. Let n and it be prime elements of K, and let e E £n, e E £X be Lubin-Tate series with coefficients in OK. Let 7r = u n, u E UK, and u = &-1, s e U. Then there is a uniquely determined power series 8 (X) E oK [[X]] such that 8 (X) = eX mod deg 2 and eo0=9 oe. Furthermore, there is a uniquely determined power series [u] (X) E OK [[X]] such that [u](X) = uX mod deg2 and eo[u]=[u]oe. They satisfy 8`0=9o[u]. Proof: Putting L (X) = EX, we have irL (X) = it L`O(X) and the first claim follows immediately from (2.2). In the same way, with the linear form L (X) = uX, one obtains the existence and uniqueness of the power series [u](X) E OK[[X]]. Finally, defining 01 = 8'P-' o [u], we get e o 81 = (e o 8)w-' o [u] = (9' o e)S'o [u] _(0`1o [u])5' o e = 04, o e, and thus 81 = 0 because of uniqueness. Hence B5' = 0 o [u]. theorem isbasedonthefactthatfieldE'canbeexplicitlygeneratedby ramified becausefljK=dK(a)1bychap.IV,(4.5).Theproofofthe element suchthatdK(a)=1and&IL6.ThefixedfieldEofaistotally we viewcasanelementofG(LIK).Thena=crcPLEFrob(LIK)is Since defined by Let K=Qp,L Proof: AsNisdenseinZp,wemayassumethatuEN,(u,p)=1. unity. Thenonehas (2.4) Theorem.Leta=up"P(a)EQ*,andlet § 2.TheNormResidueSymboloverQ, i.e., 7rEEX.Wewillshowthat Indeed, [u](A°)_(1+A°)"-1 of 1by Substituting theprimeelementA such that (1+X)"P-1 =[u](e(X)).By(2.3),thereisapowerseries8(X)EOK[1X]1 as wellthepolynomial[u](X)=(1+X)"-1.Thene([u](X))_ to thecompletionLofL,andconsidertwoLubin-Tatepolynomials a primeelementnEwhichisgivenbythepowerseries0of(2.3). we have Since en-1(X)P(X), sothat XP"-'(P-') modp.Finally,e'(X) by Eisenstein'scriterion,ase(X)=XPmodp,and so e'(X) =e(ei-1(X)).P(X)ismonicofdegreepn`1(p- 1)andirreducible is theminimalpolynomialof7r_r,wheree`(X)defined bye°(X)=Xand In ordertodothis,assumeaandcp=IPLhavebeenextendedcontinuously e` (7rz)=e'(O(),))09`(e` (A))=0'((i+A)P`-1)ef' en(9rL) =0,en-1(7rL) CIO e(X)=upX+XP ande(X)=(1+X)P-1 r,^ jr' =0 °°' is totallyramified,wehaveG(LIK)-G(LIK),and eoB=8'°oe o°, P(X) =en-'(X)P-1+upEZP[X] P(nE')en-1(rj) = "''y `-h and letorEG(LIK)betheautomorphism YrE =0(A). ° =u-t 0 0,andthus e([u]O°)) =0(1)7rE> III III G]. and = en-1(X) 1 ofL,weobtainaprimeelement 0°=Bo[u]. acv "-1 P(irz) =0. be aprimitivepn-throotof 1 =A,andtherefore (up +en-1(X)P-1)= en-I (X)p-1 w°4 tai 331 1), ... some p"-throotofunity. Exercise 4.Deducefromexercises2and3the formula (u,Qfor Hint: Seechap.IV,§5,exercise1. Putting v=]l;onehas(x,LIK)v-t. Exercise 3(TheoremofDwoRK).LetLIKbeatotallyramified abelianextensionof y=fiz°'-',v; EG(LIK). Exercise 1.Thep-classfieldtheory(d:GQ-- rLIK(u) asumodNLIKL*,andso(u,LIK)=or. element 7r read §4.Letusnotethatonewouldgetadirect,purelyalgebraicproof,if required. and chooseaEG(LIK)suchthat p-adic numberfields.LetXEK*andYL*suchthatNE, k(y)=x.Letz;EV Show thatNtIRL*=K*,andeveryyEL*withNE,k(y)1isoftheform the completionofmaximalunramifiedextensionL(resp.K)K). Exercise 2.LetLIKbeatotallyramifiedGaloisextension,andlet(resp.K) Hint: Showthatthisstatementisequivalenttoformula(2.4):(u,Qp symbol (,LIK). the cyclotomic7LP-extensionofQp(see§1,exercise7)yieldsamenormresidue 7LP-extension ofQp,andthep-classfieldtheory(d:GQ-*7Lp,DQp-->-ZP)for and everyzeroofP(X)=e"(X)/e"-t(X)wouldhavetobeaprime & =urpL.ThissplittingfieldwouldthenhavetobeequaltheE the polynomiale"(X)isabelian,andthatitselementsareallfixedunder one couldshowwithoutusingthepowerseries0thatsplittingfieldof and thereforerLIK(v)=umodNLIKL*,i.e.,(u,LIK)(a,LIK)or,as we obtain 332 In ordertoreallyunderstandthisproofoftheorem(2.4),onehas Observing thatNLIK Boa !}' l-, NEIK(rE) _(-I)dp(o)= E suchthatNZIK(7rj)asumodNLIKL*,inwhichcase vii 'L7 ?c. -1) _(-1)dp,d=[L:K](seechap.II,(7.13)), ywrc-1 = IIz (_l)dpu Chapter V.LocalClassFieldTheory 7Ly, v:Q*->.Z)fortheunramified ' . as umodNLIKL* Q p). °(a hence (a,K( the Hilbertsymbol,wehave Proof: TheimageofaundertheisomorphismK*/K*" - G(LJK)ofclass character Xb:G(LIK)-+Angivenby(r)=t/ '.Bydefinitionof the isomorphismK*/K*"-Hom(G(LIK),µn)ofKummer theoryisthe field theoryisthenormresiduesymbola=(a,LIK).The imageofbunder by (3.1) Proposition.Fora,bEK*,theHilbertsymbol (a b)Eg,isgiven the followingproposition. symbol. Itsrelationtothenormresiduesymbolisdescribedexplicitlyin (bilinear inthemultiplicativesense).ThispairingiscalledHilbert therefore definesanondegeneratebilinearpairing The bilinearmap On theotherhand,Kummertheorygivescanonicalisomorphism and classfieldtheorygivesusthecanonicalisomorphism By (1.5),wethenhave residues. "Hilbert symbol".Thisisahighlyremarkablephenomenonwhichwilllead theory. Itistheinterplaybetweenboththeories,whichgivesriseto one hand,Kummertheory(seechap.IV,§3),andontheother,classfield char(K) =0).OversuchafieldKwethenhaveatourdisposal,onthe is relativelyprimetothecharacteristicofK(i.e.,ncanbearbitraryif the groupµofn-throotsunity,wherenisa'naturalnumberwhich us toageneralizationoftheclassicalreciprocitylawGauss,n-thpower § 3.TheHilbertSymbol Let L=K("*)bethemaximalabelianextensionofexponentn. Let Kbealocalfield,or=IR,C.Weassumethatcontains G(LIK) xHom(G(LIK),1-An)--*µn, BCD )IK) (a, K( Hom(G(L IK),µn)=K*/K*n a § 3.TheHilbertSymbol =(a,LIK)=(ab) b G(LIK) Q--'K*IK*". CO. )=Xb(o)=Q K*/K*n NLIKL* =K*n, )IK) X K*/K*n = ( _ ) :a) / fb (cr,X) / An 211 . X(Or), i1+ 333 (ii) (1) (3.2) Proposition. 334 is cyclicofdegreem,andtheconjugatesx-i that yd=bhasasolutioninK,andletndm.ThentheextensionK(48)I for someprimitiven-throotofunity.Letdbethegreatestdivisornsuch (vi) reformulatesthenondegeneratenessofHilbertsymbol. Proof: (i)and(ii)areclearfromthedefinition,(iii)follows(3.1), (vi) (v) (iv) (iii) follows from Choosing x=1,b1-a,and0,-athenyield (v).(iv)finally Hence xn-bisanormfromK()IK,i.e., such thatj=imodd.Wemaythereforewrite The Hilbertsymbolhasthefollowingfundamentalproperties: If bEK*andXKsuchthatxn-b (a (a'1 (a pb) If (ab)=1forallbEK*,thenaEK*n. (alb) _ (alb) P (aPb)(bPa) b) a)=1and(a' a)=1, = 1 (aPb) (aP1), (apb) (ap,11), xn (bra)-t, xn -b=1JNK(#)IK(x a isnormfromtheextensionK(')IK, - b=fl (a, a, (xn P - n-1 i=0 ab)(b, dab)-(ab,pab a)(aPb)(bPa)(b, i=0 d-t pb,b)=1. Chapter V.LocalClassFieldTheory 0, then - f"=b, are theelementsx-J P b r-, )=1 D because (a,lR(/)IR)=1forb>0,and(-1)T course, § 3.TheHilbertSymbol K' bethesplittingfieldofpolynomialX"-umodpoverresidue n Iq-1inthatcase.Firstweestablishthe call thisthecaseoftameHilbertsymbol.Sinceµ"cµq-1onehas a localfield(¢l[8,cC)whoseresiduecharacteristicpdoesnotdividen.We the letterpsymbolicallystandsforaninfiniteplace. nIr, i.e.,7trEK*",andthusxUKK*". Jr isaprimeelementofK.ThenvL("i L =K("x)isunramifiedoverK,andletxuJrr,whereuEUK into linearfactors,so field K'(seechap.II,§9,p.173).ByHensel'slemma,X"-usplitsover class fieldK,andletK'IKbetheunramifiedextensionwithresidue Proof: Letx=uy'withuEUK,YK*,sothatK(im)K("u). unramified ifandonlyxEUKK*'. (3.3) Lemma.Let(n,p)=1andxEK*.TheextensionK(/)IKis (3.4) Proposition.If(n,p)= 1anda,bEK*,then will nowprovethe with CO(u)Eµq-1andUKu=a)modp.With thisnotationwe where a=VK(a),.vK(b). In thecaseK=Rwehaven1or2.Foronefinds,of Next wedeterminetheHilbertsymbolexplicitlyincasewhereKis Since UK=µq_1XUKI,everyunituehasaunique decomposition (a'bb) P fib = 1,andforn2wehave l ( a a b P P b l )=(-1) = C K'isunramified.Assumeconverselythat u =co(u)(u) ]"m cv((-1)a sgna-1 2 0 ba \4-1)l ) =nVL(Jr')EZ,hence sgnb-1 2 " sgna-I for b<0.Here 335 (3.5) Proposition.Let(n,p) =1anduEUK names arejustifiedbythe We callittheLegendresymbol,orn-thpowerresidue symbol.Both it. Wemaythereforeput hence ("u)_(7r,u),becauseµq_1ismappedisomorphicallyontoK*by By (3.3),weseethatK'IKisunramifiedandbychap.IV,(5.7),(7r,K'IK) is obviouslybilinear(inthemultiplicativesense).Wemaythereforeassume Proof: Thefunction 336 (in thecase(n,p)=1)isindependentofchoiceprimeelement UK --)-1C*. is theFrobeniusautomorphismV='PK' y = (7r, -7r)_( that aandbareprimeelements:=it,-7ru,uEUK.Sinceclearly The propositionshowsinparticularthattheHilbertsymbol is therootofunitydeterminedby 7C, 1t P and K'=K(y).Thenwehave (X, u)= (U) 7r, -n 'PY Y = - Yq-1(q-1)ln w(u)and ) =1,wemayrestricttothecasea7r,bu.Let (up) (a, b):=w((-1)afi p ll := = u Jr, ll (----) u u isann-thpowermodpK. (q-1)/n I K. for (n> K'IK)Y mod pK. w(u) a- / ba III Chapter V.LocalClassFieldTheory (u)(q-1)ln Consequently, l U EUK. . (q-1)ln (q-1)1n - Then onehas = ( 7r u\ (n, u)mod III P Y . . 337 and (5, - 1} a2 if u is a unit. . p . mod pK ? w(u) _ (c)' ? w(u) _ III a21 b21 2 2 a',b'EUQ and ( P P ) E(a1a2) as E(a1) + E(a2) mod 2. _ (-1) (-1)(p-1)/2 (-1)(a2_1)/8 l and w(u) E ant U = W(u) P:---Ic, aAby. b2a) b=pflb', ( \ also in the case p In. Let us look at the case also in the case p In. (-1)a = (-1)', = If a E Z2, then (-1)a means If a E Z2, then (-1)a b) . 1 = P l 22 a, b ( \a2 ) a=Pea', =!D(u)n= = 1 =!D(u)n= be a primitive (q - 1)-th root of unity, and let m = g and let m unity, root of (q - 1)-th primitive be a An elementary computation shows that 1. 2 2, then It is an important, but in general difficult, problem to find explicit formulae in general difficult, problem to find It is an important, but n(ala2) = q(a1) + 77(a2) mod 2 § 3. The Hilbert Symbol Hilbert The § 3. Proof: Let Proof: root of unity, and a primitive m-th Then a is (2 for the Hilbert symbol = Q where n = 2 and K integer - a mod 2. where r is a rational Q , we write (3.6) Theorem. Let n = 2. For a, b e lip I n particular, one has (L' P) _ Ifp=2anda,bEUQ2,then p = 2. We put i7(a) = and will be left to the reader. So let E(a) = a Proof: The claim for the case p # 2 is an immediate consequence of (3.4), Proof: The claim for the case p # 2 is it Thus both sides of the equations we have to prove are multiplicative and is enough to check the claim for a set of generators of is such a set. We postpone this for the moment and define (a, b) _ (alb) residue characteristicpofK. BRUCKNER. Sincetheresulthasnotpreviouslybeenpublishedinaneasily by-land5. defines anisomorphismU(2) for n>1.Sincear-+2adefinesanisomorphism2222-+2322,xNX2 those of i.e., 2wouldbeasquareinQ therefore todetermine(2,5).5)=1wouldimplyx)forallx, case. Thereforewehave(-1,-1)=-1. i.e.,x=y22+z2,y,zEQ2.Since 5=4+1and2=1+1,wefindthat 338 if and denote theresidueofdifferentialfdX, where rpistheFrobeniusautomorphismofW.Further, letRes(fdX)EW f P(X)denotetheseries with aLaurentseriesf(X)EW((X)). x EKcanbewrittenintheform ring ofWinvectorsovertheresidueclassfieldK).Theneveryelement of integersthemaximalunramifiedsubextensionTKIQ accessible place,westateitherewithoutproofforthecasen=p"ofodd general case.Itwasdiscoveredonlyin1964bythemathematicianHELMUT (1, -1,5,5)isasystemofrepresentativesU/U(3),U/U2generated UQ2, Ui"l=U.Bychap.II,(5.5),exp:2"Z2-+U(°)isanisomorphism (-1,x) =1forallx,i.e.,-1wouldbeasquareinQ2,whichisnotthe (-1, 2)=5)1.Ifwehad-1)1,thenitwouldfollowthat f E1+pW[[X]] It remainstoshowthat By directverificationoneseesthatthevalueswejustfoundcoincidewith We have(2,2)=(2,-1)1and(5,5)(5,1.Itremains We have(-l,x)=1ifandonlyxisanormfromQ2( For anarbitraryLaurentseriesf(X) So letgpPCK,chooseaprimeelementrofandWbethering It ismuchmoredifficulttodeterminethen-thHilbertsymbolin resp. (-l)E(°)Elbl,intherespectivecases. log f:_ N,.4 f'(X) =rawX'P d logf:= --). , whichisnotthecase.Hence(2,5)=-1. x =PIT), 00 `--' UM. ItfollowsthatU(3)CU2.Since (-1)t+t ..r Tom is generatedby{5,-1}.WesetU= t f Chapter V.LocalClassFieldTheory dX, (f CD, ai X` Tam E W((X)),let p (i.e., the )1Q2, and Let 339 is a prime urn ai = 0. i-*-oo d log f P) mod p". P 9P P 1 log gP l a;X`, ai E W, W ((X)) * such that f (7r) = x and 00 i=-CC =;.w(x,Y) , g E d log g- _ P P f P s=r(7r), f (x,y) log P (1 and thus but it is much more complicated. A more recent treatment but it is much more complicated. A be a primitive p°-th root of unity. Then 1 - Then of unity. p°-th root primitive be a 1 + (1 - Xe7l(X))P° 1 - (1 - Xe?1(X))P° 1 2 Or; is a primitive p" -th root of unity (p $ 2). Putting 7r = 1 - a,) om' For the proof of this theorem, we have to refer to [20] (see also For the proof of this theorem, we these formulae the following It would be interesting to deduce from Now let Now h(X) = field where denoting by S the trace map from ' to Q p, we obtain for the p ° -th Hilbert symbol (x-y) of the field 0v the (3.7) Theorem. If x, y E K* and f where w (x, y) = TrWIZ Res h. deduced an explicit formula for the [69] and [135]). BRUCKNER has also case n = 2 the case n = 2', has been given by of the theorem, which also includes G. HENNIART [69]. and HASSE (see [9]) relative to the classical result of IwAsAwA [80], ARTIN n(X) E W[[X]] be a power series such that be a power series n(X) E W[[X]] series and let h (X) be the for the p°-th Hilbert can now state BRUCKNER's formula With this notation we char(m) # 2. symbol (x'2) , p = g(7r)=y,then for some unit s of K, where e is the ramification index of KIQ e is the ramification unit s of K, where for some § 3. The Hilbert Symbol Hilbert The § 3. Q element of 'C3 such thatp=x.Showxra X,Lhaskernel where onehasX.,(a)=a-z; for allorEG(LIK),withanarbitrary extension LIKofexponentna surjectivehomomorphism Exercise 4.AbstractKummertheory (chap.IV,(3.3))yieldsforthemaximalabelian that onehasker(p)=W(]FP). can p :W.(,Z)- closure, andletW(K)betheringofWittvectorslength n,withtheoperator Exercise 3.LetKbealocalfieldofcharacteristicp,let kbeitsseparable Exercise 2.Deduceproposition(3.8)fromtheorem(3.7). Exercise 1.Forn=2theHilbertsymbolhasfollowingconcrete meaning: a formalresidueResn-IlogDb.Astothesupplementarytheorems, one mayalsointerpretthe-exponentinformulae(1)-(3)as(p-1)- Since instance, onecanindeedobtaintheformulaefromBRUCKNER'Stheorem(3.7). in thecasev=1,andbyIwnsAwA[80]general.Infor The formula(1)wasprovedindependentlybyARTIN[10]andHASSE[61] representation ofbasanintegralpowerseriesinitwithcoefficientsZp. one hastodefinealsoDlog st coefficientofait-adicexpansionlogDb.Inthiswayitappearsas (3) (2) where Dlogbdenotestheformallogarithmicderivativeinitofanarbitrary (1) (3.8) Proposition.ForaEUn 340 The supplementarytheorems(2)and(3)gobacktoARTINHASSE[9]. For aEU p (-) =1 a, b P ) Sao one hasfurthermorethetwosupplementarytheorems Omodp, ..^ pa =Fa-a(seechap.IV,§3,exercises2and3).Show ax' +bye-z2=0 7r P can ( 'al a,bl Hom(G(LI _ S( r ip-1, i=p-1 000 ) andbEfivonehas S(loga)/p° togaDlogb)/p° D logrr=7t' has anontrivialsolutioninK. a.) Chapter V.LocalClassFieldTheory and loga-0modp2, x F-->X.=, E W,,(L) .L' do forthefieldQp. "division points"thatdothe sameforthefieldKasp"-throotsofunity bitrary localfieldKbyintroducing anewkindofrootsunitywhichare group allowsustoconstruct suchanexplicitcyclotomictheoryoverar- tensions Q tered forthecaseofcyclotomicfieldsoverfieldQp, i.e.,withtheex- Remark: Suchaformulacanalsobegivenforn>1(P.KOLCZE [88]). Exercise 7.Showthatinthecasen=1symbol[x,a)isgivenby does notdependonthechoiceofprimeelement7r. For everyfEKonehas be thecanonicalmaptodifferentialmoduleofKI(seechap.III,§2,p.200). with coefficientsinK.Showthatforco where f,istheformalderivativeoffinexpansionaccordingtopowers7r K =K((7r)).Let Exercise 6.LetKbetheresidueclassfieldofand7raprimeelementsuchthat (vi) [x,a)=0forallx (v) [x,a)=0forallaEK* that p=x. (iv) [x,a)=0 (iii) (ii) [x+y,a)=[x,[y,a). (i) [x,a)=(a, where (,LIK)isthenormresiduesymbol.Show: Exercise 5.Define,forXEW,(K)andaK*,thesymbol[x,a)W,,(F,)by §4. FormalGroups The mostexplicitrealizationoflocalclassfieldtheorywe haveencoun- [x, ab)_[.x,a)+b). p (1;)IQ,where C/] I K)-4,if a ENK(t)IKK(i)*,where § 4.FormalGroups d:K -+S2KIK, [x, a)=TrKI]FnRes(x [x, a)=X.,((a,LIK)), is ap'-throotofunity.The notionofformal X EpW,(K). Res w:=a_, df =f,'d7r, E W,,(K)with9z=x. a EK*P". fF- df, da) a a;7r')dlr, theresidue E . is anelementsuch 341 342 Chapter V. Local Class Field Theoryy (4.1) Definiton. A (1-dimensional, commutative) formal group over a ring o is a formal power series F (X,Y) E o[[X,Y]] with the following properties: (i) F (X, Y) = X + Y mod deg 2, (ii) F(X,Y) = F(Y,X) "commutativity", (iii)F(X, F(Y, Z)) = F(F(X,Y), Z) "associativity". From a formal group one gets an ordinary group by evaluating in a domain where the power series converge. If for instance o is a complete valuation ring and p its maximal ideal, then the operation x+y := F(x,y) F defines a new structure of abelian group on the set p. Examples : 1. Ga (X, Y) = X + Y (the formal additive group). 2. G,,, (X, Y) = X + Y + XY (the formal multiplicative group). Since X + Y + XY = (1 + X)(1 + Y) - 1, we have (x+y)+1=(x+1).(y+ 1). Gn, So the new operation + is obtained from multiplicationvia the translation Gm X H X + 1. 3. A power series f (X) = a1X + a2X2 + E o[[X]] whose first coefficient a1 is a unit admits an "inverse", i.e., there exists a power series f-1(X) =a, 1X+ E o[[X]], such that f (f (X )) = f (f -1(X)) = X. For every such power series, F(X,Y) = f-1(f(X)+f(Y)) is a formal group. (4.2) Definition. A homomorphism f : F G between two formal groups is a power series f (X) = a1X + a2 X2 + E o[[X]] such that f(F(X,Y)) = G(f (X), f (Y)) . §4. Formal Groups 343 In example 3, for instance, the power series f is a homomorphism of the formal group F to the additive group Ga. It is called the logarithm of F. A homomorphism f : F -4 G is an isomorphism if a1 = f'(0) is a unit, i.e., if there is a homomorphism g = f -1 : G -+ F such that f(g(X)) = g(f(X)) = X. If the power series f (X) = a1X + a2 X2 + satisfies the equation f (F (X, Y)) = G(f (X), f (Y)), but its coefficients belong to an extension ring o', then we call this a homomorphism defined over o'. The following proposition is immediately evident. (4.3) Proposition. The homomorphisms f : F --> F of a formal group F over o form a ring End, (F) in which addition and multiplication are defined by (fg)(X) = F(f (X), g(X)) ,(f o g)(X) = f (g(X)). F (4.4) Definition. A formal o-module is a formal group F over o together with a ring homomorphism o ---; Endo(F),a [a]F(X), such that [a]F (X) - aX mod deg 2. A homomorphism (over o' D o) between formal o-modules F, G is a homomorphism f : F -3 G of formal groups (over o') in the sense of (4.2) such that f([a]F(X)) = [a]G(f(X)) for alla E o. Now let o = OK be the valuation ring of a local field K, and write q = (OK : PK). We consider the following special formal OK -modules. (4.5) Definition. A Lubin-Tate module over OK for the prime element 7r is a formal OK -module F such that [ir]F(X) = Xq modjr. of formalc-modules,anditis anisomorphismifaisunit. (ii) ForeveryaEOKthepower series[a]e, (4.6) Theorem.(i)TheLubin-Tatemodulesfor7rare precisely theseries If e(X)=wesimplywrite[a]e,e(X)_[a]e(X) Thus FisaLubin-Tatemoduleiftheendomorphismdefinedbyprime I~-, Fe (X,Y),withtheformalOK-modulestructuregivenby that (a EOK)bethepowerseries(uniquelydeterminedaccordingto(2.2))such series fortheprimeelementitofK,andlet totality ofallLubin-Tatemodules.Lete(X),e(X)E0K[[X]]be Gm isaLubin-Tatemodulefortheprimeelementpbecause respect tothemapping Example :Theformalmultiplicativegroup element .7rgivesviareductiontheFrobeniusendomorphismofF. f (X) mod 7rof[7r]F(X)isanendomorphismF.Butontheotherhand, obtain aformalgroupF(X,Y)overtheresidueclassfieldIFq.Thereduction fact, ifwereducethecoefficientsofsomeformalc-moduleFmodulozr, theory, totheeffectthatprimeelementscorrespondFrobeniuselements.In 344 The followingtheoremgivesacompleteandexplicitoverallviewofthe This definitionreflectsoncemorethedominatingprincipleofclassfield [a],,e(X) 7Lp -±EndZp(,Gi) Fe(X,Y) =X+Y = X9isclearlyanendomorphismofF,itsFrobeniusendomorphism. Fe(X,Y) EQK[[X,Y]l = aXmoddeg2, ICI [p]Gm(X)=(1+X)"-1-Xpmodp. OK -*Endo,e(Fe), III .-r , mod deg2, C." a 1-* [a]e,e:Fe>Fe [a]G,,, (X)=(1+X and ,A. e([ale,e(X)) = e(Fe(X,Y)) =Fe(e(X),e(Y)) a H[a]e(X) III e (X) Chapter V.LocalClassFieldTheory [a]e,e(X) EOK[[X]l is ahomomorphism is aformalZp-modulewith )a - 1= ,A.[ale,e(e(X)) )=1 00 (v)x , defines aLubin-Tatemodulewith logarithmloge=f- X +7r-1XQ7r-ZXQZ- Exercise 4.Letnbeaprime elementofthelocalfieldK,andletf(X)_ Exercise 3.logG,n(X)_ such that/r(X)Fl(X,0)=1.ThenlogeX+°_inX" doeswhatwewant. Hint: LetF1=aF/aY.DifferentiatingF(F(X,Y),Z)F(X, F(Y,Z))yields such thatloge(X) Fl (X,0)=1moddeg1.Let>/r(X)+F°°a,,XRE[[X]] bethepower over R,thereexistsauniqueisomorphism Exercise 2.LetRbeacommutativeQ-algebra.Thenforeveryformal groupF(X,Y) Exercise 1.End,(G.)consistsofallaXsuchthataEo. the conditione([rr]e(X))_[Tr]e(e(X)),resp.e(e(X))=e(e(X)). instance, bothpowerseriescommencewiththelinearformirXandsatisfy and thendeducestheirequalityfromtheuniquenessstatement.In(6)for that bothsidesofeachformulaaresolutionsthesameproblem(2.2), (6) showsthatFeisaLubin-Tatemodule. is ahomomorphismofrings,i.e.,thatFeformaloK-module,and (1) and(2)showthatFeisaformalgroup.(3),(4),(5) (6) [7r]e(X)=e(X) (5) [ab]e uniqueness statementof(2.2).Fortheotherclaimstheoremonehas [a]e,e isahomomorphismofformaloK-modulesfromFFtoFe.Finally, (4) [a+b]e,e(X)=FeQa],,e(X),[b]e,e(X)), (3) [a]e,e(Fe(X,Y))=Fe([a]e,e(X),[a]e,e(Y)), (2) Fe(X,Fe(Y,Z))=Fe(Fe(X,Y),Z), (1) Fe(X,Y)=Fe(Y,X), to showthefollowingformulae. and F=Febecausee(F(X,Y))F(e(X),e(Y)),ofthe Proof: IfFisanyLubin-Tatemodule,thene(X):=[7r]F(X)E6, §4. FormalGroups The proofsoftheseformulaeallfollowthesamepattern.Onechecks F(X,Y)=f-'(f(X)+f(Y)), e III (X) =[a]e,e([b]ee,=(X))1 X moddeg2,thelogarithmofF. OK n=1 00 (-1)"+1 Then ) EndoK(Fe), loge :F nn h'' = log(1+X). [a]F(X)=f-'(af(X)), G, 0 a I->[a]., , ae K, -'] series 345 The formalgroupsfurnishageneralizationofthenotionpn-throot us toconstructaperfectanalogueofthetheorypn-thcyclotomic Hint: ThepowerseriesBof(2.3)yieldsanisomorphism0:Fe+Fe. extension kIK. become isomorphicovero,,whereKisthecompletionofmaximalunramified Exercise 6.TwoLubin-TatemodulesFeandforprimeelementsTrfalways but fordifferentprimeelementsnand5T,areneverisomorphic. Exercise 5.TwoLubin-TatemodulesoverthevaluationringOKofalocalfieldK, 346 (5.1) Proposition.LetFbe a formalOK-module.ThenthesetFwith we thereforeobtainimmediatelythe in F.Fromthedefinitionofformalo-modulesandtheir homomorphisms G (x1,...,x,)convergesinthecompletefieldK(xt series withconstantcoefficient0,andifxt, the givenlocalfieldK.IfG(X1, this themaximalidealFofthe-valuationringalgebraic closurekof the powerseriesoveradomaininwhichtheyconverge. We nowchoosefor corresponding extensions. unity, andprovideanexplicitversionofthelocalreciprocitylawin (see chap.I(7.13)),replacingQpbyanarbitrarylocalgroundfieldK. field Q( operations v,' A formaloK-modulegivesrisetoanordinaryOK ifweread Formal groupsarerelevantforlocalclassfieldtheoryinthattheyallow a ) overQp,withitsfundamentalisomorphism E OK,isanOK-moduleinthe usualsense.WedenoteitbyFF. § 5.GeneralizedCyclotomicTheory x+y=F(x,y) Q.. G(Q (oIQ)-+(Z/pnZ)* ..., Xn)EOK[[Xl,X,11isapower and Mme. Chapter V.LocalClassFieldTheory' ... ,xnE , ... , F, thentheseries v-, xn) toanelement cod § 5. Generalized Cyclotomic Theory 347 If f : F -> G is a homomorphism (isomorphism) of formal OK -modules, then f:pF - +pG,x1) f(x), is a homomorphism (isomorphism) of ordinary OK -modules. The operations in pF, and particularly scalar multiplication a x = [a]F(x), must of course not be confused with the usual operations in the field K. We now consider a Lubin-Tate module F for the prime element 7r of OK. We define the group of 7r"-division points by F(n) = J X EFI 7r' - X = 01 = ker([7r"]F) This is an OK-module, and an oK/7r'oK-module because itis killed by 7r'ioK. (5.2) Proposition. F (n) is a free OK /7r'oK -module of rank 1. Proof: An isomorphism f : F G of Lubin-Tate modules obviously induces isomorphisms f : pF - pG and f :F(n) -* G(n) of OK- modules. By (4.6), Lubin-Tate modules for the same prime element 7r are all isomorphic. We may therefore assume that F = Fe, with e(X) = Xq-{-7rX = [7r]F(X). F(n) then consists of the q'2 zeroes of the iterated polynomial e°(X) = (e o ... o e) (X) = [7r"] F M, which is easily shown, by induction on n, to be separable. Now if )E F(n) N F(n - 1), then OK --> F(n),a f-r a A, is a homomorphism of OK-modules with kernel7r'0K. It induces a bijective homomorphism oK/7r'0K -* F(n) because both sides are of order q'. (5.3) Corollary. Associating a [a]F we obtain canonical isomorphisms oK/7r'0K -- EndoK (F(n))and UK/UK Aut0K (F (n)). Proof: The map on the left is an isomorphism since oK/7r'ioK - F(n) and EndOK(oK/7rnoK) = 007r' OK. The one on the right is obtained by taking the unit groups of these rings. prime elementofL,i.e.,L" =K(X,),and is itsminimalpolynomial.In particularonehasNLIK(-X.)=7r to thepolynomiale(X)EE,andlet,l"F(n) Furthermore thefollowingistrue:letFbeLubin-Tatemodule Feassociated such that i.e., foreveryUEG(L"IK)thereisauniqueclassumod UK),withuE cyclotomic fields. following theoremshowsthecompleteanalogyofLubin-Tateextensionswith of unity.LnIKisthereforethep"-thcyclotomicextensionQ So G,,(n)consistsoftheelements-1,where Example: IfOK=ZpandFistheLubin-TatemoduleG,,, splitting fieldofthen-folditeration a Lubin-Tatepolynomiale(X)Eg,,,then=[7rIF(X)andL,iKisthe hence K(G(n))=K(F(n)).IfFistheLubin-TatemoduleFebelongingto on theprimeelement7r,notLubin-TatemoduleF.ForifGis These fieldsarealsocalledtheLubin-Tateextensions.Theyonlydepend Since F(n)S-=F(n+1)wegetatoweroffields the fieldofir"-divisionpointsby 348 1) withGaloisgroup (5.4) Theorem.L"KisatotalIyramifiedabelianextension ofdegreeqi-t(q- f :F-+G,EOK[[X]]suchthatG(n)=(F(n))CK(F(n)),and another Lubin-Tatemodulefor7r,thenby(4.6),thereisanisomorphism Given aLubin-TatemoduleFfortheprimeelement7r,wenowdefine On (X)= G(L"IK) =AutOK(F(n))-UK/UK), e" (X)_[P"]Gm(X)=(1+X)pn-1. e" (X)=(eo...e)_[7r"]F KCLICL2C...CL, =UL,,. ee' A =[u]F(X) - tXX) - L"=K(F(n)). .-. = X4"-1(4-1) + for Chapter V.LocalClassFieldTheory' X EF(n). ... +7rEOK[X] varies overthep"-throots n_t F(n -1).Then?"isa p , (p. pn)IQp.The then [I':K]=[L,, :K]=[La:K]. degree the fixedfieldof6.SincefE1K =dK(Q)1,EIKistotallyramified.Ithas view vasanautomorphism ofthecompletionf,=L"KL.Let1be Let &beanelementinFrob(L" IK)suchthator_=&L.anddK(Q)=1.We automorphism suchthat ''' Proof :Theproofisthesameasthatof(2.4).LetaE G(LIK)bethe uirVK(a) EK*,uUK,onehas (5.5) Theorem.ForthefieldLnIKofIra-divisionpoints andfora norm residuesymboloftheLubin-Tateextensions. Q p(µp,")(see(2.4)),weobtainthefollowingexplicitformulafor This provesthetheorem. It isinjectivebecauseLageneratedbyF(n),anditsurjective induces anautomorphismofF(n).Wethereforeobtainahomomorphism ramified extensionK(a.")Iofdegreeqn-1(q-1).EachaEG(LK) of thisEisensteinpolynomial,andisthereforeaprimeelementthetotally module associatedtoe,andXaEF(n)NF(n-1),then.Lnisclearlyazero is anEisensteinpolynomialofdegreeqn-1(q-1).IfFtheLubin-Tate is aLubin-Tatepolynomial,then Proof :If § 5.GeneralizedCyclotomicTheory Generalizing theexplicitnormresiduesymbolofcyclotomicfields On (X) = qn-1 (q #G(L1IK) ?[K(A,):K]=qn-1(q t17 = en-1(g) e(X) =Xq+rr(aq_1Xq-1-a2X2)+7rX - 1)becauseznK=and E=E'KL.Consequently en-1(X)q-1 +,,r(aq-tea-1(X e"(X) G(L,,IK) -jAut,K(F(n))-UK/U(nl. (a, LJK),l=[u-1]F(X), ,fl Xor =[u-1]F(X), ,-r X, EF(n). )q-2 +...a2en-1(X))+g ), EF(n). - 1) =#UK/Utai. -I' ` 349 NL,IK(-An) ENL,IKLn,weget division pointsofFeiandNEIK(-7rE)=7runby(5.4).Sincen we have7rEEFe(n)NFe(n-1).Hence=K(7rE)isthefieldof7r"- Since e'(9(,ln))=00 is aprimeelementofEbecause Let ,l,EF(n)N(n-1).a.,,isaprimeelementofL,and o [[X]],withsEUk,suchthat and letF=F.By(2.3),thereexistsapowerseries9(X)EX+ 350 Proof :Fora=u7r°K(a)wehave to thegroup(7r)x (5.6) Corollary.ThefieldLIKof7r'-divisionpointsistheclassrelative and thus (5.7) Corollary.Themaximal abelianextensionofKisthecomposite generalization ofthelocalKronecker-Webertheorem(1.9): a where L,ristheunionUn°1 LnofthefieldsLof7rn-divisionpoints. Now leteEE,,J£nbeLubin-TateseriesoverOK,where7r=un, For themaximalabelianextensionKabIK,thisgives thefollowing CJ' ill (a, LnIK)=(7r rL,IK(6) =NEIK(-7rE)=7rmumod 9'=9o[u]F 7rE =9`P(X)9,([u-'IF(xn))0(Xn)nE [u-1 ]F=idF(n) (a,LnIK)=1 UK) CK*. vK(a), and Ln IK)(u,=a. Kab =KL,, nE =B(in) = 0forin,and#n-1, [u-1]F(?)=X Chapter V.LocalClassFieldTheory u1 E ICI (cc=coK). UKl 11' - aE +2i forallXEF(n) (7r) XUK1 E _ 351 UK) (n) form a basis (n) form a C NLIKL* for a suitable C NLIKL* n > 0. F A E F(1), F UK) F pn+i for all .LEF(T) AEF(1) r[ h(X+A), n>0. F zEF(n) g=ho7r. - N" (h) mod h1(X)= fl h(X+A) CKLn=Kab. N (h) E X' S*. F N(h) o [7r] = rj h(X +A) . Nn+1(h) g(X +A) = g(X) N"(h)o[7rn]= F h E X' S* for i> III If h EX'S*, i > 0, then N"+1(h)/Nn (h) E S* and Let g E S be a power series such that Let g E S be a power If g E S is a power series such that g(F(1)) = 0, then g is divisible by [7r], i.e., such that g(F(1)) = 0, then g is divisible If g E S is a power series (vi) (i) N(h1h2) = N(h1)N(h2). (ii) N(h) - h mod p. (iii) mod (iv) h - 1 mod p' for i-> 1 ==> N(h) - 1 = N(Nn-1(h)), one has (v) For the operators N°(h) = h, Nn(h) Show: (i) h(X) E S. g(X) = [7r](X)h(X), (ii) ((7r) x UK 1) fl ((irf) x UK). The class field of (7r) x UK 1 is L, and that (7r) x UK 1 is L, The class field of 1) fl ((irf) x UK). ((7r) x UK exists a unique power series h (X) in where we write X +X = F (X , A). Then there S such that then the power series Exercise 2. If h(X) is a power series in S, = h1(X) for all A E F(1). also belongs to S, and one has h1(X+A) (uniquely determined by exercise 1 Exercise 3. Let N(h) E S be the power series and exercise 2) such that norm operator. Show: The mapping N : S -+ S is called Coleman's Exercise 1. Let F = Fe be the Lubin-Tate module for the Lubin-Tate series e E E, be the Lubin-Tate module for the Lubin-Tate Exercise 1. Let F = Fe (g E S I g(O) E UK). [a] = [a]e. Let S = oK[[X]] and S* = with the endomorphisms § 5. Generalized Cyclotomic Theory Cyclotomic Generalized § 5. 7rf E NLIKL* we have Then extension. abelian be a finite Let LIK Proof: since the UK is open in K*, and f . Since NLIKL* for suitable have (7rf) x of 1, we of neighbourhoods x the group (7rf) the class field of L is contained in n. Hence that of degree f . It follows extension K f UK is the unramified of (7r f) x LCKfL,, X33 aEOL] 0 < i < n. For every II. for all dx 0 < i < n, Chapter V. Local Class Field Theory Field Class V. Local Chapter h; g; is a solution. (Go: Gs) for s for 0 + . K* D UK = UK°) D UM D UK) D .. § 6. Higher Ramification Groups G(LIK) 2 G°(LIK) R G'(LIK) D_ G2(LIK) 2 G(LIK) 2 G°(LIK) R G'(LIK) D_ G2(LIK) G1(LIK)={o EG(LIK)I vL(aa-a)>i+l let Pi E Jr"-'piOj+t, 0 < i < n. Then there exists a power series h(X) E S such power series h(X) E Then there exists a 0 < i < n. let Pi E Jr"-'piOj+t, Considering the homomorphism 352 0 < F(i +1) for '](A) E A; _ [Jr" n > 0, and -, F(n), A E F(n+1) 4. Let Exercise + 1)), extension L;+1 = K(F(i of the Lubin-Tate A; is a prime element i < n. Then = A; o;+t maximal ideal p;+t ring of L;+i, with oK [A; ] is the valuation and o;+t = Show: that put, for 0 < i < n: with h;(X) E o[X] and Hint: Write ,8; = 7r'-`Aoh;(A;), [rr'+t ] Then h = g; (X) = [rr"+t ] [,r' ]/ F(n + 1) -, F(n) and A; Exercise 5. Let A E a power series h(X) E o[[X]] such that u E UL,,,, there exists from L,,+, to Li+,. where N; is the norm E S*. Show that ht(X) E o[X], and put h2 = N"(ht) Hint: Write u = ht(A), is a power series E Jr"-' p t o;+i . Then by exercise 4 there ,B; = Nn,; (u) - h2 (A;) = h2 +h3 works. that $ = h3 (4), 0 < i < n. Show that h h3(X) E o[[X]] such are discussed in detail in [79], 5.2. Remark: The solutions of these exercises K of local fields by the norm residue defined for an abelian extension L I descending chain in the group K* on the left we have the (*) (**) symbol, it is striking that both groups are equipped with a canonical filtration: symbol, it is striking that both groups are right there is the descending chain of higher unit groups UK), and on the upper numbering (see chap. H, § 10). of ramification groups G` (L I K) in the groups in the lower numbering The latter arose from the ramification via the strictly increasing function 353 _ 0 chap. U, A_"__ remarkable extensions. , Y]]. Thus Ln , we may write an isomorphism (UK),L,=IK). We and a 0 1, then a E O n I K) of G (L" I K). = X, F(0,Y) = Y, u E UK ), and this points of a Lubin-Tate we have necessarily Y) E OK [[X < i < qk - 1. - 1). Then A is a prime , < i < qk - ifm .'~ because (, L I K) : UK/UK) -+ X° - A = F (X tea) UK) § 6. Higher Ramification Groups Ramification Higher § 6. by the rule by the the inverse function where 1 is arithmetic fact that the arithmetic in a precise (*) and (**) 1) the higher § 10, exercise (6.1) Proposition. Let L I (6.1) Proposition. element module for the prime Proof: By (5.4) and (5.5), Proof: By (5.4) and for UK/UK) -+ G(LkIK) that therefore have to show Let U E G u E subgroup UK) / UK) onto the Let u = 1 + E7rn, K element of L" and from (5.4) If m>n,then a=1,so that [Jrr]F(A) is a prime element of [E]F (An-m). As L,7I Ln_n, is totally [E7rm]F(A) = we have F (X By chap. IT, §10, i iL IK (Q) > qk >_ i + 1, and (UK ),L"IK) c G1(LnIK). If conversely iL IK (6) = qm > i > qk-i shows the inclusion Gi (Ln I K) ered themaintheoremofhigherramificationtheory. 354 may evenassumethatL=L,,,.By(6.1),thenormresiduesymbolmaps Lubin-Tate modulefor,r.Inviewofchap.II,(10.9),andIV,(6.4),we which, by(5.6),isequaltothefieldL,,,of7r°1-divisionpointssome sufficiently big.ThereforeLIKiscontainedintheclassfieldof(7r)xUK'), n =NLIK(r)isaprimeelementofand(Jr)xUKc_L*form so wemayreplaceLIKbyLIL°. chap. IV,(6.4),andV,((1.2),wehave >'LILo(*LOIK(s)) =*LILO(s)(seechap.II,(10.8)).Ontheotherhand,by hand G"(LIK)=G"(LIL°)because'i/iLOIK(s)sand1/fLIK(s) the maximalunramifiedsubextensionofLIK,thenwehaveonone Proof :WemayassumethatLIKistotallyramified.ForifL°IK maps thegroup symbol (6.2) Theorem.IfLIKisafiniteabelianextension,thenthenormresidue other words,tisajumpiffor all£>0,onehas call thesenumbersthejumps ofthefiltration{Gt(LIK)}t>_1G(LIK).In numbers t>.-I.Thuswemay askforwhichnumberstheychange.We G" (LIK). q'-I. ThisyieldsrILIK(q"-1)=nandthus(UKl,LIK) =Gqn with gi.=#Gi(LIK)=#G(LmILn)_ But wehave(seechap.II,§10) group UK)ontothe From thispropositionwegetthefollowingresult,whichmaybeconsid- If nowLIKistotallyramifiedandnLaprimeelementofL,then Higher ramificationgroupsGt(LIK)wereintroducedfor arbitraryreal AID (Uio,LIL°) G(Lm ILn)=Gi(LmK) UK (n) r1LIK(q"-1)- onto thegroupG"(LIK),forn>0. ( = Gt(LIK) ,LIK):K* _)G(LIK) (NLOIKUL ),LIK) 1 Gt+E(LIK). for g°(g1+...+ggn_i) Ire Chapter V.LocalClassFieldTheory (qm-1 _qn-1)(q-1)for qn-1 (6.3) Proposition (HASSE -ARF). For a finite abelian extension L I K, the jumps of the filtration (G' (L I K))t,_i of G (L I K) are rational integers. Proof: As in the proof of (6.2), we may assume (since Gr (L I K) = Gt (L I L°)) that L I K is totally ramified and contained in a Lubin-Tate extension L,,,IK. If now t is a jump of {G'(LIK)}, then by chap. II (10.9), t is also a jump of {Gt (L,,, I K)}. Since by (6.1), the jumps of {G,s (L,,, I K)} are the numbers q" - 1 , for n = 0, ... , m - 1 (q = 2 is an exception: 0 is not a jump), the jumps of {Gt (L", I K)} are the numbers TILm IK (q" - 1) =n, forn=0, ...,m-1. The theorem of HASSE-ARF has an important application to Artin L -series, which we will study in chap. VII (see chap. VII, (11.4)). product of elementsupEKpwhereapisaunitintheringop integersofKp,for Thus anideaeisafamily almost allp.InanalogywithAK,wewritetheidelegroup astherestricted Op CKp. is calledthe"restrictedproduct"ofKpwithrespect tothesubrings Addition andmultiplicationaredefinedcomponentwise.Thiskindofproduct in Kpforalmostallp.Theadelesformaring,whichisdenotedby of elementsupEKpwhereprunsthroughallprimesK,andisintegral family second syllable,isderivedfromtheoriginalterm"additiveideae"-a for theprinciplewhichreducesproblemsconcerninganumberfieldKto to providingasuitablebasisfortheimportantlocal-to-globalprinciple,i.e., by theFrenchmathematicianCLAUDECHEVALLEY(1909-1984)withaview "ideal element",whichwasabbreviatedasid.el. analogous problemsforthevariouscompletionsKp.CHEVALLEYusedterm notion ofideleisamodificationtheideal.Itwasintroduced base fieldistakeninglobalclasstheorybytheideaegroup.The The idelegroupofKisdefinedtobetheunit An adeleofK-thiscuriousexpression,whichhasthestresson The roleheldinlocalclassfieldtheorybythemultiplicativegroupof Global ClassFieldTheory § 1.IdelesandIdeleClasses Chapter VI AK =IIKp. IK =IIKp IK =AK. a =(ap) a =(ap) p Op. p .^. coo fin' 358 Chapter VI. Global Class Field Theory with respect to the unit groups op. For every finite set of primes S, IK contains the subgroup IK=r[KpxF1 Up pes p¢S of S-ideles, where Up = Kp for p infinite complex, and Up = R+ for p infinite real. One clearly has IK = U IS, s if S varies over all finite sets of primes of K. The inclusions K C Kp allow us to define the diagonal embedding K*) IK, which associates to a E K* the ideae a E IK whose p-th component is the element a in K. We thus view K* as a subgroup of IK and we call the elements of K* in IK principal idi les. The intersection KS=K*nIK consists of the numbers a E K* which are units at all primes p S, p {' oo, and which are positive in Kp = R for all real infinite places p 0 S. They are called S-units. In particular, for the set Scc, of infinite places, Ks- is the unit group oK of og. We get the following generalization of Dirichlet's unit theorem. (1.1) Proposition. If S contains all infinite places, then the homomorphism A:KS X(a) = (log lalp)pES' pES has kernel A(K), and its image is a complete lattice in the (s -1) -dimensional trace-zero space H = { (xp) E [IPES R I EpES xp = 01 , S = #S. Proof: For the set S,,,, = {p I oo}, this is the claim of chap. I, (7.1) and (7.3). Let Sf = S N Sam, and let J(Sf) be the subgroup of JK generated by the prime ideals p E Sf. Associating to every a E KS the principal ideal is = '(a) E J(Sf), we obtain the commutative diagram 1 oK K S ) J (Sf) I)L, Iz Ix// PESO. peS pESf § 1. Ideles and Idele Classes 359 with exact rows. The map X" on the right is given by x"( II pup) F1 vp log M(P) PESf pESf (observe that Jalp = M(p)-Up(a)), and maps_J(Sf) isomorphically onto the complete lattice spanned by the vectors ep = (0, ...,0, log R(p),0, ...,0), for p E Sf. It follows that ker(a.) = ker(A.') = µ(K), and we obtain the exact sequence () 0 ---im().') -) im(),) im()"), where the groups on the left and on the right are lattices. This implies that the group in the middle is also a lattice. For if x E im(A), and U is a neighbourhood of i (x) which contains no other point of im(),"), then i -1(U) contains the coset x + im(?'), and no other. It is discrete since im(X') is discrete. For every p E Sf, if h is the class number of K, then ph belongs to i (KS), i.e., J(Sf)h C i(KS) C J(Sf). The groups on the left and on the right have rank #Sf, hence so does i (KS). In the sequence (*), the image of i therefore has rank #Sf, and the kernel has rank #S,, -1. Hence im(A) is a lattice of rank #S,, -1-F#S f = #S - 1. It lies in the (#S-1)-dimensional trace-zero space H, since FI,ES- Ia (p = fp Ia Ip = 1 for a E KS. (1.2) Definition. The elements of the subgroup K* of IK are called principal ideles and the quotient group CK = IK/K* is called the idele class group of K. The relation between the ideal class group CIK and the idele class group CK is as follows. There is a surjective homomorphism (): IK - JK fl pVp(«,) pt- from the idele group IK to the ideal group JK. Its kernel is Is-_ flK*x flUp. P1- pt00 a'=as-'E Is,andthusaEIKK*. in ailieS,wehavevp(a'p) =0,i.e.,apEUpforallp¢S.Hence by IK-+JKtotheidealai= i.e., (a)=aiforsomeprincipal ideal(a).Theidelea'=as-Iismapped then thecorrespondingideal(a)=rjpta,p"p(1p)belongs to someclassaiPK, finite setofplacescontainingtheseprimesandthe atinfinity,then one hasIK=IKK*. are composedofafinitenumberprimeidealspt,... , p,NowifSisany Proof: Leta,a, (1.4) Proposition.IK=IKK*,i.e.,CKIKK*/K*,if Sisasufficiently the finitenessofformerisreflectedintermslatterasfollows. the onto therepleteidealclassgroup,withkernelIKK*/K*.Wethereforehave induces asurjectivehomomorphism (see chap.III,§1).Ittakesprincipalidelestorepleteidealsand onto therepleteidealgroupJ(a).Itskernelis with kernelIKK*/K*.Wemayalsoconsiderthesurjectivehomomorphism It inducesasurjectivehomomorphism 360 big finitesetofplacesK. (1.3) Proposition.C1K=IK/IK°°K*,andPic(o)IK/IKK*. -,' In ordertoseethis,weusetheisomorphismIK/IK°° =JK.IfaEIK, 0=fib In contrasttotheidealclassgroup,ideaegroupisnotfinite.But c., ,- ah beidealsrepresentingthehclassesofJKIPK.They IK -fJ(Q), :°. IK={(ap)E'KI CK -Pic(o) F.. ti' 'p, HPt,,,, can p°p(an)Since theprimeideals occurring ... IX H Chapter VI.GlobalClassFieldTheory = 1forallp} P ttpvp(ap) If XEK*isaprincipalidele, thenwefindbychap.111, topological groupcarriesovertotheideleclass CK=IK/K*.For the absolutenormofatoberealnumber any idelea=(ap)EIK,itsclassinCKwillbedenoted by[a].Wedefine deduces x1x21=v1v21E1.1,acontradiction. one xEK*.Indeed,fromx1=yv1,x2yv2K*,with x1 That thesubgroupisclosedfollowsforacompletelygeneralreason:since from 1,thenwegetthecontradiction is suchaneighbourhood.ForifwehadprincipalidelexEitdifferent no otherprincipalidelebesides1. Proof: Itisenoughtoshowthat1EIKhasaneighbourhoodwhichcontains Tt(x) _flpIxI VV-1 Cit.ForeveryyEIK,theneighbourhoodVthen containsatmost (x, y)Hxy-1iscontinuous,thereaneighbourhood Vof1suchthat (1.5) Proposition.K*isadiscrete,andthereforeclosed,subgroupof'K. topological group. of 1inIKwhoseclosureiscompact.Thereforealocallycompact the Wp,forpIoo,arebounded,then11 are compactforp¢S.ThereforethesameistrueofgroupIIpVSUp.If and W.CKpisabasicsystemofneighbourhoods1EK.ThegroupsUp where SrunsthroughthefinitesetsofplacesKwhichcontainallp100, system ofneighbourhoods1EIKisgivenbythesets § 1.IdelesandIdeleClasses As K*isclosedinIK,thefactthatIKalocallycompact Hausdorff The idelegroupcomesequippedwithacanonicaltopology.Abasic .1,t=taEIKI IapIp=1forp{oo,Iap _ ..0' 1=IIlx-llp= IIIX-11p-HIX-11P 1 'r7 < r[ix-1Iprjmax[Ixlp,1}=1. = 1.Wethushaveacontinuous homomorphism m(a) =1L pfi- p pES HWpX HUpcIK, f7t :CK p mar, ,-T pt- .-7 pf- p¢S pcS 1(p)"o(ao) + R. Wp xFjp = _1lap(p p 0'" - llp ) 1 ,y+ (1.9). We mayalsowriteitasarepleteideal number fieldK.Insteadofpn,wetakeanyintegralidealm=plp'P It comesequippedwithacollectionofcanonicalsubgroupswhicharetobe field KasthemultiplicativegroupKpdoesforap-adicnumberK. formed fromtheideaegroup (1.7) Definition.Thegroup condition. real place,itsymbolizespositivity,andfora,complexplaceistheempty For afiniteprimepandnp>0thismeanstheusualcongruence;for for np>0.GivenupEK*wewrite every placepofKweputUP=Up,and with np=0forpIoo,andwetreatitinwhatfollowsasamoduleofK.For viewed asanaloguesofthehigherunitgroups § 1.IdelesandIdeleClasses, simplify matters.Mostofall, itwasmadesubstantiallybecauseofthechoice fewer congruencesubgroups thanintheothertexts.Thischoicedoesnotonly up ofidelesainIKarealways positiveatallrealplacesp,sowehavehere from thatgivenin[107]bythe author:inthepresentbook,components "Zahlbericht" [53].Itdiffers from thosefoundinmodemtextbooks,andalso classical one,asgiven(intheideal-theoreticversion)for instanceinHasse's Remark :Thisdefinitionoftherayclassgroupdoes correspondtothe is calledtherayclassgroupmodm. is calledthecongruencesubgroupmodm,andquotient groupCK/CK The ideaeclassgroupCKplaysasimilarroleforthealgebraicnumber UpnP) :- up =-1modpn0 C* =K* R* CK; t]. r., 1+pnP, CK =IKK*lK*, IK = m = I1 p llpna ly , p , U( v) p if piscomplex, if pisreal, if pfoo, n up E UpnP) Up(n) = 1+p'ofap-adic 363 364 Chapter VI. Global Class Field Theory of the canonical metric (, ) on the Minkowski space KR (see chap. I, § 5). In fact, we saw in chap. III, § 3, that this choice forces the extension C I R to be unramified. We will explain in § 6 below how to interpret this situation, and how to reconcile it with the definition of ray classes in other texts. The significance of the congruence subgroups lies in that they provide an overview over all closed subgroups of finite index in CK. More precisely, we have the (1.8) Proposition. The closed subgroups of finite index of CK are precisely those subgroups that contain a congruence subgroup C. Proof: CI Kis open in CK because IK = fp Upnplis open in IK. IK is contained in the group Is' = Fl. K* xl lt.Up, and since (CK : IK K*/K*) = #ClK = h < oo, the index (CK : CK) = h(IK K* IKK*) < h(IK : IK) Upnp) = h fl (UpUnp) ) fl (Kp : ) pt* pl- is finite. Being the complement of the nontrivial open cosets, which are finite in number, Cg is closed of finite index. Consequently, every group containing CK is also closed of finite index, for it is the union of finitely many cosets of C. Conversely, let N be an arbitrary closed subgroup of finite index. Then N is also open, being the complement of a finite number of closed cosets. Thus the preimage J of N in IK is also open, and it thus contains a subset of the form W=j(Wpx j-jUp, pES p¢S where S is a finite set of places of K containing the infinite ones, and Wp is an open neighbourhood of 1 E K*.p If P E S isfinite, we are liable to choose Wp = Upn° c Kp form a basic system of neighbourhoods of 1 E K. If p E S is real, we may choose Wp c R. The open set W, will then generate the group R+, resp. K* in the case of a complex place p. The subgroup of J generated by W is therefore of the form 1K, so N contains the congruence subgroup C. mapping, andtheproposition isproved. [a] _[,B]EIKK*/K*=C. ThereforeCKisthekernelofabove p (a) =(a).Thecomponents oftheideae,B=as-1satisfyPpEUpfor to thekernel,thenthereis an(a)EPK,witha1 in thekernel.Conversely,ifclass[a]representedby aEIK)belongs Since (a)=1foraEIK,thegroupCKIKK*/K*iscertainly contained surjective homomorphism in PK.Thereforethecorrespondencea The elementsaEIKlnK*arepreciselythosegenerating principalideals theorem, thereexistsanac=K*suchthatapaas1 mod pnpforplm, Then IK=1K Proof :Letm= induces anisomorphism We thenhavethe b/c oftwointegersrelativelyprimetomsuchthatbascmodm.This out tobepositive.Thecongruenceaas1modmmeansthatisthequotient The latterconditionmeansthat,foreveryrealembeddingK-+IR,aturns and apa>0forpreal.Thus (1.9) Proposition.Thehomomorphism all pim=flpfp"p.Weput is tantamounttosayingthataas1modp"pinKp,i.e.,EUp"plfor and letPKbethegroupofallprincipalideals(a)Esuchthat description. LetJKbethegroupofallfractionalidealsrelativelyprimetom, § 1.Id8lesandIdeleClasses The rayclassgroupscanbegiventhefollowingpurelyideal-theoretic moo, andP.E CK =IKK*/K*IK/1KnK*-JKIPK ( ):IK-->JK,aF(a)=flpvp(a.) (")K*, becauseforeveryaEIK,bytheapproximation l p a as1modmand Up"p> forpImoo,inotherwords, 13E1K,andhence p'1', andlet CK/CK =C1K CIK _ (apa)E = Ji-IPK .Q. coo a totallypositive. IK 1,sothata IK1aEIKIapEUp"forpImoo}. (a) =llpfoopvp("p)definesa . P.. pfoo = pa-'E 1 nK*,suchthat acv IK )K*. coo 365 .°c These lattergroupsmaybeviewedastherayclassoffieldQ: ideal classgroupsontheonehand,and(Z/mZ)*other. introduced byHEINRICHWEBER(1842-1913)asacommongeneralizationof 366 ray classfields,whichsatisfycanonically be GaloisextensionsK'"IKgeneralizingthecyclotomicfields:so-called important fact.ForallmodulesmofanarbitrarynumberfieldK,therewill It isclassfieldtheory,whichprovidesafar-reachinggeneralizationofthis canonical isomorphism G(Q(µm)IQ) ofthem-thcyclotomicfield ideals (a)suchthata-1modp'pforpImoo,i.e.,thekernelofq. which haveapositivegenerator-1modm.Butthesearepreciselythe homomorphism J,-+(Z/mZ)*whosekernelconsistsofallideals(a) the positivegeneratorontoresidueclassmodm,wegetasurjective Proof: Everyideal(a)EJEhastwogenerators,aand-a.Mapping (1.10) Proposition.Foranymodulem=(m)ofthefieldQ,onehas We thereforeobtainanexact sequence IK/IK°OK*, whereIK Proof: OnehasC1K (1.11) Proposition.Thereisanexactsequence general notarayclassgroup-asfollows. to theidealclassgroupClK-whichaccordingourdefinition here,isin (see §6).Therayclassgroupmod1isofparticularinterest here.Itisrelated where o+isthegroupoftotallypositiveunitsK. The rayclassgroupsintheideal-theoreticversionClK=JK/PKwere The group(Z/mZ)*iscanonicallyisomorphictotheGalois 1R*/18+-*CIK-+CIK1, f-" 1 ,-Z+ U00 ioo O., IK K*/IKK* a'__ III CQ/C =Cl'(Z/mZ)*. L]. G(Q(hm)IQ) =CQ/Cm a.. .y. III ,al = fpUpandIs-_fpt Upxjlp,,,.K*. G(KmIK) =CKICK p real CK/CK =1K/IKK*and, by (1.3),C1K III 111 cc' CK/CK -+ClK 211 Chapter VI.GlobalClassFieldTheory 51- We thereforeobtaina ".: 1. ,°, But IK°°nK*=n*,IKo*,andIK°°/IKFpI1K*/Up For thegrouponleftwehaveexactsequence almost allprimeidealssplitcompletely,thenL=K. (ii) OtherwiseonehasatleastthataE (i) If of primes.LetaEK*andKp"',forallp¢S.Show: Exercise 2.LetKbeanumberfield,m=2°m'(m'odd),andletSfiniteset unique solution,foreverynENandyA0/7L. (iii) AQ/7Lisarbitrarilyanduniquelydivisible,i.e.,theequationnx=yhasa (ii) ThequotientgroupAQ/Ziscompactandconnected. Exercise 1.(i)AQ=(Z(9zQ)xR. § 1.IdelesandIdeleClasses e1, ...,s,bea7L-basisofthegrouptotallypositiveunits K. Exercise 6.Letpi,...,p,,bethecomplexprimesofK.For YER,letMy)be a Hsz,insuchwaythatOt(s'')=1. Z -.IK,nF->a',bycontinuitytoanexponentiationx l Exercise 5.LetsEoKbetotallypositive,i.e.,I'.Extend theexponentiation implies x;=0,i1,...,t. i.e., anyrelation are thenindependentunitswithrespecttotheexponentiationelementsof7L, Exercise 4.Letsi,...,e,EoKbeindependentunits.TheimagesE1,E,in1/ with xEZ. taking integerpowersofideaesaEIfextendsbycontinuitytoexponentiationax Exercise 3.WriteIK=Ifx1,,withr[p{Up,1. Hint: Usethefollowingfact,provedin(3.8):ifLIKisafiniteextensionwhich into thegroupCKO=([a]eCK I defines acontinuoushomomorphism Exercise 7.Sending (ii) aisprincipalidealifandonlyA;EZC7LxRy, EZCR. form agroup,andhaveabsolutenormOZ(a)=1. (i) Theideaesoftheform the idelehavingcomponente2-'YatPk,andcomponents1 allotherplaces.Let 17 !p realR*/R-{-' 1 is,* nK*/IIK* a =e is cyclic,where2"a (AI, ...,A,,Y1,...,ys)HEll...el'0l(Y1)...0,(Ys) ). I ... f: (7Lx1R)'xl[ls-tCK El' Tt`r=1,x;EZ, d11(Yl) Tt([a]) =1),withkernelVx V. > Is/IK-IK°°K/IKK* s (Ys), primitive rimitive 2°-throotofunity,thenaEK*m. .O. '17 X, E7LxR,y,l[8, -* IK=I'xI., Up. Showthat 1. 367 0 W.5 whenever 93and93'lieabove thesameplacepofK. components aqbelongtoKp(j3lp),andifonehasfurthermore ap=api element a=(aq3)EILthereforebelongstothegroupIK ifandonlyits which willalwaysbetacitlyusedtoconsiderIKasasubgroup ofIL.An In thiswayweobtainaninjectivehomomorphism pass fromafieldKtoanextensionL.SoletLIbefiniteof Exercise 14.ComputethekernelsofC1K-C/KandClK--->CI,form`jm. W.5 whose componentsa'aregivenby group ILofLbysendinganidelea=(up)EIKtothe idelea'=(a')EIL algebraic numberfields.WeembedtheidelegroupIK ofKintotheidele positive units-1modm. where o+,resp.isthegroupoftotallypositiveunitso, Exercise 13.Foreverymodulem,onehasanexactsequence positive numberinowhichisprimetoanarbitraryfixedideal. Exercise 12.Leto=OK.Everyclassin(o/m)*canberepresentedbyatotally integral idealwhichisprimetoanarbitraryfixedideal. Exercise 11.EveryidealclassoftheraygroupC1Kcanberepresentedbyan line. direct productoftcopiesthe"solenoid"(2xR)/Z,scirclesR/Z,andareal Exercise 10.TheconnectedcomponentDKof1intheideleclassgroupCKis in CK,anditistheconnectedcomponentofICK. Exercise 9.ThegroupDKistheintersectionofallclosedsubgroupsfiniteindex (ii) fyieldsatopologicalisomorphism Exercise 8.(i)TheimageDKoffiscompact,connectedandarbitrarilydivisible. 368 Every isomorphismo We shallnowstudythebehaviourofidelesandideleclasseswhenwe III I -->0+*/0+(o/m)'- § 2.IdelesinFieldExtensions f :((7Gx1[1)/76)' a' =apEKpCL*for am. coy : L C'S a:IL IK -)IL, o Linducesanisomorphism r-. >IaL (R/Z)S Chapter VI.GlobalClassFieldTheory 't1 C1K 13Ip. C11 -+1, DK bra . (]. II- .., § 2. Id6les in Field Extensions 369 like this. For each place3 of L, a induces an isomorphism or :LT -*(6L)QT. For if we have a = T-lim aj, for some sequence a; E L, then the sequence vat e oiL converges with respect toI,T in (oL)vp, and the isomorphism is given by a = 3-lim al I-+ va = cr 3-lim o'a;. For an idele a E IL, we then define o'a E IL to be the idele with components (oa)o = o-ap E (QL),T. If L I K is a Galois extension with Galois group G = G (L I K),then every a E G yields an automorphism or: IL - IL, i.e., IL is turned into an G -module. As to the fixed module IL _ (a E IL I au = a for all a EG), we have the (2.1) Proposition. If L I K is a Galois extension with Galois group G, then ILG =IK. Proof: Let a E IK C IL. For o' E G, the induced map a : LT -* LQT is a Kp-isomorphism, if 93Ip. Therefore (aa)ap = as .T = a'p =.aQp, so that as = a, and therefore Of E 1LG. If converselya = (a,) E IL , then (aa)'T = aE G. In particular, if or belongs to the decomposition group Gqj = G(Lq3lKp), then aq3 = 93 and aap = ap so that ap E K. If a- E G is arbitrary, then or: LT -* L,T induces the identity on Kp, and we get aT = vat = aQp for any two places3 and o r93 above p. This shows that a EIK. The idele group IL is the unit group of the ring of adeles AL of L. It is convenient to write this ring as AL = II Lp, p where Lp = fl LT- Tip Lp is Lq3. Therefore Tip ) Upx, X A,=IL Chapter VI. Global Class Field Theory Field Class VI. Global Chapter AL, fl11 NL,IKp(a,,). Lp -* K* TI p TIP Lp, Kp -->Lp, AK NLIK:IL -kIK and as in the case of a field extension, we define and as in the case of NLPIKP NL,IKp (ail) = det(ail) . IK=A NLIK(a)p = up : Lp .fl NL,Ixp(ap) = det(ail) _ fl det(ail) _ fl NLTIKp(ap). The idele norm enjoys the following properties. Every up e Lp defines an automorphism Every up e Lp defines (2.2) Proposition. If L I K is a finite extension and a = (aT) E IL, the local (2.2) Proposition. If L I K is a finite extension given by components of the idele NLIK(a) are In this way we obtain a homomorphism It induces a norm homomorphism and IK = L[K*. Explicitly the norm between the idele groups IL = IIpL* of an idele is given by the following proposition. Kp-automorphism Up : Lp Proof: Putting up = (acp)pIp E Lp, the ail : LT the direct product of the Kp-automorphisms [L : K]. These embeddings yield the embedding embeddings yield [L : K]. These whose restriction considered above. turns out to be the inclusion L of the Kp-vector space the norm of at,, by 370 E Lp Up of elements families (ail) of all consists product IIpLp The restricted Via the diagonal embedding for almost all p. Up E Cep = fIp CST such that : Kp] _ degree E,,Ip[LT Kr-algebra of Lp is a commutative the factor +-' O." 371 aa - ! ) ap,(ra). - 4-4 TIP AL = AK ®K L, cps r.. NLIK (CO = detAK (a) or :AL =AK OK L-AK OK aL=AQL "ti ,ate Kp®KLLp= fLT, ap®ar NLIK o NMIL NLIK = `ti If L I K is embedded into the Galois extension MI K and if G = G (M I K ) K and if G = G (M Galois extension MI embedded into the If L I K is Here are consequences of the preceding investigations for the idele class In this way the inclusion by components IK c IL is simply given by the In this way the inclusion by components The proofs of (i), (ii), (iii) are literally the same as for the norm in a field (iii) are literally the same as for the The proofs of (i), (ii), via or (a (& a) = a ® aa, and the norm of an L-idele a E A* is simply via or (a (& a) = a ® aa, and the norm the determinant which a induces on the finite AK - of the endomorphism a : AL -* AL algebra AL = AK ®K L. groups. Lp = L ®K Kp (see chap. II, (8.3)), the Kp-automorphism fY : Lp -* Lp, chap. II, (8.3)), the Kp-automorphism Lp = L ®K Kp (see tensoring with K. the K -automorphism x : L -+ L by y H xy, arises from Hence det(ff) = det(x). practical reasons, it is convenient to Remark : For fundamental as well as considerations which allows us to adopt a formal point of view for the above ideles and their components. This avoid the constant back and foiih between the ring of adeles AL of L as point of view is based on identifying (see chap. H, (8.3)) which results from the canonical isomorphisms rp : L -+ LT. Here rp denotes the canonical embedding by K c_ L. An isomorphism embedding AK - AL, a H a ® 1, induced L -* oL then yields the isomorphism § 2. Ideles in Field Extensions Field in Ideles § 2. M we have C L C fields K of For a tower (i) Proposition. (2.3) NMIK (ii) = has fora E IL: NLIK(a) then one and H = G(MIL), E IK. = a1L:K] fora (iii) NLIK(a) idele of K the principal idele x E L* is the principal (iv) The norm of norm NLIK (x). defined by the usual ''' once we identify I, § 2). (iv) follows from the fact that, extension (see chap. surjectivity. the formxa=ay/yforsome ycL*.Puttinga'=ay-1yieldsa'L*aL* Noether's version(seechap. IV,(3.8))suchacrossedhomomorphismisof Indeed, for somexaEL*.Thisisa"crossedhomomorphism", i.e.,wehave aL* ECG The kernelofIL_-_>.CLisnL*=IKnL* K*=L*G.The This givesusanexactsequenceofG-modules we havetoshowthatIKnL*=K*.LetMIKbeafiniteGaloisextension Proof: TheinjectionIK--fILclearlymapsK*intoL*.Fortheinjectivity, (2.4) Proposition.IfLIKisafiniteextension,thenthehomomorphism 372 and aa'=aaay-1axaay-1 =ay-1a',hencea'EI.Thisproves surjectivity ofIL___>CLisnotaltogetherstraightforward. Toproveit,let deduced fromthefirstisstillexact.TheinjectivityofL*G -ItGistrivial. We claimthatthesequence or r=Ginducesanautomorphism Proof: TheG-moduleILcontainsL*asaG-submodule.Henceevery is canonicallyaG-moduleandCL=CK. (2.5) Proposition.IfLIKisaGaloisextensionandG=(LK),thenCL Galois descentfortheideleclassgroup: class aL*hasarepresentativea'inIK.Itisimportanttoknowthatwehave subgroup ofCL:anelementaL*ECL(aIL)liesinCKifandonlythe with GaloisgroupGcontainingL.ThenwehaveIKcILSI",and IK -ILinducesaninjectionofideleclassgroups IKnL*cIKnM*c(IKnM*)G=IKnM*G=IKnK*=K*. Via theembeddingCK-->CL,ideleclassgroupbecomesa x at= a=Qa'=a(.)a fir, L. ForeveryaEG,one ara Fhb .., 1L*G CL I)L*ILCL1. CK -*CL, ara Q714 CL, xar =xaQxr. as then hasa(aL*)=aL*,i.e.,asaxa aL* H(aa)L* IL aK* i ra as Chapter VI.GlobalClassFieldTheory CL =axrxa.ByHIlbert 90in aL*. . 1 field axiomofchap.IV,(6.1). TodothiswewillfirstcomputeitsHerbrand usual traceTrLIK(x). (v) ThetraceofaprincipaladelexEListhe in AKdefinedbythe (iv) TrLIK(a)=[L:K]a (iii) IfLIKisembeddedintotheGaloisextensionMK,and ifG=(MIK)and (ii) ForatoweroffieldsKCLM,onehasTrMIK=TrLIK 0TrMIL (i) TrLIK(a)p=Y_TIPTrL.IKp(as.) endomorphism xr-paxoftheAK-algebraAL,andshow: Galois descent.Moreprecisely,foraextensionLIK,thehomomorphism Exercise 3.Thecorrespondencebetweenidelesandideals,a norm NLIK:IL-+IK: resp. L,behavesasfollowsundertheinclusioniLIK by (2.3).Hencewegetanormmapalsofortheid6leclassgroupCL, H =G(MIL),thenonehasforaEAL,TrLIK(a)F,eG/x aa. Exercise 5.DefinethetraceTrLIK CIK Exercise 4.Theidealclassgroup,unliketheid6ledoesnothave (For thenormonideals,seechap.III,§1.) following rule,inthecaseofaGaloisextensionL(K, Exercise 2.LetLIKbeafiniteextension.Theabsolutenorm91ofidelesK, where op,resp.Op,isthevaluationringofK,Lp L (&KKp=n Exercise idele group. It enjoysthesameproperties(2.3),(i),(ii),(iii),asnormmapon § 3.TheHerbrandQuotientoftheIdeleClassGroup Our goalnowistoshowthat theid6leclassgroupsatisfies The normmapNLIK:IL--)IKsendsprincipalidelestoideaes § 3.TheHerbrandQuotient oftheIdeleClassGroup c1L(LIK° isingeneralneitherinjectivenorsurjective. v,' 1. Let lp L co1, ...,mbe '71(NLIK (a))_01(a) O1(iLJK(a)) induces, foralmostallprimeidealspofK,anisomorphism for aEAK. (NLIK(a)) =NLIK(((X)). NLIK :CL-*CK. ...ED a : AL---*AKbyTrLIK(a)=traceofthe basis ='Rt(a)EL:Kl . of LIK.Thenthe for for TIP flop, a EIL. a EIK, : IK-*IL,resp.underthe (a), satisfiesthe 1 isomorphism 373 reduces ustolocalfields.ForeveryplacepofKweput IL maybedescribedinthefollowingsimplemanner,whichimmediately group first. idele group,andbythatoftheunitgrouponother.Westudy quotient. ItisconstitutedontheonehandbyHerbrandquotientof 374 its decompositiongroup.Aso-variesoverasystemofrepresentatives Choose aplace93ofLabovep,andletGT=G(LtpIKp)cGbe where therestrictedproductistakenwithrespecttosubgroupsUL,pcL. decomposition groups LPandUL,pareG-modules,wehaveforthe-moduleIL Since theautomorphismsaEGpermuteplacesofLabovep, above theplacesofS.ForILwehavetheer77G-moduledecomposition then defineIS=IL (3.1) Proposition.LandUL,paretheinducedG-modules get thefollowing In termsofthenotioninducedmoduleintroducedinchap.IV,§7,wethus G/Gp, a93runsthroughthevariousplacesofLabovep,andweget ramified inL,thenwehave fori=0,-1that (3.2) Proposition.IfLIKisacyclicextension,andifS contains allprimes and (3.1)givesthe where foreachp,q3isachosen primeofLabovep. H' (G,Is)=®(Go,L*) Let LIKbeafiniteGaloisextensionwithgroupG.TheG-module Now letSbeafinitesetofplacesKcontainingtheinfinite places.We Lp =FILQfla(L*), LP =IndGW(L), peS Lp=nL* andUL,p=fJUc. , where TIP a IG= S denotesthesetofallplacesLwhichlie pES IL =IIL* 11 Lpx 11UL,p, and p UL,p =FjUa'fa(UU). UL,p =IndGT(Uq3). Chapter VI.GlobalClassFieldTheory p¢S H' (G,IL)=®H(GT,L a '°- `SIP a P an isomorphism Proof: ThedecompositionIL primes, then of proposition. Thesecondisanimmediateconsequence: H'(GT, UT)=1,bychap.V,(1.2).Thisshowsthefirstclaimof H' (G,UL,p)=(GT,UT).Forp (7.4), wehavetheisomorphismsH'(G,Lp)- and aninjectionH'(G,V)-*flposUL,p).By(3.1)chap.IV, § 3_TheHerbrandQuotientoftheIdeleClassGroup By localclassfieldtheory,wefind#H°(Gp,L*)_ (Kp Proof: WehaveH(G,IL) (3.3) Proposition.IfLIKisacyclicextensionandifScontainsallramified locally everywhere,i.e.,ifeverycomponentapisthenormofanelement where q3isachosenplaceabovep.Inotherwords: For thisweneedthefollowing general = np.Hence where np=[LT:Kr]. H` (G,IL)=lilyH An ideleaEIKisnormofanLifandonlyit The propositionsaysthatonehasH-'(Q,IL)=[I),becauseH-1(GT,L* As fortheHerbrandquotienth(G,IL)weobtainresult: Next wedeterminetheHerbrand quotientoftheG-moduleLS=LnIL. 1) byHilbert90.Furtheritsaysthat coo 93* S H`(G, Is_®H`(G,Lp)®H'(G,V), h(G,IL) = IKINLIKIL =®KpINLTIK,L H°(G, IL)=flH°(Ge h(G,IL) =rj PES = fpES = (®pES #H-1(G, IL) #H°(G,IL) S PES p H-1(Gq3, L*)=1and pES Ly) ®V,V= peS H` (GT,L*) S, LtpIKpisunramified.Hence np, _ F1np. , L*). pES 93, = H'(GT,L*)and l ip¢S ®H` (GT,L*). P UL, p,givesus : 375 El s) contradiction provesthelemma. If twaschosensufficientlylarge, thenzcannotbewritteninthisway.This with IzI coy I bethesup-normwithrespecttocoordinatesofbasis o(1)=i twi = wi= Fay, 0 = `ti ... z =tnjvj+Etcinivi. F,=Zw1+ ...+ZwS, i=1 Q(1)=i i=L s s or (1)=i ci wi=0, ci wi=t Itvl -yI h(G,LS) = 1 f np, ®C7 log Iaalpep _ slog IaIQ_ipaeQ-ip T ES e p = v(Elog IaIQ_iTe,-LT) = aA(a). '43 E 9)f be the standard basis of the vector space '43 E 9)f be the standard 0 -* 7Ze0 - P -) 1 /7Leo -* 0, F'=®Zwq =®ED 7Lwp=®rp X:LS TIP A(aa) R. By (1.1), the homomorphism rp = ®Zwq3 = ® a(7Lweo) = IndG°(7Lwpo), ?Es h(G, LS) = h(G,A(LS)) = Now let L I K be a cyclic extension of degree n with Galois group with Galois degree n of extension be a cyclic let L I K Now § 3. The Herbrand Quotient of the Idele Class Group Class Idele the of Quotient Herbrand The § 3. denote the places of S. We of L that lie above be the set of places and let S where np = [LT : Kp}. G = G(LIK), let S be a finite set of places containing the infinite places, the infinite set of places containing let S be a finite G = G(LIK), by LS. LS of S-units simply the group G -module LS satisfies quotient of the The Herbrand (3.5) Proposition. Proof: Let {ep I V = f has kernel A(L) and its image is an (s - 1) -dimensional lattice, s = #S. We has kernel A(L) and its image is an (s make G operate on V via we have, for a E G, Then A is a G-homomorphism because Therefore eo = F to 7L, the exact sequence lattice 1' in V. Since 7Le0 is G-isomorphic yields the identities together with the fact that F/7Leo = A(LS), accordance with lemma (3.4). Then We now choose in F a sublattice r', in we have and aw p = wok. This identifies F as the induced G-module chap. IV,(7.4),weconcludethat lattice T'hasthesamerankasF,soisthereforeoffiniteindexinF.From where 93oisachosenplaceabovep,andGpitsdecompositiongroup.The 378 and from(3.3)(3.5)weobtainthe IL =ILL*.Suchasetexistsby(1.4).Fromtheexactsequence places ScontainingallinfiniteonesandprimesramifiedinL,suchthat Herbrand quotientoftheideleclassgroupCL.Todoitchooseafiniteset Thus wedofindthath(G,LS)=1r1pesno,wherenp#Gp[LT:Kp]. In particular(CK:NLIKCL)>n. G =G(LIK),then (3.6) Theorem.IfLIKisacyclicextensionofdegreenwithGaloisgroup arises theidentity thus contradicting(3.6). p GofLIKisdifferentfrom G(LIK).Hence GpgG(LIM).Thereforeevery subextension ofLIKdegree p.Foreveryp0S,thedecompositiongroup Proof :AssumethatthesetS ofnonsplitprimeswerefinite.LetMIKbethe then thereareinfinitelymanyplacesofKwhichdonot split inL. (3.7) Corollary.IfLIKiscyclicofprimepowerdegree n=p"(v>0), From theHerbrandquotientofILandLSweimmediatelyget From thisresultwededucethefollowinginterestingconsequence. S splitscompletelyinM.We deducefromthisthatNMIKCM=CK, h(G, Ls) .v. = = n 1 n 1-BLS-kIL 1 h(G,1') = h(G, CL)=IL)h(G,LS)-I, 11 h(Gp,Z). h(G,CL) = .Z' cad f1, rah f1, _ n 1 #H_1(G,CL) PCs fl h(G,r,')= #H°(G,CL) Chapter VI.GlobalClassFieldTheory ILL- IL boo = n. n p fl I h(Gp, Zwcpo) w&0 CK =NMIKCM. the ideaeas-1isanormofsomeidelefiIM,i.e.,=(NMIKf)aE in NMs,IK,M*becauseMT=Kp.Since NM,PIKPM', forallpES.IfP there existsanaEK*suchthatapa-tiscontainedintheopensubgroup § 3.TheHerbrandQuotientoftheIdeleClassGroup Exercise 1.IftheGaloisextensionLIKisnotcyclic,thenthere areatmostfinitely contradicts (3.7). completely splitinL,thenthesamewouldholdforprimesp'ofK.This of primeorder,withfixedfieldK'.IfalmostallprimespKwere is tantamounttoGT=1,andhencethefactthatpsplitscompletelyinM. Hence psplitscompletelyinLif#H\G/Gp=[L:K]#H\G.Butthis equals thenumber#H\G/GtpofdoublecosetsHaGIinG(seechap.I,§9). let GTbeitsdecompositiongroup.ThenthenumberofplacesLabovep H =G(MIL).AlsoletpbeaplaceofK,TMabovep,and fact, letMIKbethenormalclosureofLK,andwriteG=(MK) Proof :WemayassumewithoutlossofgeneralitythatLIKisGalois.In If almostallprimesofKsplitcompletelyinL,thenL=K. (3.8) Corollary.LetLIKbeafiniteextensionofalgebraicnumberfields. NMIKIMK*. ThisshowsthattheclassofabelongstoNMIKCM,so completely inL;,fori>1,but whicharenonsplitinLI. L; flLj=Kfori0j.Thenthere areinfinitelymanyprimespofKwhichsplit Exercise 4.LetLi, such thatK*DisdenseinIKandDCNLIKL*.ThenL=K. Exercise 3.LetLIKbeafiniteabelianextension,andletD beasubgroupofIK unramified overK. generated bytheFrobeniusautomorphismsV Exercise 2.IfLIKisafiniteGaloisextension,thenthe groupG(LIK)is many primesofKwhichdonotsplitinL. Indeed, letaEIK.Bytheapproximationtheoremofchap.II,(3.4), So assumeLIKisGalois,;K,andletaEG(LK)beanelement IKINMIKIM =®KpINM,IK,M. ... ,Lr1Kbecyclicextensions ofprimedegreepsuchthat - S, thenapa-tisautomaticallycontained p of allprimeideals ,ti one 1 ofLwhichare fl, rig ,fl L3. t3. 379 cyclic groupµ(K)whoseorder isdivisiblebyn.Therefore By (1.1),KSistheproduct ofafreegroupranks-1andthe KS. ByKummertheory,chap.IV,(3.6),wehave so thatxz-nEA.ThisshowsD=4K*n,andthus L=K(Z). z EK*.Then an ideley=(yp)whichcanbewrittenasaproduct =azwithaeIK, write x=upyp,withupEUp,ypK.Putting1 forpES,weget the saidkernel.Bychap.IV,(3.6),wecertainlyhavethat L=K Proof: WeshowfirstthatL=K We chooseafinitesetofplacesScontainingtheramifiedplaces,thosethat S containstheplacesramifiedinL.Bychap.V,(3.3), wemaytherefore D =L*nnK.IfxED,thenKO(/)IKpisunramifiedfor allp¢Sbecause where 4isthekernelofmapKS of KthatdonotbelongtoSsuch (4.1) Proposition.Onehass>r,andthereexistsasetTof-rprimes again KS=IKnK*forthegroupofS-units,andweputs#S. divide n,andtheinfiniteones,whichissuchthatIK=ISK*.Wewrite group oftheform is afixedprimepower,andletLIKbeGaloisextensionwith down explicitly,andthisallowsustocomputetheindex(CK:NLIKCL). extension. ForsuchanextensionthenormgroupNLIKCLcanbewritten with thesecond.WewillreduceproblemtocaseofaKummer identity iscuriouslyinaccessiblebywayofdirectattack.Wearethusstuck either H-t(G,CL)=1orH°(G,CL)(CK n =[L:K]ofthecyclicextensionLIK,itwillnowbeenoughtoshow 380 The fieldN=K(nKS)containstheLbecause4 =L*nnKSc_ Having determinedtheHerbrandquotienth(G,CL)tobedegree So letKbeanumberfieldthatcontainsthen-throotsofunity,wheren xz_n G(NIK) =Hom(KS/(KS)n,Z/nZ) = upapEUpforallp¢S,i.e., § 4.TheClassFieldAxiom G(LIK) =(Z/nZ)' if 4=L*`:nKS,andthenthatis Chapter VI.GlobalClassFieldTheory 1 1peTKp/K**nk ,-' : NLIKCL)=n.Thefirst xz_n . E IKnK*=KS, with KS/(Ks)n general. Remark: Itwillfollowfrom (5.5)thatAIKCL=CK(S,T)alsoholdsin In particular,ifLIKiscyclic,thenNLIKCL=CK(S,T). Then onehas where (4.2) Theorem.LetTbeasetofplacesasin(4.1),andlet This showsthatAisthekernelofmapKs therefore have N IKinwhichtheprimespl, has orderdividingn.ThisshowsthatNi=Zi. is thereforecyclic,andnecessarilyofordernsinceeachelementG(NIK) extension K(nu_),uES.ThedecompositiongroupG(NIZi)QNi) is unramifiedinN,becausebychap.V,(3.3),itevery in Nibecause'li Ti, fori=1,...,s-r.Indeed,thisdecompositionfieldZiiscontained the groupA=L*nflKsaskernelof-- is possibleby(3.7).WenowshowthatthesetT=(pi of KlyingbelowX31, a prime¶3iofNiwhichisnonsplitinNsuchthattheprimesp1,...,p,s_r i =1,...,s-r.ThenLnsi-iNi.Forevery 0 i, rank rsothat NLIKCL ?CK(S,T) be aZ/nZ-basisofG(NIL),andletNithefixedfield6i, I K(S,T)=flKp'1xF1KpfiUp. is nonsplitinN:-Ontheotherhand,primepi 0 ... ,93,-,-arealldistinct,anddonotbelongtoS.This CK(S,T) =IK(S,T)K*/K*, pCS ... ,ps_,- and two XEKpjt, i=1,...,s-r. Kp;(f/x)=Kp, i=1,...,s-r, (CK :CK(S,T))_(LK]. °,' III peT split completely.ForXEKswe p¢SUT 1 lpeT fl;_ Kit/Kp*n ..., s-rwechoose K;/Kpn. ,. . . ,ps_r) realizes - 381 11 The map for then(3.6)impliesM=K,henceyEK*"nIK(S,T)c IK(S,T) nK*,andM=K("y).ItsufficestoshowthatNMIKCMCK, Proof: Theinclusion (4.3) Lemma.IK(S,T)nK*=(KSUT)R 382 Proof oftheorem(4.2):Theidentity(CK:CK(S,T))=[LK]follows unramified (seechap.V,(3.3)).Thisiswhy[a]ENMIKCM,q.e.d. is an-thpower.Forp holds becauseyEK*".HencewehaveMT=KpforpcTsincea'u' checking thateverycomponenta'isanormfromMTIKp.ForPESthis class:as a,andweshowthata'ENMIKIM.By(3.2),thisamountsto cep =xup,upEUp,forPT.Theidelea'ax-1belongstothesame #U pIU=nsincepfn.WethusfindanelementxeKSsuchthat This isalsotheorderofproductbecausebychap.II,(5.8),wehave get and dK*n/K*n=d/(KS)".From(1.1)Kummertheory,wetherefore is surjective.Forif. [a] ECK=IKK*/K*,andletaIKberepresentativeoftheclass[a]. because #(SUT)=2s-r, and t The orderofthegrouponleftis Since IKuTK*=IK,theorderofgrouponright is from theexactsequence order ofthegroupinmiddle is (IKUT :IK(S,T)) 1 - (IKUT nK*:IK(S,T)K*)=(KSUT(KSUT)n)n2s-r (IKUT IKUT nK*/IK(S,T)nK* #(K /d) K* :IK(S,T)K*)=(IKK*/K*IK(S,T)K*/K*) s denotes itskernel,thenobviouslyK*"nA=(KS)n, = pfl(Kr: P-1 (KSUT)n __ S UTitholdsbecauseapisaunitandMTIK. #KS/(KS)n #d/(KS)" KS = (CK:CK(S,T)) 02,04, C- IK(S,T)nK*is Kin) =pfl c KSUTInviewofchap.II, (5.8),the fl U,/Un -> Chapter VI.GlobalClassFieldTheory IKUT K*11K jKUTIIK(S,T) #G(LIK) - g`' nS 2 . (S, T)K*. n n InIpt trivial. LetyE 00" (KSUT)n Let = .-r n'S. 1. §4. The Class Field Axiom 383 Altogether this gives n2s (CK T)) nr = [L : K]. : CK (S, =nzs = We now show the inclusion CK (S, T) c NLIKCL. Let a E IK (S, T). In order to show that a E NLI KIL all we have to check, by (3.2), is again that every component ap is a norm from LplKp. For p e S this is true because ap E K*" is an n-th power, hence a norm from Kp ("p) (see chap. V, (1.5)), so in particular also from LpIKp. For P E T it holds because (4.1) gives A C Kp", and thus Lq3 = K. Finally, it holds for p 0 S U T since ap is a unit and LT I Kp is unramified (see chap. V, (3.3)). We therefore have IK(S>T) c NLIKIL, i.e., CK(S,T) C NLIKCL. Now if L IK is cyclic, i.e., if r = 1, then from (3.6), [L:K] Now that we have an explicit picture in the case of a Kummer field, the result we want follows also in complete generality: (4.4) Theorem (Global Class Field Axiom). If L IK is a cyclic extension of algebraic number fields, then [L : K]fori =-0, #H`(G(LIK),CL) 1 fori =-1. Proof: Since h(G(LIK),CL) _ [L : K], it is clearly enough to show that #H°(G(LIK),CL) I [L : K]. We will prove this by induction on the degree n = [L : K]. We write for short H°(L IK) instead of H°(G(L IK), CL). Let M I K be a subextension of prime degree p. We consider the exact sequence CM/NLIMCL N"'I CK/NLIKCL CK/NMIKCM -- 1, i.e., the exact sequence H°(LIM) H°(LIK) --* H°(MIK) 1. If p < n, then #H°(LIM) I [L : MI, #H°(MIK) I [M : K] by the induction hypothesis, hence #H°(LIK) I [L : M][M : K] = [L : K].- Now let p = n. We put K' = K(µp) and L' = L(ip). Since d = [K' : K] I p - 1, we have G(LIK) = G(L'IK'). L'IK' is a cyclic 384 Chapter VI. Global Class Field Theory Kummer extension, so by (4.2), #H°(L' I K') = [L' : K'] = p. It therefore suffices to show that the homomorphism (*) H°(L I K) H°(L' I K') induced by the inclusion CL CLis injective. H°(L I K) has expo- nent p, because for x E CK we always have xP = NLIK (x). Taking d = [K: K]-th powers on H°(L I K) is therefore an isomorphism. Now let x = x mod NLIKCL belong to the kernel of (*). We write x = yd, for some y = y mod NLIKCL. Then y also is in the kernel of (*), i.e., y = NL'IK'(z'), Z' E CL', and we find: Yd = NK'IK(Y) = NL'IK(Z') = NLIK(NL'IL(Z')) E NLIKCL. Hencex=yd=1. An immediate consequence of the theorem we have just proved is the famous Hasse Norm Theorem: (4.5) Corollary. Let L I K be a cyclic extension. An element x E K* is a norm if and only if it is a norm locally everywhere, i.e., a norm in every completion LTIK, (gllp). Proof: Let G = G(L J K) and G = G (LT I K,). The exact sequence 1-*L"-AIL-) CL±1 of G -modules gives, by chap. IV, (7.1), an exact sequence H-'(G, CL) ) H((G,L*) ) H°(G,IL). By (4.4), we have H-t(G,CL) = 1, and from (3.2)it follows that H°(G, IL) _ ®P H°(G p, L*). Therefore the homomorphism K*INLIKL* K*1NL1,IK,L- P is injective. But this is the claim of the corollary. It should be noted that cyclicity is crucial for Hasse's norm theorem. In fact, whereas it is true by (3.2) that an element x E K* which is everywhere locally a norm, is always the norm of some idele a of L, this need not be by any means a principal idele, not even in the case of arbitrary abelian extensions. § 5. The Global Reciprocity Law 385 Exercise 1. Determine the norm group NLIKCL for an arbitrary Kummer extension in a way analogous to the case treated in (4.2) where G (L I K) - (7L/ p°7G)T . Exercise 2. Let c be a primitive m-th root of unity. Show that the norm group NQ(r)IQCQ equals the ray class group mod m= (m) in CQ. Exercise 3. An equation x2 - ay2 = b, a, b E K*, has a solution in K if and only if it is solvable everywhere locally, i.e., in each completion K. Hint: x2 - ay2 = NK(.,)IK(x -,`y) if a V K*2. Exercise 4. If a quadratic form aixi + represents zero over a field K with more than five elements (i.e., atx + 0 has a nontrivial solution in K), then there is a representation of zero in which all x; # 0. If Hint: If 0, b # 0, then there are non-zero elements a and such that aa2 + b82 = A. To prove this, multiply the identity (t - 1)2 4t (t+1)2 + (t+1)2 1 by A and insert t = bye/a, for some element y 0 0 such that t±1. Use this to prove the claim by induction. Exercise 5. A quadratic form ax2 + by2 + cz2, a, b, c c K*, represents zero if and only if it represents zero everywhere locally. Remark: In complete generality, one has the following "local-to-global principle": Theorem of Minkowski-Hasse: A quadratic form over a number field K represents zero if and only if it represents zero over every completion K. The proof follows from the result stated in exercise 5 by pure algebra (see [113]). § 5. The Global Reciprocity Law Now that we know that the idele class group satisfies the class field axiom, we proceed to determine a pair of homomorphisms (GQ-') Z, CQ-v) 2) obeying the rules of abstract class field theory as developed in chap. IV, § 4. For the Z -extension of Q given by d, we have only one choice. It is described in the following (5.1) Proposition. Let Q IQ be the field obtained by adjoining all roots of unity, and let T be the torsion subgroup of G(SQ IK) (i.e., the group of all elements of finite order). Then the fixed field QIQ of T is a i-extension. 2 and the group isomorphic is = Zp contains (Z/p°7L)* Q(P) (7G/127L)* _ *. (7G/127L)* latter P02 L) n T = fl 7L/(p - 1)Z x 7L/2Z. T = fl 7L/(p - 1)Z = 7Lp Chapter VI. Global Class Field Theory Field Class VI. Global Chapter (E~ ("' d:GK where = 7L p x 7L/(p - 1)Z for p 7L/(p - 1)Z for = 7L p x 1 d:GQ -+Z G(Q(P)IQ) Q as our base field. If K IQ is a finite extension, Q as our base field. If K IQ is a finite G(Q(tn)IQ) = G(Q(tn)IQ) 211,, dK= n T of G(521Q) torsion subgroup V is the fixed field of T, this implies that G(41Q) is the fixed field 7L1, x 7L/(p - 1)Z for p 54 2 and Z2 = Z2 x 7L/27L. The torsion 7L1, x 7L/(p - 1)Z for p 54 2 and Z2 = G(S2IQ) = G(S2IQ) 0.j G(SpIQ) = IFm the torsion subgroup of T. Since the the torsion subgroup G(521Q) = 7L* = 7L x f, G(521Q) = 7L* = There is no canonical choice as in We fix anisomorphism G (Q I Q) = Z. Another description of the 7L-extension 5IQ is obtained in the following of the 7L-extension 5IQ is obtained Another description the case of local fields. However, the reciprocity law will not depend on the the case of local fields. However, the continuous surjective homomorphism choice. Via G (Q Q) = Z, we obtain a G(QIQ). With this we continue as in of the absolute Galois group GQ = chap. IV, § 4, choosing k a surjective homomorphism then we put fK = [K fl Q : Q and get which defines the Z-extension K = KQ of K. K I K is called the cyclotomic Z-extension of K. We denote again by coK the element of G (K I K) which is to of T is isomorphic ® 7L/27L, we see that the closure T (@p't27L/(p - 1)7L to T. Now, if G(S21Q)/T - 7L. field obtained by prime number p, let S2p IQ be the manner. For every order. Then adjoining all roots of unity of p-power and Z , - 1)7L, resp. 7L/27Z, and taking its subgroup of ZP is isomorphic to 7L/(p Galois group fixed field gives an extension (P)IQ with Q = j1P The 2-extension Q IQ is then the composite 386 we find Q(µn), = U,,,1 Since S2 Proof: WON FI, Zp, and ZP But 7L = x Z/2Z. Consequently, 7G2 = Z2 the This shows that § 5. The Global Reciprocity Law 387 mapped to 1 by the isomorphism G(K I K) = Z, and by coLIK the restriction cPK I L if L I K is a subextension of K 1 K. The automorphism cOL I K must not be confused with the Frobenius automorphism corresponding to a prime ideal of L (see § 7). For the GQ -module A, we choose the union of the idele class groups CK of all finite extensions K IQ. Thus AK = CK. The henselian valuation v : CQ -+ i will be obtained as the composite [ ,O 2, CQ G(&IQ) where the mapping [,d IQ] will later turn out to be the norm residue symbol (,d IQ) of global class field theory (see (5.7)). For the moment we merely define it as follows. For an arbitrary finite abelian extension L I K, we define the homomor- phism [ LIK] : IK - G(LIK) by [a, L I K] = flap, Lp I Kp), p where Lp denotes the completion of L with respect to a place P3Ip,. and (ap, L p I Kp) is the norm residue symbol of local class field theory. Note that almost all factors in the product are 1 because almost all extensions L p I Kp are unramified and almost all ap are units. (5.2) Proposition. If L I K and L' 1 K' are two abelian extensions of finite algebraic number fields such that K C_ K' and L C L', then we have the commutative diagram IK' [ L G(L'IK') I [ ,LK])G(LIK). Proof: For an id6le a = (asp) E IK, of K', we find by chap. IV, (6.4), that (aT,L' I K') IL,= (NK/IK,(aT),LpIKp) ,(1Ip), and L'offinitealgebraicnumber fieldsKandK'. This showsthat[a,LIK]isthe onlyaccumulationpointofthefamily{oaL'). for allL'DNbecause neighbourhoods (i.e.,NIKisoneofthefinitesubextensions ofLIK),then They convergeto[a,LIK],forifLIK]G(LN)isone ofthefundamental all finite,sothatwemaywritedownthefiniteproducts finite subextensionsofLIK,thenthesetsSL'_[pI profinite grouptotheelement[a,LIK].Indeed,ifL'Kvariesover G (LpIKp)_c(LIK),andtheproductfJp(ap,Lconvergesin L' IKpofallfinitesubextensions(seechap.II,§8).ThenLpIKpisGalois, above p,butratherthelocalization,i.e.,unionofcompletions where LPdoesnotdenotethecompletionofLwithrespecttoaplace one hastheequation precisely thiselement,onceweidentifyG(LIK)= by (5.2),anelementoftheprojectivelimitl,imG(L'IK),and[a,LIKIis of LIK.Inotherwords,ifaEIK,thentheelements[a,L'IK]define, by itsrestrictions[ homomorphism 388 and (2.2)implies [NK'IK(a),LIK] It isclearthatproposition(5.2) remainstrueforinfiniteextensionsL If LIKisanabelianextensionofinfinitedegree,thenwedefinethe QL' IN=JJ(ap,NpIKp)[a,NIK][a,LIK] Obi 141 ` 011 =fl(ap,LpIKp)EG(LIK). fl(NK'IK(a)r,LpIKp) =ffl(NK'IK,(ap),LpIKp) p , LIK]L':=[ p (aT,L'I K')IL=[a,L'IK']IL. [a, LIK]=flap,LpIKp), [ UL' E[a,LIK]G(LIN) rESL' LIK] :IK-±G(LIK) p , L' IK]tothefinitesubextensions Chapter VI.GlobalClassFieldTheory L' p TIP (up, L'IKp) G (L'IK).Again 11 are § 5. The Global Reciprocity Law 389 (5.3) Proposition. For every root of unityand every principal idele a c K one has [a, 1. Proof: By (5.2), we have Hence we may assume that K = Q. Likewise we may assume that has prime power order 2. Now let a E Q*, let vP be the normalized exponential valuation of Q for p oo and write a = uppvP(). For p ; Z,oo, Qp I Qp is unramified and (p, Q p I Q p) is the Frobenius automorphism cop: -a P. From chap. V, (2.4), we thus get pvP(a) forp # t, oo, `P withnp = up for p = f, sgn(a) for p = oo. Hence P i where a flnP = sgn(a) 11PvP(a)ug = sgn(a) npuP(a)a-l = 1. P pol, oo P__ - Since the extension KIK is contained in the field of all roots of unity over K, the proposition implies [a, K I K] = 1 for all a E K*. The homomorphism KIK ]: IK -f G (KIK) therefore induces a homomorphism [ ,KIK] :CK -- G(KIK), and we consider its composite VK :CK ) Z with dK : G (KIK) -* Z. The pair (dK, vK) is then a class field theory, for we have the (5.4) Proposition. The map vK : CK -7L is surjective and is a henselian valuation with respect to dK. x ER+,then[x,kIK]L=1foreveryfinitesubextension that M=K,andso[IK,LIK]G(LK).Thisyieldsfurthermore completely inthefixedfieldMof[IK,LIK].By(3.8),thisimplies is surjective.Indeed,,since(,LpIKp) then themap Proof :Wefirstshowsurjectivity.IfLIKisafinitesubextensionofkK, 390 because foreveryfiniteextensionLIKwehavetheidentity (i) issatisfiedbecauseVK(CK)=Z,andcondition(ii)followsfrom(5.2) surjectivity ofVK=dKo[ subgroup ofG(KIK)andthereforeequaltoG(KIK).Thisprovesthe and n=[L:K].Therefore[CK,KIK]isaclosed,dense LIK ofKIK,becausewemayalwayswritex=y°withyER+ completion K.ThusCK=CoxR.Now,[I[8+,KIK]1,forif if weidentifyIR+withthegroupofpositiverealnumbersinanyinfinite (see §1)thegroupCKiscompactby(1.6),andweobtainasplitting, UK, KIK]=[CK,isdenseinG(KIK).Intheexactsequence [IK,LIK] containsalldecompositiongroupsG(LpIKp).Thussplit reciprocity law: abstract theoryinchap.IV,(4.7). and isthereforepreciselythe inducedhomomorphisminthesenseof satisfies theformula homomorphism VK=dKo[ constitutes aclassfieldtheory,the"globaltheory". Theabove axiom, thepair In thedefinitionofahenselianvaluationgiveninchap.IV,(4.6),condition As themainresultofglobal classfieldtheorywenowobtaintheArtin In viewofthefactthatideleclassgroupCKsatisfies theclassfield VK(NLIKCL) =VK(NLIKIL)dK[NLIKIL,KIK] VK [ ,LIK]=fl(,LVIKV):IK->G(LIK) fKdQ0[ QIQ]oNKIQ = .fLIKdL[IL,LIL]fLIKVL(CL).fLIK2. (dQ:GQ-+Z, vQ:CQ-+7L) , KIK]. C"0 p , KIK]:CK r," Chapter VI.GlobalClassFieldTheory Kp l fKVQoNKIQ -±1 for finiteextensionsKIQ, G(LPIKp) issurjective, symbol. WeviewitalsoasahomomorphismIK--*G(LIflab. with kernelNLIKCL.Themap( fields wehaveacanonicalisomorphism (5.5) Theorem.ForeveryGaloisextensionLIKoffinitealgebraicnumber § 5.TheGlobalReciprocityLaw (5.6) Proposition.IfLIKisanabelianextensionandpaplaceofK, field theory,asfollows. These homomorphismsexpressthecompatibilityoflocalandglobalclass which sendsapEK*totheclassofidele G (LpIKp)-+K),andontheothercanonicalinjection In particular,forapEKpwe havetheidentity Thus, ifLIKisasubextensionofkand=(ap)E IK, then KIK_, orifL=K(i),i Proof: WefirstshowthatthepropositionholdsifLIKis asubextensionof is commutative. then thediagram subextensions ofKIK. which showsthatthediagram iscommutativewhenrestrictedtothefinite ( , The inversemapofrLIKyieldsasurjectivehomomorphism For everyplacepofK,wehaveontheonehandembedding KIK) :IK-3G(KIK)agreebecausefromchap.IV,(6.5), wehave (a, LJK)=[a,K]flap,Lp(Kp). dK c(,KIK)=vK=dK0[,KIK]. rLIK : (ap) _(...,1,1,1,ap,1,1,1,...). ( ((ap),LIK) =(ap,LpIKp), CK Kp T<.z G(LIK)ab -+CKINLIKCL , LIK): ( ( , and pIoo.Indeed,thetwomaps[,KIK LIK) iscalledtheglobalnormresidue CK -+G(LIK)ab ... G(LpIKp) G(LIK) p i I 391 resp. LIL,andconsiderthe field diagram order. WedenotebyKI R.3 G(L,f K,)_+G(LOIKp)issurjective. to L'.ThenL'=K(i)and Kp=R,LPC,sothemapping and chooseforK'thefixedfieldofrestrictioncomplex conjugation we mayassumemoreoverthatG(LIK)isgeneratedby a. power order,becausetheygeneratethegroup.Passingto thefixedfieldofa image ofG(L'IKp).Itisevensufficienttodothisonly forallUofprime every aEG(LpIKp),somespecialextensionL'K'such thataliesinthe G(L' IKp) and hencealsothefrontone,forallelementsofG(LOI Kp) intheimageof the propositionisalreadyestablished,thenbackdiagram iscommutative, for trivialreasons.IfnowL'1K'isoneofthespecialextensions forwhich bottom arecommutativebychap.IV,(6.4),andthesides where L.=KpL,KpKOK',L'KL'.Inthisdiagram,thetopand the diagram This wouldmeanthata=NLqKq(aq)forq class ofCL,i.e.,(-1)a=NLK(a)forsomeaEK*andanideleIL. If wehad((-1),LIK)=1,thentheclassof(-1)wouldbenorma R+ isthekernelof(,LpIKp),and(-1,Kp)complexconjugationin L' JK'beanabelianextension,sothatKCK',LL.Wethenconsider therefore -1ENL,IKp(L'P)=NCIRC*R+,acontradiction. 1 =[a,LIK]fJq(a,LqIKq)(a,LpIKp),sothat(-1,LpIKp)1,and (a,L41K4) =1forq G(LpIKp) =G(CIR).Thus,allwehavetoshowisthat((-1),LIK) 392 When ptoo,wefindtheextension L'IK'asfollows.Letabeofp-power When pIooandL We nowreducethegeneralcasetothesespecialcasesasfollows.Let On theotherhand,letL=K(i),pIoo,andLp0Kp.ThenKpI[8*, G(Lp IKp) G(L IK) G(L0IKp). Thismakesitclearthatisenoughtofind, for G(L' IKp) G(L' IK') Kp,i.e.,Kp=11,Lp=C,weputL'=L(i)9U, p and(-a,LpIKp)=1.By(5.3),wewouldhave , resp.LtheZr-extension l 114 ' CK/NLJKCL '-n Chapter Vl.GlobalClassFieldTheory Kp INLplKpLp try p and-a=NLpIKp(ap),i.e., - CK-/NL,IK,CL, Kp*/NLpIK' Lp contained inKI, CAD X20 ca) 1. on theid6leclassamodK". The productformulaisaconsequence ofthefactthat(a,LIK)dependsonly a =(ap), ideles. Butthisisexactlythestatementof(5.6): Proof: SinceIKistopologicallygeneratedbytheideles oftheform In particular,foraprincipalid6leEK'wehavetheproduct formula (5.7) Corollary.IfLIKisanabelianextensionanda= (ap)EIK,then of alLpunderG(L'IK)-+(LKp).Thisfinishestheproof. and K'=L.L'Iisthereforeasubextensionofk'K,theimage in chap.IV^(4.5),conditions(ii)and(iii),itthenfollowsthat[K':K] We nowtakethefixedfieldK'ofaIL,andextensionL'=K'L.As yam (a,LIK) =((ap),LIK)(ap,LIKp) = :-4, ap EKp, ... it isenoughtoprovethefirstformulaforthese (a,LIK) =II(ap,LpIKp) K L /LP fJ(a,LpIKp) =1. p I p IK of FrobeniuscpKEG(KIK).As =kVIfIK. K /LP °a' k v q 1, andthusis f(aq,LgIKq). 393 inclusion also a norm the a Zp -module of rank proves is is represented on the one is represented on PIP Chapter VI. Global Class Field Theory Field Class VI. Global Chapter PIP rp(E)=r+s-1, $.0 NCLnK*=NPL;. rankz G(MpIK) = s + 1. GIMP I H) - rI Uptt/(II Up't n f), ff E NCL n Kp K. Then ii ff E NCL n Kp K. 1 - G(MpIH) -* G(MIK) -3 CIK(p) -+ 1, 1 - G(MpIH) -* G(MIK) -3 CIK(p) -+ (Xp), xp E K. This gives (xp)a = NP with a E K*. gives (xp)a = NP with a E (Xp), xp E K. This p I Kp , .4) 4. The group f (p) := E fl H PIP Up1t Identifying Kp with its image in CK under the map ap I-> (ap), we obtain ap I-> (ap), the map under image in CK with its Kp Identifying where k is the closure of the (diagonally embedded) unit group E = oK in fPIP Up. where k is the closure of the (diagonally unit rank. where r, resp. s, is the number of all real, resp. complex, places; in other words, The Leopoldt conjecture was proved for abelian number fields K IQ by the American mathematician ARMAND BRUMER [22]. The general case is still open to date. K°nIK is the maximal abelian extension of K, then CK/DK = G(K°bIK). K°nIK is the maximal abelian extension Hint: Use (5.6) and (5.8). let MP I K be the maximal abelian p- Exercise 3. Let p be a prime number, and let H I K be the maximal unramified extension unramified outside of (p I p). Further, places split completely. Then there is an subextension of Mp I K in which the infinite exact sequence where C1K(p) is the p-Sylow subgroup of the ideal class group CIK, and there is a where C1K(p) is the p-Sylow subgroup of canonical isomorphism Exercise G (Mp I K). rp (E) is called the p-adic rp (E) := rankzp (9(p)) = [K : Q] - rankZP the famous Leopoldt conjecture: Problem: For the p-adic unit rank, one has Exercise 1. If DK is the connected component of the unit element of CK, and if Exercise 1. If DK is the connected component Kpb = K°bK Exercise 2. For every place p of K one has (5.8) Corollary. For every finite abelian extension one has abelian extension For every finite (5.8) Corollary. = (xp,LpIKp) we see from (5.6) that ((xp),LIK) Proof: For Xp E NpL NpL* C NCL. of (xp) is contained in NCL. Therefore = 1. Thus the class Conversely, let the other hand ideae a = NP, P E IL, and on hand by a norm by an idele LglKq for every shows that a is a norm from Passing to components a is formula (5.7) shows that q 0 p, and the product xp E NFL*, and this from LpIKp. Therefore NCL n K* S; NpL*. 394 = NLIK abbreviations N where we use the further corollary, the following NL and NP = the valuationsofvariouscompletionsKpimpressuponit(see§1). group CKasatopologicalgroup,equippedwithitsnaturaltopologywhich algebraic numberfieldK.Forthisitisnecessarytoviewtheideleclass class fieldtheoryacompleteclassificationofallabelianextensionsfinite § 6.GlobalClassFields norm groupNLIKCLandisthereforeoffiniteindex,becausefrom(5.5), which areopeninthenormtopologypreciselyclosedsubgroupsof Proof: Bychap.IV,(6.7),allwehavetoshowisthatthesubgroupsNofCK field ofAl.Itsatisfies closed subgroupsoffiniteindexinCK.Moreoveronehas: is a1-1-correspondencebetweenthefiniteabelianextensionsLIKand (6.1) Theorem.Themap Al = to showthatAlisopeninthe normtopology,i.e.,containsagroup closed. SinceFKisclearlyalsoclosedinCK,wegetthat NLIKCLisclosed. The normmapiscontinuous,andCLcompactby(1.6). HenceNLIKCLis homomorphism IR:CL CK =xI'K.Bythesametoken,1'Kisagroupofrepresentatives forthe for thehomomorphismfl under themapping denote byI'Ktheimageofsubgrouppositivereal numbersinK. show thatNLIKCLis.Forthis,wechooseaninfinite placepofKand (CK :NLIKCL)=#G(LIK)ab.ToshowthatAliscloseditenoughto finite indexforthenaturaltopology. The fieldLIKcorrespondingtothesubgroupAlofCKiscalledclass LI CL2(AML,_ML2, n =p NL IKCL.Forthiswemay assumethattheindexnisaprimepower.Forif As inlocalclassfieldtheory,thereciprocitylawprovidesalsoglobal If thesubgroupAlisopeninnormtopology,thenitcontainsa Conversely letAlbeaclosed subgroupofCKfiniteindex.Wehave NLIKCL =XNLIKFKxfK XTK. M, andifthe.Nareopenin thenormtopology,thensoisM. pr' , and N cCKisthegroupcontaining Alofindexp°' § 6.GlobalClassFields K* L h-±NL=NLIKCL R. Wethereforeget G(LIK) =CK/M. : CK-+R ML1L2 =ML,nNL2, CK. ThenI'Kisagroupofrepresentatives with kernelCK(see§1),i.e., ' NL,nL2 =NL,NL2. , then .'h pro 395 ray classgroupmodm: prompt thefollowingdefinition. bring inthecongruencesubgroupsCKofcorresponding tothemodules called therayclassfieldmodm. (6.2) Definition.TheclassfieldKtIKforthecongruence subgroupCKis m =fl, theorem givesaclearoverviewofalltheabelianextensions ofKoncewe NLIKCL =N.ThisextensionListheclassfieldfor N.Theexistence subgroup NoffiniteindexinCK,anabelianextension LIKsuchthat theory becauseitsmainassertionistheexistence,for anygivenclosed This finishestheproof. NK'IK (IK'(S'))cIK(S),sothat the aboveargument.Usingchap.V,(1.5),thisgivesonotherhandthat then, IK'=IK'K'*andCK'(S')NL'IK'CL',withL'K'(/k),by of primesK'lyingaboveinS.IfSwaschosensufficientlylarge, to K,thenweadjointhemandobtainanextensionK'IK.LetS'betheset L =K(RKS),becauseoftheremarkfollowing(4.2).Iftheydonotbelong group. Ifthen-throotsofunitybelongtoK,thenCK(S)=NLIKCLWith Thus itisenoughtoshowthatCK(S)=IK(S)K*/K*CNcontainsanorm the group infinite onesandthoseprimesthatdividen,suchIK=IKK*.Since where SisasufficientlybigfinitesetofplacesKcontainingthe topology). ThereforeJcontainsagroup Then JisopeninIKbecauseNCK(withrespecttothenatural 396 (IK The Galoisgroupoftheray classfieldiscanonicallyisomorphictothe The abovetheoremiscalledthe"existencetheorem"ofglobalclassfield Now letJbethepreimageofNwithrespecttoprojectionIK-CK. : J)=n,JalsocontainsthegroupHpEsKp"xfl NL'IKCL' =NK'IK(NLIIK'CL,)NK'JK(CK'(S'))CCK(S). p";, (see(1.7)).Theyareclosedoffiniteindexby(1.8),and they °q, ..' Ins IK (S)=rjKp'2xflUp. G(K'IK) =CK/Cg. UK=11{1}x 11Up, afro cad pES pES Chapter VI.GlobalClassFieldTheory pS p¢S ESC (1), andhence CM CJVI(aEIK It followsfromtheidentityAlflKp=NL*(see(5.8)): AUK So toprovef=r[pfp,wehaveshowtheequivalence for poo).Onethenhas Proof: LetAl=NLIKCL,andletmfpp"Pbea module(np=0 is theconductoroflocalextensionLpIKp,then (6.5) Proposition.Iffistheconductor-of-theabelianextensionLIK,andp obtain the place pwedefinef=1.Thenviewastherepleteidealfl be thesmallestpowerfp=p"suchthatUK_CNLPIKPL*.Foraninfinite defined theconductorfpofa-adicextensionL0IKpforfiniteplacep,to not trueingeneralthatmistheconductorofKmIK.Inchap.V,(1.6),we NL KCL.TheconductorfofLI(orAlL)isthegcdallmodulesm (6.4) Definition.LetLKbeafiniteabelianextension,andletNL= CK ,wegetfrom(6.1)the because clearlyCK2Cr'.Sincetheclosedsubgroupsoffiniteindexin One has such thatLCKm(i.e.,CKNL). class fieldKmIK. (6.3) Corollary.EveryfiniteabelianextensionLIKiscontainedinaray are by(1.8)preciselythosesubgroupscontainingacongruencesubgroup § 6.GlobalClassFields K fIisthereforethe (up EUp (up-1modp'P=(ap)ENfKp=NpL*) forallp in CK (RP) mom smallest ii EJV) = apENpLp) and f =JIfp. ray classfieldcontainingLIK.Butitis for ofElK p Km C p w forallp. Up00 C NpLp forallp. , p° and 397 fplpn°. case, aswedidinchap.M.Then(6.5)yieldsthe an infiniteplacep,providedwecalltheextensionL0IKpunramifiedinthis ramified ifandonlyitsconductorfpis541.Thiscontinuestoholdalsofor 398 familiar cyclotomicfields: Then: (6.6) Corollary.LetLIKbeafiniteabelianextensionandfitsconductor. so thatC'=NCQ(N,m),andthisprovestheclaim. other hand,CQ/C=(Z/mZ)*by(1.10),andtherefore id6le inI of Qp(Iip,,p)IQp,accordingtochap.V,(1.8).Thismeans,by§3,thatevery of theunramifiedextensionQ m =m'p'P. Proof: Letm=FjpgPoop"p.Then16 of m-throotsunity. class fieldmodmofQisthe (6.7) Proposition.Letmbeanaturalnumberand=W.Thentheray theorem ofKroneckerand Weber (seechap.V,(1.10))totheeffectthat global classfields,whichwill bediscussedattheendofthissection. This localdiscoverydoes,however, originatefromtheproblemofgenerating points offormalgroups-a fact thatcarriesalongway(seechap.V,§5). were theLubin-Tateextensionswhichcouldbegenerated bythedivision local fieldsthingsweredifferent.Theretheanaloguesof therayclassfields about themisthattheyexist,butnothowtogeneratethem. Inthecaseof are notmadetotakeovertheimportantroleoflatterbecause allweknow KmIK asanaloguesofthecyclotomicfieldsQ(µ,,,)IQ. Nonetheless,they :". By chap.V,(1.7),thelocalextensionLpIKp,forafiniteprimep,is In thecaseofbasefieldQ,rayclassfieldsarenothingbut Note inpassingthattheabove propositiongivesanotherproofofthe According tothisproposition,onemayviewthegeneral rayclassfields is anormofsomeideleQ(µm).ThusC'CNCQ(N,,,).Onthe (CQ :Cn) Then 0211 UJ p isramifiedinL np) is Qm p (µ,,,,)IQp,butalso certainly containedinthenormgroup = Q(Am) ... I"' Chapter VI.GlobalClassFieldTheory (CQ :NCQ(un,)) = HPMAPoo gyp! P If CAD in thenormgroup Upnp) Lei , x R+.Let .07 ... J D By (1.11),thegroupC11 special place.ItiscalledthebigHilbertclassfieldandhasGaloisgroup m =(m),sothatLCQ(µm). by (1.8)thenormgroupNLIQCLliesinsomecongruencesubgroupCQ, ,fl every finiteabelianextensionLIQiscontainedinafield0(µ,,,)Q,because § 6.GlobalClassFields It satisfiesthe Hilbert classfield",isdefinedtobethemaximalunramifiedabelianextension ideals areunramified.TheHilbertclassfield,ormoreprecisely,the"small characterized by(6.6)inthefollowingway. The bigHilbertclassfieldhasconductor exact sequence place p,thecommutativediagram(see(5.6)) Proof: WeconsiderthebigHilbertclassfieldK11and, foreveryinfinite In particular,thedegree[H:K]isclassnumberhK ofK. canonically isomorphictotheidealclassgroup: (6.9) Proposition.TheGaloisgroupofthesmallHilbert classfieldHIKis H IKinwhichallinfiniteplacessplitcompletely,i.e.,therealstayreal. abelian extensionofK. (6.8) Proposition.ThebigHiIbertclassfieldisthemaximalunramified Among allabelianextensionsofK,therayclassfieldmod1occupiesa Since theinfiniteplacesarealwaysunramified,thismeansthatallprime r-+ IK/1KK*( p * is linkedtotheordinaryidealclassgroupby p real G(K'IK) -Cl' G(HIK) R*/R* -CIK ,K''IK,) CIK > G(KIpIK G(K11K). . . 'K 1 andmaythereforebe 1. 399 400 Chapter VI. Global Class Field Theory The small Hilbert class field HIK is the fixed field of the subgroup G, generated by all G (K' I Kp), p I oo.Under (, K t I K) this is the image of fl Kp) I' K*/Ij K* = IK K*/IKK*, plc where Is' = j-jpl. Kp x Flpl Up. Therefore by (1.3), G(H I K) = G(K' I K)/G, = 1K/IK°°K* = ClK . El Remark: The small Hilbert class field is in general not a ray class field in terms of the theory developed here. But it is in many other textbooks where ray class groups and ray class fields are defined differently (see for instance [1071). This other theory is obtained by equipping all number fields with the Minkowski metric (x,Y)K=>2a,XVYr (tEHom(K,C)), ar = 1 if t = i, a, = i if r T. A ray class group can then be attached to any replete module m=flpnv, p where n V Z, n > 0, and n 0 or = 1 ifoo. ThegroupsU('attached p p- , p= pI p to the metrized number field (K, (,)K) are defined by 1+p11p, for np>0,and Up for np=0,if pfoo, U(nv) R*, if p is real and np = 0, p - ][8+, ifpisreal andnp=1, C* = Kp, if p is complex. The congruence subgroupmod m of (K, (, ) K) is then the subgroup CK = IKK*/K* of CK formed with the group IK=Iju(nc) p p and the factor group CK/CK is the ray class group mod in. The ray class field mod m of (K, (, )K) is again the class field of K corresponding to the group CK C CK. As explained in chap. III, § 3, the infinite places p have to be considered as ramified in an extension L IK if Lp ; K. Likewise, where F'=F-,{0}.p(z)is ameromorphicdoublyperiodicfunction,i.e., an algebraiccurveviatheWeierstrassp-function lattice F=Zw1+7Gur,inC.Thisisatoruswhichreceives thestructureof knowledge ofthetheoryellipticcurves,wehavetorefer to[96]and[28]. briefly describethemhere.Fortheproofs,whichpresuppose anin-depth may constructtheabelianextensionsofanarbitrarynumberfieldina of Kronecker'sJugendtraum(Kronecker'sdreamhis youth).Wewill quadratic fieldK.Theresultsforthiscasearesubsumed underthename answer tothisquestionhasbeengivenonlyinthecase ofanimaginary the historicoriginofnotionclassfield.Acompletely satisfactory similarly concreteway,viaspecialvaluesofanalyticfunctions.Thiswas function e2"`Z.Thequestionsuggestedbythisobservationiswhetherone to (6.7),bythem-throotsofunity,i.e.,specialvaluesexponential Riemann-Roch theoryofchap.III,andhastheadvantagebeingsimpler. metric (x,y) treatment ofthisbookisforceduponusalreadybythechoicestandard with theMinkowskimetric.Thetheoryofrayclassfieldsaccordingto finds inthetextbooksalludedtoabove.Itcorrespondsnumberfields Q( +-1),whichwasnotarayclassfieldbefore.Thisisthetheoryone ray classfieldforthemodulembecomesmaximalrealsubextension of unityistherayclassfieldmodmpg,wherep,,,infiniteplace.The m =]-[p,,.p.InthecaseofbasefieldQ, at allplaces.ThebigHilbertclassfieldistherayformodule class field.ItisnowthemaximalabelianextensionofKwhichunramified class fieldsareconcerned.Therayfieldmod1isthesmallHilbert ramified inLifandonlypoccurstheconductorf. and np=1ifLp where nowforaninfiniteplacep,wehavefp=pnPwithnp0ifLpKp, m =fIpp"PsuchthatCKCNLIKCL,istherepleteideal the conductorofanabelianextensionLIK,i.e.,gcdallmodules § 6.GlobalClassFields An ellipticcurveisgivenasthequotientE=C/PofC byacomplete Over thefieldQ,rayclassmod(m)canbegenerated,according This entailsthefollowingmodificationsofabovetheory,asfarray P W=bar(z) p (z+w)=(z) xTy ronKRtakeninchap.I,§5.Itiscompatiblewiththe c., Kp. Corollary(6.6)thencontinuestohold:aplacepis '.y 'C7 - "., coo f=fIfp, + wE yaw p for all L (z 0°7 i w Er, cot ]' °.° of m-throots °°. 401 The constants92,93onlydependonthelatticeP.andaregivenby 402 curve EoverCofgenus1.Animportantroleisplayedbythej-invariant cC/f NS-->{(x,y)EC21y2=4x3-g2x-$3}, S ccC/f'ofpoles,onegetsabijection thus beinterpretedasfunctionsoncC/f'.Ifonetakesawaythefiniteset 92 =(r)60 and itsatisfies,alongwithitsderivativep'(z),anidentity plane 1111,thenj(E)becomesthevaluej(z)ofamodularfunction,i.e., W1,602 offinsuchanorderthatr=&)I/o).2 It determinestheellipticcurveEuptoisomorphism.Writinggenerators 4x3 -g2x$3.ThisgivesthetoruscC/Tstructureofanalgebraic onto the curves C/aareobviouslyofthiskind. then End(E)®Qisnecessarilyanimaginaryquadratic numberfieldK, such thatzfc_1'.Generically,onehasEnd(E)=Z.Ifthis isnotthecase, elliptic curveE=C/fisgivenasmultiplicationbya complex numberz complex multiplication.Thismeansthefollowing.Anendomorphism ofan does aswell.ThetoriC/aconstructedinthiswayareellipticcurveswith OK ofintegersformsalatticeinC,andmoregenerally,anyideal of aholomorphicfunctionjonIwhichisinvariantunderthesubstitution (ii) Everyj(a)generatesthe HilbertclassfieldoverK. on theidealclass.fiofa,and willthereforebedenotedbyj(R). (i) ideal ofOK.Thenonehas: (6.10) Theorem.LetKbeanimaginaryquadraticnumber fieldandaan are thefollowing. and onesaysthatthisisanellipticcurvewithcomplex multiplication. The H Now letKc_cCbeanimaginaryquadraticnumberfield.Thenthering The consequencesoftheseanalyticinvestigationsforclass fieldtheory The j-invariant(a)ofC/a is analgebraicintegerwhichdependsonly x.25 ar+d affine algebraiccurveinC2givenbytheequationy2= j(E)=j(r`)= for everymatrix p'(z)2 1a , 93= = 4do(z)3-g2b'a(z)93 263383 3 (a A d ) (f') =140 E SL2(Z). Chapter VI.GlobalClassFieldTheory with 4=g2-27g3. lies intheupperhalf- 3 z r-(p(z),bo'(z)), - .pandp'may 2 numbers j(a1)areconjugatetooneanotheroverK. (iii) § 6.GlobalClassFields m 01,weform,foranylatticeFCC,theWeberfunction i.e., therayclassfieldmod1,tofieldsforarbitrarymodules big andsmallHilbertclassfield.Inordertogobeyondthefield, of K(j(a))abovep. where rppEG(K(j(a))IK)istheFrobeniusautomorphismofaprimeideal93 (iv) ForalmostallprimeidealspofKonehas Hilbert classfieldKI=(j(.fit)). With theseconventionswethenhavethe denoted by only dependsontheclassA*,notchoiceofa, banda.Itwillbe integral idealinA*.Thenabm-1=(a)isaprincipalideal.Thevaluera(a) are senttotheidealclass(m)A-1.LetabeaninA,andletb classes intherayclassgroupC1=JK/Pwhichunderhomomorphism Let AEC1Kbeanidealclasschosenonceandforall.Wedenoteby.fi.*the (ii) Foranarbitrary.fi*,thefield K(j(.Ft),t(.*))istherayclassfieldmodm class S,aredistinctalgebraicnumberswhichconjugate overthe (6.11) Theorem.(i)Theinvariantsr(.),r(.q),..., forafixedideal over K: Note thatforatotallyimaginaryfieldKthereisnodifferencebetween can If a1,...,aharerepresentativesoftheidealclassgroupCIK,then rr(z) _ Km =K(j(J),r(A*)) 2834 -2936 aPr(z), -235 9293 pp.l (a)=I(p-la), -c (A*)=ra(a)- Clg -CIK COI a L2 Pr (z), vet 2 &r(z), if 92=0, if $293,0, if 93=0 . . y{. , poi , 403 tea) II CK/(CK)", and one -> Z/nZ. For X = (Xp) in the first and For X = (Xp) in the ,, Z/nZ) Chapter VI. Global Class Field Theory Field Class VI. Global Chapter p p pFS PVS p p P (X,a) = H' (K, w») -* II H' (K,, !-L") H'(K,Nn) -+ Il H'(Kp,A,,) H' (K, Z/nZ) - ]J H' (Kp, Z/n7Z) L1. H' (K, Z/nZ) -- II H' (K P The cokernel of the second map is EtH'(Kp,Z/nZ) x IIH1(Kpjt,,) --> Z/nZ EtH'(Kp,Z/nZ) x IIH1(Kpjt,,) p IIH'(Kp,Z/nZ) x J]H'(Kp,A,,) L[H'(LT,Z/n7G) x is a finite extension, then one has a commutative diagram is a finite extension, (iii): (*) 404 Then field. Hilbert class the small H I K the big, and K' K be 1. Let Exercise (o` : o+). places, and 2' _ r is the number of real (Z /2Z)'-', where G(KIIH) = a totally Q(,/d-). Let s be and K = 2. Let d > 0 be squarefree, Exercise as : HI = 1 or = 2, according K. Then one has [K' unit of positive fundamental NKIQ(s)=-for=1. groups of the norm (IK)"K*/K* is the intersection The group (CK)" = Exercise 3. extensions L I K of exponent n. NLIKCL of all abelian V, § 1, exercise 2) field K, local Tate duality (see chap. Exercise 4. (i) For a number pairing yields a non-degenerate taken with respect to where the restricted products are of locally compact groups, resp. the subgroups H,,.(Kp,Z/nZ), product, it is given by a = (a',) in the second (ii) If (iii) The images of and respect to the pairing (*). are mutual orthogonal complements with of Z/nZ, and A' = Hom(A,K") instead an arbitrary finite GK -module A instead Hint for where LIK is the maximal abelian has H'(K,7G/nZ) = Hom(G(LIK),Z/nZ), extension of exponent n. that the statements of exercise 4 extend to Exercise 5 (Global Tate Duality). Show of p.". and exercise 4 of chap. V, § 1. Hint: Use exercises 4-8 of chap. IV, § 3, K, then the map Exercise 6. If S is a finite set of places of is surjective if and only if the map Hilbert classfield.Moreover, hestatedthehypothesisthateveryabelian as theidealclassgroupCIK anditssubgroupsdidforthesubfieldsof of a"rayclassfield"K'nIK (whichatfirstwasonlypostulatedtoexist) Clm, alongwithitssubgroups, woulddothesameforsubextensions with a=1modin,andtotally positive.Heconjecturedthatthisgroup 'ate of allidealsrelativelyprimetoagivenmodulem,bythe principalideals(a) group a rayclassgroup,whichhedefinedinanideal-theoretic way asthequotient with agrainofsalt-onlydifferentinstancescommon concept,thatof realized, aswasalreadymentioned,thatthegroupsCIK and(Z/nIZ)*are- their familiarisomorphismsG(Q(µm)IQ)=(Z/mZ)*.HEINRICH WEBER But overQ,wenaturallyencounterthecyclotomicfields Q(µ,,,)IQwith radical, forQhasnounramifiedextensionsatallbyMinkowski's theorem. abelian extensionsofK.IfthebasefieldisQ,thisrestriction isofcourse restriction tothesubfieldsofHilbertclassfield,i.e.,unramified the insightsthatwehavegainedinprecedingsection,thismeans class groupCIKathandtodothis,alongwithitssubgroups.Intermsof But atfirst,insteadoftheideaeclassgroupCK,therewasonlyideal guided bythedesiretoclassifyallabelianextensionsofanumberfieldK. had beencompletedinthelanguageofideals.Fromverystart,itwas given extensionsLplKpascompletions(see[109]). one handhasGaloisgroupisomorphictoG,andwhichontheother can beembeddedintoG.ThenthereexistsaGaloisextensionLIKwhichonthe places, andletLpIKp,cS,begivenGaloisextensionswhosegroupsG. Note: (see also[10],chap.X,§2). which satisfiestheidentityofdegrees C>11 exists acyclicextensionLIKwhichhaspKpascompletionforES,and for thetriple(K,n,S),then,givencyclicextensionsL,,IKpPES,therealways Exercise 7(TheoremofGRUNWALD).Ifthelastconditionexercise6issatisfied K(I.c2") (see§1,exercise2). n =2"m,(m,2)1,orifSdoesnotcontainallplacesp12whicharenonsplitin is injective.ThisthecaseinparticularifeitherextensionK(97V)Icyclic, § 7.TheIdeal-TheoreticVersionofClassFieldTheory Class fieldtheoryhasfounditsideae-theoreticformulationonlyafterit cad W-' § 7.TheIdeal-TheoreticVersionofClassFieldTheory Let Gbeafinitegroupoforderprimeto#ji(K),letSset [L :K]=scm{[LpKp]] CIK - JK/UK '.1 405 .fl TAKAGI (1875-1960),andcastbyEMILARTIN(1898-1962)intoadefinite, After theseminalworkofAustrianmathematicianPHrrrrPFURTWANGLER by thecasewherebasefieldisQ,whoseabelianextensionsareall extension oughttobecapturedbysucharayclassfield,aswassuggested 406 to thegroupsbetweenCKandCK,hence,inviewofisomorphism fields K'IK.Theirsubfieldscorrespond,accordingtothenewpointofview, by lookingatcongruencesubgroupsCKinCK,whichdefinetherayclass point ofviewcanbevindicatedintermstheid6le-theoreticversion field KTtIIK,andtherebymakeitamenabletoclasstheory.Theclassical such anextensionwasgiven,inordertoaccommodateitintotherayclass extensions ofLIKatonce,avoidingchoosingamodule'meverytime simplification thattheideleclassgroupCKencapsulatedallabelian canonical form. [44], theseconjectureswereconfirmedbytheJapanesearithmeticianTErn contained incyclotomicfieldsQ(µ,,,)IQbytheKronecker-Webertheorem. obviously characterizedbythecongruence where 7rp automorphism group G(LTIKp)cG(LIK)isthengeneratedbytheclassical Frobenius ideal ofKandj3aprimeLlyingabovep.The decomposition accessible idealgroups. the numerousapplicationsofmoreelementaryandimmediately obligation towardshistory,butafactualnecessitythatisforceduponusby of globalclassfieldtheoryfromtheid81e-theoreticone.Thisisnotonlyan to thesubgroupsofrayclassgroupCl. where qisthenumberofelements intheresidueclassfieldofp.Weput The idele-theoreticlanguageintroducedbyCHEVALLEYbroughtthe Let LIKbeanabelianextension,andletpunramified prime In whatfollows,wewanttodeducetheclassical,ideal-theoreticversion is aprimeelementofKp.AsanautomorphismL, cpp rppa -aqmod3 DOA (pp =(gyp,Lq3IKp), oho CKICK =ClK IPp =: ( .-+ LIK for all p Chapter VI.GlobalClassFieldTheory K ). , !fl a EoL '4' ...777 'C1 ,-.n is putting, foranyideala=flpp'pp: from thegroupJKofallidealsKwhicharerelativelyprimetomby prime idealpfmisunramifiedinL,wegetacanonicalhomomorphism Such amoduleiscalledanofdefinitionforL.Sinceby(6.6)each Now letmbeamoduleofKsuchthatLliesintherayclassfieldmodin. of theArtinreciprocitylawisnowprovidedbyfollowingtheorem. if (nrp)ECKdenotestheclassofideae(...,1,7rp,...). element ofKp,thenclearly § 7.TheIdeal-TheoreticVersionofClassFieldTheory isomorphism yieldsacommutativediagram by sendinganidelea=(ap)totheideal(a)Hpf,,pLp("p). This Proof: In§1,weobtainedtheisomorphism() commutative diagram with kernelH'/PP,whereH°`=(NLIKJL)PK,andwehaveanexact definition forit.ThentheArtinsymbolinducesasurjectivehomomorphism (7.1) Theorem.LetLIKbeanabelianextension,andletmamoduleof ( LIK) and weshowthatfisgiven by theArtinsymbol. The relationbetweentheidele-theoreticandideal-theoreticformulation is calledtheArtinsymbol.IfPEJPaprimeidealandirp 1 1 -+NLIKCL Hm/PK I CK/CK (LIK) (LIK) ()I (LI ( LIK p a K) ) : CIK- )=((np),LIK), ClK : JP-)G(LIK) CK - ( ,LIK)) f 1 Fj , 1 \ (.LIK) (LIK) > G(LIK), G(LIK) G(LIK) )up. lid : CK/CK-+CIK=JK/PK , G(LIK)-)1 G(LIK) -1. id 0f- 407 / lImoo}. The for As 7 3{co W. NLIK(ILm))K*/K* for a E JK, only depends on NLIK( II f3Uv(a )). r-. , Chapter VI. Global Class Field Theory Field Class VI. Global Chapter Tip v 3euIn. As in the proof of (1.9), we then get 3euIn. As in the proof : JK/Hm -, G(LIK). the group H'IP'. We view the module the group H'IP'. , and that it is surjective. pfoo Tip NLIK(a)p = 11 NL,IKP(ap) NLIKCL = bin (LIK) to show that the image of NLIKCL under the map to show that the f((c)) =(c,LIK)=((np),LIK) =(). f((c)) =(c,LIK)=((np),LIK) pnp as a module of L by substituting for each prime ideal p L by substituting for each prime pnp as a module of ,0 (NLIK(a)) = 11 11 p1 JK -* G (LIK) remains : CK -> JP/PP is : CK -> JP/PP Let p be a prime ideal not dividing m, zr, a prime element of Kp, element zr, a prime m, not dividing ideal be a prime Let p It ) the "ideal group defined mod m" The group Hm = (NLIK JL)PK is called The most important consequence of theorem (7.1) is a precise analysis and c E CKICK the class of the ideae (irp) _ (... , 1, 1,Trp, 1, 1, ...). Then (... , 1, 1,Trp, 1, 1, the ideae (irp) _ the class of and c E CKICK PK and (c) = p mod ( L j K) : ( the class a mod PK. It defines an isomorphism 408 symbol by the Artin --) G(LIK) is induced that f : JK I PK This shows m = fl p = FIT,, of K the product where IL) = (a E IL I Up E CL = ILm)L*IL*, elements of ideles NLIK(a), for a E IL are the classes of norm = f, Ipvs (asp) (see chap. III, (1.2)), (see (2.2)), and since vp(NLV IK,(ap)) the ideal the ideae NL I K (a) is mapped by () to the homomorphism () : CK -f Therefore the image of NLIKCL under )PK/PK, q.e.d. JK/PK is precisely the group (NLIKJL (7.2) Corollary. The Artin symbol (LnK) between subextensions of the ray that the correspondence L r- HI is 1-1 belonging to the extension LIK. From the existence theorem (6.1), we see belonging to the extension LIK. From JK containing P. class field mod m and subgroups of of the kind of decomposition of any unramified prime ideal p in an abelian extension L I K K. It can be immediately read off the ideal group H m C JK which determines the field L as class field. class fieldarepreciselytheprincipal primeideals. (7.4) Corollary.Theprimeideals ofKwhich.splitcompletelyintheHilbert (see (6.9)).Thisgivesusthe strikinglysimple corresponding idealgroupHC_JK=isthePK ofprincipalideals ray classfield rational primespfm,thelawwhichwehadalreadydeduced inchap.I, are characterizedby (10.4), andinparticularthefactthatprimenumberswhich splitcompletely JQ /PQ=(Z/mZ)*(see(1.10)),weobtainforthe decomposition of and theidealgroupcorrespondingtoQ(/.s,,,,)inJJis thegroup at thecyclotomicfieldQ(p,,,,)IQ,conductorismodulem=(m), the conductorofLIK. completely arepreciselythosecontainedintheidealgroupHt,iffis finishes theproof. of theFrobeniusautomorphismV.=(Lp) degree istheorderofdecompositiongroup43joverK,i.e., p isunramified,theq3iarealldistinctandhavesamedegreef.This Proof: Letp=T, positive integersuchthat JK/Hm =G(LIK),thisisalsotheorderofpmodHminJP/H'.This of r=n1fdistinct then pdecomposesinLintoaproduct let Hmbethecorrespondingidealgroup. definition forLIKthatisnotdivisiblebyp(forinstancetheconductor),and degree n,andletpbeanunramifiedprimeideal.Letmamoduleof (7.3) Theorem(DecompositionLaw).LetLIKbeanabelianextensionof § 7.TheIdeal-TheoreticVersionofClassFieldTheory In thecaseofHilbertclassfieldLIK,i.e., fieldinsidethe Let ushighlighttwospecialcases.IfthebasefieldisK=Qandwelook The theoremshowsinparticularthattheprimeidealswhichsplit If fistheorderofpmodH'inclassgroupJK/H',i.e.,smallest BCDmod 1inwhichtheinfiniteplacessplitcompletely, prime ideals . q3,betheprimedecompositionofpinL.Since p-1modm. pf EHm, III of degreefoverp. p=T,...93,. . In view oftheisomorphism 409 As Proof: LetK1IKbetheHilbertclassfieldofKandletK21 principal ideal. by thefollowingtheorem,knownasprincipalidealtheorem. 410 reduced tothefollowingpurelygroup-theoreticresult. is thetrivialhomomorphism.SinceKIIKmaximalunramifiedabelian show thatthetransfer where i is trivial.Bychap.IV,(5.9),wehaveacommutativediagram class fieldofK1.Wehavetoshowthatthecanonicalhomomorphism (7.5) Theorem.IntheHilbertclassfieldeveryidealaofKbecomes independent interestinthatit givesanadditiveinterpretationofthetransfer. is aZ-basisofIH.Wefirst establish thefollowinglemma,whichalsohas For everysubgroupHofG, wehaveIHCIG,and{r-1ITEH,t ideal IG,whichisbydefinitionthekernelofringhomomorphism group ringZ[Q]={EveGn,o-IEZ),weconsider theaugmentation is thetrivialhomomorphism. subgroup, andG"thecommutatorsubgroupofG.If(G :G') )-F, Chapter VI.GlobalClassFieldTheory Y'na. Cr G(KI IK), I Ver 1} 2 411 IG/Ic. 8 r P P.r H/H' , pER pER (IH + IGIH)IIGIH Ver S (IH + IGIH)IIGIH, pSr=Sr+SpSr /I IH+IGIH -* H/H'. p,r H/H' Ver(Q mod G') = f j orp mod H', S : IGG - (IH + IGIH)/IGIH S(x mod IG) = x > p mod IGIH, S(x mod IG) = x GIG' 1-i P,r 0 = r np,rpSr = r np,r(Pr - P) = i np,rpr - F_ (Enp,r)P, 0 = r np,rpSr = r np,r(Pr - P) = The transfer is now obtained as (*) = 1 mod H' because 8(pr)8r = It sends S(pr')Sr E IGIH to r'rr'-tr-' homomorphism which is inverse to pS(r'r) - PST' - Sr. It thus induces a the isomorphism GIG' (*). In particular, if H = G, we obtain where vp E H is defined by 6p = p'Up, p' E R. Ver thus induces the homomorphism where the homomorphisms 8 are induced by or i-k SQ = Q - 1, and the 8 are induced by or i-k SQ where the homomorphisms given by homomorphism S is G/H. of the left cosets R E) 1 of for a system of representatives that the homomorphism Proof : We first show The elements pSr, r E H, induced by r H Sr = r - 1 has an inverse. IGIH. Indeed, it follows from r ; 1, p E R, form a 7G-basis of IH + that they generate IH + IG IH, and if the pr, p are pairwise distinct. then we conclude that np, r = 0 because a surjective homomorphism Mapping pSr to r mod H', we now have § 7. The Ideal-Theoretic Version of Class Field Theory Field Class of Version Ideal-Theoretic The § 7. has a G, one index in of finite H subgroup For every Lemma. (7.7) diagram commutative For ifweputµ=> be theadjointmatrixof(µik).Then 412 so that(5U)µ-0modIGZ[G]IG'=IGIG'forallU.This yields which givesameaningtothedeterminantµ=det(µik)E 7L[G/G']. Let(a.kj) Qi and where ,uik-mikmodIG.Infact,thetkareproductsofcommutators that The formulae8(xy)=3x+6y+3x5y,3(x-1)_-(.x)x-1yieldbyiteration Consequently, where fisgivenbyannxn-matrix(mik)withdet(mik)=(G: ei =(0, of leftcosetsG/G',andletU1,...,UnbegeneratorsG.Mapping G" ={1},i.e.,thatG'isabelian.LetRE)1beasystemofrepresentatives Proof oftheorem(7.6):ReplacingGbyGIG",wemayassumethat Since p'runsthroughthesetRifpdoes,wegetasclaimed the identity given byS(SUmodIG)=EpERSUPIGIH.FromUpp'Upfollows Qi-1. Weview(µik)asamatrixoverthecommutativering S(Sp modlG) III . Mob . . , 0,1,...0)E (Svj)µ =>(SUi)µikXkj0modIGZ[G]IG', 0)Z f)Z" III Sp+(SU)p=SQp+Sp'+Sp'SUp. PER S i-1 1 7R 7L[G/G'] =Z[G]/Z[G]IG', i=1 11 u1 UI.c=rnpup=Fnpp. i,k n PER npp, where;5=pmodG', then forallvEGIG', µ =rpmodZ[G]IG'. cr likTk=1 m,k Sup= F 7L" toUi,wegetanexactsequence pER rk) =F-(30ri)1-Lik0, III P pER i=1 with n (SU)p =-SUEpmodIGIH- Chapter VI.GlobalClassFieldTheory )GIG' P Tic cG'. pER . 1, a°) G'). 413 ,.o fl[ III to an integral basis w1, ... , co" of Ok is III IGIG'. p = (8Q)µ - 0 mod pER O., III h = (93, ... f3r)e, G .c, K=KocK1 cK2CK3C..., t]. µ-det(mik)-(G:G')-m(G:G')mod 'G, µ-det(mik)-(G:G')-m(G:G')mod S(Sv mod I) - 80- A problem which is closely related to the principal ideal theorem and is closely related to the principal ideal A problem which If e = (Hp : Hf) and pf is the smallest power of p which belongs to Hp, then If e = (Hp : Hf) and pf is the smallest power The following exercises 2-6 concern a non-abelian example of E. ART/N. The following exercises 2-6 concern a non-abelian § 7. The Ideal-Theoretic Version of Class Field Theory Field Class of Version Ideal-Theoretic The § 7. as transfer is we see that the now lemma (7.7), m = 1. Applying we even have which was first put forward by PWLJPP FuR7wANGLER is the problem of the by PWLJPP FuR7wANGLER is which was first put forward tower is the question whether the class field class field tower. This This implies that all n p are equal, hence µ = m >pER p mod 7L[G]IG', and 7L[G]IG', p mod = m >pER hence µ are equal, all n p implies that This homomorphism since the trivial a finite number of Hilbert class field of Ki, stops after where Ki+t is the the last field in the would have the implication that steps. A positive answer not only the ideals of K, but in fact all tower had class number 1 so that in it naturally generated the greatest its ideals become principal. This perspective for a long time all attempts to interest. But the problem, after withstanding by the Russian mathematicians solve it, was finally decided in the negative (see [48], [24]). E.S. GOLOD and I.R. SAFAREVt6 in 1964 prime ideals p which are ramified in an Exercise 1. The decomposition law for the like this. Let f be the conductor of L K, abelian extension L I K can be formulated the smallest ideal group containing H of H f c JK the ideal group for L, and Hp conductor prime to p. r = of , n = [L : K]. where the 93i are of degree f over K, and field above p. Hint: The class field for H. is the inertia .... .-. X + I is irreducible. The discriminant of Exercise 2. The polynomial f (X) = X5 - d = 19. 151. a root a (i.e., the discriminant of 9L [a]) is aX + b is 55b4 + 28a5. Hint: The discriminant of a root of X5 + ring ok of integers of k. Exercise 3. Let k = Q(a). Then Z[a] is the Hint: The discriminant of Z[a] equals the discriminant of Ok because on the one hand, both differ only by a square, and on the other hand, it is squarefree. The transition matrix from 1, a, ..., &' therefore invertible over Z. Exercise 6.ShowthatKIQ(.,/19151)isa(non-abelian!)unramifiedextension. ideal awith071(a)<4.IfzA1,thenhastobeprimepsuchthat Hint: Show,usingchap.I,§6,exercise3,thateveryidealclassofKcontainsan Exercise 5.Khasclassnumber1. group C75,i.e.,itisofdegree120. Exercise 4.ThedecompositionfieldKIQoff(X)hasasGaloisgroupthesymmetric 414 These symbolsallfittogetherinthefollowingproductformula. It isgivenviathenormresiduesymbolby is generatedbyelementsofprimepowerorder. show thatacanbechosensuchLflK'=Kuse(3.7),andthefactG(LIK) chap. II,§6,exercise2.PutK'=K(a)andshowthatLIisunramified.To close toa,,whenembeddedintoKp.ThenKp(ap)cKp(a)byKrasner'slemma, approximation theorem,chooseanalgebraicnumberawhich,foreverypES,is Hint: LetSbethesetofplacesramifiedinLIK,andletLO=Kp(ap).By L K'Iisunramified. exist infinitelymanyfiniteextensionsK'suchthatLfl=K,and Exercise 7.ForeveryGaloisextensionLIKoffinitealgebraicnumberfields,there which isnotthecase. 01(p) =2or3.HenceOk/p7L/27L7L/37Z,sofhasarootmod3, (8.1) Theorem.Fora,bEK* onehas for everyplacepofK,then-thHilbertsymbol containing thegroup/L,,ofn-throotsunity.Inchap.V,§3,weintroduced, definite formulation.Letnbeapositiveinteger>2andKnumberfield In classfieldtheoryGauss'sreciprocitylawmeetsitsmostgeneraland eon ... tr. § 8.TheReciprocityLawofthePowerResidues (a, K,( ".y C-P .0+ )IKp) KP*xK** fl(a,b)=1. Chapter VI.GlobalClassFieldTheory a, b) P An. fro and hencethetheorem. Proof: From(5.7),wefind § 8.TheReciprocityLawofthePowerResidues number aprimetob,wedefinethen-thpowerresiduesymbolby and moregenerally choice oftheprimeelement7randthatonehas element ofKp.Wehaveseenthatthisdefinitiondoesnotdependonthe where pisaprimeidealofKnotdividingn,EUp,and7r Hilbert symbol: We nowprovethegeneralreciprocitylawforn-th powerresidues. arguments. Ifbisaprincipalideal(b),wewritefor short Here(a) (8.2) Definition.Foreveryidealb=flp°nprimeton,and (8.3) Theorem.Ifa,b(2K* areprimetoeachotherandn,then {fl(a b)]./[fl(a,Kp('v/)IKp)]/=(a,K( In chap.V,§3,wedefinedthen-thpowerresiduesymbolintermsof The powerresiduesymbol(a) r-, I when (a)=1 ()\a)-1 b (a) (9-')I' 0. pl mod p, a -cr"modp, is obviouslymultiplicativeinboth II pal' (ab) q =IR(p) =j(7)v,. )IK)Y= b 415 b , and forpositiveoddintegersb,wehavethetwo"supplementary theorems" (8.4) Gauss'sReciprocityLaw.LetK=Q,n2,and letaandbbe explicit description(chap.V,(3.6))oftheHilbertsymbol(!-'-b)forp=2 numbers p,1,iscontainedinthegeneralreciprocitylaw(8.3)asaspecial the theoryofGausssumsinchap.I,(8.6),casetwooddprime Here pI(b)meansthatoccursintheprimedecompositionof(b). because u,bcU.Forthesamereason,wefind where zrisaprimeelementofKp.Forifweput=u7rvp(a),then(pb)1 Proof: Ifpisprimetobnoo,thenwehave 416 odd, relativelyprimeintegers.Thenonehas chap. I,(8.6). and p=oo,weobtainthefollowingtheorem,whichismoregeneralthan case. Forifwesubstitute,inthecaseK=Q,n2,intoformula(8.3) (8.1) thengives (b) ( For thelastequationweneed againtheproductformula: Gauss's reciprocitylaw,forwhichwegaveanelementaryproofusing b 2 ) (ab n12 (b)(a) pl 1 T7 al H p b) l 2)vP(b) n(b ( (byt,(a) a,b V (a l ( bpa) :.y )Up(b)pfj (b ) - = 1 (-1) = (b1)=(-1)bz (7rb)Up(Q) ptH (bpa) for pprimetoabnoo. z a-I b-1 p 2) `CD (b) Chapter VI.GlobalClassFieldTheory )-Lp(Q) a)._`" - _pH sgna-I sgnb-I (apb) pH (apb (a) pH(a Ij 2 =(226)(200b -t7 ) ,". ) = (-1) bzz lj )-1 i1 . chap. V,(3.7). finally completedinthe1960sbymathematicianHELMUrBRUCKNER,see explicit computationoftheHilbertsymbols(apb)forpmnoo.Thiswas to reallysettletheoriginalproblem,classfieldtheorywasstilllacking appears asasimpleandspecialconsequenceofArtin'sreciprocitylaw.But reciprocity law.Theabovelaw(8.3)ofthepowerresiduesnow for alongtime,andtheall-embracinganswerwasfinallyfoundinArtin's for asimilarlawthen-thpowerresiduesdominatednumbertheory by iterationthequadraticresiduesymbol(6) example: residue modulob). symbol (a)=1isnolongerequivalenttotheconditionthataquadratic residue symbol(although,forbnotaprimenumber,theconditionthat § 8.TheReciprocityLawofthePowerResidues C Class fieldtheoryoriginatedfromGauss'sreciprocitylaw.Thequest 40077 In theaboveformulation,reciprocitylawallowsustocomputesimply The symbol(b)iscalledtheJacobisymbol,oralsoquadratic 6365 /- / =\40077I 1/ (6365) / -\112 1\ \ 1I=( 40067 7 3 887 /-\1887)1887I 704 / -\40077 11 3 `-/ 4 3 2 -\31 , asisshowninthefollowing 11 .P. 11 6365 - 3 2 (111 (40077) WIN 6365 417 Proof: ForRe(s)=Q>1+6,theseriesE' identity represents ananalyticfunctioninthehalf-planeRe(s)>1.OnehasEuler's first. where sisacomplexvariable.Ittothisimportantfunctionthatweturn prototype ofsuchafunctionisRiemann'szeta are morethaneverthefocusofnumber-theoreticresearch.Thefundamental have increasinglymoltedtotheforegroundofarithmeticscene,andtoday J'. This iswhyzetafunctions,gswellastheirgeneralizations,theL-series, be rewardedbytherevelationofsomesurprisingandsignificantrelationship. that wesucceedinstealingoneofthesewell-guardedtruths,mayexpectto a simpleshape,butitisunwillingtoyielditsmysteries.Eachtime,however, hidden withinasingleanalyticfunction,itszetafunction.Thisfunctionhas fact thatagreatnumberofdeeparithmeticpropertiesfieldare log denotestheprincipalbranch ofthelogarithm(see[2],chap.V,2.2).The limit. Thisisthecaseifandonlyseries>01loganconverges, where converge ifthesequenceofpartialproductsPn=a1. ourselves thataninfiniteproductfn° convergent inthisdomain.InordertoproveEuler'sidentity, weremind the convergentmajorantE',1/n1+a,i.e.,c(s)isabsolutely anduniformly convergent inthedomainRe(s)>1+S,foreveryS0.Ittherefore (1.1) Proposition.Theseries(s) where prunsthroughtheprimenumbers. One ofthemostastoundingphenomenainnumbertheoryconsists .L" Zeta FunctionsandL-series § 1.TheRiemannZetaFunction (S) =II 5'M CDR Chapter VII P 1_p_S n=1 n 1 a,, co 1 of complexnumbersanissaidto 1 n isabsolutelyanduniformly 1 I 1/ns=F'11/n°admits an hasanonzero We obtaintheseries of itstermsa,,. In thiscasetheproductconvergestosamelimitevenafterareordering product iscalledabsolutelyconvergentiftheseriesconvergesabsolutely. 420 In thisproductwenowexpandtheoffactors This impliestheabsoluteconvergenceofproduct p(1+8)n, It convergesabsolutelyforRe(s)=a>1+S.Infact,sincepns corresponding toalln .ti p" ::.se Re(s) >1.Itdoes,however,admitananalyticcontinuationtothewhole closely. Byitsdefinition,thefunctionisonlygivenonhalf-plane significance ofthezetafunction.Itchallengesustostudyitspropertiesmore numbers inasingleequation.Thisalreadydemonstratesthenumber-theoretic § 1.TheRiemannZetaFunction latter isdefinedforRe(s)>0bytheabsolutelyconvergentintegral zeta function facts willbeprovednext.Theproofhingesonanintegralformulaforthe equation whichrelatestheargumentsto1-s.Thesecrucial complex plane,withthepoints=1removed,anditsatisfiesafunctional residues (-1)'/n!.Therearenopolesanywhereelse. (ii) meromorphic continuationtoallofC. (1.2) Proposition'.(i) and obeysthefollowingrules(see[34],vol.I,chap.I). k=0,1,2,.... (iv) 3) r(s)r(s+1)=2LF(2s)(Legendre'sduplicationformula). 2) r(s)r(1-s)= 1) r(s+=sr(s), (iii) Now sumoverallnENandget y Hirn2y,whichgivestheequation Euler's identityexpressesthelawofuniqueprimefactorizationnatural To relatethegammafunctiontozetafunction,startwith substitution It isnowherezeroandhassimplepolesats=-n,n0,1,2,...,with It hasthespecialvaluesr(1/2) It satisfiesthefunctionalequations (s) whicharisesfromthewell-knowngammafunction.This sin 7rs n'-sI'(s) n The gammafunction r(s) n 2s 1 = = f 0 J f 00 f 0 J 00 0 e-yy 00 / n0=0 Y' e-Rn2yys dY e-an2YYS COI dY r(1) =1,r(k+1)k!, is y analytic andadmitsa f14 d Y 421 422 Chapter VII. Zeta Functions and L-series Observe that it is legal to interchange the sum and the integral because 00 00 7rn2yys,dy dY E fl e = f n=1J Y n=1 Y 0 0 < co. Now the series under the integral, g(y) _ e-"n2y, n=1 arises from Jacobi's classical theta series en'in2z enin2z, 6(z) _ =1+200 nEZ n=1 i.e., we have g(y) =(0(iy) - 1). The function 2 Z(s) = is called the completed zeta function. We obtain the (1.3) Proposition. The completed zeta function Z(s) admits the integral representation co Z(S)= 2 f(o(iy) - 1)y2. Y 0 The proof of the functional equation for the function Z(s) is based on the following general principle. For a continuous function f : R+ - C on the group R+ of positive real numbers, we define the Mellin transform to be the improper integral 00 L(f,s)=f (f(Y)-f(00))Ysdy 0 provided the limit f (oo) = limy-,,,, f (y) and the integral exist. The following theorem is of pivotal importance, also for later applications. We will often refer to it as the Mellin principle. subset. Thesameholdsforf°°(g(y) - of s.Itthereforeconvergesabsolutely anduniformly,forallsinthecompact therefore admitstheconvergent majorantfloo for ally>1,withconstantsB,B'.Theintegralfl'(f (y)- bound fortheintegrandofMellinintegralL(f,s). Therefore theconditionf(y)=ao+0(e-cy°)gives following upper a =Re(s),isboundedfory>1byconstantwhichindependent ofa. Proof: IfsvariesoveracompactsubsetofC,thenthefunction resp. bo Remark 2:Condition(ii)istobeunderstoodsaythattherenopole if ao=0,resp.bo0.Butthereisapole,whichsimple,# limit inquestion,soourcase,asy-->oo. rp(y) =c(y)*(y),forsomefunctionc(y)whichstaysboundedunderthe Remark 1:ThesymbolV(y)=0(+/r(y))means,asusual,thatonehas (iii) (ii) admit holomorphiccontinuationstoC-,(0,k). varies inanarbitrarycompactdomaincontained(sECI (i) for somerealnumberk>0andcomplexC#0,thenonehas: equation for y-+oo,withpositiveconstantsc,a.Ifthesefunctionssatisfythe (1.4) Theorem.Letf,g:R+-+Cbecontinuousfunctionssuchthat They arethereforeholomorphicfunctionson{sECI § 1.TheRiemannZetaFunction The integralsL(f,s)and(g,convergeabsolutelyuniformlyifs They havesimplepolesats=0andkwithresidues They satisfythefunctionalequation Res3=0 L(g,s)=-bo, Ress_o L(f,s)=-ao, 0. f(y) =ao+O(e-`y°) 1 (f(y)-ao)ys-tl two f (y)=Cykg(y), Ress_kL(g,s) =C-lao. Res5_k L(f,s)=Cbo , g(y) =bo+O(e-cy°), bo)ys-i dy. 't7 Y T_, B d ywhichisindependent ..D Re(s) >k}.They resp. Re(s) >k). ao)y'-t dy e'cyaya, 423 I For thesecondintegral,substitutionyt-+ and (1,oo)write 424 where therefore obtain By theabove,italsoconvergesabsolutelyanduniformlyforRe(s) f (1/Y)=CYkg(Y)give: following property. function Z(s).Infact,Jacobi's thetafunction6(z)ischaracterizedbythe This finishestheproofoftheorem. have beencontinuedtoallofC'.{0,k}andweL(f functions, andoneobviouslyhasF(s)=CG(k-s).ThusL (f the wholecomplexplane,aswesawabove.Sotheyrepresent holornorphic The integralsF(s)andGconvergeabsolutelylocally uniformlyon where Swapping fandg,weseefromg(1/y)=C-1yk(y)that: Now letRe(s)>k.Wecuttheintervalofintegration(0,oo)into1] The resultcannowbeapplied totheintegral(1.3)representing 0 f 1 L (f,s)= G(s) _ F (s)= f[(g(y) -bo)Ys+C-1(f(Y) 00 f[fY) f (f 1 00 00 L(f,s) (f(Y)-ao)YsdY=-ao L(g,s)=- (Y) -ao)YS - ao)ys+C(g(Y) ao S s + f bo ao s s C o+f + + dy Chapter VII.ZetaFunctionsandL-series Y 00 C_'ao s -k s-k Cbo 1 (g(Y)-bo)Y--dy-k----s 00 + ITV f 0 f (f ( 1 t Y)Y_S +G(s) (Y) -ao)YS - bo)Yk-s] ao)Yk-s] 1/y andtheequation k s +F(s), d Y , s)=CL(g,k-s). t dY Y , s)andL(g, dy dy Y Y . Cbo k. We The squareroot half-plane IHI={zECIIm(z)>0},andsatisfiesthetransformationformula for everyE>0.Itthereforerepresentsananalyticfunctionontheupper converges absolutelyanduniformlyinthedomain(zECI (1.5) Proposition.Theseries § 1.TheRiemannZetaFunction uniquely bytheconditions where logindicatestheprincipalbranchoflogarithm.Itisdetermined so wetakeitforgrantedhere.Observethatifzliesinill(thendoes-1/z. i.e., Z(2s)istheMellintransform Proof: By(1.3),wehave equation and s=1withresidues-11,respectively,satisfiesthefunctional admits ananalyticcontinuationtoC (1.6) Theorem.Thecompletedzetafunction of thefunctionf(y)=26(iy). Since We willprovethispropositioninmuchgreatergenerality§3(see(3.6)), h(z)2=z/i 6(iy) =1+2e-"y(1+E .N. z/i isunderstoodtobetheholomorphicfunction Z(2s) = Z(s) 0(-l/z) = NIA and 6(z) _ Z(s) =Z(1-s). h(z) =e21°g2/' Z(2s) =L(f,s) 2 h(iy)=fy- >0foryElR+. 0 J (9(iy)-1)yS 0 nGZ z/i 6(z). e7rin2z (0, 11,hassimplepolesats=0 n=2 e-n(n -1)y) 'y Y 2 Im(z) >8}, 425 Substituting (1-s)/2,resp.s/2,intotheformulae(1.2),(iii), 2)and3)gives (*) The equationZ(1-s)=Z(s)translatesinto Proof: Z(s)= and sodoest(s),asr(1/2)=v6F.Theresiduecomesout tobe 1' (s/2).Hence(s)hasnopole.Ats=1,however,Z(s) asimplepole, satisfies thefunctionalequation continuation toCN(1),hasasimplepoleats=1withresidueand (1.7) Corollary.TheRiemannzetafunction(s)admitsananalytic the functionalequation and simplepolesats=0,1withresidues-11,respectively,satisfies Accordingly, Z(s)=L(f the functionalequation poles ats=0,1/2withresidues-1/2and1/2,respectively,itsatisfies By (1.4),L(f,s)hasaholomorphiccontinuationtoC 426 one hasf(y)=1+O(e-"y).From(1.5),wegetthetransformationformula For theRiemannzetafunctionitself,theoremgives f (1/Y)= ResS_1 c(s)=7r1/2r(1/2)-1Ress=tZ(s)1. (1 -s)=2(27r)-'I(s)cos (1 1-s r(s)r(l+s) 2 L(f,s) =L(f, , s/2)hasa 2 2O(-1/iy) =2Y1/20(iy)YI,2f(Y) s ) r( s) 1+s has asimplepoleats=0,butsodoes 2 2 =n7_S holomorphic continuationtoC Z(s)=L(f,2)=L(f,2-2)=Z(1-s). Chapter VII.ZetaFunctionsandL-series r(`2s) 1 r(2) = 2 cos(Trs/2) ' -s). s 2s (s)- Tt r(s)' (0, 1/2)andsimple . {0,1} »'D El (1.8) Theorem.Foreveryinteger k>0onehas now provetheremarkable more naturalandbettersuitedforthefurtherdevelopment of thetheory.We except forBL,whereonefinds-1insteadofi.Buttheabove definitionis the Bernoullinumbers.AsF(t)= In generalonehasB21+1=0forv>1,becauseF(-t) = F(t)-t.Inthe The firstBernoullinumbersare classical literature,itisusuallythefunction Their relationtothezetafunctiongivesthemaspecialarithmeticsignificance. and aredefinedbytheseriesexpansion from thefunction at thenegativeoddintegersaregivenbyBernoullinumbers.Thesearise functional equationbythefactthatvaluesofRiemannzetafunction function atthepointss=2k,kcN.Thephenomenonisexplainedvia These areexplicitevaluationsofthespecialvaluesRiemannzeta It isonlythebeginningofasequence: has thesuprisetodiscoverremarkableformula Inserting thisinto(*)nowyieldsthefunctionalequationclaimed. and aftertakingthequotient, § 1.TheRiemannZetaFunction At somepointduringthefirstmonthsofstudieseverymathematicsstudent BO =1,BL f-+ coo r(2)/r(12 2 1 n4 , B2= C7' 90 1 6 1 F (t)_YBk- n , B3=0,B4 F(t) = -in 4 .-. , .--1 R'7 sl 2 =-Ir er k=0 E °° = t 00 et-1 2S (1-k)=-Bk HIV t et n6 6 1 1 + t,thisdoesnotchangeanything k. tk e' III 2 t cos 945 1 30 1 2 F(s). which servesfordefining 7r , B5=0,B6 Ow' 6 , `"i etc. 42 ,...SIN 1 427 from stoa: and aE[e,co],weconsiderthepath 428 Proof: Theintegrationdoesnotactuallytakeplaceinthecomplexplanebut integrable on(0;a).Thenonehas and alsotheinteriorofK.LetG(z)beaholornorphicfunctiononUN{0} (1.9) Lemma.LetUbeanopensubsetofCthatcontainsthepathCs,a KE ={zI which firstfollowsthehalf-linefromatoe,thencircumference in X.Wenowhave and z and on theuniversalcoveringofC*, with apoleofordermat0,andletG(t)t"-'-1(nEN),forRe(s)> where's We preparetheproofproperbyafunction-theoreticlemma.Fore>0 Ce,a zs-1 G+- J a =(a,Elx is thepath f are holomorphicfunctionsonX,namely I z=s)inthenegativedirection,andfinallyhalf-line CE, a 'T7 f X ={(x,a)EC*xRIargx-amod27r}. z(x a)=x G(z)zns-1dz= e27rinsf G(z)z)u-1 dz {0}, Ke={seit Ce, a=(a,e]+Ke[e,a), G(Z)zns-1 dz Ce,a =1ea+Ke+lea = _ zs-1 (x,a)=e(s-1)(10AIx!+ia), = ! G(t)tns-1dt (e2nins 9 a Chapter VII.ZetaFunctionsandL-series J w _ 1) t e[0,2'r}},'a=[s,a)x{2rr} G(t)tns-ldt, 0 J /a a G(t)tns-1 dt. be and theclaimreducestoidentity z =27riv,vEZ, is ameromorphicfunctionofthecomplexvariablez,withpolesonlyat Proof of(1.8):Thefunction passing tothelimitass-+0. and sincetheintegralonleftisindependentofe,lemmafollowsby as s-30.Infact,onehaslim Since Re(s)> § 1.TheRiemannZetaFunction place ontheuniversalcovering X=((x,a)EC*xRIargx-amod2ir} the half-linefrom-eto-oo.Infact,integralsover (-cc,-s]and which tracesthehalf-linefrom-ooto-s,followedby circumference We mayreplaceitwiththepath-Ce=(-oo,-s]+Ke +[-s,-oo), for 0 Resz.o -,% CE, Q Ke f J Izl =s)inthepositivedirection,from-eto-e,andfinally ... i.e., Re(ns-m)>0,thelastintegralI(s)tendstozero -.4 G(z)zns-l dz F(z)z-k-I G(z)zns-l dz F(z)= 0. Bk/kistheresidueof(k-1)!F(z)z-k-Iat0, H(s) =I = -if = -i e->0 = = (e2Jrins_1)fG(t)t"s-'dt+I(E), 2.7ri 1 0 J 0 2n ez_1 -Ce N_. I z=sistakeninthepositiveorientation. z ez Izl=e l r f F(z)z-k-'dz= F(z)zs-' s"sG(ee-`t) =>Bk- k=0 e CO a = 0.Thisgives 'CS G(ee-it)Ens-te-it(ns-I)ee-itdt dz z z zk _ -1(1 -k) (k -1)! III ITS 8 429 we substituteti-+ntandget gamma integral The integralontherightwillnowberelatedtozeta function.Inthe has asimplepoleatz=0sothat,forRe(s)>1,(1.9)yields from stooo.Thefunction circumference KEinnegativedirectionfromstos,andfinallythehalf-line where thepathCE=rp-to(-C6)followshalf-linefrom00tos,then we obtain Since zorp=-zand transformation Now substitutezH-z,ormoreprecisely,applythebiholomorphic uniformly forallsEC.ItthusdefinesaholomorphicfunctiononC,andwe Zs-t (x,a)= of C*,asin(1.9),andz,zs'-taretheholomorphicfunctionsz(x,a)=x, 430 find that H(s) _-e-"is G(z) = (zs t o W)(x,a)=zs-t(-x,a-7r) e(s-t)(log Ixl+ia)Theintegral enis rp:X H(s) __e-irrs CE - CC. f G(z)zs-tdz e-nis) JG(t)ts F(-z)z-t = _ -e P(s) 0 1 -e-z 00 Rest=oF(z)z-k-t e-z (x,a)i 0 00 CE 0 f f f -isszS-i(x,a), o= f = Chapter VII.ZetaFunctionsandL-series e-tts e-ntts dt - F(-z)zs-t - dt t 1 27r z 1 -e-z converges absolutelyandlocally > (-x,a-7r). - -2i dt t .n. H(1 -k). 1 `-G e(s-t)(Oog Ixl+ia-i7r) sinus dz - 1=>2e-nz z 0 f/ G(t)ts- 00 n=1 00 dt t of R*/Q*,andonehas regulator. Itiswell-determined uptoarationalfactor,i.e.,itisanelement The imageR2kofanonzeroelementinK4k_I(Z)®zQis calledthe2k-th mysterious canonicalisomorphism elucidated onlyrecently.Surprisinglyenough,itisthehigher K-groups Ki (Z)fromalgebraicK-theory,whichtakethelead.In fact,onehasa k =1,2,3,..., (1.10) Corollary.Thevaluesofc(s)atthepositiveevenintegerss=2k, which goesbacktoEuL.ER. s =1-kweobtain,sinceF(k)(k1)!, Since bothsidesareholomorphiconallofC,thisholdsforsEC.Putting From thisand(1.2),2),weget r(2k) _(2k-1)!,theprecedingtheoremgivesfollowingcorollary, The interchangeofsummationandintegrationisagainjustifiedbecause Summing thisoverallnENyields § 1.TheRiemannZetaFunction The values(2k-1),k>1,atthepositiveoddintegers havebeen Applying thefunctionalequation(1.7)for(s)andobservingthat Rest=o H(s) =-2isin2rsP(s)(s) are givenby F(z)z_k-I (2k) =(-1)k_i r (2k-1)_-R2kmodQ*. n=1 = K4k-I(Z)®R -R. 0 1 2rri I 00 le nttsl H(1 -k)= Z 0 J 00 G(t)ts 2(2k) ! ( dt 2n 2k t ) < 00. r(1 -s) dt B2k (1 -k) (k -1)! 27ri 'r. ' (s), q.e.d. 431 SPENCER BLOCHandKAZUYAKArohavefoundacompletedescriptionofthe most generalkind.Theseinsightsaresummarizedwithinthecomprehensive deep insightsintothearithmeticnatureofzetafunctionsandL-series tremendous influenceonfurtherarithmeticalresearch,andhasopenedup This discoveryoftheSwissmathematicianARMANDBORELhashada 432 functional equationZ(s)=Z(1-s),i.e., function I'(s)isnowhere0andhassimplepolesats=0,-1,2, Euler's identity(1.1)showsthatonehas(s)0forRe(s)>1.Thegamma theory oftheTamagawameasure. special zetavalues(2k-1)(i.e.,notjustadescriptionmodQ*)vianew Beilinson conjecture(see[117]).Inthemeantime,mathematicians may bewritten,accordingtoR/EMANN,astheseries the naturalnumbers.Thedistributionfunction consequences fortheproblemofdistributionprimenumbers withinall Re(s) =t Riemann Hypothesis: famous, stillunproven 0 30 000 / -I- 2z. Then for all matrices k =F(t)ea` = Bk(x)-1, k +3 k e2 k=0 -I- (Bk+i(n) - Bk+1(0)) k + 1 20 000 Bm (x) _ L (.f , s/b) = ba''1bL (g, s) sk(n) = lk +2 e' - 1 0(2z) = t e('+i)l sk(n) = 1'0(4)={(a 10 000 6 000 t Exercise 4. For the power sum one has Exercise 5. Let in the group Exercise 1. Let a, b be positive real numbers. Then the Mellin transforms of the Exercise 1. Let a, b be positive real numbers. are defined by Exercise 2. The Bernoulli polynomials Bk(x) so that Bk = Bk(0). Show that Exercise 3. Bk(x) - Bk(x - 1) = krk-l. On this matter, we urge the reader to consult the essay [142] by DON ZAG/ER. On this matter, we urge the reader to consult functions f (y) and g(y) = f (ayb) satisfy: § 1. The Riemann Zeta Function Zeta Riemann The § 1. mathematics: in mysteries of the biggest poses one which (n, m) The trivialcharacterX°mod m,X°(n)=1for(n,m)1,0 the multiplicativefunctionX:Z-*CCby always inducedfromaprimitivecharacterX'modf.Given X,wedefine case, thegcdofallsuchdivisorsiscalledconductor fofX.SoXis of aDirichletcharacterX'modm'foranyproperdivisorI m.Inthegeneral It iscalledprimitiveifitdoesnotariseasthecomposite A Dirichletcharactermodmisbydefinitiona Dirichlet L-series.Theyaredefinedasfollows.Letmbeanaturalnumber. of thebasicandseminalprinciplescurrentnumber-theoreticresearch(see[1061). forms, whichwehaveintroducedinthecaseofRiemannzetafunction,isone for thegroupP0(4).TherepresentationofL-seriesasMellintransformsmodular The Legendresymbol(d)andtheconstantEdaredefinedby where one hastheformula 434 The mostimmediaterelativesoftheRiemannzetafunctionare Jacobi's thetafunctionO(z)isthusanexampleofamodularformweighti 1, playsaspecialrole.When readmod1,wedenoteitbyX=1. r-, X: (Z/mZ)*)S'={zECCIIzl=1}. .CD l (\ (Z/mZ)* d raz+bl Sd J 1(Y,z) _(d)sd'(cz+d)1/Z. § 2.DirichletL-series 10 = t X (nmodm) (1, i, (IdI)' (Idl rcl ) (Z/m'Z)* if d-3mod4. if d-lmod4, cz-}d1=I(Y,z)7A(z), Chapter VII.ZetaFunctionsandL-series III III otherwise, ifc Then therule of Xby important whenwegeneralizefurther.Wedefinetheexponent pc:(0,;1) Dirichlet charactersXmodm.Thisphenomenonwillbecome increasingly L (X,s)whichhastheeffectofrealizingitasMellin transformofa >,i theta series.Wedo,however,havetodistinguishnowbetween evenandodd study, alsocomparingitwiththeprecedingsection. special. caseoftheaboveL-seriesL(X,s).Werecommend itforcareful orientation, theproofoffunctionalequationwillbegiven hereinthe is anessentialgoalofthischapter.Inordertoprovidesome preliminary in alargerclassofL-series,theHecketreatmentwhich to theargumentI-s.Thisparticularlyimportantpropertydoesinfacthold X =X°),andtheysatisfyafunctionalequationwhichrelatesthearguments continuation tothewholecomplexplane(withapoleats=1incase may omitithere. have togiveitagaininamoregeneralsituation§8below(see(8.1)),we literally thesameasforRiemannzetafunction.Since,moreover,wewill function onthehalf-planeRe(s)>1.WehaveEuler'sidentity (2.1) Proposition.TheseriesL(X,s)convergesabsolut.,lyanduniformlyin present section. the generalL-seriesL(X,s)usingsamemethods.Thisistaskof obtained forthisspecialfunctioninthelastsectioncanbetransferredto character X=1,wegetbacktheRiemannzetafunctionc(s).Allresults where sisacomplexvariablewithRe(s)>1.Inparticular,fortheprincipal the lastsection.ForaDirichletcharacterX,weformL-series It isalsocalledtheprincipalcharacter.Consideringitintheorytobe the domainRe(s)>1+8,forany80.Itthereforerepresentsananalytic developed nowhastheeffectofsubsuminghereeverythingwehavedonein § 2.DirichletL-series The proofagainhingesonanintegralrepresentationof thefunction Like theRiemannzetafunction,DirichletL-seriesalsoadmitananalytic In viewofthemultiplicativityXandsince E.' tea. ..d L(X,s) 'L7 X ((n))=(n) X(-i) =(-1)px(1). L(Xs) P 'y' I -X(p)p-s n=1 0o x(n) (In n n, ,°7 n coo C'+ I X(n)<1,theproofis 'on 5'S ii-!- 435 (see §7). play theleadingpartwhenweconsiderhigheralgebraicnumberfields These Grbj3encharakterearecapableofsubstantialgeneralizationandwill relatively primetom.ThisfunctioniscalledaGr6f3encharaktermod defines amultiplicativefunctiononthesemigroupofallideals(n)whichare 436 because Here, swappingtheorderofsummationandintegrationisagainjustified, We multiplythisbyX(n),sumoverallnEN,andget Substituting y1-+rrn2y/m,weobtain arises fromthethetaseries The seriesundertheintegral(*), (*) X(n)nP =X(-n)(-n)Pimplies that where weadopttheconventionthat00=1incasen0, p=0.indeed, We nowconsiderthegammaintegral (1) 2 n=t }IN 00 s±e m 0 00 T' (X,s)L(X,s)= T'(X,s)=Ts+p If )2L 8(X, z)=X(0)-}-2r-(n)nP e'in2z/m (m 1r) r(X,s)n = 8(X,z) _ (Re(s)+P)/2r Re(s)+p g(y) =EX(n)nP IX(n)nPe-irn2Y/my(s+P)/2I d -1 n=1 2 nEZ f 0 00 JnPe_rn2Y/rny(s+P)/2.;)±. 0 J n=1 00 E X(n)nP n=1 00 Chapter VII.ZetaFunctionsandI-series, 0 2 00 e-nn2Yl ni X(n)nPen'in2z/m e-Yy(s+P)12d Y e-sn2Y/my(s+p)!2 d Y 7'f Y .S' oo. y °8I' .dab the factor which isassociatedwithRiemann'szetafunctionaswesawin§1.Weview function character 1,andx(0)=0otherwise.WhenmthisisJacobi'stheta so thatg(y)= § 2.DirichletL-series For thisfunction(*)givesusthe define thecompletedL-seriesofcharacterx: Lp(s) =1/(1-x(p)p-')oftheproductrepresentation(2.1)L(X,s)to in (*)asthe"Eulerfactor"atinfiniteprime.ItjoinswithEulerfactors formula asassumedintheorem(1.4).Todothis,weusethe following So wehavetoshowthatthethetaseries9(x,iy)satisfies a transformation the thetaseries. This iswhythefactornphadtobeincludedinorderlinkL-series the naturalnumbers,whereasinthetaserieswesumoverallintegers. where c(X)=(ZL)p/2. (2.2) Proposition.ThefunctionA(X,s)admitstheintegralrepresentation and forzEH,onehasthetransformationformula converges absolutelyanduniformlyinthedomainIm(z)>3, forevery3>0, (2.3) Proposition.Leta,b,ILberealnumbers,p>0.Then theseries Let usemphasizethefactthatsummationinL-seriesisonlyover We wanttoapplytheMellinprincipleaboveintegralrepresentation. A(x,s)= 0,,(a, b,- A(X,s) =L.(X,s)L(X,s)', z (0(x, iy)-x(0))withx(0)=1,ifxisthetrivial eµ(a, b,z)=T,e'ri(a+g)2z+27ribg L.(X,s) = c(2/ 11z) 8(Z) =Te7rl 0 f 00 e-27riab gEµZ nEz (-1 ml n s/2 n2z z/t r(x,s) B !1' Re(s) >1. (8(x,iy)-x(o))y(s+p)l2 yy b az). 437 a, z). I'D . P urn 91/u(-b, a, z). Ir(X)I= / . p e'i(a+s)2z+2rribg and Chapter VII. Zeta Functions and L-series and Functions VII. Zeta Chapter u=O N0) m-I 8 E(hZ riPe2niab ]-1(z/i)P+70 µ(-b, 0,,(a, b, z) = (27ri)PzP0, (a, b, z) 0,,(a, b, z) = (27ri)PzP0, r(X,n) Oil,(-b, a, z) = (21ri) e- Oil,(-b, a, z) = (21ri) dP daP 27riab r(X,n)=X(n)r(X) 9, (a, b, z) = F (a + g)P 9, (a, b, z) e dP daP 9, (a b, - 1/z) = This proposition will be proved in § 3 in much greater generality (see (3.6)), (see greater generality 3 in much in § will be proved This proposition formula for the theta This corollary gives us the required transformation so we take it for granted here. The series BP,(a, b, z) is locally uniformly The series BP,(a, it for granted here. so we take in § 3. Differentiating This will also be shown in the variables a, b. convergent function a, we obtain the = 0, 1) in the variable p times (p have More precisely, we an. > 0, one has the transformation (2.4) Corollary. For a, b, tc E II8, µ 438 (2.3), we dP/ daP to the transformation formula Applying the differentiation get the formula sums which are defined as follows. series 6 (X, a), if we introduce the Gauss sum r(X,n) associated to the (2.5) Definition. For n E Z, the Gauss to be the complex number Dirichlet character X mod m is defined Forn = 1, we write r(X) = r(X,1). character x mod m, one has (2.6) Proposition. For a primitive Dirichlet 0. we 439 e2ni v(/.c-1)/n, eni(a+g)Zz/m v=O m-I 'r(X,v)e-2niv/m v=o // M- X lit) pm(0,a,mz), m-1 u=o gEmZ x(v)e-2niv/n: = E X (a) F (a + g)P a=0 m-I 1.rP (X) m (mz/i)P+z6 i z-1 r' X(a)em(a,0,z/m). v=o a=0 m-+I W=1 -Xnt-1+...+X+1. r(x)ny e2ni vlt/n: a-2ni v/m = r e2ni vlt/n: a-2ni v/m enin2z/m = X-1 (X, Z) = Xm-1 X (1 ) - 1/mz) = an m-th root of unity ; 1, hence a root of the polynomial an m-th root of unity 6(x, - 1/z) = V=OA--O nEZ m-Im-I Bm(a,0, _ e2'r`(i`-I)/m is r(x) 2= r(x)r(x) = r(x) 2= r(x)r(x) We now obtain the following result for the theta series 9(x, z) . We now obtain the following result for B(X, z) = L X (n)nP = § 2. Dirichlet L-series Dirichlet § 2. = X (vn) : from = 1 follows (n, m) in the case identity The first Proof: l, X zero. Indeed, since $ 1, both sides are When d = (n,m) X(n)x(v). that = 1 mod m/d such case choose an a we may in this is primitive, that by x (a) and observing r (X, n) m and x (a) 56 1. Multiplying a # 1 mod so that r(X,n) X(a)r(X,n) = r(x,n), = e2nivn/m gives e2nivan/n: have Further, we character mod m, then (2.7) Proposition. If x is a primitive Dirichlet to x, i.e., its inverse. where X is the complex conjugate character The last sum equals m for tL = 1. For j. 0 1, it vanishes because then m for tL = 1. For j. 0 1, it vanishes The last sum equals =MX(1) = m. Therefore I r(X)12 have the transformation formula according to the classes a mod m, Proof: We split up the series 0(x, z) a=0,1, ...,m-1, and obtain hence By (2.4), one has 440 Chapter VU. Zeta Functions and L-series and this gives 8pO, a mZ = gP eir:&2 +2aiag = 1 e2nian/''np enin2z/m I mpnEZ 8E,Z Multiplying this by X (a), then summing over a, and observing that r(X,n) = X (n)r(X), we find: 1 1 m-1 B(X, - 1/z) =ipm(mz/i)p+2 r X(a)8jm(0,a,mz) a=0 1 t z iPpZP+1(mZ/i)p+z z/n: nEZ a=0 1 (zl i)p+2i(X) E X (n)npeain2z/m iP naZ i (z/i)p+1 PW 0(X, Z). The analytic continuation and functional equation for the function A(X , s) now falls out immediately. We may restrict ourselves to the case of a primitive character mod m. For X is always induced by a primitive character X' mod f, where f is the conductor of X (see p. 434), and we clearly have L(X,s) (1 -X(P)P-s)L(X,s), PI- Pff so that the analytic continuation and functional equation of A(X, s) follows from the one for A(X', s). We may further exclude the case m = 1 (this is not really necessary, just to make life easy), this being the case of the Riemann zeta function which was settled in § 1. The poles in this case are different. (2.8) Theorem. If X is a nontrivial primitive Dirichlet character, then the completed L-series A(x,s) admits an analytic continuation to the whole complex plane C and satisfies the functional equation A(X,s) =W(X)A(y, 1 - s) with the factor W (X) =iplw. This factor has absolute value 1. formula (2.7)gives We thereforeobtainA(X,s)andsimilarlyalsoasMellintransforms By (2.2),onehas and thereforef(y)=O(e-"yln')`,likewiseg(y)O(e-"y/nm). We haveX(0)=0,sothat Proof: Letf( § 2.DirichletL-series These arealgebraicnumberswhichlieinthefieldQ(X) generated bythe nontrivial primitiveDirichletcharactersXmodm,thegeneralized Bernoulli holds withW(X)=tpl to allofCandthattheequation Theorem (1.4)thereforetellsusthatA(X,s)admitsananalyticcontinuation f of thefunctionsf(y)andg(y)atpoints'=SS if pE{0,1}isdefinedbyX(-1) =(-1)PX(1). we find(-1)kBk,x=X(-1)Bk, x,sothat values ofX.Since numbers Bk,xdefinedbytheformula zeta functiongeneralizestotheDirichletL-series(X,s)if weintroduce,for ( 1l The behaviourofthespecialvaluesatintegerargumentsRiemann Y = W(X)A(X,1-s) C 2 Fx(-t) _EX(-1)X(m-a) B(X, -1/IY)= y A(X,s) =L(f,s') ) =c1 Fx (t)_>X(a) n(X,s) = 2 0(X, iy)=2EX(n)nPa-nn a=1 n: Bk,x =0fork#pmod2, 1B( =W(X)L(g,p+2 a=1 . By(2.6),wehaveIW(X)=1. X, m t., i y c(Z) f0(X,iy)yS2 c(X)r(X) ) 2iP ana n=1 °O e ..J 0 l and mt - 00 t eat g ( y YP+se(x>IY) = t e(m-a)t ) emt -1 1 A(X,s) =L(g,s') = _ EBk,x - s+p) c k=O oo )B( z yln', = X(-1)Fx(t), X, dy y P. i k. y), c(X)=( tk W(X)L(g,±) The transformation . r(X) i ) p/2 441 . The functionalequationalsogives the we deducefork>1that character 1.Fromthefunctionalequation(2.8)andfact that L(X,k) p E(0,1),x(-I)=(-1)PX(l),providedthatXisnot theprincipal From theequations(2)and(3)onededucesequation(1)in exactlythesame .-. manner asin(1.8). with thefunction (2) by X(n),andsummingoveralln,yields Multiplying theequation (1) that has polesatmostz= the meromorphicfunction Proof: TheproofisthesameasforRiemannzetafunction(see(1.8)): (2.9) Theorem.Foranyintegerk>1,onehas 442 (3) The theoremimmediatelygivesthat coo Gx (z)_EX(n)e'2 .Or L(X,1-k)=- - L(X,1 -k) n=1 Fx (z)= co L(X,1-k)=0 fork#pmod2, r(k) r(s)L(X, s)= F(S) a=1 L(X,1-k)=-Bk,x m °D = residueofFX z v EZ.Theclaimthereforereducestoshowing X(a) Bk = EX(a) ns 1 kx¢0 fork-pmod2. a=1 m - J e mz Z eaz 0 00 Chapter VII.ZetaFunctionsandL-series 0 CO .-- - 1 e-ntts dt fGx(t)t2 1 -e- = e-az `.b k=0 t 00 (z)z-k mz dt Bk, t atz=0. 1C' = Fx(-z)z1. x III k. zk 0, complete generality. not haveanyrelationwithnumber fieldsanddeservetobeintroducedin a higherdimensionalanalogue oftheupperhalf-planeH.Aprioritheydo series. Butnowhigherdimensionalthetaseriesarerequired whichliveon which allowsustoviewtheL-seriesinquestionasintegrals overtheta with thesamemethods.Inparticular,Mellinprinciple appliesagain, results obtainedin§1and2extendtothesegeneralizations inthesameway, field Q.Theyhaveanaloguesforanyalgebraicnumber field K,andthe Exercise 4.ForaprimitiveoddcharacterXonehas Exercise 3.ForthenumbersSk,x(v)_Fo=iX(a)ak,k>0,onehas Exercise 2.Bk, Thus Bk,x(0)=Bk,.Showthat Exercise 1.LetFx(t,x)_aiX(a)en mathematician A.A.BEILINSONcanbefoundin[110]. asssociated totheDirichletcharacterXaredefinedby Minkowski space.AdetailedtreatmentofthisdeepresulttheRussian certain "regulators"definedviacanonicalmapsfromhigherK-groupsinto function atthepoints2k.Uptounknownalgebraicfactors,thesevaluesare similar remarksapplyastheoneswemadein§1aboutRiemannzeta (2.10) Corollary.Fork=pmod2,k>1,onehas § 3.ThetaSeries Riemann's zetafunctionandDirichlet'sL-seriesareattached tothe tai For thevaluesL(X,k)atpositiveintegerargumentsk#pmod2, x (x)- L(X,k) _ Sk'x(v"`) Bk, x (x- Bk,x(x) =F Fx(t,x) _ § 3.ThetaSeries k m) =kFa a=i 1 X(a)a#0. (-1)1+(k-P)/2 i=0 k=O k (a+x)t 1(Bk+1,x(vm)-Bk+t,x(0)) - l (k)B.,Xxk_i Bk,x(x)-. i X(a)(a r (X) 2iP The BernoullipolynomialsBk,x(x) k. tk _ 2n kBk,- m + x-m)k-',k>0. k X . 443 C -algebra involution rH(rEX),andletn=#X.Weconsiderthen-dimensional analogues asfollows.LetXbeafiniteG(CIR)-set,i.e.,setwithan 444 leave allnumber-theoreticaspectsaside. The number-theoreticapplicationswilloccurthere.Butforthemomentwe Minkowski spaceKR( forms ann-dimensionalcommutativeR-algebra,andC=®RC. One clearlyhasz=*z*.Theset the involutionszi-+z*and*zgivenby We calltheinvolutionz the followingcomponents: multiplication. Ifz=(zr)EC,thentheelementCisdefinedtohave of alltuplesz=(zt)ZEX,ztEC,withcomponentwiseadditionand If zEC,then*zistheadjoint elementwithrespectto( It isinvariantunderconjugation,(x,y)=y),andrestricting ityields scalar product endomorphism C-+C,xHzx.Furthermorewehaveon Cthehermitian Here Tr(z),resp.N(z),denotesthetrace,determinant. ofthe homomorphism a scalarproduct( The familiarobjectsC,R,R*,IEII,11,log,findtheirhigherdimensional For theadditive,resp.multiplicative,groupC,C*,wehave If KisanumberfieldofdegreenandX=Hom(K,C),thenRthe Tr:C-*C, N:C*-+ C*,N(z)=11zr. , R=[fC]+={zECI z=Z} ), i.e.,aeuclideanmetric,on theIR-vectorspaceR. zi =zt, (x, Y)_ y'^ CD' K ®(QIR)whichwasintroducedinchapterI,§5. z theconjugationonC.Inaddition,wehave (xz, Y)=(x,*zy) C= (Z)i =FF. resp. Tr(z)_T, zr, xrYz =Tr(x*y) rEX Chapter VII.ZetaFunctionsandL-series C *Zr =zt. r resp. , ), i.e., Z.) homomorphisms multiplicative group If SER,wesimplywritex>8tosignifythatxrforallr.The Thus wefindforthecomponentsofx=(xr)ER±thatxtxrR. In R,weconsiderthesubspace § 3.ThetaSeries Im(-1/z) >0,sincez2Im(-1/z)=-Im(z-tzz)Im(z)0. If zliesinH,thensodoes-1/z,becausezzER+,andIm(z)>0implies Putting Re(z)=2(z+z),Im(z) We finallydefinetheupperhalf-spaceassociatedtoG(CIR)-setXby of positiverealnumbersxrsuchthatxT= will playaparticularlyimportantpart.Itconsistsofthetuplesx=(xr) will beusedconstantlyinwhat followswithoutspecialcross-reference.This case n=1.Werecommendthatthereadermemorizethem well,forthey shows theanalogyofnotionsintroducedwithfamiliar onesinthe the zrmoveonlyinplanecutalongnegativerealaxis. Thetable if weagreetotaketheprincipalbranchoflogarithm andassumethat is well-definedby also includesthenotation H CCDR=RDR+, H CJRDR±2R+, For twotuplesz=(Zr),p(pr)EC,thepower log :R+R±, R* and R+={xER±Ix>0} =[fR+]+ Rt= {xERIx=x*}=[FIR]+. z, z*,*z,Tr,N,( H={zECIz=z*, Im(z)>0}. NIA o°° R+, H=R}+iR+. ZPr =eprlogzr, zP =(zpr)EC z x =(xr) x=(xr)H Ixl=(Ixrl), 1 1:R*-+R*, 2r I z Tao (z -z),wemayalsowrite , ), x>8,z°. , anditoccursinthetwo r r log x=(logxr). log :R+--Rf, log: R+ R 445 446 Chapter VII. Zeta Functions and L-series The functional equations we are envisaging originate in a general formula from functional analysis, the Poisson summation formula. It will be proved first. A Schwartz function (or rapidly decreasing function) on a euclidean vector space R is by definition a C°°-function f : R --* C which tends to zero as x oo, even if multiplied by an arbitrary power 1jxilm, m > 0, and which shares this behaviour with all its derivatives. For every Schwartz function f, one forms the Fourier transform e .f (Y)= I.f (x) e-vri (x. Y) dx R where dx is the Haar measure on R associated to(, ) which ascribes the volume 1 to the cube spanned by an orthonormal basis, i.e., it is the Haar measure which is selfdual with respect to (,). The improper integral converges absolutely and uniformly and gives again a Schwartz function f. This is easily proved by elementary analytical techniques; we refer also to [98], chap. XIV. The prototype of a Schwartz function is the function h(x) = e-7r(C'X) All functional equations we are going to prove depend, in the final analysis, on the special property of this function of being its own Fourier transform: (3.1) Proposition. (i) The function h(x) = e-'(",') is its own Fourier transform. (ii)If f is an arbitrary Schwartz function and A is a linear transformation of R, then the function fA(x) = f (Ax) has Fourier transform f(A-'y), TA(Y) = IdetAl where to is the adjoint transformation of A. Proof: (i) We identify the euclidean vector space R with R" via some isometry. Then the Haar measure dx turns into the Lebesgue measure dxl ... dx". Since h(x) e-"?,we have h = r["(e-"X+ so we may assume n = 1. Differentiating 00 h (Y) = f h (x) e-2;r'xy dx -00 §3. Theta Series 447 in y under the integral, we find by partial integration that 00 d - 27rix h(y) = -27ri f xh(x)e- ydx = -2rryl2(y). dy 0 This implies that h y) = C e-may2for some constant C. Putting y = 0 yields C = 1, since it is well-known that fe-lrx2dx = 1. (ii) Substituting x H Ax gives the Fourier transform of fA(x) as: f f(Ax)e-zn'(x,y)dx=rf(x)e-vri (Ax.y)I detAI`'dx 1 J Jfe_2A_'y7ri dx =(x,'1T (A- y) I de t A l I det A I From the proposition ensues the following result, which will be crucial for the sequel. (3.2) Poisson Summation Formula. Let P be a complete lattice in R and let T'=Ig'E RI (g,g') EZforallgEP) be the lattice dual to r. Then for any Schwartz function f, one has: 1 E f (g) = .(g') Er vol(r)8,', where vol(r) is the volume of a fundamental mesh of r. Proof: We identify as before R with the euclidean vector space R" via some isometry. This turns the measure dx into the Lebesgue measure dxl dz". Let A be an invertible n x n-matrix which maps the lattice Z" onto F. Hence F = AZ" and vol(t) = I det Al. The lattice Z" is dual to itself, and we get F' = A*7G" where A* = `A-', as g'EP't= `(An)g'=`n`Ag'EZ for aim E Z" Ag'E7L" g'E'A'Z". Substituting the equations P = AZ", TA (y) 1 .f (A* y) P' = A*Zn,.fA(x) _ .f (Ax), =vol(t) In ordertoprovethis,letuswritefinsteadoffAandtaketheseries into theidentitywewanttoprove,gives 448 whose Fouriercoefficientsaregivenbythewell-knownformula partial derivativesoff.Sog(x)isaC°°-function.Itclearlyperiodic, the convergentseries for almostallkEV.Henceg(x)ismajorizedbyaconstantmultipleof function, wehave,ifxvariesinacompactdomain, It convergesabsolutelyandlocallyuniformly.ForsincefisaSchwartz It followsthat Swapping summationandintegrationgives and thereforeadmitsaFourierexpansion p Er[iZwillhenceforthbecalled admissible. such thatpiE{0,1)ift= z,andpip?=0ifr with theparametersa,bER andatuplep=(pr)ofnonnegativeintegers, We applythePoissonsummationformulatofunctions an = _ .f(n) 0 f f nEZ" E .f(n)=g(0)anf(n), ...J g(x)e-27r"nxdx=Ef-..Jf(x+k)e-2,ritnx ff(a, b,x)=N((x+a)P)e- 0 an = g (x+n)=(x) 0 nEZ" If(x+k)I Ilkll"+l n c=7L", 0 . i. Suchanelement .,, .ti q.e.d. transform andonehas for somepolynomialP(x). Proof: Itisclearthatff(a,b,x)aSchwartzfunction,because on R.ItsFouriertransformis (3.3) Proposition.Thefunctionf(x)=fp(a,b,x)isaSchwartz § 3.ThetaSeries For anarbitraryadmissiblep,wegettheformulabydifferentiatingptimes We themforeobtain We nowdifferentiatepptimesbothsidesof(*)inthereal variableap,for representatives oftheconjugationclasses(r,t)suchthat r the elementsofXsuchthatp=;5,andletorrunthrough asystemof exists acoupler0T.Wethereforeproceedasfollows. Let pvaryover components aTofa,noraretheseindependenteachother, whenthere in thevariablea.Nowfunctionsareneitheranalytic theindividual (*) the identity all p,andapplypatimesthedifferential operator Pr pt=0,wemaychooseainsuchwaythatp-0.Then onehas Let p=0.By(3.1),thefunctionh(x)e-a(x,x)equalsitsownFourier (a+x,a+x)=F_(ap+xP)2+2E(aa+xa)(aa+xQ). iii f(Y) =J f(x) =fo(a,b,x) f(Y) = I fp(a,b,x)I `FL fo(a,b,Y) = = e-27ri(a,b)fo(-b,a,y). = e2'b)e-Jr(Y-b,y-b)+27ri(y,a) = e2si(y-b,a)h(y-b) f h(x)e-2;ri a RR= P 1 e-2ni(a, b)fo('-b,a,Y) P(x)I a [iTi(P)e2rri(a,b)1-1 (y-b,x-a) h(a+x)e27ri(b,x)Cbri(x,y)dx fp(-b, a,Y) e-r(a+x,a+x), a a dx h(a+x)e2n:(b,x). 'CI T. Since 449 450 Chapter VII. Zeta Functions and L-series for all or.. Here we consider ay = o + i no as a function in the real variables a , nQ ("Wirtinger calculus"). On the left-hand side fo(a, b, y) e-n(a+x,a+x)+2jri(b, x) e-2ni(x'Y)dx, > we may differentiate under the integral. Then, observing that pQ = 0 and ((aa +xa)(a& +xa)) = (aa +xa), we obtain aao fl(-27r(ap + xp)) P' J f p F I( -27r (aa + xa)) PO e-' (a+, a+x)+2ai (b, x)-2ni (x, y) dx a =N((-27r)P) N((a+x)P)e-1r(a+x,a+x)+21ri(b,x)e-21ri(x,Y)dx J =N((-2,-r)P) fp(a,b,y). The right-hand side of (*), e-2ni(a,b)-1r(-b+y,-b+y)+27ri(a,y) = e21ri(a,-b+y)-1r(-b+y,-b+Y) in view of (a, -b+y) a0(-by+yp)+E(aa(-ba+yo)+aa(-ba+ya)), P a and as pa = 0, becomes accordingly e-21ri(a,b)fo(-b, N ((27ri)P) N((-b + y)P) a, y) = N ((27ri )P)e-21ri (a, b) fp(-b, a, y) Hence fp(a,b,y) = N(i-P)e-2si(a,b).fp(-b,a, y). We now create our general theta series on the upper half-space H={zECIz=z*,Im(z)>0} =R±+iR+. (3.4) Definition. For every complete lattice I' of R, we define the theta series or (Z) _ z E H. ger More generally, for a, b E R and any admissible p E F1, Z, we put 9j (a,b z) = F_ N((a+g)P) erz`((a+g)z,a+g)+21ri(b,g) ger i 451 The xEK IIgli 9' Eµ2, 2 (a E R, g E F). y Seµz (xx), If Ilgll > 4supllxll t'O ?n2> N((a+g)")I e-" (0+9, 0+9) N((a+g)")I > IIg112 -211ahl smallest eigenvalue of the matrix i=1 1 Al = (2µr 1)" - (24 - 1)a. 2 1lgll)2 xEK 00 E P (A)Aq+ e- e-as(a+g, a+g) F- IfgIK < 00, t ger ll1g112> 1s 2 (Ilall - tip , gi) yi yj is the N((a+g)1)e";l(a+g)Z,a+g)+27ri(6. 1)I N((a+g)1)e";l(a+g)Z,a+g)+27ri(6. I,fg lK Z-basis of r, and for g = Ei 1 m; gi E F, let Z-basis of r, and gE j" IN((a+g)P)I <µg+1 forallaEK, E" j(gi 1 °_° w (a+g,a+g) fg (a) = N ((a + g)P) ... , g. be a I P N((a + >migi)P) is a polynomial of degree q in the mi, (q = Tr(p)), N((a + >migi)P) is a polynomial of From the Poisson summation formula we now get the general § 3. Theta Series Theta § 3. (3.5) Proposition. The series Br (a, b, z) converges absolutely and uniformly and absolutely b, z) converges Br (a, The series Proposition. (3.5) x R x H. subset of R on every compact Im(z) > 8, we find all z E H such that S E ]R, 8 > 0. For Proof: Let ((gi, gj)). Let show that put I fg I K = sup I fg (x) 1. We have to For K C_ R compact, Let gl, define Ilxll = µg=mWmi1. Furthermore, then for all a E K: where e = Einf functions of a. It follows that the coefficients of which are continuous finds a subset I" C_ F with provided µg is sufficiently big. One therefore finite complement such that series on the right is clearly convergent. 452 Chapter VII. Zeta Functions and L-series (3.6) Theta Transformation Formula. One has Brp,(a,b, - 1/z) _ [iTr(P)e2n;(a,b)vo1(I')]-'N((z/i)p+z)8,(-b, a, z). In particular, one has for the function Or (z) = Br°, (0, 0, z) : Br(-1/z) =N(zlt)Br (z). vol(e) Proof: Both sides of the transformation formula are holomorphic in z by (3.5). Therefore it suffices to check the identity for z = iy, with y E R. Put t = y-1 "2, so that 1 z = i;7, and - 1/z = it2. Observing that t = t* =* t, so that (fit, r7) *t77) tr7), we obtain Br(a,b, - 1/z) = N(t-p) N((ta +tg)p) e-n(ta+tg,ta+tg)+2ni(t-1b,tg) gEr Let a = ta, ,B = t- lb. We consider the function fp(a 6,x) = N( (a +x)') e-n(a+x,a+x)+2ni(j3,x) and put (Pt (a, A, x) = fp (a, 9, W. This gives (1) Brp.(a b,- 1/z) = N(t-p) tot(a,f, g) gEr and similarly z = i gives that (2) Bp, (-b, a, z) = N(tp) E 9,-1 (-P, a, g). g'Er, Now apply the Poisson summation formula 1 (3) f (g) = f (g1) gE vol(I)g,E, to the function f (x) = tot (a, ,l, x) = fp (a, f, tx). Its Fourier transform is computed as follows. Let h (x) = fp (a, so that f (x) = h (tx) = ht (x). The transformation A : x i-+ tx of R is self-adjoint and has determinant N(t). Thus (3.1), (ii), gives The Fouriertransformhhasbeencomputedin(3.3).Thisyields the gammafunction (and Riemann'szetafunction). without proofforprovingthefunctionalequationofDirichletL-series mation formulasought. Since t=(z/i)-1/2,i.e., Substituting thisinto(3)andmultiplyingbyN(t-p)gives,(1)(2): §4. TheHigher-dimensionalGammaFunction' We defineisomorphisms where depending whether#p=1or2.Wethenhave the lastsection.FirstwefixaHaarmeasureonmultiplicative groupR+: gamma functionforeveryfiniteG(CIR)-setX,buildinguponthenotationof In ordertogeneralizethisprocess,wenowintroduceahigher-dimensional by y-ry,resp.(y,y)F->y2, and obtainanisomorphism Let p={t,YJbetheconjugationclassesinX.Wecallreal orcomplex, The passagefromthetaseriestoL-seriesin§1and2wasaffordedby For n=1,weobtainproposition(2.3),whichatthetimewasused R+p =R+' 9 (a,b,-1/z)_ § 4.TheHigher-dimensionalGammaFunction f(Y) = resp. F(s) =Je-YY R+p =[R+x R* R+p -R+ R+ --}r}R+. + - (t2p+1)-1 =(z/i)p+71 0 o p R* +p' [N(ipt2p+t)e2'rt(a.b)vol(T')]-1O d +={(Y,Y)IYER+}. , this isindeedthetransfor- -'Op,(-b, C,, Y) a C3. 453 where Tr(s)=si+si. factors aregivenexplicitlyby (4.2) Proposition.DecomposingtheG(CIR)-setXinto itsconjugation gamma functionasfollows. and theconvergenceofintegralcanbereducedtocaseordinary where s,=s,forp={r),resp.sp=(sr,sr)forp=(r,=t),r classes p,onehas gamma functionassociatedtothe (4.1) Definition.Fors=(Sr)ECsuchthatRe(sz)>0,wedefinethe on flPR. XP rixp,resp.(xp,xp)H2xp,correspondstotheLebesguemeasure it ismappedtotheHaarmeasuredxonR±whichunderisomorphism is calledthecanonicalmeasureonR.Underlogarithm where product measure We nowdenotebyyytheHaarmeasureonR+whichcorrespondsto 454 The integrandiswell-defined,accordingtoourconventionsfromp.445, dt t is theusualHaarmeasureonR+.Thethusdefined rp(sp) 1 2'-Tr(sP)r(Tr(sp)), rx(S) = R± =jlR±p r(sp), rx(S) =FTrp(sp), log P R' fN(e_Yy2. G (C JR)-setXby P Chapter VII.ZetaFunctionsandL-series fJ R, P if preal, ifp complex, I_. F. The i. 455 -p.. t t . -)(sr,sr)) if p real, if p complex. p 3';, .". l dt N(7r-s/2)1-X(s/2) (f, VL p jp R+p, fJ 2T (Tr(sp)/2), 00 f e-2 (v)/ 2(27x)-Tr' S f N(e 0 fj FX(S) = I'x(Sl) LX(s) = 1 t Lp(sp) , Lx(s) = y dy -Sp/2r(sp/2), \R*, Yy f N(e-yys)y d= 21-Tr'(s)r(Tl.(s)) f 7r J R+ N(e-yys) f f R` Lp(s) T. The factors L;(sp) are given explicitly, by (4.2), as T. The factors L;(sp) are given explicitly, The proposition shows that the gamma integral r(s) converges for The proposition shows that the gamma We call the function For a single complex variable s E C, we put §4. The Higher-dimensional Gamma Function' Gamma Higher-dimensional The §4. Proof: The first statement is clear in view of the product decomposition the product in view of is clear statement The first Proof: has only one conjugation (C R) -set X which is relative to a G The second r let X = {r, i.`}, fX(s) = F(s). So = 1, then trivially class. If #X Mapping one then gets and, since d(t/2)2/(t/2)2 = 2dt/t, the substitution t H (t/2)2 yields = 2dt/t, the substitution t H (t/2)2 and, since d(t/2)2/(t/2)2 analytic continuation to all of C, s = (si) with Re(sr) > 0, and admits an obvious way by the ordinary gamma except for poles at points dictated in the function r(s). X into the conjugation the L-function of the G(CI R) -set X. Decomposing classes p, yields = {r} and sp = (s, s7) for p = {r, z where as before we write sp = st for p r number ofreal,resp.complex,conjugationclassesX,wefind where 1=(1,...,1)istheunitelementofC.Denotingrl,resp.r2, 456 With thisnotation,(1.2)impliesthe Then wehave,foranarbitraryG(CR)-setX: and inparticular L-function Lx(s): (iv) LR(s)LR(s+1)=Lr(s)(Legendre'sduplicationformula). (iii) LR(1-s)LR(1+s)=cos7rs/2' (ii) LR(s+2)=27rLR(s),Lc(s+1)=zLc(s) (4.3) Proposition.(i)LR(1)=1,Lc(1)n and ontheother Proof :Ontheonehandwehave (4.4) Proposition.LX(s)=A(s)Lx(1-s)withthefactor The propositionthereforeresults fromtheidentityLx(s)=La(s)r'LC(s))2. In thesamewayweput As aconsequenceweobtainthefollowingfunctionalequationfor LR(s) =Lx(s)n-s/2r(s/2), Lc (s)=LX2(27r)-sr(s), LR(1 -s) LR(S) LC(1 -s) LX (s)=Lx(s1)7r Lc (s) A(s) =(cossrs/2)r'+r2(sin7rs/2)1'2LC(s)'. rx(s) =2('-2s)r2r(s)r'r(2s)r2. = cosTrs/2sinlrs/2LC _ LX(s) =Lnz(s)"Lc LR(1 -s)LR(1+s) LR(s)LR(1 +s) LC (1-s)Lc(s) Lc (S)2 -ns12rx Chapter VII.ZetaFunctionsandL-series Lc(s)Lc(1 -s)_ if X=[r,T),t#T. if (s12), =cosYrs/2 X =(r}, - sinnsL,c(s)2 2 (S)2. fir, n =#X, LC (s), sinus' 2 457 . i is associated with the field Q i is associated with 1 1 t core --+ 01(a)S 1 - ` 1(P)-S P (s) _ Yk .hi K (S) = > K (S) = II § 5. The Dedekind Zeta Function Dedekind Zeta § 5. The Just like the Riemann zeta function, the Dedekind zeta function also Just like the Riemann zeta function, the The proof proceeds in the same way as for the Riemann zeta function The proof proceeds in the same way as The Riemann zeta function This concludes the purely function-theoretic preparations. They will now They preparations. function-theoretic the purely concludes This admits an analytic continuation to the complex plane with 1 removed, and admits an analytic continuation to the complex the argument s to 1 - s. This is it satisfies a functional equation relating what we are now going to prove. The argument will turn out to be a higher dimensional generalization of the one used in § 1 for the Riemann zeta function. where a varies over the integral ideals of K, and 0(a) denotes their absolute where a varies over the integral ideals of absolutely and uniformly in (5.2) Proposition. The series K (s) converge 0, and one has the domain Re(s) > 1 + S for every S > K. where p runs through the prime ideals of 01(a) is multiplicative. We do not (see (1.1)), because the absolute norm that also applies to Hecke go into it here, because it is the same argument §8 as a common generalization of L-series, which will be introduced in zeta function. Dirichlet L -series and of the Dedekind of rational numbers. It generalizes in the following way to an arbitrary It generalizes in the following way of rational numbers. field K is Dedekind zeta function of the number (5.1) Definition. The defined by the series norm. § 5. The Dedekind Zeta Function Dedekind The § 5. to number theory. be applied n = [K : Q]. number field K of degree fundamental meshhasvolume (5.4) Proposition.(51,s)=O't(a)tF yields the where NdenotesthenormonR*(seechap.I,§5).The lemma therefore The fieldKmaybeembeddedintoK.Thenonefindsfor aEK*that one hasb=as1withaEabca. as 1=ba1,then(a)(b),sothatab-1Eo*.Thisshowstheinjectivity by a*/o*thesetoforbits,i.e.,classesnon-zeroassociated of themapping.Butitissurjectiveaswell,sinceforeveryintegralbES, Proof :IfaEa*,thena-1=(a)a-1isanintegralidealinSt,andif a-'. Thenthereisabijection (5.3) Lemma.LetabeanintegralidealofKandAtheclass elements ina. then theunitgroupo*ofooperatesonseta*=a'.{0),andwedenote The integralidealsinriaredescribedasfollows.Ifaisfractionalideal, The functionalequationisthenprovedfortheindividualfunctions(A,s). so that ideal classgroupC1K=J/PofK,intothepartialzetafunctions 458 By chap.I, To theG(1CJR)-setX=Hom(K,1C)correspondsMinkowskispace First wesplituptheseriesK(s),accordingtoclassesAofusual a*/o* -14{bEAIintegralI, (5.2), theidealaformscompletelatticeinRwhose Jt((a)) =INKIQ(a)IIN(a)I, KR =R=[fJc]t OK (S)_ vol(a) = CEa*/o* integral aEJ Chapter VII.ZetaFunctionsandL-series z da, 'R(a)s 1 N(a1 d f---+b=aaC1. )1, coy CAD 459 dy. dy , (.fi, s) (see § 4, p. 455) d s) we associate the theta s) we associate a) = Idxis/2Lx(s) erzi (az/dog", ncay/d."" e-7r(sy/d;1,r,a)N(Y)s R* aEa R R+ a R', (xri) H (Ix= I). We then obtain R', (xri) H (Ix= I). We s a) y H irIa12y/da/n IN( IdKis/27r-ns/2rK(s/2) Z(S,s) = Z.(sM.t,s), g(Y)e j'K(S) = rx(S) = 1 N(e-yys) j'K(S) = rx(S) = 1 O (a, z) = Ba(Z/dl/n) _ O (a, z) = (A, s) via the gamma integral associated to the G (C I R) -set associated to the G (C the gamma integral (A, s) via I dK I sn-ns FK (s) (A, 2s) = f g(y)N(Y)s IdKls7r-ns1'K(S) Z.(s) = .v. c." I denoting the map R* § 5. The Dedekind Zeta Function Dedekind The § 5. a, of the discriminant value of absolute the I denotes da = 9t(a)2IdK where K. To the series the discriminant of and dK is series to It is related (t;), X = Hom(K, > 0 (see (4.1)). In the integral, we substitute where s E C, Re(s) of a*/o*, yields Summing this over a full system 9t of representatives with I with the series legal, for the same reason as in the Swapping summation and integration is 422). We view the function case of the Riemann zeta function (see p. function as the "Ruler factor at infinity" of the zeta and define hinted at in the case of thej a*/o*. This difficulty - which was already The desire to realize this function as an integral over the theta series 0 (a, s) is The desire to realize this function as an integral we sum over all a E a, whereas) frustrated by the fact that in the theta series a system of representatives of, summation in the series g(y) is only over Riemann zeta function - will now be overcome in the general case as follows. is containedinthenorm-onehypersurface 460 Let d*xbetheuniqueHaarmeasureonmultiplicativegroupSsuchthat we obtainadirectdecomposition Writing everyyeR+intheform fundamental meshofthelattice2G.Anysuchchoicesatisfies (Dirichlet's unittheorem).ChooseFtobethepreimageofanarbitrary Tr(x) =0),andthegroup10*I takes thenorm-onehypersurfaceStotrace-zerospaceH=IxERtI We willnotneedanymoreexplicitdescriptionofd*x. the canonicalHaarmeasuredy/yonR+becomesproduct Proof: DecomposingR+=SxR*,wefind where w=#p(K)denotesthenumberofrootsunityin K. of thefunction (5.5) Proposition.ThefunctionZ(A,2s)istheMellintransform 10*12 The imageIo*oftheunitgroupundermapping( We nowchooseafundamentaldomainFfortheactionofgroup IE12 1sEo*}onSasfollows.Thelogarithmmap Z(A, 2s) f(t) =fF(a,t) Y = log :R.-3R±, S={x ER+IN(x)=1}. = Z(S,2s) =L(f,s) 0 ff dy 00 y R+=SxR+. x =N(y)tN(y), s = d*xx .'3 aEOt is takentoacompletelatticeGinH T e-n(axt',a)d*xts W 1 Chapter VII.ZetaFunctionsandL-series (xt) F---;(logx.t), F J B(a,ixt'/")d*x, dt t dt t : R*-3R+ maps Ftorl2F,sothat The transformationxr+r12xofSleavestheHaarmeasured*xinvariantand hypersurface Sintothedisjointunion with t'=(t/da)t/".ThefundamentaldomainFcutsupthenorm-one § 5.TheDedekindZetaFunction finally thatf(oo)= furthermore thatasrunsthroughtheseta*=a,(0}exactlyonce,and Observe herethatwehavetodividebyw=#p(K),because11(K)isjust to d*x,andthelatticewhichisdualainR.This achieved bythe to computethevolumevol(F)offundamentaldomain Fwithrespect transformation formula(3.6).Inordertofindthepreciseequation, wehave for thefunctionfF(a,t),whichinturnderivesfrom generaltheta follows viatheMellinprinciplefromacorrespondingtransformation formula show that the kernelofo*-ao*I(seechap.I,(7.1)),henceEIEI chap. I,(7.5)). (5.6) Lemma.ThefundamentaldomainFofShasthefollowing volume following twolemmas. where risthenumberofinfinite placesandRistheregulatorofK(see with respecttod*x: s f Ye- Using thisproposition,thefunctionalequationforfunction Z(.J,s) aE91 *xt', a)d*x= .^' Z(F,2s)=f(f(t) w fF d*x,as0(a,ixoo)=1.Thisresultdoesindeed 0 00 nEIO*I Iixt= w 1 vol(F) =2r-1R, S =U112F F F f(&a EEO* aE9t n 2F nEIo*I - 0 f aE9R f(c)o))tsdt e-yr (axt',a)d*x e-tr(aext',ae)d*x ) -1)dx=f(t)-f(oo). . .17 t =L(f,s). = w TeObserve `ti 461 regulator R.Thusweget Adding thefirstr-1linestolastone,allentriesof lastlinebecome zeroes hastheabsolutevalueofitsdeterminantbydefinition equaltothe zero, exceptthelastone,whichisn= of thedeterminant span thefundamentalmesh0.Itsvolumeisn2r-1times absolutevalue the parallelepipedspannedbyvectors2e1,...,2er-1,,:e,ifet, H =((xi)E][fir where 0denotesafundamentalmeshoftheunitlatticeGintrace-zerospace whether piisrealorcomplex.BydefinitionofF,wealsohave with thevectore=(epi,...ep,)ER',ep;1,resp.2,depending Let uscomputetheimage*a(FxI).1=(1,...,1)ES.Thenwefind (see §4,p.454)transformsdy/yintotheLebesguemeasureof]Rr, isomorphisms i.e., thevolumeofa(FxI)withrespecttody/y.Thecomposite quantity vol(F)isalsothevolumeofFxIwithrespecttod*xdt/t, Since I=(tER+ measure d*xxdt/tbytheisomorphism Proof: Thecanonicalmeasuredy/yonR+istransformedintotheproduct 462 I Exi=0).Thisgives a :SxR*+ Vra((1,t)) =elogtII' det I *a (Fx(11)=20, log eir ell vol(F) =2r_1R. > R,-+F1IR=R'_ ...... ,.] R+, Chapter VII.ZetaFunctionsandL-series er-1,r er-i, 1 (x, t)1axt pl- ep,. Thematrixabovethese ep, ep1 n n 1 C3' elogt lh' . ' ofthe , er-I 0 E 463 xa c o-I', implies that (xt) on Ka and 0 the (xt) on Ka for t-*oo,c>0. Br (z) (*gz, *g) = (gz, g) ) O(e-cth/h N(z/i) .\i vol(F) TrKIQ(xao) c Z for all a E a R + 1 ='JT(a)-2,R(0)-2IdK I = 1/(O(a)2IdK I) = 1/da. Br(-1/z) = *F'{xEKI Tr(xa)c7L} w fF {a, t) = t lie fF-i (at) 2 +Xnan, with x; E R, then Tr(xai) _ >t x; Tr(a;ai) = ni E 7Z' then Tr(xai) _ >t x; Tr(a;ai) = ni E +Xnan, with x; E R, d(a.D)_1 +. fF(a, t) = X E *F'={*gERI (g,a)EzforallaEa)={xERI Tr(xa)c7G} *F'={*gERI (g,a)EzforallaEa)={xERI § 5. The Dedekind Zeta Function Dedekind The § 5. involution (xt) r->. asterisk denotes the where the (5.7) Lemma. The lattice F' in R which is dual to the lattice r = a is r = a the lattice is dual to R which F' in The lattice Lemma. (5.7) given by K IQ. different of we have Proof: As (x, y) = Tr(*xy), -basis of a and immediately x E K, for if a 1, ... , a is a 7L Tr(x a) c 7L implies IQ (alai) E Q equations with coefficients Tr(a; ai) = TrK is a system of linear x E K. It follows that so all xl E Q, and thus t) satisfy the transformation formula (5.S) Proposition. The functions fF (a, and one has x = x1a1 we obtain the, i-1 = {x E K I TrK,Q(xo) c 7L}, and By definition we have equivalences x E* T' Proof: We make use of formula (3.6) mesh has volume; for the lattice F = a in R, whose fundamental I" dual to r is given by (5.7) vol(I) = Yt(a) I dK I t/2. The lattice as *T' _ (ac0)-1. The compatibility Br, (z) = 0*r, (z). Furthermore we have The transformation x H x-' of the multiplicative group S fixes the Haar, measure d*x (in the same way as x H -x fixes a Haar measure on R")I 464 Chapter VII. Zeta Functions and L-series and maps the fundamental domain F onto the fundamental domain F-1, whose image log(F-1) is again a fundamental mesh of the lattice 2 log I o* j. Observing that N(x(tda)'1n) =rtda for x E S, we obtain fF(a, 1 = 1 J ea(ix tda) d*x t w F 1 ( Bal-1/ix n tda) d*x W F_i ( 1 tda)1/2 f0(aa)_l (ix n tda) d*x w vol(a)J F-I t1/2 w [ 0(ac0)-1 (ix n tld(a)-, ) d*x F -I = t'/2 f _i ( (act)-' , t) . This shows the first formula. To prove the second, we write w1 1 vol(F) fF(a,t) = d*x+ f (B(a,ixt' n)- 1) d*x = +r(t). w w F F The function r (t)satisfies r (t) = O (e-lt'I°) ,c > 0,t -> co, as the summands of 6(a, ixt'I") - 1 are of the form n , e-"(°x'Q) a E a, a 0, t' = t/da. The point x = (xi) varies in the compact closure F C { R+] + of F. Hence xT > 8 > 0 for all r, i.e., (ax,a) _ IrQ12xT > 8(a,a) and so vol(F) r(t) n w Writing m=min{(a,a)IaEa,a¢0)and M=#(aEaI(a,am),it follows that O(e-ctI/ t') - 1 = e-n8F71 t'(M + 0a(is ` (a,a)>nv where c = We thus get as claimed yol(F) _c,,/n 2r-1 't,/,. fF(a,t)= +O(e )= R -{-O(e- ). 0 -w w 465 C9; w 2r ' s w 2r 2 -s) 0T(6) to C an analytic continuation ensures the analytic continuation of (0, 11 with simple poles at s = 0 and g(t) =ao+O(e-`t), resp. 2ao = - R E resp. -R. integral w 2r =ao+O(e-ctl/'), w f (. ) = t ll2g(t) 2r Z(ft's)=L`f,2) (q,s) = --R, L(f,s) =L(g, and .l! correspond to each other via .fiR' = [D]. It and .l! correspond to Z(.P, S) = Z(.',1 - s), Z(.P, S) = Z(.',1 - N . -2ao = - - R, Z(A, s) = Is/2Yr-ns/21K (s/2), ad-nits Is/2Yr-ns/21K (s/2), t R. Proposition (1.4) thus .f (t) 2w _ JdK This last proposition now enables us to apply the Mellin principle (1.4) to principle the Mellin to apply enables us now last proposition This admits an analytic continuation to C s = 1 and residues with simple poles of L (f , s) at s = 0 and s = 2 with residues -ao, resp. ao. with simple poles of L (f , s) at s = 0 and Therefore {0, 1) and satisfies the functional equation {0, 1) and satisfies the where the ideal classes ((aD)-l, t). Then (5.8) implies Proof: Let f (t) = fF (a, t) and g(t) = fF_, and with ao = functional equation the Mellin transforms of f and g, and the this yields the following result, where the notations dK, R, w, and r signify dK, R, w, and r where the notations the following result, this yields function (5.9) Theorem. The Z.(s) residues has simple poles at s = 0 and s = 1 with § 5. The Dedekind Zeta Function Dedekind The § 5. partial zeta functions fF (a, t). For the the functions of unity, and the regulator, the number of roots as before the discriminant, places, respectively. the number of infinite F>~ 466 Chapter VII. Zeta Functions and L-series and satisfies the functional equation Z(.fi,s)=L(f,2)=L(g,12s)=Z(.kt',1-s). This theorem about the partial zeta functions immediately implies an analogous result for the completed zeta function of the number field K, ZK(S) = Z(-R,s) (5.10) Corollary. The completed zeta function ZK (s) admits an analytic continuation to C [0, 11 and satisfies the functional equation ZK(S)=ZK(1-S). It has simple poles at s = 0 and s = 1 with residues 2rhR 2rhR resp. w w where h is the class number of K. The last result can be immediately generalized as follows. For every character X:J/P) S' of the ideal class group, one may form the zeta function Z(X,s) = where X (a) it integral91(a)3 and X (a) denotes the value X (Ft) of the class 51 = [a] of an ideal a. Then clearly Z (x, s) _ E X (A) Z (A, S), and in view of 1t' = .At`1 [c'], we obtain from (5.9) the functional equation Z (X , s) = X (D)Z (X ,1 -S). If X E 1, then Z(X, s) is holomorphic on all of C, as Er X (S) = 0. 467; c (s) )r2L / hR 2 Re(s) > 1. . As dK 11/2 w IdK 1112 C.. 2' (27r)r2 wldx11/2hR =hR/e5. w l 21" (27r)r2 w(a)s > log g 2r1 (27r)r2 q-, 7 (cos 2 K(1 -s) = K Ress=1 x (s) = Ws) _ .--I = IdKIs Zoo(s) =IdKIS/2Lx(s) = IdKISI2LR(s)r1Lc(s)r2, Zoo(s) =IdKIS/2Lx(s) A(s) K(s) = Zoo(S)ZK(S) = IdKI-S/2Lx(S)-1ZK(s), K(s) = Zoo(S)ZK(S) It satisfies the functional equation We now conclude with the original Dedekind zeta function Dedekind original with the conclude We now The proof of the analytic continuation and functional equation of the The proof of the analytic continuation The formula for the residue § 5. The Dedekind Zeta Function Dedekind The § 5. The Euler factor at infinity, Z., (s), is given explicitly by § 4 as (s), is given explicitly factor at infinity, Z., The Euler places. By denotes the number of real, resp. complex, where r1, resp. r2, = IdK 11/2/7,r2 (4.3), (i), one has Z,(1) the we obtain from (4.4) has an analytic The Dedekind zeta function K (s) (5.11) Corollary. (i) { I). continuation to C '. III, (3.5)). the genus of the number field K (see chap. (ii) At s = 1 it has a simple pole with residue (ii) At s = 1 it has a Here h denotes the class number and (iii) Dedekind zeta function was first given by the mathematician ERICH HECKE Dedekind zeta function was first given by we have presented here, albeit in a (1887-1947), along the same general lines the theory we are about to develop somewhat different formulation. Further, goes back to HECKE in the following sections § § 6-8 also substantially function K(s)forthefieldK=Q(lan,)ofm-throotsunity,andisbased hands ontheEulerproductandthuscomputezetafunction. law forthedecompositionofprimesinthisfieldsufficientlywelltolayour us todeterminetheclassnumberhoffieldK,providedweknow is commonlyknownastheanalyticclassnumberformula.Itdoesallow 468 this is1.Soletp{m.Bychap.I,(10.3),ftheorder ofpmodmin contains thefactor p inK,andletfbethedegreeofp;,i.e.,T(p;)=pf.ThenOK(s) the fieldK.Letp=(pt Proof: Theproofhingesonthelawofdecompositionprimenumberspin (5.12) Proposition.IfK=Q(Am)isthefieldofm-throotsunity,then on the (see §2)ishighlyremarkable.ItresultsfromstudyingtheDedekindzeta Finally, takingtheproductover allp,wegetOK(s)=G(s)J]L(X,s). Gp) elementsinthepreimageofX(p).Itfollowsthat where' indicatescharactergroups.Wethereforefindr= #(G/G,,)=(G X '-a(p)definesanisomorphismGP=ILf,andgivesthe exactsequence index ofthesubgroupGpgeneratedbypinG=(Z/mZ)*. Associating (Z/mZ)* ande=1.Sinceefrcp(m),thequotientr = rp(m)/fisthe On theotherhand,L-seriesgivefactor]lx(1-x(p)p-s)-t where XvariesoverallDirichletcharactersmodnz,and .'YThe followingapplicationofcorollary(5.11)toDirichletL-seriesL(X,s) II(1 - x I PIP fJ(1 X(P)P-s)-t = G(s) =TI(1- ... pr)ebethedecompositionofprimenumber K(s) =G(s)IIL(X,s), G/GPGµf1, = fl(1-`n(p)-s)-t ova 5EAf pInt PIP II (1- -'71(p)-s)-t Chapter VII.ZetaFunctionsandL-series = (1- x p-s)-r =(1-p-fs)-r 01(p)-s)-i. p-fs)-r. . ForpIm § 5. The Dedekind Zeta Function 4(9 For the trivial character X° mod m, we have L(X°,s) = fpl,,,(1 - p- ) c(s), so that OK(s)=G(s) fl(1-p(s) fl L(X,s). PI- X#XI Since c(s) and OK (s) both have a simple pole at s = 1, we obtain the (5.13) Proposition. For every non-trivial Dirichiet character X, one has L(X, 1) 0 0. This innocuous looking result is in fact rather profound, and yields as 1 concrete consequence (5.14) Dirichlet's Prime Number Theorem. Every arithmetic progression 1 a, a±m, a±2m, a±3m, ..., with (a, m) = 1, i.e., every class a mod m, contains infinitely many prime numbers. Proof: Let X be a Dirichlet character mod m. Then one has, for Re(s) > 1, logL(X,s) = ->2log(1-X(p)p_S) = >2 >2X(Pms)= >X(P)+gx(s), P P m=1 MP P P where gx (s) is holomorphic for Re(s) > 1 - this follows from a trivia estimate. Multiplying by X(a-1) and summing over all characters mod m yields EX (a-') logL(X, s) _ EEX (a 1P) Ps + g(s) x xP >2X(a_1b) is _ >2 > +g(s) b=1 x p=b(m) P om-f-g(s). p=a(m) Ps Note here that 0, if a=Ab, >X(a-1 b) _ x (P(m) _ #(Z/mZ)*,if a = b. When we pass tothe limit s--f 1 (sreal > 1), logL(X,s) stays bounded for X X ° because L(, 1) 0 0, whereas log L (X °, s) = for everyalgebraicintegeraE orelativelyprimetom. such that for whichthereexistsapairofcharacters (6.1) Definition.AGrofiencharaktermodmisacharacter X:Jm-+S' define asfollows. equation, HECKEwasledtothenotionofGriJ3encharaktere, whichwe Here avariesoverallintegralidealsofK,andonedefinesX (a)=0whenever which thecorrespondingL-seriescouldbeshowntohave afunctional (a, m)01.Searchingforthemostcomprehensiveclassof charactersXfor as welltheDedekindzetafunction,L-series we mayassociatetoit,asacommongeneralizationoftheDirichletL-series of allidealsKwhicharerelativelyprimetom.Givenanycharacter in§13. the classesamodm.Wewillcomebacktothisinmoregeneralcontext method givessharperresultsonthedistributionofprimenumbersamong case DirichletwasledtothestudyofL-seriesL(X,s).Thisanalytic algebra (seechap.I,§10,exercise1).Searchingforaproofinthegeneral proved. Thus thesumcannotconsistofonlyfinitelymanyterms,andtheoremis holomorphic ats=1,wefind left-hand sideoftheaboveequationthereforetendstooo,andsinceg(s)is 470 EP,,, log(1-p'S)+log(s)tendstooobecausec(s)hasapole.The Let mbeanintegralidealofthenumberfieldK,andletJ'°group For a=1,Dirichlet'sprimenumbertheoremmaybeprovedbypure Xf:(o/m)*)S', Xco:R*S', fro X:Jm --*S' § 6.HeckeCharacters S-1 p=a(m) lim X((a)) =Xf(a)Xc(a) (X, S)_ or. fz ccIzI=I), (P m) Chapter VII.ZetaFunctionsandL-series a Ps gZ(a)s X(a) °n. = oo. . this section. abstract interpretationinwhat follows,butitwillbeexplainedattheendof the ideleclassgroupof numberfieldK.Wewillnotusethismore one tointroduceideaes.Infact,allGr6f3encharakterearise ascharactersof a Gr5j3encharakterX(exercise5). and itcanbeshownthateverysuchpairofcharacters(Xf, X,,)comesfrom In fact,forEo'wehaveXf(s)=1,andthus a =b/c,fortwointegersb,crelativelyprimetom,suchthatbmodm for aEK.Letusrecallthatthecongruence=_1modmsignifies associated withaGrbj3encharakterXsatisfytherelation or, equivalently,aE is denseinR*,bytheapproximationtheorem,andonehasXc"(a)=X((a)) the group character XF, a well-definedclassin(o/m)*.ThecharacterX,,.,andthusalsothe of allfractionsrelativelyprimetom,becauseeveryaEKdetermines it extendsuniquelytothegroup ideals breaksupintoafiniteandaninfinitepart.From underlines thefactthatrestrictionofaGrOj3encharaktertoprincipal for allalgebraicintegersaEorelativelyprimetom.Thislastidentity for allaEosuchthat=1modm.Forifthisisthecase,thenrule a characterX,),,ofR*suchthat Xf(a) =X((a))X.(a)-1definesacharacterXfof(o/m)*whichsatisfies § 6_HeckeCharacters (E)) =1.ThetwocharactersXfandX,,,, The attempttounderstandGrdj3encharaktereinaconceptual wayleads The characterX,,.factorsautomaticallythroughR*/om,where 'r1 A characterXofJ'°isaGr5f3encharaktermodmassoonthereexists are determineduniquelybytheGr6f3encharakterX,since Xf(E)Xo(E) =1 Up"Plc KpforpIm,ifm OmE E0*IE=lmodm}. X((a)) =Xf(a)X.(a) x((a)) =X.(a) K(m)={aeK*I(a,m)=1} for allEo*, o(m)laecI(a,m)1) ... = fpp"P. of (o/m)*,resp.R*/Om, ml `y. Xf(E)Xoo(E) III 471 -oc-.,, S1. S1. In every c78 R*/om through R*/o"`'). , (o/m')* . We Put to J"' is the Gr6f3encharakter X xf(a-lal)Xoo(a-la1)Xf(a)X,,.(a) R*/om , 00 Chapter VII. Zeta Functions and L-series and Functions VII. Zeta Chapter resp. S', 00 x' X'(a) = X(a)Xf(a)X.(a) Let X be the restriction of the GrOJ3encharakter Let X be the restriction = X(at)Xf (al)x.(al) E J"''/P"'', there is an ideal a E J"' which is relatively E J"''/P"'', there is an ideal a E J"' which I obi (o/m')* - X ((a)) = X'((a)) = Xf (a)X' (a) = Xf(a)&(a), X ((a)) = X'((a)) = S', and let Xf, X' be the pair of characters associated with X'. the pair of characters associated with S', and let Xf, X' be (ab) = Xf(a')X"o(a'). = Xf(a)X.(a)Xf (b)Xao(b) = Xf (ab)x W. Let Xf be the composite of (o/m)* alai, al EJm,(al)EP',then onehas(aal1)EJm,and Xf factors through (o/m')*. Xf factors X is the restriction of a GroJ3encharakter X': J"'' -+ S1 mod m'. X': J"'' of a GroJ3encharakter X is the restriction X (a)x'(a)xon(a) = X (a)x ((aal 1)) X'((a')) = X ((a)) X'((b)) = X ((a)) X'(b)xoo(b) The Grof3encharakter X mod m is called primitive if it is not the (o/m)* (b) E P'"e, then we have The restriction of the character X' from J'' The restriction of the character X' from prime to m' and a' = ab, (a) E J'", of J"', and if (a) is a principal ideal corresponding pair of characters Thus X' is a Grbjiencharakter mod m' with restriction of a Gr6J3encharakter X' mod m' for any proper divisor m'Im. divisor of m. Then the following conditions are equivalent. m. Then the following divisor of (i) (ii) Proof: (i) = (ii). X' : J'"' the composite of Let Xf, resp. X,,o, be O(m) C o('"') : We then find for a E determined X,) = X, because Xf and X,, are uniquely so that Xf = Xf and through (o/m')* (and by X. Thus Xf factors (ii) class a' mod P'' E P"' prime to m, i.e., a' = as for some (a) choice of the ideal a E J', for if This definition does not depend on the 472 a let m' be in, and mod be a Grofiencharakter Let X Proposition. (6.2) is theconductorofXf,i.e.,smallestdivisormsuchthatXffactor: such thatXistherestrictionofaGr6f3encharaktermodf.By(6.2), proper divisorm'Im.TheconductorofXisthesmallestfin is primitiveinthesensethatitdoesnotfactorizethrough(o/m')*forany According to(6.2),thisisthecaseifandonlycharacterXfof(o/m)' § 6.HeckeCharacters x'-Xmodm,then x'y-xyEmmsir-l=i-'={aEKlTr(a)EZ}, is thedifferentofKIQ.ThenwedefineGausssumXftobe (6.3) Definition.LetXfbeacharacterof(o/m)*andyEm-iD-1,whereV character Xoo. through (o/f)*. Gauss sumswillplayacrucialrole. a viewtoprovingfunctionalequations.Forthis,thefollowing propertiesof define thetaseriesandL-seriesattachedtoHecke'sGrbj3encharaktere with the Gausssumintroducedin(2.5)byr(Xf,n)=rm(Xf,m}. Wewillhaveto on theo/m-module rm(Xf, y)dependsonlyonthecosety+zr',i.e., and therefore so that where xvariesoverasystemofrepresentatives(o/m)*. and furthermore and aEo.Thenonehas (6.4) Theorem.LetXfbeaprimitivecharacterof(o/m)*, letyEm-tD-1 Let usnowhaveacloserlookatthecharacterXf,andthen III The Gausssumdoesnotdependonthechoiceofrepresentativesx,forif ='o rm(Xf, ay)= e2ai Tr(x'y)=Tr(xy) I rm(Xf,Y)= rm(Xf, y)=FXf(x)e2ai Tr(x'y) -Tr(xy)mod7G 0, Xf(a)rm(Xf, y), xmod m (.r, M)=I ` t(m), In thecaseK=Q,m(m),wegetback III The sameargumentshowsthat if (YMD,m)=1. '17 `'C if (a,m)01, if (a,m)=1, Ti(xy) it definesafunction 47= j and foreverydivisoraim, (6.6) Lemma.Onehas find the x' -xmodm,then(x'x)yED-1,henceTr(x'y)Tr(xy) modZ.We These sumsdonotdependonthechoiceofrepresentatives x,forif the sums Proof: Ifa=pi` For thisfunctionwehavethe (6.5) Proposition.Ifa01,thenrpc(b)=0. consider theMibiusfunction make thefollowingpreparations.Forintegralidealsa=pv' 474 Now, foryEm-10-1andeveryintegraldivisoraofm,welookat The mostdiffcultpartofthetheoremislastclaim.Toproveit,we III bin Ta(Y) AM w(1)+E/4pi)+Eµ( CMG = 1+1)+121(-1)2+...+\r)(-1)r x modm (x, m)=a lu(a)= 1+(-1))r=0. Sa(Y) =S p° r,vi>1,then e2niTr(xy) i Ti (Y)_Eg(a)S.(Y), 0, (-1)r, 1, (0l(a), 0, bra aim and il Proof: In view of (6.5), we have µ(a)S.(y) _ E pt (a) E Tb(Y) _ F Tb(y) E u(a) = T1(y) aim aim b bim alb album If y E a 1t-1 and a I x, then xy E t-1, so that Tr(xy) E Z, i.e., all summands of Sa are 1 and there are #(a/m) = M(11) of them. If on the other hand y 0 a1-0-1, thenwe can find in a/m a class z mod m such that zy D_1, i.e., Tr(zy) 0 7G, so that e27riTr-(zy) 54 1, andwe obtain 21r i Tr(zy) e Sa(Y) _ E e27riTr((x+z)y) - Sa(Y), xmodm aix since x + z varies over all the classes of a/m as x does, so that we do find sa(y) = 0- Proof of Theorem (6.4): Let a E o, (a, m) = 1. As x runs through a system of representatives of (o/m)*, so does xa. We get e27r1 Tr(xay) tm(Xf, ay) = >2 Xf(x) x mod m (x, m)='1 = X f(a) F Xf(xa) e27r1Tr(xay) x mod m (x, m)=1 = Xf(a)tm(Xf, y) Let (a, m) = ml 1.Since Xfisprimitive, we can find a class b mod in E (o/m)* such that m Xf(b) 1 andb - lmod - MI As a consequence, ab - a mod m, so that aby - ay r= D-1, and by what we have just shown, Xf(b)tm(Xf, ay) = rm(Xf, bay) = tm(Xf, ay) Finally, in view of Xf(b) # 1, we find t,,,(Xf, ay) = 0. As for the absolute value of the Gauss sum, we see from (6.6) that ate. 476 We nowmakeuseofthecondition(ymD,m)=1.Itimpliesthat nonzero onthesubgroupofzmodinE(o/m)*suchthat z-1modm/a: For a01,thelastcharactersumvanishessinceXfisprimitive, andtherefore and thusy(z-1) vp(y) _-vp(m)-vp(c))and prime divisorpof other handz#1moda Indeed, ifz-1Eaim,theny(z1)m-1-0-'a-1m=a-1-0-1.Ifonthe phisms (6.7) Proposition.Thecharacters XofR*,i.e.,thecontinuoushomomor- X,,) ofR*.Theyaregivenexplicitlyasfollows. statements ofthetheorem. character. Sowefinallyhavethat the sumreproducesitselfundermultiplicationwithavalue Xf(x) vp(y(z -1)) X(Pm) =1. I I x1,wefinallyobtainthecharacters),ofR*as X(x) mo=o a -1modmandtotallypositive. boo = N(( 1XX X ((a))= Xoo(x) = X :Jm/Pm-S1 I m =Jp',, )PIxIcg) O-S Xf(a)N((lal)P)' "CI pt- =°c Chapter VII.ZetaFunctionsandL-series = N(XPIxI-P+iq N(xP XI-P+iq) /J call p-iq,theexponent J'" -+S'suchthat S1 is R. El GrOJ3encharaktere Xmod1satisfying Xc:=1. i.e., thecharactersX (6.10) Corollary.Thecharacters oftheidealclassgroupC1K=JIP, through Jf Jf'/Pf' of (o/m)*-(o/f')*-+S'(see(6.2)).Bytheabove,X"gives a is therestrictionofaGr6j3encharakterX":Jf' This impliesthatf'If.Ontheotherhand,Graj3encharakter XJ'-->S' is theninducedbyacharacterX':Jf/Pf---)-S',sotheGrOJ3encharakter ... conductor ofthecorrespondingGrbj3encharaktermodm. X:J'"/P'-S;' the typeclaimed. gives usacharacterof(o/m)*.ThereforeXisindeedGroJiencharakter for aEKm,and P, E(0,11forrreal,andpz=0complex.WehaveX((a))Xoo(a) because defines acharacterofR*/R*+).ItisinducedbyX,,,,R*which- Then thecomposite an isomorphism totally positive)andR*)_((xt)ER*Ixr>0forrreal).Thenwehave such thatX(Pm)=1.LetKm{aEK*Ia1modm),K.(=-K' mod m. Therefore XfactorizesthroughJm/Pm,andisthusaDirichletcharacter have Xf(a)=1,andx.(a)thenX((a))Xf(a)X,,I. complex. Proof: LetXbeaGroJ3encharaktermodmwithcorrespondingcharacters X :Jm-3S'modmistherestrictionofGroJ3encharakter X': Xf, X.of(o/m)*,R*/om,suchthatX...istype(p,0)withpi=0forr § 6.HeckeCharacters Let fbetheconductorofDirichletcharacterXmodm,andlet Conversely, letXbeaDirichletcharactermodm,i.e.,ofJm ' /Pf S' suchthattheDirichletcharacterX:Jm/Pm-*factors ' . Hence Km/K+ 1 -isoftheformX00(x)= x" : J Vv:< Km/K+ f If',sothat=. Xf(a) =X((a))XCO(a) E osuchthata S' suchthatX(P)=1,are precisely the t) R*/R(+) =II(±1). J°'/P- -4S' p real 1 modm,wethenobviously ' S' , soXf )P) withp=(pr), is thecomposite Jf -+S'. character gy! 479 0 o.. t5" trivial, so p100 I, = f1 Kp . , as the multiplicative group R* of pt- K. Observe that I7II differs slightly Chapter VII. Zeta Functions and L-series and Functions VII. Zeta Chapter If = I1 Up"Pi Al X(If)=1. where mod 1. The associated character Xf is associated character mod 1. The X((a)) = Xf(a)X.(a) = Xf(a) = 1, Xf(a)X.(a) = Xf(a) X((a)) = 1 "` = If" x I. fpt,,. Up -+ S' is a compact and totally disconnected subgroup of We call m a module of definition for the Hecke character X if We call m a module of definition for the : In order to deal with Hecke characters concretely, consider an integral In order to deal with Hecke characters To conclude this section, let us study the relation of Gr5f3encharaktere to let us study the relation of Gr5f3encharaktere To conclude this section, Up if n = 0, and the n-th group p { oo, then Up(') is the group of units Every Hecke character admits a module of definition, since the image of X (6.11) Definition. A Hecke character is a character of the id81e class group Hecke character is a character of the id81e (6.11) Definition. A X(K*) = 1. of the ideae group I = L[pK* such that this ideal the subgroup I'° of I, If of higher units for n > 1. We interpret 1 the R-algebra R = K ®Q I[8 = Ilpl introduced in chap. Vl, § 1, in from the congruence subgroup I' _ Flp Up"PI = R instead of the component K. that, for real p, we have the factor Up°l class group J'°/P'° mod. m, but The effect is that I/IrK* is not the ray group P'° of all principal ideals (a) isomorphic to the quotient J°'/P'" by the in chap. VI, (1.9). We will refer such that a = 1 mod m - this is seen as to J'IP' as the small ray class group. X.(a) = Xf(a)-'X((a)) = 1, and thus X,,. = 1, because K* is dense 1, because K* is and thus X,,. = = 1, X.(a) = Xf(a)-'X((a)) 1, then 1 satisfying X', = mod X is a Gr6f3encharakter in R*. If conversely the ideal class X(P) = 1, and X is a character of for a E K*. Therefore group. class group. characters of the ideae field K, i.e., a continuous homomorphism C = I /K* of the number no = 0 for p I oo. We associate to ideal m = IIp p"P of K, i.e., no > 0 and 480 is of J/P X A character = (1). have (o/m)* = 1 we For m Proof: a Gr5f3encharakter and 481 is the (7rp) 1. 1, ) .Y' I/ImK* 1rK*/IfK* I/If K* 1 m n K* = om, If fl K* = 1 * C(m)-4St JmC C:Jm-pC(m) X:C(m)-fS1 C(m) = I/IfK*. = 0 for almost all p. For it we can take the ideal all p. For it = 0 for almost tiw as a module of definition. as a module -> J'n/Pm _ 1. 1 -) R*/Om -) C(m) 1m fl K*/If fl K* -+ 1 m/If 1 -) IrK*/ImK* will not in general factor through the small ray class group factor through the small ray class will not in general 1 Up' where np Up' where Given a Hecke character X with module of definition m, we may now Given a Hecke character X with module Since X (If) = 1, the character X : C = I/ K * - St induces a character X : C = I/ K * - = 1, the character Since X (If) (.... 1, 1, ir,,1,1, . . .). This mapping does not depend on the choice of (.... 1, 1, ir,,1,1, . . .). This mapping (up), up c Up, for p fi in, lie in I. the prime elements, since the ideles Hecke characters with module of yields a 1-1 correspondence between definition m and Grbj3encharaktere mod in. The reason for this following construct a Gr5f3encharakter mod m as follows. For every p { oo, we choose construct a Gr5f3encharakter mod m as a homomorphism a fixed prime element 7rp of Kp and obtain the class of the idele which maps a prime ideal p t m to Taking the composite map I/I'"K* = Jm/Pm (see chap. VI, (1.7), (1.9)), which bears the following chap. VI, (1.7), (1.9)), which bears I/I'"K* = Jm/Pm (see There is an exact sequence (6.12) Proposition. from the two exact sequences Proof: The claim follows immediately In the second one, one _has § 6. Hecke Characters Hecke § 6. of the form a subgroup has to contain the kernel and so finite, S1, hence p' m = fp},,. of the group But it relation to C(m). 1°1/If=1,),, = R*, and soImK*/IfK*=R*/om. 1, ) C(m) modifK*=1, C(m) ] indicates taking classes. and x E K*. E If c/r : R*/o'n _ VI(b) = b-1 mod If K*, VI(b) = b-1 mod If Chapter VII. Zeta Functions and L-series and Functions VII. Zeta Chapter a E I be the idele with components ap = a a E I be the idele with with is considered as an idele in I. For a E 0, is considered as an idele in I. For , let J' x (o/m)* x R*/O' * C(m) J' x (o/m)* c(a) = y mod IfK* c((a)) =a modifK*. Then f : J'n x (o/m)* x R*/o'n S(a) _ ((a)-1, a mod m, a mod o'n) a mod m, a mod S(a) _ ((a)-1, : (o/m)* -> C(m), rp f ((a, a mod m, b mod O'n)) = c(a)rp(a)i/r(b), yaaO,)b-1 = l;x K(m)/om f ((a, a mod m, b mod Om)) = c(a)cp(a)*(b) = 1, f ((a, a mod m, b mod Om)) = c(a)cp(a)*(b) sp(a) = aa,, mod If K*, sp(a) = aa,, mod If 1 .r., f(S(a)) =c((a))-lfp(a),/r(a)=a 482 sequence exact is a canonical There Proposition. (6.13) given by where S is Proof : For every a E K () Proof : For every a 1 for p I moo. It is then obvious that for p t moo and ap = components in the principal idele a according to its Let us decompose a = afaw, and define the homomorphisms I = If x I... as a product by where every b E R* = I,,, I, so we get in C(m) the equation a = 1 mod m, we have afa-1 E If C [ V(a) = aa.] = [afa.] _ [a] = 1, where every s E om, one has sf E If, so This shows that fp is well-defined. For thus *(s,,) = 1. Consequently ,/r is [sue] = [s.sf] = [s] = 1 in C(m), and well-defined. We now define the homomorphism by is exact. The homomorphism S is and we show that the resulting sequence so that f o S = 1. Conversely, let clearly injective. For a E K ('n) one has and let a = yp = jr' pfor p t moo, and yp = 1 for some idele y with components for p I moo. This yields an identity Gr6f3encharaktere modm. X ofC(m),i.e.,Heckecharacters withmoduleofdefinitionm,and (6.14) Corollary.Thecorrespondence XHocis1-1betweencharacters Xoo areuniquelydeterminedbyX,weobtainthe for aEK("').ThismakesXGr6f3encharaktermodm, and sinceXf that the triplesX,Xf,XooofcharactersJ°',resp.(o/m)*, R*/Om,such the charactersofJ`°x(o/m)*R*/o"'thatvanishonS(K(m)/o"'), i.e.,to 112 b = p moo,andyp=1forImoo.Thisgivesya-lacEIf,ifwedefine, where theideleyhascomponentsyp= for pIm.Leta= idele a,multiplyingitbyasuitablexEK*,insuchwaythatapUpn°) in C(m).Bytheapproximationtheorem,wemaymodifyrepresenting: and thisshowstheexactnessofoursequenceinmiddle. So wehave b =a)x-1,andthus Finally, forpInowefind(yaa,,,,b'1)p=abp1xinKp,sothat As XE For ptmooonehas and also0=vpvp(a-lx)sinceaisrelativelyprimetom.Thisgives vp =vp(alx).ForVImonehas § 6.HeckeCharacters By theprecedingproposition,charactersofC(m)correspond 1-1to The surjectivityoffisprovedasfollows.LetamodIfK*beclass! (a, amodm,bo')_((ax-1),ax-1ow) then f((a,1modm,bom))= Uphl°), one hasx=1modm,hence x ((a))-1Xf(amodm)X,,.(aom)=1 l p"°(a°). Thenwehave c(a)=ymodI K*, III (yaa,,.b-1)p =zrp°apxinKp,andso *(ax-1) _*(b) (P(ax-1) _(P(a) .-. a =(ax-1). nv°p(a°) yb-1 1 =1=px,sothatxEUp" = ya,=-amodIfK*. = Epap,EpEUp,for , 483 0 484 Chapter VII. Zeta Functions and L-series Exercise 1. Let at = fi=i in, be a decomposition of m into integral ideals which are pairwise relatively prime. Then one has the decompositions and Let Xf be a character of (o/m)*, and let Xfi be the characters of (o/m;)* defined by Xf. If Y E m7'?-'/cr-1, and if yi E m are the components of y with respect to the above decomposition, then rm(Xf, y) = f 1 rm, (Xfi, yr) i=l Exercise 2. Prove the Mobius inversion formula: let f (a) be any function of integral ideals a with values in an additive abelian group, and let g(a) _ f(b) bin Then one has f(a) = Fu- ag(b) bin b Exercise 3. Which of the characters ),(x) = N (xP Ix I -P+"?) of R* are characters of R*/o'°? Exercise 4. The characters of the "small ray class group" J'°/P'° mod m are the Grof3encharaktere mod m such that X= = 1. Exercise5. Show thateverypair of charactersXf : (o/m)* --.>. S1and X : R*/o' --> S1 such that Xf(E)X.(e) = Ifor allE E o* comes from a Grofiencharakter mod m. Exercise 6. Show that the homomorphism c : JC (m) is injective. § 7. Theta Series of Algebraic Number Fields The group P of fractional principal ideals (a) is constituted from the elements a E K*, and it sits in the exact sequence 1-± - +K* -) P-) 1. In order to form the theta series we will need, let us now extend K* to a group K* whose elements represent all fractional ideals a E J. Proof: LettheidealclassgroupJ/Pbegivenbyabasis[b1], for allaEK*. (7.1) Proposition.Thereisacommutativeexactdiagram § 7.ThetaSeriesofAlgebraicNumberFields and weconsiderthemapping Each class[b]EJ/Pcontainsauniquelydeterminedidealoftheform be thesubgroupofC*generatedbyK*andelementsb;=(b;r)EC. in Csuchawaythatbtt=btrwheneverriscomplex.WedefineK*to bh' = v; modhiareuniquelydetermined,h;beingtheorderof[b,]inJ/P.Let where aEK*iswell-determineduptounitso*,andtheexponents fractional idealacJcanbewrittenintheform choose, foreveryoneofthesebasicclasses,anidealbl,...,b,..Then with asubgroupK*CC*containingsuchthatjaiER+,and the classDID],and v; +vt=tt;)1h1,0 - h,-betheclassnumberof K.Thenwehavefor III f([b][b'])=bi'...RAI ... b°r)(btbr') "+o IR((a)) a=abi' -.ba b17 =h'rb, bl1...bwfb;I...bar > K* f ([b])=bl''...b'. with 0 jig =[MD], I-+ a, i.e.,b=asforsomeaEK*, a modmIEdtmi} ) J/P--)1. a, andthus 1, and 487 '; ; come fromaGr5f3encharaktermodm,butthetreatmentofthetaseries and putX(a)=0foraEosuchthat(a,m)1.Intheapplications,xwill ideal numbersm,d.Form=olet1.Wenowstudycharacters where tisthedifferentofKIQ.Letm=(m)andD(d),withsomefixed 488 because aandzareintegral.Ontheotherhand,ifmodmisfixed In fact,ontheonehandwehave class W.Inparticular,weputr(X)=r(x,1). Then wedefinetheGausssum (7.4) Definition.LetaEnbeanidealinteger,andlet.theclassof(a). is independentofsuchanoriginX. representatives z,andtheorem(6.4)yieldsatoncethe with y=X/md,whichsatisfies(ymD,m)1since yrnD _(z)and and inparticular by xxmodm,withxinvaryingovertheclassesof(o/m)*. Therefore class of(o/m)*whichmapstoif,then,inview(7.3),we gettheothers .L'm-1D-1, soyEK*,andonefinds since theclassofideal(y)=(a)(x)(m)-1(d)-1isprincipal where xmodmrunsthroughtheclassesof(o/m)*whicharemappedto and Ir(x)I=.fs7t(m). (7.5) Proposition.ForaprimitivecharacterXof(o/m)*, one has ((x ),m)=1.Consequently,r(X,a)doesnotdependon thechoiceof The Gausssumr(X,a)reducesimmediatelytotheoneconsideredin§6, (-e C3. c°) wad r (X,a)_ y E(y)_ rm(X,y) = bbl) r(x,a) =X(x)rm(X,Y), y =za/mdE r(X) =X(x)rm(X,Y) X :(o/m)* r(X,a) =X(a)r(X) xmod m (a)m'?' Cm1D-1 x modm (x, -)=1 L. X (X) Chapter VII.ZetaFunctionsandL-series X(x)e2" ) C*, e2ni Ti(za/md) , C3. term ofthethetaseriesisX(0)N(0P)=1,whereasit0inallothercases. m =1if1.Thecase1,p0isexceptionalinthattheconstant where m,darefixedidealnumberssuchthat(m)=mand(d)D.Wetake §3, p.448)andacharacterXof(o/m)*,weformtheHeckethetaseries an arbitrarynumberfieldK.GivenanyadmissibleelementpEF1,Z(see are attachedtothefieldQ.Wenowhavefindtheiranaloguesrelative § 7.ThetaSeriesofAlgebraicNumberFields have thegeneraltransformationformula(3.6)atourdisposal. and tothisendwedecomposethemfurtherintothetaseriesforwhich For thesepartialthetaseries,wewanttodeduceatransformationformula, corresponds totheidealclassAunderisomorphismK*/K*-J/P. where avariesoverallidealintegersintheclassAEK*/K*which into partialHeckethetaseries Proof: Inthethetaseries8P(A,X,z),itsufficestosumover theelementsof classes of(o/m)*,thenonehas (7.6) Lemma.Assumethatm class A,andletaEo(')beanidealnumbersuchthat(a)=a. the classes sequence (7.3) is eitherdisjointfromAorelseitcontainedinA.Inview oftheexact A no(m)becauseXiszeroontheothers.Everyclasszmod mE(o/m)* where Pisthelatticem/aCRand The thetaseries0(X,z)usedin§2thetreatmentofDirichletL-series Let usdecomposethethetaseriesaccordingtoidealclassesAEJ/P Let abeanintegralidealrelativelyprimetomwhichbelongsthe 8P(A>X,z) =X(a)N(a") 8p(A, X,Z)_ Br(x,0,z) _E 81(X,z) 1-;(o/m)*-f(o/m)*--) J/P-)1, r`4 ax modm=a(x+a1m) aE(.flo)U(0} aEOU(o) gEr x modm ''' 1 orp;0.Ifxmodmvariesoverthe X (a)N(ap)eni(azllmdl,a) X(x)8r(x,0,zIa2/mdl) , CAD X(a)N(aP)en'(az/in:dl,a) N((x+g)P)e'r'((x+g)z,x+g) 489 (2) From (3.6)wenowget mesh ofFisbychap.I,(5.2), respect to( (1) to (5.7),byT'=a/m1.(Hereasin§4,theasterisksignifiesadjunctionwith Proof: ThelatticeF'dualtother=m/aCRisgiven,according This factorhasabsolutevalueIW(X,p)=1. with theconstant (7.7) Theorem.ForaprimitivecharacterXof(o/m)*,onehasthe for theseriesOrandproposition(7.5)onGausssums,wenowobtain element withcomponentsj5,=pT.Fromthetransformationformula(3.6) transformation formula are thedifferentresidueclassesof(o/m)*containedinkThisgives 490 OP (A,X,z)= For anyadmissibleelementp=(pr),wewillwriteforthe the 9r(0,x,zlmd/a2I) _ series A(z) 9rP,(x,O, OI(A,X, -1/z)=W(X,P)N((zli)P+.I)6P(S',X,z) vol(j') =M(m/a) = X(a)N(aP)EX(x)6r(x,0,zla2/mdl). = X(a)N(aP)Ex(x) , x modmgEr ), i.e.,(x,ay)_(*ax,y).)Thevolumeofthefundamental W(X,P) = factor - 1/Imd/a2Iz) = x modm xmod m X [ITr(P)N\\ (ax)N((ax +ag)P)eni(a(x+g)z/Imdl,a(x+g)) 9 'Er' I dKl=N(Im/al)(Id1)1/2 (( gEr EN((x +g)P) Imdl Chapter VII.ZetaFunctionsandL-series and )P)I -' N(gP)e27ri(x,g')ezri(g'zjmd/aZI.g') - r(X) rn(m) .In eni((x+g)zla2/mdl,x+g) . o--.7 representatives of(o/m)*,then axvariesoverasystemofrepresentatives Now considerthesuminparentheses.Ifxvariesover asystemof class A.Furthermore,(g)is an integralidealintheclassA',andsince.S of thoseclasses(o/m)*which aremappedunder(-/m)*-->JIPtothe with thefactor (3) into(1)andformula(7.6),with-1/zinsteadof z,weobtain Equation (1)thusbecomes Let usnowconsiderfirstthespecialcasem=1,p0(whichwas tag IgEK,(ag)Co)=as1aT'.Consequently essentially treatedalreadyin§5).Inthiscase,wehave(.ktflo)U{0} Substituting allthisinto(2)yields the set and N((*g)P)=N(gP).Ifg'variesoverthelatticeT",theng (3) Writing g'= § 7.ThetaSeriesofAlgebraicNumberFields Now assumem BP (A, 01(A, X,z) =B(z) O ,(0x,zlmd/a2I) 0P(A, X,-1/z)=N(aP)EX(ax)O(x,0,1/zlmd/a2I) (g'lmdla2lz,S) _(*gzlmdla2Illmd1a12,g) X NR. gek'n0 z) Y_ (md/a)*I" =(md/a)a(mD)-1(A'flo)U{0}. *(nad/a)' = N(( *g 0P(A,X, -1/z)=N(z/i)7BP(A',7,z) gE(k'no)U(o) N(gP)( 1 orp e'i(agzlldl,ag) the rulesstatedin§3give )P/ B(z) =A(z) x modm 't? (x, g')=Tr(axg/md), 9E(. 'fl(;)U{O) 0. ThenwehaveX(0)N(0P)=Substituting j x modm X(QX)Q21iTr(axg/md) N((md/a)P) gef N(aP) erri(gzIa2/dl,g) Br (zld/a21). 4L, N(g")e-'Tr(axglmd)eni(gzllmdl,g) eni(gz/Imdl.g) = BI (gz/Imd1,g) lam" \.J Of if bears thesamerelationfi.'.fi=[mO]toSzas.fidoes.t3',werecognize 492 we finallyarriveattheidentity Substituting innowtheresult(7.5), sum inquestionastheGauss we haveIW(X,p)l=1. where onehastoobservethatTr(p)=Tr(p),aP*aP,a*aIal2,and with thefactor (4) contemplated this,thereader should reflectupontheadmirablesimplicityof number fieldKhavebeenused.Firsttheyservedtobreak upthetheta peculiar wayinwhichalmostallfundamentalarithmetic properties ofthe himself amomentofcontemplation.Lookingback,he willrealizethe (7.8) Corollary.8P(X,-1/z)=W(x,P)N((z/i)P+z)BP(X, z). and (7.7)yieldsthe ImdlP =(*Imdl)PImdl'becauselmdlER.Sincelr(X)l_ the numberfield. the thetaformulawhichencapsulates alltheseaspectsofthearithmetic law, butintheendtheyarereassembledtoformanewtheta series.Having series, thentheseconstituentswerereshuffledbytheanalytic transformation We recommendthatthereaderwhohasstudiedabove proofallow If m01orp,-f0,wefindforthespecialthetaseries: BP(X,z)= E W(X,p) 01(A,X, -1/z)=W(X,))N((zli)P+z)0z) r(X,g) _ aEO [iTr(P) 1Ti'(P) r(X) 0t (m)LL r(X) T(X,g) =X(8)r(X), sn(m) x modm q't(m)]-1N( Imd/a2lP) riTr(P)N N X(ax)e27riTi(axg/md) X(a)N(aP)e'i(az/jmdl,a)EBP(A,X,z), ((Imdj )P) Chapter VII.ZetaFunctionsandL-series \\ and I 0.M ImdI J and Coo \ p\-1 /I N k ale I al2P/ N((md/a)P) (aP*aPl a P z(X `n(m), El Furthermore, wehavex a, b,thenthereexistelementsx,yE holds ink*. Exercise 1.Defineidealprimenumbersandshowthatuniquefactorization next section. play anessentialrolewhenwenowpassfromthetaseriestoL-seriesinthe does notenterintothisformula,thatis,Dirichlet'sunittheorem.Thiswill § 8.HeckeL-series such thata=1modm,onehastheexactsequence Exercise 6.ForthefactorgroupJm/PmbyF'ofallprincipalideals(a) an idealintegerprimetoa. Exercise 5.Ifa,mEo,thenthereexistsineveryresidueclassmodprimetom, such awaythattheindividualsolutionsareequivalent(withrespectto^-), moduli issimultaneouslysolvableifeverycongruenceindividuallyin Exercise 4.Asystemoffinitelymanycongruenceswithpairwiserelativelyprime and onlyif(a,m)Ib.Thissolutionisuniquemodm,provided(a,n2)=1. Exercise 3.Thecongruenceax=bmodmhasasolutioninowithintegralxif notation a Exercise 2.Letobethesemigroupofallidealintegers.Ifit=(a,b)isgcd K' =(aEK*Ia=_1modm).Thenonehas(/m)*K(m)/Km.' Exercise where 0m=(sE0*1s=1modm). this character,weformtheL-series be acharacterofthegroupidealsrelativelyprimetoM. Withrespectto (a, m)01.Thenthefollowing propositionholdsincompletegenerality. where avariesovertheintegral idealsofKandweputX(a)=0whenever CU- There ishoweveroneimportantfundamentallawofnumbertheorywhich Let inbeagainanintegralidealofthenumberfieldKand let 7. fi meansUP-1EK*. Let k(m)bethepreimageofJmunderK* 1 -+O*/O'ni(/m)* III d/a, resp.y- § 8.HeckeL-series L(X,s)_> m.$ X:J, Q.. d=xa+yb. F,; U (0)suchthat d/b, unlessx=0,resp.y0.Herethe a <`< ) S' 9q(a)s X(a) ' I0. pro J, and gyp-, 493 let the termssuchthat01(a) 1 1 1 1 -X(P)t(P) L(X,s)=F1 p n r log E(s)=F_E E(s)=j1 npn(1+S) -qtr r-1 (Pi)_S 1 _ _ _ P P 9t( vt,...,vr-0 1 + =dlogi;(1+8). P 1 -x(P)91(p)-S 1 - 00 n=1 nM(p)ns Chapter VII.ZetaFunctionsandL-series sa(p)s +`n(p)IS X(a) = exp\ X (p) N O'1(a)S + 1 X (a) rat 1 X (`R(pl)°t ...Mpr)ur)S ( (p)n X X (p)2 ( X(p)`RW-S 9q(a)>N (pl)"I... X(pr)Ur n'R(p)s , X (p)n .. . p1+8and since $R(a)S X (a) 495 is monotone 1 X (a) N EN w(a)s 'F(a)t+a N )(t+S)u 1 pifa 1 (91(1 71(a)>N 1 :C I npn(1+S) co X00 : R* - S1. 1 °O log((1 - S7t(pi)-(1+S)) -1) r ) S1, p X,s) i=1 i=1 n=1 rL Fd m(a)t+a p ,n and L( < = = d log(w + S)) . with the series L(X, s), we get L(X, s), we the series with i=1 - S 1 ) X : J, 1 X((a)) = Xf(a)Xoo(a) 1 N m(a)t+a 1 - X(Pi)'N(Pi)-s - 1 - X(Pi)'N(Pi)-s fit( r7 it 1 i=1 log( fi (1 - 2(pi-(1+S)) -1) We now face the task of analytically continuing the L -series L (X, s) We now face the task of analytically continuing § 8. Hecke L-series Hecke § 8. Comparing now in (*) the sum in (*) the now Comparing term of side tends to zero, as it is the remainder For N --* oo the right-hand the sequence (E,n(a) Lx(s) =N(n-512)I'x(s/2) N(n-s/2)JN(e-yyS/2) coo (S n"off/o*--4{aESIintegral} LM X,s)=> L (A,X,s)_r L(X,s)L(A,X,s) COQ Xf :(o/m)* QE91 IN(a)IS integral Chapter VII.ZetaFunctionsandL-series 0.E. ) X((a)) J`(a)S S1 tYx (a) R* , a H(a), 51, aJy "'1 (m> o;~ prime § 8. Hecke L-series 49, which has been attached to the G(CIR)-set X = Hom(K, C). The character xof R* corresponding to X is given by (6.7) as Xoo(x) = N(XPIXI-p+iq)for an admissible p E ft 7L and a q E R±. We put s = s1 + p - iq, where s E C is a single complex variable, and Lw(X,s)=Lx(s)=Lx(s1+p-iq). In the integral 1'x (s/2) = J N (e-y ys/2) Y, we make the substitution y r- + jrIaI2y/Imdl(a E 91), where m, d E o are fixed ideal numbers such that (m) = m and (d) = D is the different of K IQ. We then obtain T'x(s/2) = N((Iandl)SI2)N(IaIs) f e-"(ay/Iandl,a)N(ys/2) dY and, since N(ImdIts/2)= (IdKIM(m))s/2 yY dKI` t(m))s/2L.(X, s)N(laIS) =c(X)fe -"(aY/Imdl,a)N(ys/2) wherec(X) = N(Imdl-P+`q)h/2.Multiplying this byXf(a)N(aP)and summing over a E 9 t yields, in view of Xf(a)N(aP)_ Xf(a)N(aPIaI-P+iq) _x((a)) N(Ials) N(lals) IN(a)Is the equation (IdK I`n(m))s/2Loo(X, s)L(A, X, s) = c(X)f g(Y)N(ys/2)d R* with the series g(y) = Xf(a)N(aP)e-R(aYllmdl,a) a E9i We now consider the completed L-series A(A, X, s) = (IdK I`n(m))s/2Loo (X , s)L (A, X, s) t dt grr(Isl2x,t). Xf(a)N(aP)e'i(az/Imdl,a) . £EO* _ t dt- Xf(a)N(aP)e-7f(axtImd1,a) t=N(Y), N(Y)tln, N(X(P-iq)I2)t-j'(s+Tr(P-iq)ln) but over all a e St fl o. We have but over all a e St fl 'CS =d*xx aEAno y Chapter VII. Zeta Functions and L-series and Functions VII. Zeta Chapter dy f g(Y)N(YS/2) yY f g(Y)N(YS/2) R* F ae9t 00 0s = 3 o.(E)N(EP) = Xf(E)N(EP). N(x(P-iq)/2) N(x(P-iq)I2)( 0(. ,X,ixtIln)-s(X)) R+=Sx][8+, A(A, X, S) = c(X) A(A, X, S) y=Xttln, x= N(EIP-iq) A(A, x. s) = c(x) ff N(X(P-iq)I2)g(xtI1n)d*xts' g9t(x,t) = N(YS/2) = N(xs/2)N(ts/2n) = 9(,,z) := 0P(., Xf, z) = s(x) + E 0P(., Xf, z) = 9(,,z) := We now want to write this function as an integral over the series as an integral to write this function We now want of g(y) - over a is extended not only - as in the case where the summation 9t of (An system of representatives will proceed in p = 0, and s(X) = 0 otherwise. We s(X) = 1 if m = 1 and Just as we did the Dedekind zeta function (see (5.5)). the same way as with there, using (*) 498 we get Then decompose with n = [K : Q], we Then, observing that we obtain the identity under the second integral will with s' = 1(s + Tr(p - iq)/n). The function be denoted by is constructed as follows. From it, the theta series 0(A, X, ixtI /') (8.2) Lemma. Xoo(E)Xf(s) = X((s)) = 1, so that Proof: For every unit s E o*, one has we get . t) 499 = gEst(x, : Q], c(X)I = x, ixtll")d*x . e'r(Eat,sa) 12x, t) gs (I s - s(X)) J F Xf(Ea)N( (sa)p) Xf(Ea)N( nElo*I EEO* F w ixtl/") S= U q2F. aE9t c(w) f I o* 1 '. F is mapped by log : R+ - - R± to a I o* 1 '. F is mapped x , C(X)E(X) r N(x(p-iq)l2)d*x. A(A,X,s) =L(f,s') N(x(p-iq)/2)(e(.q, ) = N(x(p-iq)l2) ge`R(x, t) = xt 1 /'/ I and I and obtain /'/ I and I = xt 1 E E N(x(p-iq)l2)Xf(sa)N((ea)p) f( EEo* aEE9t EEO* u s9t, we get u s9t, we EEO* _ A(Jq,X,s) = (s + Tr(p - iq)/n). Here we have set n = [K (s + Tr(p - iq)/n). Here we have set (CJ z f(t) = fF(A, x, t) = From this lemma we now obtain the desired integral representation of now obtain the desired integral representation From this lemma we gyt( Is IZx, t) = gyt( Is IZx, § 8. Hecke L-series Hecke § 8. for short We put Since S fl the function A (A, X, s). We choose as in § 5 a fundamental domain F of S s). We choose as in § 5 a fundamental the function A (A, X, group for the action of the fundamental mesh of the lattice 2 log I o* 1. This means that we have fundamental mesh of the lattice 2 log I o* (8.3) Proposition. The function of the function is the Mellin transform at s' = of roots of unity in K. N(Imdl-p+iq)1/2, and w denotes the number Proof: One has where We haveseenbeforethat 500 The factthatwemayswapsummationandintegrationisjustifiedinexactly F In eachoneoftheintegralsonright,wemaketransformation Since S= (*) Using (8.2),thisgives #(rE0*1 Isl=n1=w,sothatweget where tz(K)denotes the exactsequence the samewayasforcaseofDirichietL-seriesin§2,p.436.Inview the hypothesesofMellinprinciple. 0P(Ft, Xf,z)whichguarantees thatthefunctionsf(t)=fF(A,X,t)satisfy This togetherwith(*)yieldstheclaimofproposition. It isnowthetransformationformula (7.7)forthethetaseries6(A,),,,z)= n2 F,xHn2x,andobtain A(r) = =f(t)-f(oo) = c(X) c(X) UgElo*I772F, A(A, X,s) l) µ(K)_+o*_to*I-#1, w fo(r) =c(X) w f-+ fo(r) =c(X) IE1=71 F fo(r) =c(X) F f f one has the groupofroots gtt(IE12x,t) =wgot(n2x,t) N(x(P-iq)l2)(9(_q,X,ixt1/n)-E(X))d*x EEO* Sot(Iel2x, t)d*x 0 f F f 00 pEIO*I nHo*1 f S Chapter VII.ZetaFunctionsandL-series I? F got (x,t)d*x. z fgEt(x,t)d*x. go't(n2x,t)d*x. CIO fo(r)ts, at = L(f,s') of unity (7p l-. in K,onehas II- joy because Xoo(x)=N(xPIXI-P+iq)N((*x)PIxl-P-'q) second, wemakeuseofformula(7.7).Itgivesus Proof: Thefirststatementfollowsexactlyasintheproofof(5.8).For where .'=[mO],andtheconstantfactorisgivenby if m=1andp0,ao0otherwise.Furthermorewehave and (8.4) Proposition.WehavefF(Ft,X,t)=ao+O(e-`t'In)forsomec>0, § 8.HeckeL-series We haveusedinthiscalculationthatN(xt/2)=N(x)1/2 =1 F-1, (7.7)yieldsforz=ixtlln: d*x invariantandtakesthefundamentaldomainFto Observing thefactthattransformationxf-1x-1leavesHaarmeasure conjugate ofX,,.,isgivenby N(xP) =N((*x)P)N(xP), andthatthecharacterX,,,complex B(S, X = = W(X)t = , - l/z) t c(X)W(x) c(x)W(x) C(X) fF(R,X, W(X) = w F-1 f N(x- = W(X)N((z1i)P+)0(-',X,z), = BP(.R,Xf,-1/z)W(X)N((z/i)P+2)Oi(.', 2+Tr(P)/n.fF-r (.', f w J ao= F-1 F-1 F f J f N(x(P-iq)/2)g(.,X,ix/tl/n)d*x r Xoo(x) =N(xPIx L IITr(P)N(( and)P)1-1 N(IdI`g12) (P-iq)/2 N(x- ( w )e(St, X, Ire X, t) +P+Z)N(t(P+2)ln)9(A',X,ixtl/n)d*x Imd) F J rN(x-iq/2)d*x =W(X)t21Tr(P)lnfFt) - 1/ixt I-P-iq) J . 11n r(Xf) - )d *x (m) ixt 1/n) 81k dx N(xPIxI-P-iq) Xf, z) 501 and _ f (t) From Re(s) > 1, I-R(m) [(Xf) , s) and L (g, s) can be +Tr(p)ing(t), Chapter VII. Zeta Functions and L-series and Functions VII. Zeta Chapter IandI -.N +Tr(p)ln- 2(s+Tr(p-iq)ln)) 2 O(e-ct'1'1) and = W (X)t (s + Tr(p - iq)l n)) iTr(P)N(( and );5)1-1 t I 2 (1) f = bo + A(A,X,s)=W(X)A(i',X,1-s) =W(X)L(g, =W(X)L(g, 2(1-s+Tr(5 +iq)ln)) = W(X)A(. ', X,1 - s), W(X) = O4. X, t). f (t) = fF(.., X, t) and g(t) = fF-1(S', A(. ,X,s) = (IdKIOl(m))s/ZL.(X,s)L(. ,X,s), A(. ,X,s) = (IdKIOl(m))s/ZL.(X,s)L(. A(A, X, s) = L (f, O(e'ct'I"), g(t) From this proposition and (1.4), we now finally get our main result. our main finally get we now (1.4), and this proposition From A (.fi, X , s) is holomorphic except for poles of order at most one at A (.fi, X , s) is holomorphic except for 502 that mod in, i.e., GrOj3encharacter that X is a primitive We may assume (see §6, p.472). Xf of (o/m)* is primitive character the corresponding of the from the L-series character differs of an arbitrary The L-series So many Euler factors. only by finitely primitive character corresponding of the follow from those equation of one and functional analytic continuation other. m. Then the x be a primitive Groj3encharacter mod (8.5) Theorem. Let function and satisfies the continuation to the complex plane C has a meromorphic functional equation and the constant factor is given by where J W = [mZZ], A(.., X, s) is holomorphic on all of C. It has absolute value I W (X) I = 1. iq)/n. In the case m 0 1 or p ¢ 0, s = Tr(-p + iq)/n and s = l +Tr(p + Proof: Let ao + L (f it follows by (1.4) that the Mellin transforms we get meromorphically continued, and from (8.3) where we have to take into account again that X,,,(x) = N(xp Ixff-i4). theory [94]providesanillustrative complement. which cannotbeimprovedupon. InadditionSERGELANG'saccountofthe reason forthisistheclarityand concisenessofTate'soriginalpaper[24], modern conceptualapproaches,westilldecidednottoinclude ithere.The Tate's thesis.Eventhoughitdoesmeetthegoalofthisbook ofpresenting American mathematicianJOHNTATE,andiscommonlyknown forshortas number fieldsandtheirideaeclassgroup.Thistheorywas developed bythe principle ofalgebraicnumbertheoryandontheFourieranalysis ofp-adic The proofofthefunctionalequationisthenbasedon local-to-global the representation(8.1)ofcorrespondingL-seriesasan Eulerproduct. this section.Itstartsfromacharacteroftheideleclass groupandfrom Remark 2:Thereisanimportantalternativeapproach to theresultsof without decomposingtheidealgroupatall, Dedekind zetafunctionbyusingidealnumbers,imitatingtheaboveproof, in [52].Ontheotherhand,onemayprovefunctionalequationfor zeta function.TheGausssumstobeusedthenarethosetreatedbyHnssE J°'/P`° intoitsclasses,Ft,andthenproceedingexactlyasfortheDedekind can beprovedwithoutusingidealnumbers,bysplittingtherayclassgroup Remark 1:ForaDirichletcharacterXmodm,thefunctionalequation It isholomorphiconallofC,ifm01orp and satisfiesthefunctionalequation to (8.6) Corollary.TheL-seriesA(X we deriveimmediatelyfromthetheorem or p simple poleats=Tr(-p+iq)/nand1+Tr(piq)/n.Ifm and s=1+Tr(p)/n,i.e.,A(A,X,s)L(f, §8. HeckeL-series For thecompletedHeckeL-series According to(1.4),inthecaseao c°.) 0, thenao=i.e.,A(.q,X,s)isholomorphiconallofC. A(X,s) =(IdKI C N{Tr(-p+iq)/n,1+Tr(p+iq)/n} WOO A(X, s)=W(X)A(X,1-s). .., (m))s12L.(X,s)L(X,s) _ Ss, , s)admitsaholomorphic 0, L(f,s)hasasimplepoleats=d 0. 2 (s +Tr(p-iq)/n))hasa' A(J.,X,s) cad continuation 503 1 1' o4, E R+1 .:, tt (1 - k) and L (X, 1 - k) of the .S2 Chapter VII. Zeta Functions and L-series and Functions VII. Zeta Chapter 42° C(v1, ...,vd) = {t1v1 +...+tdvd I ... , vd are linearly independent vectors in Vk. A finite disjoint ... , vd are linearly independent vectors § 9. Values of Dirichlet L-series at Integer Points § 9. Values of Dirichlet L-series We first prove a new kind of unit theorem for which we need the following We first prove a new kind of unit theorem Thus instead of idly copying this theory, we have chosen to provide a to provide have chosen theory, we this idly copying instead of Thus the analytic reasons that we have proved It was for pedagogical .°. The results of § 1 and § 2 on the values a subfield of R, and Vk a fixed k-structure of V, i.e., a k-subspace such a subfield of R, and Vk a fixed k-structure k-rational simplicial cone of that V = Vk ®k R. By definition, an (open) dimension d is a subset of the form where v1, called a k-rational polyhedric cone. union of k-rational simplicial cones is We call a linear form L on V k-rational if its coefficients with respect to a k-basis of Vk lie in k. L-series, and the very theory of theta series has to be seen as an important theory of theta series has to be seen L-series, and the very in its own right. arithmetic accomplishment over: for the equation of L-series four times continuation and functional Dedekind zeta for the Dirichlet L-series, for the Riemann zeta function, the number for general Hecke L -series. This explains function, and finally the expose the general case directly would shrink of pages needed. Attacking to be said that the size of Tate's thesis. Still, it has to little more than theory at fundamental importance for number Tate's theory has acquired large through its far reaching generalizations. L-series will now be extended to Riemann zeta function and the Dirichlet real number field. We do this generalized Dirichlet L-series over a totally TAKURO SHINTANI (who using a method devised by the Japanese mathematician [128]). died an early and tragic death) (see [127], an n-dimensional R-vector space, k notions from linear algebra. Let V be 504 proof of Hecke's original modem treatment framework and a conceptual out that Hecke's approach to fathom. It turns difficult which is somewhat of even claim a number of its own, and can to have a relevance continues equation of the Riemann For the functional over Tate's theory. advantages out of it would be L-series, for example, and the Dirichlet zeta function since its p-adic expense, formalism with all to develop Tate's proportion used here. Also, at a beginner's level with the method they can be settled simplicial conesof-':mension r3' ... ,Lt P={xEVI Li(x)>0,Mj(x)>0} P N(0)=UCjCj(u)R+u. r-, a., 0. Forn=1and2thelemmaisobvious.We jEJ vd3 E jEJ . (0)isadisjointunionofk-rational Vk, dj U '°. A It' o... coo ... ,Li 505 this actionhasafundamental domain whichisaQ-rationalpolyhedriccone: of totallypositiveunitsactson R*)viamultiplication,andwewillshowthat Since R=K®QR,thefieldisaQ-structureofR.The group (Observe thatonehasR*+)=R+onlyinthecasewhereK istotallyreal.) n =[K:Q],andletR[[(TC]+bethecorrespondingMinkowski space SrnM}1M in1979,fallsintothiscategory.LetKbeanumber fieldofdegree foundations ofalgebraicnumbertheory.Thefollowingtheorem, provedby union ofthesetsCi, If wedefine,foreachi=1,...,n-d, (r EHom(K,C)).Define are eitheremptyork-rationalpolyhedriccones.ThereforeCNC'isalso. then wefind,ascanbecheckedimmediately,thatCNC'isthedisjoint and foreachj=1, linear formsL1,. rational cones.LetdbethedimensionofC.Thentherearenk-rational Proof: WemayassumewithoutlossofgeneralitythatCandC'arek- is alsoak-rationalpolyhedriccone. (9.2) Corollary.IfCandC'arek-rationalpolyhedriccones,thenN finitely manyk-rationalsimplicialcones,thensoisP. As P=NU; cone inPhasnonemptyintersectionwith8j,thenitiscontaineda1. For everyj=1,...,in,letajP(xEPIMM(x)0).Ifasimplicial is adisjointunionoffinitenumberk-rationalsimplicialconesand(0}. 506 C'=(XEVI LI(x)=...=L,,-d(x)=0,MI(x)>0,.,Md(x)>0}. It isarareandspecialeventifnewsubstantialinsight addedtothe C, =(XECI C1= xEC R*+)=((xt)ER*Ixt>0 forallreal-r . . 1 , L,-d,M1,---,Mdsuchthat ... , BjP, weseethatsinceP ... , Li(x) _...=Ln_d(x)0, M1(x) >0,...,Mj_1(x)0,Mj(x)<0 d, Cn o+ =onR1+1 CDR d Ct A ... ,Cn-d' Chapter VII.ZetaFunctionsandL-series `+a C1, ...,Cd.By(9.1),these ;40 (0) isadisjointunionof ±Li(x)>0}, cot bra '-e D°, 10. C1, ...,C,"inR'+1suchthat converge to1asp"N(zv)=N(zv),i.e.,x(limsv)x.This wouldmean PuZv =Evzu,Pu),v/,k.zandz'wouldconvergetox ;nowpvwould z,,, zvEUi/v(x),andEvE, If not,thenwewouldfindsequencesA,z,,,A;,z'vEC11,,, A,,,AuER+, a betheclosureof45inH,andputF=tSince. (2) that limsv=1,whichisimpossible,sinceEdiscretein R. small sothat form y=AzwithAER+andZU5(x).Wemaynowchoose Ssufficiently -rational simplicialconeinR*)withxECa,andeveryyCSisofthe even choosethevitolieinKnU8(x).ThenCS=C(v1,...,v")isa with tj>0.SinceKisdenseinRbytheapproximationtheorem,wemay is clearlyabasisv1.....u,,EU&)ofRsuchthatx=t1v1+---+t"v" Let XEFandUs(x)={yRI closed, soisF.Itthereforecompact,andwehave onto mapping unit theorem,E(beingasubgroupoffiniteindexino*)ismappedbythe positive scalarelement.Indeed,x=IN(x)I'1"(x/I'"").ByDirichlet's Every XER{+)isinauniquewaytheproductofanelementSand (1) { xE[ftR]+I Proof: WeconsiderinR*+)thenorm-onehypersurface (9.3) Shintani'sUnitTheorem.IfEisasubgroupoffiniteindexino+, then thereexistsaQ-rationalpolyhedricconePsuchthat § 9.ValuesofDirichletL-seriesatIntegerPoints F beingcompact,wethusfindafinitenumberofQ-rational cones m $I-' a complete f :S-.[fR]+, C3nEC5=O forallEEE,s¢l. Tr(x) =0} Ri+)= UsP lattice S=}xER*(+) IIN(x)I=1}. EEE r F =U(C;nF) . y'' S =UsF. (1. r Let 0beafundamentalmeshofr,let i=1 IIx -yII<81(::R*),S>0.Thenthere m EEE of 1, suchthatAvzv= (xr) h--> (disjoint union). the trace-zero log Ixrl) , is boundedand space and thus H 507 andCinEC1=0forallseE,a 508 we obtain 0-rational polyhedriccone.ObservingthatCinsCi=0foreEE,s EC1 andCiaredisjointforalmostallsEE.Hence,by(9.2),C In ordertoturnthisunionintoadisjointone,weputC(])=C1and and (2),wededucethat (ii) (i) Ci following properties: -rational polyhedricconesCi°), and ECil)nC;1)=0forallsEEandi=2,...,m. Cim-t) which enjoysproperties(i),(ii),and(iii)withv+1insteadof v. Consequently, Then Ci°+]) We putCt°+1)=C,")fori c Ci, 5'E' -t3 i=1 LEE , Cmm-1) C` v+1)=Ci) R(*+) =U , CP) =Ci M `'. is asystemofQ-rationalpolyhedricconessuchthat 1=1 SEE is afinitesystemofQ-rationalpolyhedriccones Rw Rc+)=U UsCi. L(X,s) _ N UsC1, * =U LEE E& 1) LEE = U ECv+1 i=1EEE .. ,C,v=1,...m-2 i=1 LEE m Chapter VII.ZetaFunctionsandL-series a 1, andalli=1,...,m.From(1) sp(a)s sCi X(a) i =2,...,m. (disjoint union). for c]) i >v+2. j. satisfying the is a El then wehave the associatedDirichletL-series.If.variesoverclassesofJ°t/P'n § 9.ValuesofDirichletL-seriesatIntegerPoints in 1+am.Thegroup Let .beafixedclass,andanintegralidealin.t.Furthermorelet with thepartialzetafunctions It integral. ThereforeaHasgivesusmapping in A.Furthermore,wehaveasCall+a-1m)=amo,sothatis m arerelativelyprime,wegeta-1Em,i.e.,(a)P().Henceaslies Proof: LetaE(1+1m)+.Thenwehave(a-i)aCm,andsince onto theset.dintofintegralidealsinA. (9.4) Lemma.Thereisabijection acts on(1+a1m)+,andwehavethe (1 +a-1m)+_a-im)flR*+)bethesetofalltotallypositiveelements belong tothesameclassunderactionofE. that EE,i.e.,aandbhaveexactlythesameimageif and onlyifthey if (a)=(b),sothatabewithEo*.Since(1+ lm)+, itfollows a E(1+a-1m)+.Fora,blm)+,wehaveas= baifandonly (a -1)aCmam,sothataE1+1m,andalso ER*),andso (A, s) The lemmaimpliesthefollowingformulaforpartial zetafunction is surjective, forifaa,aEP',isanintegralidealin A,then E=o+= IsE0*1Elmodm,eER*I) (1 +a1m)+/E-4.dint, L(X,s) _Xs) Q.. (1 +a-'M)+- '.' s) = M(a)S .-. a integral it 1 aESt Y, aE g(a)s Aint IN(a)is 1 a 1 aa, 509 every aE(1+-1m)flCicanbewrittenuniquelyintheform we haveExevie1+a with rationalnumbersye>0.Putting by. by 1.Itisthereforefinite.SinceCiCR*)thesimplicial conegenerated Proof :R(.fi,Ci)isaboundedsubsetofthelattice1minR,translated Then wehavethe w herz=(z1,...,zd,)variesoveralldi-tuplesofnonnegativeintegers. with thezetafunctions (9.5) Proposition.ThesetsR(.ft,C1)arefinite,andonehas and positive integer,wemayassumethatallvielieinm.Let system of.generatorsofCi.Multiplyingifnecessarybyaconvenienttotally cones Ci.Foreveryi=1,...m,letv be adisjointdecompositionofR*)intofinitelymanyQ-rationalsimplicial we nowapplyShintani'sunittheorem.Let where 91runsthroughasystemofrepresentatives(1+a-1m)+/E.Tothis 510 vi 1, . . . vid, Em,everya(1+1m)flCicanbewrittenuniquely as (Ci,x,s) Ye=xe+zt, 0 n coo A w(a sp(a)s ajiti X(a) 1 x,.) ofpositiverealnumbers,wewritethe Zl,.., Zr-oi=1 ... ti-Iti+l ... ,m,xthroughtheelementsofR(A't,Ci), i=1 EEE 00 i=1 xCR(SC,) 1 and Y- n i=1 m i=1 xER(F,C,) m > Bk(A,x)('), n L!(z1, H L;(z+x)-s. n F- ('(Ci,x,s) ... ,zr) tn)k-1 j=1 E ajizj r 511 and A = a, and A = ... dtn, 0, and observing that `N. tn)S-1 dt1 q=1 Bk(A,x) (zi +xj)Ll(t), x,.), is easily proved. (-1)n(k-t)+rBk(A,x), Chapter VII. Zeta Functions and L-series and Functions VII. Zeta Chapter ,j=1 ... , Zr), zi E Z, zi > fl e-k (t1 ... i=1 tiLi(z+x)](t,...tn)s-1dtl...dtn. . . . , 1 - +x)-S exp(uxjLj (t)) exp(uxjLj used +s by the same arguments that we have n exp(uLi (t)) - 1 exp(uLi (t)) , 00 ti -- L; (z + x)t,, 0 i=1 '' FI ... i 00 ttL; (z +x) = ... , t,-1, ti+1, ... , t, . For r = n = 1 ... , t, . For r ... , t,-1, ti+1, f f n 0 , n i=] n Bk(A, 1 -x) _ JT L (z i=1 w 0 j (s)n = 00 0 r(s)n Summing this over all z = (z1, we substitute and obtain where 1 - x signifies (1 - X 1 where 1 - x signifies forRe(s) > series i; (A, x, s) is absolutely convergent (9.7) Proposition. The complex plane. continued to the whole r/n, and it can be meromorphically s = 1 - k, k = 1, 2, ..., are given by Its values at the points Re(s) > r/n is deduced from the Proof: The absolute convergence for convergence of a series En t remainder of the proof is similar repeatedly. It will be left to the reader. The to that of (1.8). In the gamma function 512 function at 0 of the Laurent expansion in the u, t 1 in the variables Bk(x) polynomial with the usual Bernoulli = ak-1Bk(x), we have Bk(a,x) The equation (see § 1, exercise 2). We cutupthespaceRnintosubsets with thefunction yields theequation § 9.ValuesofDirichletL-seriesatIntegerPoints For 0 (A,x, 1-k)_(-1)n(k-l)k-nEBk(A,-x)(i) (Ci,x,s) r(k)" A(s) n 'r. (-1)n(k-1)n ((-l)n(k-1)F(k)"1n) _ (22ri) (27ri)" at s=1-k.Thusexpression(2)turnsinto 1 1 rlN(x +z1vil Z R identifiestheMinkowskispacewithI8", n Kf f f[g(uy)um(1__1(flyt)_kfldy]du, (ebrins -1)(e2Tris1)-1enyris Ki-1 times thecoefficientof E ..+ Chapter VII.ZetaFunctionsandL-series Bk(A, 1-x) r(1 -s)" j=1 .°; ... ,ti_1ti+1,tointhedirect Tr7 n i=1 kn +... +zdiVidi)I-s exp(u(1 n exp(uLj(Y)) (-1)n(k-1)+rBk(A,x) mentioned , . -xj)Lj(Y)) . , xn)withpositive ti. u"(k-1) (fe#iye)k-i - 1 ... IfK 50] ,^i yt=l (9.9) Corollary(SIEGEL-KL!NGEN). Thevaluesofthepartialzetafunction s=-1,-2, -3,... If thenumberfieldKisnottotallyreal,thenwehave (s)=0for,all we deducethatK(1-k)=0foroddk>1,anditis# forevenk>1. rational numbers.Fromthefunctionalequation(5.11), generated bythevaluesofcharacterXf.TheK (1-k)areeven for k>1,arealgebraicnumberswhichalllieinthecyclotomic fieldQ(Xf) zeta functionK(s).Thetheoremsaysinparticularthatthevalues L(X,1=k), Here aisanintegralidealintheclass.fiofJ'/P' points s=1-k,k1,2,3,... and thevaluesofDirichletL-seriesL(X,s)aregivenby (9.8) Theorem.Thevaluesofthepartialzetafunction(.fi,s)atintegral and, from(9.5)and(9.6),weobtainbyputtings=1-kthe For XE][8+,wethereforeget component ofz1vi+ then Ai=(ask)isa(dixn)-matrixwithpositiveentries,andthek-th If § 9.ValuesofDirichletL-seriesatIntegerPoints (A, s)atthepoints,s=0,-1, -2,...arerationalnumbers. BCD This resultabouttheDirichletL-seriesL(X,s)alsocovers the K(1 -s)=IdKIs_h12(cos L(X, 1-k) (.h, I-k)=9q(a)k_1 (Ci,x,s) _ Lk(zl, ..,zd,)= * + zd1vidibecomes z k=1 11 Lk(z1, - , 7rS are givenby 2) X(a)01(a)k_1 [(-i)d l r1+r2 i=1 m ...,zdi)-S =(Ai,x,s), l=1 d; j=1,...,d1 -1)d' (Sin aikZi O .ti xeR(..,C;) 7rS y{.,, Bk A 2)r2 .ate ( k" Bk (Ai,x) 1 x) k Obi Dedekind I `mix Now wehadBk(A,x)=nF1Bk(A,x)iji,whereBk(A,x)ljlwasthe indices {1,2,...,n}sothat closure ofKIQandCrEG(LIQ).Thenainducespermutationthe rational numbersx=(x1,...,xr).Toseethis,letLIQbethenormal enough toshowthatBk(A,x)isarationalnumberforeveryr-tupleof according tothechosennumberingofembeddingsr matrix (a1),whereajiisthei-thcomponentofaj,afteridentifying12=l[l;° Proof: Leta1,...,arbenonzeronumbersinK,andletAthe(rxn)- 516 found increasinginterest.Likeintheclassnumberformula,whichexpresses under theactionofGaloisgroupG(LIQ),andthusbelongstoQ. in LandthataBk(A,x)(j)=Bk(A,x)(Q(j)).ThereforeBk(A,x)isinvariant with Lj(t)=a1t1+ of thefunction coefficient of have mentionedbefore.Thep-adiczetafunctionofatotally realnumber and PIERETTECASSOU-NOGUEShaveusedSHINmw'sresultto prove theexistence L -serieswhichwehaveencounteredmayprovetobeimportant. of numbertheory.Onthewaytowardsit,insightsinto thenatureof (see [99],[12]).Theproofofthisconjectureisagreat,ifstill remote,goal the LichtenbaumconjectureasEulercharacteristicsinetale cohomology by astrikinglysimplegeometricinterpretation:theyappearaccordingto STEPHEN LICHTENBAUM,thesignificanceoftheseL-valuescanbeexplained to "motives".AccordingaconjectureoftheAmericanmathematician to extendanextremelywideclassofL-series,theL-seriesattached of allthespecialvaluesindicateadeeparithmeticlawwhichappears the behaviourofDedekindzetafunctionatpoints=0,properties which isrelatedtotheordinary DedekindzetafunctionK(s)by field Kisacontinuousfunction of p-adicL-series.TheseplayamajorroleinIwasawatheory, whichwe The natureofthespecialvaluesL-seriesatintegerpointshasrecently Finally wewanttomentionthattheFrenchmathematiciansDANIEL, BARSKY r-. aajj =aj(j) u'a(k-t)+r p(-n) =K(-n)Fl(1-9(p)n) (t1, ...,t1_1,tj+1,tn)k-1intheTaylorexpansion U r + aint,,.ThismakesitclearthatBk(A,x)(j)lies j=1 exp(uLj(t)-1) p :ZP\{1}-)QP, r exp(xjuLj(t)) Chapter VII.ZetaFunctionsandL-series : !S pip 'r3 -.8 , : K-+R.Itis § 10. Artin L-series 517 for all n E N such that -n = 1 mod d, where d = [K(µ2p) : K] denotes the degree of the field K(92p) of 2p-th roots of unity over K. The p-adic zeta function is uniquely determined by this relation. Its existence hinges on the fact that the rational values K (-n) are subjected to severe congruences with respect to p. § 10. Artin L-series So far, all L-series we have considered were associated to an individual number field K. With the Artin L -series, a new type of L -series enters the stage; these are derived from representations of the Galois group G(LIK) of a Galois extension L I K. This new kind of L -series is intimately related to the old ones via the main theorem of class field theory. In this way they appear as far-reaching generalizations of the old L-series. Let us explain this for the case of a Dirichlet L -series L(X X (n) s) = ns= 1 - X (p)p-s attached to a Dirichlet character X : (7L/m7L)* ) C*. Let G = G(Q(gm)IQ) be the Galois group of the field Q(ii,,,) of m-th roots of unity. The main theorem of class field theory in this particular case simply describes the familiar isomorphism (7L/m7L)* --* G, which sends the residue class p mod m of a prime number p fi in to the Frobenius automorphism (pp, which in turn is defined by cpp = p for E Un,. Using this isomorphism we may interpret X as a character of the Galois group G, or in other words, as a 1-dimensional representation of G, i.e., a homomorphism X:G) GLI(C). This interpretation describes the Dirichlet L-series in a purely Galois- theoretic fashion, L(XI s)=17 ptm 1 - X (cOp)p-s and allows us the following generalization. are rootsofunitybecausethe endomorphism qpTofVIPhasfiniteorder. L-series (see(8.1)),observingthatthee;infactorization half-plane Re(s)>1.Thisisshowninthesamewayas fortheHecke Re(s) >1+8,forany80.Itthusdefinesananalytic functiononthe coq where prunsthroughallprimeidealsofK. L-series attachedtopisdefinedbe (10.1) Definition.LetLJKbeaGaloisextensionofalgebraicnumberfields conjugates. Wethusarriveatthefollowing and IV,theFrobeniusautomorphismsVTrpT,aresimultaneous to cps,asthedecompositiongroupsGpandGp,,inertiaIT above p.Infact,adifferentchoice only dependsontheprimeidealp,notchoiceofidealT x i-+x`I,whereq=01(p).rppisanendomorphismofthemoduleVIP with GaloisgroupG,andlet(p,V)bearepresentationofG.ThentheArtin invariants. Thecharacteristicpolynomial automorphism VTwhoseimageinG(K()3)IK(p))istheq-thpowermap (9.5)). ThefactorgroupGW/IpisthereforegeneratedbytheFrobenius onto theGaloisgroupofresiduefieldextensionK(93)I(p)(seechap.I, IT theinertiagroupof93overp.Thenwehaveacanonicalisomorphism a primeidealofLlyingabovep.LetGpbethedecompositiongroupand the completeexpressionp(a)v.LetpbeaprimeidealofK,andletTIp Our shorthandnotationfortheactionofaeGonvVisav,instead finite dimensionalG-vectorspaceV,i.e.,ahomomorphism Galois groupG=(LIK).Arepresentationofisanactionona 518 The ArtinL-seriesconvergesabsolutelyanduniformlyin thehalf-plane Let LIKbeaGaloisextensionoffinitealgebraicnumberfieldswith det(1 -'PTFqq(p)-S;VIA')_ H(1-Ei`n(p)-S) ow" L(L IK,p,s)=r( X03 p:G -GL(V)=Autc(V). GTIIp -24G(K(q3)IK(p)) det(l -cpet;VIT) p -.0'd 0-.6 COD l'Ip yieldsanendomorphismconjugate Chapter VH.ZetaFunctionsandL-series i=1 d CAN 1 C7° VIP) ^'b If p The specialimportanceofcharacterscomesfromthefollowing fact: that f(vto--t)=(t)iscalledacentralfunction(orclass function) all One hasXp(1)=dimVdegree(p),andXp(Qty-1) =Xp(t)for where p,,variesoverallnon-equivalentirreduciblerepresentations ofG. then raiscalledthemultiplicityofpainp,andonewrites equivalent topreciselyramongtherepresentationsinthisdecomposition; of irreduciblerepresentations.Ifanrepresentation(p"',V«)is a directsum modules VandV'areisomorphic.Everyrepresentation(p,V)factorsinto Two representations(p,V)and(p',V')arecalledequivalentiftheG abelian groupissimplyacharacter admit anyproperG-invariantsubspace.Anirreduciblerepresentationofan, of V.TherepresentationiscalledirreducibleiftheG-moduleVdoesnot For theirproofswereferto[125]. behaviour underchangeofextensionsLIKandrepresentationsp.This study, wefirstcollectbasicfactsfromrepresentationtheoryoffinitegroups. allows todeducemanyoftheirexcellentproperties.Asapreparationforthis does notexistforgeneralArtinL-series.Buttheyexhibitaperfectlyregular to theexpansion simply theDedekindzetafunctionK(s).Anadditiveexpressionanalogous; § 10.ArtinL-series Two representationsareequivalentifandonlytheircharacters areequal!. The characterofarepresentation(p,V)isbydefinitionthe function The degreeofarepresentation(p,V)finitegroupGisthedimension For thetrivialrepresentation(p,C),p(o)=1,ArtinL-seriesis! o', reG.Ingeneral,afunctionf:G-+Cwiththe F a r,,,Pa, then Xp:G -C,Xp(a)=tracep(Q). "3- p:GC*=GLj(C). ... V=Vm®...®VS WS) XpraXp.. P-Er«P., .O+ a a `n(a)s property 'J" 519 520 Chapter VII. Zeta Functions and L-series The character of the trivial representation p : G -* GL(V), dim V = 1, p(o) = 1 for all a E G, is the constant function of value 1, and is denoted by 1G, or simply 1. The regular representation is given by the G-module V=C[G]={ F_ xtrI xt (=- C}, tEG on which the v E G act via multiplication on the left. It decomposes into the direct sum of the trivial representation Vo = CLJEG or,and the augmentation representation { ElEG X1 0'I & xa = 0}.The character associated with the regular, resp. the augmentation representation, is denoted by rG, resp. uG. We thus have rG = uG + 1G, and explicitly: rG(a) = 0 form; 1, rG (1) = g = #G. A character X is called irreducible ifit belongs to an irreducible representation. Every central function p can be written uniquely as a linear combination rP=>c5X,cx EC, of irreducible characters.is a character of a representation of G if and only if the cx are rational integers > 0. For instance, for the character rG of the regular representation we find rG =EX(1)X, where X varies over all irreducible characters of G. Given any two central functions rP and >/r of G, we put 1E 9 (a)(a) , g = #G, g aEG where is the function which is the complex conjugate of rfr. For two irreducible characters X and X', this gives 1, if X = X" i 0, ifX X' In other words, (,) is a hermitian scalar product on the space of all central functions on G, and the irreducible characters form an orthonormal basis of this hermitian space. For the representations itself,this scalar product has the following meaning. Let V=Vi ED ...ED Vr be the decomposition of a representation V with character X into the direct sum of irreducible representations V. If V' is an irreducible representation with character X', then (X, X') is the number of times that V' occurs 521 is the .ti for all rEH}, X(a), g=#G. #N t g QEG *(a)_F*(r)- VG = 1 I (co,h*(ik)) = dim of finite groups. If cp is a central G be a homomorphism (x, X') _ (X1,X')++(Xr,X'), (x, X') _ IndG(V)={f:G-*VI f(ax)=rf(x) We then write 'p instead of h*(co), and *q instead of h*(fl. One has We then write 'p instead of h*(co), and The following result is of great importance. Now let h : H two special cases. This will be applied chiefly to the following instead of h* ((p), and ** instead In this case we write rp I H or simply 'p If 'p is the character of a representation (p, V) of G, then If 'p is the character of a representation character of the representation (p o h, V). (10.2) Frobenius Reciprocity. For every central function i/r on H there is For every central function (10.2) Frobenius Reciprocity. has function h*(ik) on G such that one one and only one central 'p on G. for all central functions a) H is a subgroup of G and h is inclusion. is the character of a representation of h*(*) (the induced function). If rp of the representation (pIH,V). If * (p, V) of G, then 'pIH is the character V) of H, then i/r* is the character of is the character of a representation (p, by the induced G-module the representation (ind(p), IndG (V)) given (see chap. IV, §7). One has on which Cr E G acts by (af)(x) = f (xa) on the right of G/H, and we where r varies over a system of representatives put i/'(rar-1) = 0 if tar-1 0 H. h is the projection. b) G is a quotient group H/N of H and § 10. Artin L-series Artin § 10. of V1, then is the character For if Xi up to isomorphism. the Vi, among ++Xr, so that X = X1 V1 is or is not isomorphic depending whether (Xi, X') = 1 or 0, and we have in V' = C, we obtain trivial representation this to the to W. Applying that particular H. Conversely, h*('p) = rp o h is a central function on function on G, then proposition. one has the following (p', V') are G p/IT If (p, V), , X', then the direct sum 1 (ii) cud , each lying above the next. Let X - p(rPT)9T(p)-s; V113) Chapter VII. Zeta Functions and L-series and Functions VII. Zeta Chapter earlier. det(1 p f"+ L(LIK,1,s) = K(s). s) = 1-7 , L(L'IK,X,s) =L(LIK,X,s) X L(L I M, X , s) = L(L I K, X*, s) , L(LIK L(LIK,X +X',s) =L(LIK,X,s)L(LIK,X',s) gab Gcp, ---+ Gt, IT, -+ 1T, G j,/1 , If M is an intermediate field, L 3 M 3 K, and X is a character If M is an intermediate field, L 3 M If X, X' are two characters of G(L I K), then If X, X' are two characters of G(L I K), Note that a character of degree 1 of a group H is simply a homomorphism 1 of a group H is simply character of degree Note that a groups, we of representation theory for finite After this brief survey V1T)det(1-rppt; Vi1q'). det(1-rpc t; (V ®V')'T)=det(1-rp,pt; 522 - G is a Z group X of a finite character Every Theorem. Brauer's (10.3) 1 characters X; of degree Xi* induced from of characters linear combination G. to subgroups H; of associated X:H-*C*. V) and (p', V') L-series. Since two representations (p, now return to Artin we will only if their characters X and X' coincide, are equivalent if and henceforth write instead of L(L I K, p, s). These L-series exhibit the following functorial p, s). These L-series exhibit the following instead of L(L I K, behaviour. character X = 1, one has (10.4) Proposition. (i) For the principal (ii) K, L' D L 3 K, and a character X of (iii) For a bigger Galois extension L' I G(L I K) one has (iv) ofG(LIM), then character X + X', and (p ® p', V ® V') is a representation with Proof: We have already noted (i) X representations of G (L I K) with characters This yields (ii). I K (iii) Let q3' I3 I p be prime ideals of L' I L -module V. G (L' I K) acts on V via be the character belonging to the G (L I K) the projection G(L'IK) -+ G(LIK). It induces surjective homomorphisms It isthereforesufficienttoprove that G i-module.Hence Putting Vi=®jaijriW,weobtainadecompositionV ®iViofVasa G modH.Wethushave(seechap.IV,§5,p.297)thedecomposition immediately thatthen(Ti)isasystemofrepresentatives ontheleftof representatives ontheleft,aij,ofGImodG,nriHrt i.Onechecks and fi=(GI we obtain We reducetheproblemtocaseGI=G,i.e.,r1.Conjugatingbyii, Then X,,isthecharacterofinducedrepresentationind(p):G (pTr EGi/Ii,andtheimageofP}'inHiisFrobeniusof'p3ioverqi. Frobenius VT,EGI/11.Thenrpi=rillprGiismappedtothe and Ii=r['Ilr.LetPEG1beanelementwhichismappedtothe We chooseelementsTiEGsuchthat= is fi=(Gi:Hili),i.e., decomposition, resp.inertia,groupsof93ioverqi.Thedegreeqioverjp inertia, groupofTioverp.ThenHi=GinH,resp.1ilIiarethe of Laboveqi,i=1,...,r.LetGi,resp.Ii,bethedecomposition, V =IndG(W).Clearly,whatwehavetoshowisthat q..., (iv) LetG=G(LIK)andHG(LOM).pbeaprimeidealofK, This yields(iii). (rpp, VIT),i.e., automorphism rpgytotheFrobeniusrpcpsothat(rpp',V of thedecompositionandinertiagroups.ThelattermapsFrobenius § 10.ArtinL-series 1, Now letp:H-->GL(W)bearepresentationofwithcharacterX. q, thevariousprimeidealsofMabovep,and43ia det(1 -of'tfi;WI+)=gpfitfi;(t det(l det(1-rpt;VI')= rdet(1-cpf'tI'';WIi). : (GInrHri-1)Ii).Foreveryiwechooseasystemof det(1 -VT,t;V11y)=-(pt,V'T). - apt;Vi det(1 -ppt;VI')=jIVI'). 1')=det(l-cpfitfi; (riW)Iinr;1 Jt(gi) =`n(p)fi V =®aijriW. i,j 0"0 i=1 i=1 Ti'. ThenGi=r1Gizi abby y9- C". C'J' W)I;nr,Hr i') ') GL(V), 1,4y) ideal 523 i , it isconstantontheright(andthereforealsoleft)cosetsofGmad1, under Iifandonlyonehasf(xr)=(x)forallrEI,i.e., then V1=IndG(WtrH).Indeed,afunctionf:G We simplifythenotationbyreplacingG1G,IlI,flrjHr,-1 524 is thematrixofowithrespecttobasis{cp'wl)V.Thisgives denotes the(dxd)unitmatrix,then Let AbethematrixofcpfwithrespecttoabasisW W11H, becauserf(x) i.e., ifandonlyitisafunctiononG.Itthenautomaticallytakesvaluesin V =IndH(W),i.e.,wearereducedtothecaser1,GG. H, fibyf=(G:HI),V,andr;WW.Thenwehavestill of G(LIK).Wethereforededucefrom(10.4)the by tandaddingittothethird,etc. by tandaddingittothesecond,thenmultiplying secondcolumn as desired.Thelastidentityisobtainedbyfirstmultiplying thefirstcolumn where Xvariesoverthenontrivial irreduciblecharactersofG(LIK). (10.5) Corollary.Onehas {1) cG(LJK)isthecharacterrG=ExX(1)Xofregular representation det(l -cpt;V)=det We mayfurtherassumethat1=1.ForifweputGG/I,HH/IAH, The character1,,inducedfromthetrivial1of thesubgroup So letI=1.ThenGisgeneratedbyrp,f(G:H),andthus CL(s) =OK(s)IIL(LIK,x,s)x(l), = f(rx)(x)forrEIflH. 1E -tA 0 BCD l0 A u V=®coW. -tE L; x01 0 0 0 0 f-1 i=O Chapter VII.ZetaFunctionsandL-series C17 ...... E 0 U -tE E 0 Cry I 1 , W inVisinvariant Wd ofW.IfE W) G(L IK,X,s)definesanentirefunction. Artin Conjecture:ForeveryirreduciblecharacterX01,theL-series Corollary (10.5)showsthatthiscouldbededucedfromthefamous an entirefunction,i.e.,aholomorphicfunctiononthewholecomplexplane. question whether,foraGaloisextensionLIK,thequotientL(s)/K(s)is § 10.ArtinL-series prime tof,andPfisthegroupofprincipalideals(a)such thata=1modf from therayclassgroupJf/Pf.HereJrisoffractional ideals Artin symbol(LaK such thatLIKliesintherayclassfieldf(seechap.VI,(6.2)).The i.e., thesmallestmodule theory. LetLIKbeanabelianextension,andletftheconductorofK, case. case. Viafunctoriality(10.4)theymaythenbeextendedtothenon-abelian particular theirfunctionalequation,transfertoArtinseriesintheabelian Dirichlet L-series.ThismeansthatthepropertiesofHecke'sseries,andin L IKcoincidewithcertainHecke-series,moreprecisely,generalized one ofthebigchallengesinnumbertheory. general itisnotknown.Inviewofitsmomentousconsequences,constitutes class groupJf/Pf,i.e.,aDirichletcharactermodf.Itinduces acharacter Composing withtheArtinsymbol(&1 a homomorphism and aispositiveinKv=IRifpreal. and wehavethe By (6.9),thischaracteronideals isaGr68encharactermodfoftype(p,0), on Jf,whichwedenoteby The linkbetweenArtinandHeckeL-seriesisprovidedbyclassfield We willshownextthattheArtinL-seriesjr'caseofabelianextensions We willseepresentlythatthisconjectureholdsforabelianextensions.In The startingpointofArtin'sinvestigationsonL-serieshadbeenthe Now letXbeanirreduciblecharacteroftheabeliangroup G(LIK),i.e., Jf/P1 ` -± then givesusasurjectivehomomorphism G(LIK), f =HP°r Pt- a modPfr--> ) , thisgivesacharacteroftheray (LK a J III 525 526 Chapter VII. Zeta Functions and L-series (10.6) Theorem. Let L I K be an abelian extension, let f be the conductor of L I K, let X 1 be an irreducible character of G(L I K), and X the associated Grofiencharakter mod f. Then the Artin L -series for the character X and the Hecke L -.series for the Gr6f3encharakter X satisfy the identity 1 ,C(LIK,X,s)=PS1-X(w) t(P) -SL(X,s), where S={plfI X(Ip) = 11. Proof: The representation of G (L I K) associated to the character X is given by a 1-dimensional vector space V = C on which G(L I K) acts via multiplication by X, i.e., o v = X (a) u. Since f is the conductor of L I K, we find by chap. VI, (6.6), that p I f p is ramified IT : 1. If X (IT) 0 1, then V IT = {01, and the corresponding Euler factor does not occur in the Artin L-series. If on the other hand X(IT) = 1, then V 1 P = C, so that (p)-S det(1 - capO(p)-S ; V t ') = 1 - X . We thus have G(LIK,X,s)=fl 1 1 1 - X P11 - X ((P YR(P) pl s (p)-:, and 1 L(X s)=Tj Ptf 1 - X (PYR(p)-S For p { f, one has( ) = qpp, and so Y (p) = X (VT). This proves the P claim. Remark: If the character X : G(LIK) -)- C* is injective, then S == 0, and one has complete equality G(LIK, X, s) = L(X,s). In this case X is a primitive Grofiencharakter mod f. If on the other hand X is the trivial character 1G, then X is the trivial Dirichlet character mod f, and we have OK(s) = fl 1 L(X,s) Plf 1 - 92(p)-S every solvableextensionLIK. for L(X,s),aswasshownin(8.5).ThisalsosettlestheArtinconjecture Hence G(LIK,X;s)isholomorphiconallofC,becausethesametrue then theaboveremarkshowsthatG(LIK,X,s)=G(LxL(X,s). of XandistheGrbj3encharakterassociatedwith:G(LxIK)-4C series G(LIK,X,s)whichcorrespondtonontrivialirreduciblecharactersX § 11.TheArtinConductor product decomposition fields admitsafinestructurebasedongrouptheory.Itisexpressedby Galois-theoretic complementwhichwillbedealtwithinthenextsection. Hecke L-series.Butinordertogothewholeway,weneedanadditional for theseEulerfactors,thefirstnaturalguidelineisprovidedbycaseof Artin L-seriesbytheright"Eulerfactors"atinfiniteplaces.Inlooking already establishedforHeckeL-series.Wehoweverhavetocompletethe basis forthiswillbetheabovetheoremandfunctionalequationwehave of abelianGaloisgroupsG(LIK).ForifLxisthefixedfieldkernel needed forthis,followingessentially thetreatmentgivenbyJ.-P.SERRE play animportantroleinthefunctionalequationofArtin L-series,which ARTIN andHELMUTHASSE.Theidealsf(X)arecalledArtin conductors.They group ofLTIKp,andg;denotesitsorder.Thisdiscoverygoes backtoEMiL where VisarepresentationwithcharacterX,G1thei -thramification with G =G(LIK).Theidealsf(X)aregivenby where XvariesovertheirreduciblecharactersofGaloisgroup in [122]. we aregoingtoproveinthe nextsection.Herewecollecttheproperties Our goalnowistoproveafunctionalequationforArtinL-series.The The theoremimpliesthattheArtinconjectureholdsforallL- The discriminant-0=1LIKofaGaloisextensionLIKalgebraicnumber f/1 § 11.TheArtinConductor fp(X) f(X) =II Z) =IIf(x)x('), i>0 90 gt pf- codim VG', pfc(x) 527 , character ofGi,and(u1)#theGinducedfrom ui.Thenonehas (iii) LetG;bethei-thramificationgroupofG,ui DK'IK =p',then (ii) preceding section). o.4 have shownthis,wemayformtheidealfp(X)=pf(x),whichwillbe We maythereforewrite (11.1) Proposition.(i)IfHisanormalsubgroupofG,then the functionaGsatisfiesfollowingproperties(weusenotationof p-component oftheglobalArtinconductorthatwewant.Firstprove is toprovethatthecoefficientsf(X)arerationalintegers>0.Oncewe with XvaryingovertheirreduciblecharactersofG.Ourchiefproblem aG isacentralfunctiononG,andwehave H CG.IfLIKisunramified,theniG(a)=0forallac=G, One hasiG(rat'1)=iG(o-),andiH(a)iG(a)foreverysubgroup valuation ofL.Withthisnotationwecanwritethei-thramificationgroupas where xisanelementsuchthatoL=OK[x],andvLthenormalized In chap.II,§10,wedefined,foranyaEG, group G=G(LIK).LetffLIK[X:K]betheinertiadegreeofLIK. 528 First letusconsideraGaloisextensionL1Koflocalfields,with If HisanysubgroupofG,andifK'thefixedfieldwith discriminant .°' aG(a)= G1=[aEGI iG(a)>i+l}. (aG,1G) = aG =>2f(X)X, aG IH=vrH+fK'KaH. aG =E iG(a) =VL(aX-X), IfF-,,,IiG(7r) x aG/H =(aG)4 i=o (GoGt) O° fiG(a) #G 1 E QeG ti, Chapter VII.ZetaFunctionsandL-series f(X) EC, aG(a) =0. (ui)x force=l. fora (I" III .-4 1, the augmentation 1. Weput and g =#G.ForanycentralfunctionpofG,weput we have,inviewofaG(a-1)=aG(a),that orthogonal to1G. This impliestheidentityforcaseu=1aswell,sincebothsidesare if aVGi,and(ur').(u)=-g/gi-f.go/giEGor01, (iii) Letgi=#Gi,g#G.SinceGiisinvariantinG,wehave(ui).(a)0 Thus rH(1)_[L:K]andv=VK(tK'IK)yieldstheformula furthermore that and inthesamewayall(l)=VK,(-0LIK').Fromchap.III,(2.10),weget f(X) _(aG,X)= EaeG(L[i)*(a) =0.ForaEGkNGk+1,wethusfind SO VKoNLIK=fLIKVLgivestheidentity By chap.111,(2.9),weknow,ontheotherhand,thatDLIK=NLIKPLIK), then generatedbyg(x)=r[o#1(ax-x).Consequently, g(X) betheminimalpolynomialofxoverK.Bychap.III,(2.4),OLIKis Now letor=1,and',0LIKbethedifferentofLIK.LetoL0K[x] Since 1G(9)=iH(a)andfLIKfLIK'fK'IK,thisimplies (ii) LetaeH,;1.Then Proof: (i)followsimmediatelyfromchap.II,(10.5). § 11.TheArtinConductor aG(1) =[L:K']VK(DK'IK)+fK'IKVK'(DLIK')vrH(l)fK'IKaH(1) For thecoefficientsf(X)inlinearcombination aG(Q) =-fLIKiG(a), ULMLIK) =vL(g'(x)) aG(a)=-f(k+1)=F aG(1) =fLIKVLMLIK)VK(ZJLIK), 9(Gi) = g oEG 1 0LIK = (7) aG (a)=VrH+fK'IKaH aG(a)X(a-') =i gi QEG1 - aG=Y- f(X)X, 1 f aH(a) _-fLIK'iH(a), qp(a), (DK'IK)[L:K')NK'IK(0LIK') ' aG) (Go=Gi) iG(a) = g aEG gi =#Gi. 1 fLIK aG(a-1)X(a) aG(1) rH(a) =0. = (X,aG), 529 `ti ;;~ function inducedbyTonG,then (ii) and cp'isthecorrespondingcentralfunctiononG,then (11.2) Proposition.(i)IfrpisacentralfunctiononthequotientgroupG/H, 530 The theoremofHASSE-ARF(seechap.V,(6.3))nowgives usthefollowing which wasintroducedinchap.II,§10.Forintegersm> -1, itisgivenby Now considerthefunction and X(G;)=dimVG;,hence from (11.1),(iii). (iii) Wehave(cp,(u;),)=((pIG;,ul) (ii) Proof: (iii) ForacentralfunctionrponG,onehas and thisisarationalinteger> 0. (11.3) Proposition.LetXbeacharacterofGdegree 1.Letjbethe degree 1. integrality statementforthenumberf(X)incaseof acharacterXof 1lLIK(-1) _-1,71LIK(0)=0,and fK'IKf (co)WithV=VK(t.KIK). we have biggest integersuchthatXIGi01 If Xisthecharacterofarepresentation(p,V)G,then(1)=dimV ,.., If WisacentralfunctiononsubgroupHofG,andq0,the (i) f(cs)= (p.,aG) =((p,aGIH)v((p,rH)+ AN) =vK()K'1K)cp(1)+fK'IKf(cp) nLjK(m)=Y'- aG/H) = f (rp)=E f(X)=Tg`codimVG,. 27LiK(S)- f(X) ='7LIK(J)+1, `4' f(W)=f((P') t>o go i>o go i=1 go g- m gi Gi (aG) q)=(q/,aG)f(V)- 0 (rs(1) -p(Gt)). S (when X=1Gweputj-1).Then Chapter VII.ZetaFunctionsandL-series (Go:G.,) dx form>1. so theformulafollows theorem (chap.II,(10.7))onehas hence by(11.2),(iii),f(X)=0t?LIK(-1)+1. provided j>0.If=-1,wehaveX(1)-(Gi)0foralli0,and X (Gi)=1,andso(1)-0.From(11.2),(iii),itthusfollowsthat Proof: Ifi G'(L'IK) =G'(LIK)H/H0Gt+e(LIK)H/H=Gt+e(L'IK) f(X) _Enif(Xi*)Eni(UK(oK,IK)Xi(1)+fK,IKf(Xi)), Gj(LI K)H/H=GJ'(L'IK) f(X)=E 1, andX(Gi+s(LIK)H/H)=X(Gj+i(LIK)H/H) 3r. Gt(LIK)H/H =G'(L'IK), X=>niXi*, i=o go G') gi ?7LIK(j)+l, ni EZ, With .,y j' ='7LIL'(1) 531 central function function aG.onGtbyat,andextendittoG=G(LIK) byzero.The Gt =G(LcpIKp)thedecompositiongroupof$43overK. Wedenotethe ideal ofLlyingabovep.Let extension ofglobalfields.LetpbeaprimeidealK, q3Ipaprime that Gf(X)(LXIK)=1,i.e.,f(X)n. has an integer,andweconcludethatf(X)=t+1isthesmallest integersuch largest numbersuchthatGt(LXIK) and Gt+E(LXIK)S;GJ+1(LIK)H/H=1foralle>0. Hencet integer suchthatG(L1K)gG(LILX)=:H.Lett=77LIK(J).Thenone Proof: By(11.3),wehavef(X)=77LIK(J)+1,wherejisthelargest of X,andftheconductorL be acharacterofG(LIK)degree1.LetLXthefixedfieldkernel (11.6) Proposition.LetLIKbeaGaloisextensionoflocalfields,andletX the followingresult. smallest integern>0suchthatG"(LIK)=1.From(11.3)wethusobtain i =TJLIK(J)-seeV,(6.2).Theconductorfp"isthereforegivenbythe which mapsUY')tothehigherramificationgroupG'(LIK)=Gjwith kernel ofthenormresiduesymbol unit groupUK')iscontainedinthenormNLIKL*.Thelatter of localfieldstobethesmallestpowerp,f=p",suchthatn-thhigher 532 of G=G(LIK)tobetheideal (11.5) Definition.Wedefinethe(local)ArtinconductorofcharacterX We nowleavethelocalsituation,andsupposethatLIK isaGalois tea' In chap.V,(1.6),wedefinedtheconductorofanabelianextensionLIK G' (LxIK)=G(LK)H/HGJ(LK)H/H, ( LIK) :K* X fp(X) =pf ap = I K.Thenonehas p I KpbethecompletionofLIK,and TIP Chapter VII.ZetaFunctionsandL-series 1. BythetheoremofHASSE-ARF,tis asp ) G(LIK), (X) p.. is the § 11. The Artin Conductor 533 immediately turns out to be the function (ap),, induced by a-pIGT. If is therefore the character of a representation of G. If now X is a character of !G, then we put f(X,p) = (X,ap) = f(XIGp). Then f p (X) = pf (X. P)is the Artin conductor of the restriction of XI to Gp = G(LpIKp). In particular, we have fp(X) = 1 if p is unramified. We define the (global) Artin conductor of X to be the product f(X) = II fp(X) pfc precision is called for, we write f (L I K, X) instead of f (X). The conductorproperties (11.2) of the f numbers(X), f and(X, p) transfer we immediatelyobtain totheI the Artin Proposition. (i) f(X +X')=f(X)f(XI)f(1) = (1) j (ii)If L' I K is a Galois subextension of L I K, and X is acharacter,of G(L'IK), then f(LI K, X) = f(L'I K, X). (iii)If H is a subgroup of G with fixed field K', and if X is a character of H, then f(LIK,X*)_O Y NK'IK(f(LIK',X)). Proof: (i) and (ii) are trivial. To prove (iii), we choose a fixed prime ideal 'j3 ofL,put G=G(LIK),H=G(LIK'), G=G(LIKp), with p = j3 fl K, and consider the decomposition G = U G rH t into double cosecs. Then representation theory yields the following formula for the character X of H : (*) X* I G p _ > X* , where XT is the character XT(v) = X(t-1Qt) of Gpfl rHr 1, and X* is 'the character of Gcp induced by XT (see [119], chap. 7, prop. 22). Furthermore 137 ='3T fl K' are the different prime ideals of K' above p (see chap. I, §9, p. 55), and we have Gtr = t-1Gt = G(Lr IKp), H= = G= n H = G(L I K'). 534 Chapter VII. Zeta Functions and L-series Now let p'* be the discriminant ideal of KIT IKp, and let f,p' be the degree of over K. Thus NK,JK W) =pfgr. Since fVr(LIK',X)=T7(xlH!,r) fp(LIK,X*)=pf(x,iGT) and we have to show that f(X*IG) vgrX(1)+f rf(XIHsr), or, in view of (11.2), (ii), that (**) f(x*IGT) _ f((xIHpr)*) But HTr = r-I(GT n rHr-I)r, and xIHTr, resp. (xlHpr)*, arises by conjugation a H rar' from xt, resp. X'. Therefore f ((x I Hcpr)*) = f (X'), and (**) follows from (*). We apply (iii) to the case x = 1H, and denote the induced character x* by SG/H. Since f(x) = 1, we obtain the (11.8) Corollary. dK'IK = f(LIK,SG/H) If in particular H = {1}, then SG/H is the character rG of the regular representation. Its decomposition into irreducible characters x is given by rG = EX(1)x x This yields the (11.9) Conductor-Discriminant-Formula. For an arbitrary Galois exten- sion L I K of global fields, one has OLIK=flf(X)X(t), X where x varies over the irreducible characters of G (L I K). For an abelian extension L I K of global fields, we defined the conductor f in VI, (6.4). By chap. VI, (6.5), it is the product f=llfp p of the conductors f p of the local extensions LT I Kp. (11.6) now gives rise to the following § 12. The Functional Equation of Artin L-series 535 (11.10) Proposition. Let L I K be a Galois extension of global fields, x a character of G(L IK) of degree 1, LX the fixed field of the kernel of x, and f the conductor of LX I K. Then one has f=f(x) Now let L I K be a Galois extension of algebraic number fields. We form the ideal c(LIK,X) =ZKj"NKIQ(f(LIK,X)) of Z. The positive generator of this ideal is the integer c(LIK,X) = IdKIXl')9q(f(LIK,X)) Applying (11.7) and observing the transitivity of the discriminant (chap. III, (2.10)), we get the (11.11) Proposition. (i) c (L I K, x + X') = c (L I K, X) c (L I K, X'), c (L I K,1) = IdK1, (ii) c(LIK,X)=c(L'IK,X), (iii)c(LIK,X,,) =c(LIK',X) Here the notation is that of (11.7). § 12. The Functional Equation of Artin L-series The first task is to complete the Artin L -series 1 V's)' LL K X' s) - det(1 - qp for the character x of G = G(LI K), by the appropriate gamma factors. For every infinite place p of K we put Lc(s)x('), if p is complex, Lp(L JK, X, s) LR(s)" LR(s + 1)",if p is real, (')+z (wT) with the exponents n+ = X , n- =X(1)_ 2 Here q qis the distinguished generator of G (LT I Kr), and LR(S) = 7r_sI2r(s/2),Lc(s) = 2(2 r)-SP(s) (see § 4). For p real, the exponents n+, n- in L p (L I K, x , s) have the following meaning. 536 Chapter VII. Zeta Functions and L-series The involution op on V induces an eigenspace decomposition V = V+ ® V-, where V+={xEVlcp,pX=x},V =IxEVlcppx=-x}, and it follows from the remark in § 10, p.521, that I dimV+= (X (1) + X (VT)), dimV- = 2( X (1) - X The functions Gp (L I K, X, s) exhibit the same behaviour under change of fields and characters as the L -series and the Artin conductor. (12.1) Proposition. (i) Gp (L I K, x + x', s) = Gp (L I K, X, s),Cp (L I K, x', s). (ii)If L' I K is a Galois subextension of L I K and x a character of G (L' I K), then G,(LIK,X,s)=Lp(L'IK,X,s). (iii)If K' is an intermediate field of L I K and x a character of G (L I K'), then Gp(LI K, X.,s) = II Lq(LIK',X,s), qlp where q varies over the places of K' lying above p. Proof: (i) is trivial. (ii) If TIT'Ip are places of L J L' 3 K, each lying above the next, then cpgpis mapped under the projection G(LIK) -+ G(L'IK) to ppp'. So x (n) = X (APT,). (iii) If p is complex, then there are precisely m = [K' : K] places q above p. They are also complex, and the claim follows from X,, (1) = mx(1). Suppose p is real. Let G = G(LIK), H = G(LIK'), and let H\G/Gp be the set of double cosets HrGT with a fixed placel of L above p. Then we have a bijection H\G/GT --* {q place of K' above p),HrGT qT = rg3I K' (see chap. I, §9, p.55). qz is real if and only if cp.,T = E H, i.e., G,T rGq3r-' c H. The latter inclusion holds if and only if the double coset HrGcp consists of only one coset mod H: HrGT = (HrG+pr-')r = Hr. We thus obtain the real places among the qZ by letting r run through a system of representatives of the cosets Hr of H\G such that rtpTr-1 E H. But, for such a system, one has ,) X. (VT) =>x(rcPspr _ x(cO.'T) this into Legendre's duplicationformulaLIa(s)LR(s+1)=Lc(s)(see(4.3))turns On theotherhandwehave i.e., Putting Ll=r3,makesq=11K'runthroughtherealplacesofK'abovep, over tothefunctionA(LIK,X, s),i.e.,wehavethe the right-handside,whichwe studiedin(10.4),(11.11),andabove,carries G (LIK)isdefinedtobe (12.2) Definition.ThecompletedArtinL-seriesforthe characterxof and obtainimmediatelyfromtheabovepropositionequations § 12.TheFunctionalEquationofArtinL-series where The behaviourofthefactorsc(L IK,X),L...(LX,s),L(LIK,x,s)on We finallyput Lp(LIK, x.,s) q complex .y. rj A(LIK, X,s)=c(LIK,X)s12Loo(LIX,s)L(LIX,s), Loo(L IK,X+X',s)=L.(LIK,x,s)Loo(Ls), Lc(s)X(t) H L. (LIK,x.,s)=Lao(LIK',x,s). L.(LIK, X,s)=L.(L'IK,s), c(LIK,X) =idKIX( L.(L IK,x,s)=1ILp(LIK,s), X«(1) _E2X(1)+X = fLq(LIK',X,s) _ qlp q real X. ((PT)_ q complex LR(s)xT_ jI ploo real qlp X (t(f(LIK,X)) . q real q real dam Lte(s+1)xcl = 537 complex. ThisGr6f3encharakterwillbedenotedX. Groftencharakter modf(X)withexponentp=(pr),sothatpr0ifris Via theArtinsymbol By (11.10),wethenhave fixed fieldofthekernelX,andletf=rjpp'pbeconductorLXIK. coincides withacompletedHeckeL-series.Toseethis,letLXIKbethe then (iii) then (ii) (12.3) Proposition.(i)A(LIK,X+X',s)=X,s)A(Ls). 538 the imageq up =-1andaq1forallplaces qdifferentfromp.Bychap.VI,(5.6), Let pbearealplaceofK,and letaEIbetheidelewithcomponents I =11pK*(seechap.VI,(1.9)),andconsiderthecomposite map where If=flpUp°"iisthecongruencesubgroupmodfof theidelegroup Proof: Weconsidertheisomorphism (12.4) Lemma.ForeveryrealplacepofKonehas r :K-*C.ThenumberspphavethefollowingGalois-theoreticalmeaning. X becomesaDirichletcharacterofconductorf,i.e.,by(6.9),primitive We putpp=prifpistheplacecorrespondingtoembedding For acharacterXofdegree1,thecompletedArtinL-seriesA(LIK,X,s) If L'IKisaGaloissubextensionofLandXcharacterG(L'K), If K'isanintermediatefieldofLIKandXacharacterG(LK'), p = I/IfK* -3Jf/Pf Jf/PfG(LXIK), (a, LXIK)=(-1,LXtIKp)in G(LXJK)isageneratorof oar A(LIK, X.,s)=A(LIK',X,s). A(LI K,X,s)=A(L'Is). I/IfK* f =f(X). Chapter VII.ZetaFunctionsandL-series G(LXIK)*V*. Jf/Pf, aF--±(LXIK), pp=[LXp:Kp}-1. a 'v) in Kq,forallrealplacesq#p.Then we maychooseanaEK*suchthat=1modf,<0inKp,and> the decompositiongroupGp=G(LxtIKr).Byapproximationtheorem, § 12.TheFunctionalEquationofArtinL-series with Proof :ThecompletedHeckeL-seriesisgiven,accordingto§8,by coincide: degree IandthecompletedHeckeL-seriesforGrbj3encharakterX and rpq3;1forpp=1.Butthisisthestatementoflemma. Since a Jf/pf istheclassof(,8) As explainedintheproofofchap.VI,(1.9),imageamodIfK* (see p.454).Ontheotherhandwehave factors Lp(sp)aregivenexplicitlyby is theL-functionofG(CIR)-setX=Hom(K,C)defined in§4.The and s=sl+p,where (12.5) Proposition.ThecompletedArtinL-seriesforthecharacterXof and with (*) IQIp_ (-1)'Pn,i.e.,X(rP)= 'U" a )PP III =aaE1(f)={xEIIXpEUp"" A(LIK,X,s) =c(LIK,X)s'2G00(LIK,X,s)G(LIK,X,s) 1 modf,wehaveXf(a)=andX,,(a)N((-1)")_ C3. A(x, s)=(IdK1`n(f(x)))s/2Loo(x,s)L(x, Lp(sp) = L (LIK,X,s)=FIGp(LIK,X,s) X ((a))=Xf(a)Xoo(a)((ot) c(LIK,X) =IdKI`I(f(LIK,X)) = (a),whichthereforemapsto(pp.Consequently, A(L IK,X,s)=A(X,s). ,mob { Lx(s) =flLp(sp) L.(X,s) =Lx(s), La (s+pp), Lc (s), (_1)PP sothatpT=1forpp0, PI- P100 forpIfoo}, III if preal, if pcomplex, ?t2 L]. iff=fppnp 539 where theXi,areinducedfromcharactersXiofdegree 1onsubgroups Proof: ByBrauer'stheorem,thecharacterXisanintegrallinear combination fashion, asaconsequenceofthefunctionalequationforHeckeL-series, Hence (*)showsthatindeedLp(LIK,X,s)=Lp(sp). Using (12.4)thisgives on G(LXIK),wegetx(q)=-lifcpp (see p.537).LetcppbethegeneratorofG(LxpIKp).SinceXisinjective for pIooands=1+p.Firstly,wehaveLp(LK,X,s)(LX We arethusreducedtoproving preceding lemma(12.4),onehas Let LxbethefixedfieldofkernelX.By(11.11),(ii),andremark 540 Hi =G(LIK1).Frompropositions(12.3)and(12.5),itfollows that with aconstantW(X)ofabsolutevalue1. continuation toCandsatisfiesthefunctionalequation (12.6) Theorem.TheArtinL-seriesA(LIK,X,s)admitsameromorphic which wehavealreadyestablished. Artin L-seriesnowfollowsfromBrauer'stheorem(10.3)inapurelyformal and by(10.4),(ii),(10.6),thesubsequentremark,onehas Lp(Lx IK,x,s)= In viewofthetworesults(12.3)and(12.5),functionalequationfor c(LIK,X)=c(LxIK,X)=IdKI(f(X)), '-r L(L JK,X,s)=L(Lx A(LI K,X,s)=W(X)A(LIK,X,1- A(LIK,X,s) = LR(s +1),forprealandTcomplex,i.e.,pp=1. LR(S), LC (S), Lp(LIK,X,s) =Lp(sp) X =r-niXi., = = flA(LIKi,xi,s)R' for prealand13real,i.e.,pp=0, for pcomplex, Chapter VII.ZetaFunctionsandL-series J K,X,s) 1, andx(VT)=1ifcp,3=1. A(Xi,s)Ri = .fl L(X, s). flA(LIK,Xi*,s)'i X55 CDR X, s) way astheEulerfactors functions, behaveunderchangeoffieldsandcharactersin exactlythesame Lp (LIK,X,s)attheinfiniteplacesp,whicharemadeup outofgamma L -series,wehavemadeessentialuseofthefactthat "Euler factors" Remark: Fortheproofoffunctionalequation completed Artin not theprincipalcharacter,theyarefollowing: the zeroesoffunctionC(LIK,X,s)inhalf-planeRe(s)<0.IfXis Here thesummationsareoverrealplacespofK.Thisgivesimmediately and theexponents with thefactor explicit form,whichiseasilydeducedfrom(12.6)and(4.3): Therefore A(LIK,X,s)satisfiesthefunctionalequation functional equation where W(X)_Hi(X,)isofabsolutevalue1. L-series A(X;,s)admitmeromorphiccontinuationstoCandsatisfythe where XristheGrbj3encharakterofK,associatedtoXi.By(8.6),Hecke § 12.TheFunctionalEquationofArtinL-series The functionalequationfortheArtinL-seriesmaybegivenfollowing A(LIK,X,s)=W(X)fA(Xj,1-s)=W(X)A(LIK,X,1-s), at s=-1,-3,5,...zeroesoforderZX(l) at s= 1a) n+= 2X(1)+E2X(@Pq), A(X,s) =W(X)[IdKIX(1)T(f(LIK,X))]s 0, -2,4,...zeroesoforder Lp(LIK, X,s)=det(1- a'. .C(LIK,X, 1-s)=A(X,s)L(LIK,3C,s), A(Xj ,s)=W(Xr)A(Xil-. x (cosirs/2)"+(sinres/2)"I'c(s)"X(l) P ,>~ t n-= 2 X (1)+r-2 cpTIR(93)-s. V17')-t p p real p real I 541 D provided itexists,iscalledthe DirichletdensityofM. progression (13.1) Definition.LetMbeasetofprimeidealsK.The limit of thistheorem. L-series, wewillnowdeduceafar-reachinggeneralization andsharpening a, inEN,(a,in)=1,thereoccurinfinitelymanyprime numbers.Using DENrNGER's theoryforpresent-dayresearch. L-series hasenjoyedinthisanalogoussituationaddstotherelevanceof The strikingsuccesswhichthegeometricinterpretationandtreatmentof notion ofdeterminantforfinitedimensionalvectorspacestotheinfinite analogy forthetheoryofL-seriesalgebraicvarietiesoverfinitefields. generality, andreachesfarbeyondArtinL-series.Itsuggestsacomplete dimensional case.Thetheorybasedonthisobservationisoftheutmost and det Dirichlet's primenumbertheorem(5.14)saysthatineveryarithmetic Pa' Lp(LIK,X,s) =deter a, a±m,a±2m, d(M) = § 13.DensityTheorems r log94) \ lim G.- 27ri NAM Chapter VII.ZetaFunctionsandL-series p 9q(p) (sid-(9p); H(Xp/Lp)) 3 m,..., . , 1 . JmTHmJPmwith indexhm=(Jm:Hm). (13.2) Theorem.LetmbeamoduleofKandHmanideal groupsuch-that generalized Diriehletdensitytheorem. (see [123],p.26).Inthenotationofchap.VI,§1and§7, weprovethe of d(M),andthatonehas8(M)=d(M).Theconverseis notalwaystrue It isnotdifficulttoshowthattheexistenceof6(M)implies theexistence One frequentlyalsoconsidersthenaturaldensity anything asfarexistenceorvalueoftheDirichletdensityisconcerned. in M.Addingoromittingfinitelymanyprimeidealsalsodoesnotchange definition ofDirichletdensityonlydependsontheprimeidealsdegree1 Since thesumE'R(p)-soverallprimeidealsofdegree>1converges, So wemayalsowritetheDirichletdensityas at s=1.Furthermore,by(5.11),(ii),wehaveK(s) because thesumEd. The lattersumobviuaslydefinesananalyticfunctionats=1.Wewrite',, f (s)^-g(s)if-isananalyticfunctionats=1.Thenwehave we obtainasin§8,p.494, § 13.DensityTheorems For everyclassSEJm/Hm,thesetP(A)ofprimeidealsin .fihasdensity From theproductexpansion 109 K(s)_, log K(S) WS) =ci S(M) _lira ,gtpi,2 p"m can d (M)=lim E x-}oo d(P(f)) 1 940 1 -01(p)-1 s-* t+o mIR(p)ms 1 s1(p)-1 taken #{p EMI`.71(p) = #{p I`n(p) X01 chi logs 1 L(s) =OK(S)IIG(LIK,x,s)x(1), E X L(X s)=r[ X(. CAD 1 - ' st(p)S X (p) log K(s) t L(x,1)#0. )_ ... p X01 = 1 -X hm, 0, Chapter VII.ZetaFunctionsandL.-series W,J- P- .'EJ"/Pm X hm r 1 ifR=S. if R PEA X (-q') (p)`n(p)-s `n A, PE-W IR(p)S I 1 PER m(p)S . mod m, t density w(m)=1/#(Z/n?Z)*. in anarithmeticprogression,i.e.,aclassmodin,(a,m)=1,have the beginning,instrongerformwhichsaysthatprimenumbers and werecovertheclassicalDirichletprimenumbertheoremrecalledat and H'=Pm,wehaveJm/Pm(Z/mZ)*(seechap.VI,(1.10)), are equidistributedamongtheclasses.IncaseK=Q,m(m), in aclassofJ'/Hmisthesameforeveryclass,i.e.,primeideals § 13.DensityTheorems that thissetdependsonlyontheconjugacyclass of Lsatisfying of allunramifiedprimeidealspKsuchthatthereexistsaideal necessarily abelian).ForeveryaEG(LIK),letusconsidertheset is particularlyimportantinthatitconcernsarbitraryGaloisextensions(not density theorem.Wenowdeduceamoregeneraltheoremwhich VP = theory isomorphismJm/Pm-=G(LIK),totheFrobeniusautomorphisms every aEG,thesetPLIK(a)hasdensity,anditisgiven by (13.4) Theorem.LetLIKbeaGaloisextensionwithgroup G.Thenfor density ofthesetPLIK(a)?Theanswertothisquestionisgivenby of orandthatonehas where ( isomorphism Let .EJ'/H'betheclasscorrespondingtoelement orunderthe LIK. ThenLIKistheclassfieldofanidealgroupHmJm DHmPm Proof: WefirstassumethatGisgeneratedbyor.Letmbe theconductorof The theoremshowsinparticularthatthedensityofprimeideals Relating theprimeidealspofaclassJm/Pm,viafield (LjK density theorem. ) is theFrobeniusautomorphismcpsof gives usaGalois-theoreticinterpretationoftheDirichlet J'/H' --LG, (Cr) ={rar-'ITEG(LIK)} d(PLIK(a)) = n PLIK(a)=0if(a)#(r).Whatisthe }., =(L PL IK(CF) p i ), #(a) (LK) P 'C7 3 overK.Itisclear 545 just thesetofallprimeidealsKwhichsplitcompletely inL. In thegeneralcase,letE'befixedfieldofa.Iffisorderor, has density Then PLIK(a)consistspreciselyoftheprimeidealspwhichliein 546 G =G(NIK),HG(NIL). Then onehas prime divisor93ofdegree1overK.So,ifLIKisGalois, thenP(LIK)is P (LIK)thesetofallunramifiedprimeidealspKwhich admitinLa Furthermore, letuswriteS to indicatethatSiscontainedinTupfinitelymanyexceptional elements. of primes,thenletuswrite consequences, whichwewillnowdeduce.IfSandTareanytwosets where Z(a)=(rEGIraar)isthecentralizerofa.Soweget As PLIE(a)-P(a),weget,foreverypEPLIK(a), Now weconsiderthesurjectivemap ramified orhavedegree>1overQ,wemayomitthemandobtain that £q=Kp,qlp.SincetheremainingprimeidealsinPLIE(a)areeither bijectively tothesetPLIE(a)ofthoseprimeidealsqinsuch j3 ofLsuchthatTipEPLIK(a)and( then, aswejustsaw,d(PLIE(a))=1.LetP(a)bethesetofprimeideals class R.BytheDirichletdensitytheorem(13.2),weconcludethatPLIK(a) (13.5) Lemma.LetNIKbeaGaloisextensioncontaining L,andlet Let LIKbeafiniteextensionofalgebraicnumberfields. We denoteby N The Cebotarevdensitytheoremhasquiteanumberofsurprising d(PLIK(a)) _(Z(a)1 .CD P(LIK)- U p-'(p) ={PEP(a)13Ip}Z(a)/(a), 112 p :PLIE(a)-+PLIK(a), d(PLIK(a)) = d(PLIE(a)) = -'C C/] (o)nH#0 112 T ifSCandS. (a)) PNIK(a) SCT d(P,IE(a)) = h.. ,®w Chapter VII.ZetaFunctionsandL-series #G } =a.ThenP(a)corresponds (disjoint union). 211 q F-rflK. 17l # #G) . Via) f f 1 . #G) sothat [N:K]=1andN=L=K. hypothesis ofthecorollary,wethereforehave only ifitsplitscompletelyinNIK(seechap.I,§9,exercise 4).Underthe extension containingL.AprimeidealpofKsplitscompletely inLifand Proof: LetNIKbethenormalclosureofLIK,i.e., smallest Galois extension LIK,thenL=K. only ifH=U(a)nHOe(a)Thisimpliesthesecondclaim. case ifandonly(a)cHwhenevern:A0,sothisholds LIK isGaloisifandonlyHanormalsubgroupofG,thisthe (13.7) Corollary.Ifalmostallprimeidealssplitcompletely inthefinite Since HCU(a)nH,1e(a),itfollowsthat The Cebotarevdensitytheorem(13.4)thenyields and H=G(NIL).By(13.5),wehave Proof: LetNIKbeaGaloisextensioncontainingL,andletG=G(NIK) has densityd(P(LIK))?n.Furthermore,one some orEGsuchthat(a)nHo0. and onlyiftheconjugacyclass(aa)ofa=(%K-),forsomeprimeideal Proof :AprimeidealpofKwhichisunramifiedinNliesP(LIK)'if (13.6) Corollary.IfLIKisanextensionofdegreen,thenthesetP(LIK) § 13.DensityTheorems PIp ofN,containsanelementH,i.e.,ifandonlypEPNIK(a)for d(P(LIK)) _ 1=d(P(LIK))=d(P(NIK))= d(P(LIK)) = P(LIK)= d(P(LIK)) ? pct (a)nHOO ,-, n 1 (a)nH,oa #(G) U coo coo #G #H L IKisGalois. PNIK(v). #G n #( U(o)) I (a)nH960 [N:K]' 1 0.a 547 0 0 I 548 Chapter VII. Zeta Functions and L-series (13.8) Corollary. An extension L I K is Galois if and only if every prime ideal in P (L I K) splits completely in L. Proof: Let again N I K be the normal closure of L I K .Then P (N I K ) consists precisely of those prime ideals which split completely in L. Hence if P(NIK) = P(LIK), then by (13.6), [N:K]1 =d(P(NIK)) = d(P(LIK)) >[L:K]1 i.e., [N : K] < [L : K], so L = N is Galois. The converse is trivial. (13.9) Proposition (M. BAUER). If L I K is Galois and M I K is an arbitrary finite extension, then P(LIK)P(MIK) tom-- LcM. Proof : L C M trivially implies that P (M I K) C P (L I K ). So assume conversely that P (L I K) 2 P (M I K). Let N I K be a Galois extension containing L and M, and let G = G(NIK), H = G(NIL), H' = G(NIM). Then we have PNIK(a) c P(LIK)= U PNIK(a) P(MI K)=(v)nH'00 U (c)nH540 Let a E H. Since PNIK(a) is infinite by (13.4), there must exist some p E PNIK(a) such that p E PNIK(t) for a suitable r E G such that (r) fl H 0 0. But then a is conjugate to t, and since H is a normal subgroup of G, we find (a) = (r) C H. We therefore have H' C_ H, and hence L C M. (13.10) Corollary. A Galois extension L I K is uniquely determined by the set P (L I K) of prime ideals which split completely in it. This beautiful result is the beginning of an answer to the programme formulated by LEOPOLD KRONECKER (1821-1891), of characterizing the extensions of K, with all their algebraic and arithmetic properties, solely in terms of sets of prime ideals, "in a similar way as Cauchy's theorem determines a function by its boundary values". The result raises the question of how to characterize the sets P (L I K) of prime ideals solely in terns of the base field K. For abelian extensions, class field theory gives a concise § 13. Density Theorems 549 answer to this, in that it recognizes P(LIK) as the set of prime ideals lying in the ideal group H'° for any module of definition m (see chap. VI, (7.3)). If for instance L I K is the Hilbert class field, then P (L I K) consists precisely of the prime ideals which are principal ideals. If on the other hand K = Q and L = Q(p,,,), then P(LI K) consists of all prime numbers p = 1 mod m. In the case of nonabelian extensions L I K, a characterization of the sets P(LIK) is essentially not known. However, this problem is part of a much more general and far-reaching programme known as "Langlands philosophy", which is undergoing a rapid development at the moment. For an introduction to this circle of ideas, we refer the interested reader to [106]. [18] [17] [16] [15] Bourbaki, N. Borevicz, S.I.,Safarevic,I.R. [13] Bloch, S. Bayer, P.,Neukirch,J. Artin, E.,Whaples,G. [14] [12] [11] [10] Artin, E.,Tate,J. [9] Artin, E.,Hasse,H. [8] Artin, E. Apostol, T.M. Ahlfors, L.V. Adleman, L.M.,Heath-Brown,D.R. [7] [6] [5] [4] [3] [2] [1] Algebre commutative.Hermann, Paris1965 Topologie generale.Hermann,Paris 1961 Espaces vectorielstopologiques. Hermann,Paris1966 Algebre. Hermann,Paris1970 Number Theory.AcademicPress,NewYork1966 Algebraic cyclesandhigherK-theory.Adv.Math.61(1986)267-304 (1978) 35-64 On valuesofzetafunctionsand'-adicEulercharacteristics.Invent. math.50 Bull. Amer.Math.Soc.51(1945)469-492 Axiomatic characterizationoffieldsbytheproductformulafor Wesley 1965 im Korperder`-tenEinheitswurzeln.CollectedPapers,Nr.6.Addison- Die beidenErganzungssatzezumReziprozitatsgesetzder"-tenPotenzreste Class FieldTheory.Benjamin,NewYorkAmsterdam1967 Wesley 1965 Uber eineneueArtvonL-Reihen.In:CollectedPapers,Nr.3.Addiso- lected Papers,Nr.7.Addison-Wesley1965 Idealklassen inOberkorpernandailgemeinesReziprozitatsgesetz.In:Col- Collected Papers,Nr.9.Addison-Wesley1965 Die gruppentheoretischeStrukturderDiskriminantenalgebraischerZahlkorper. Collected Papers.Addison-Wesley1965 son-Wesley 1965 Beweis desallgemeinenReziprozitatsgesetzes.CollectedPapers,Nr.5.Addi- Berlin 1976 Introduction toAnalyticNumberTheory.Springer,NewYorkHeidelberg Complex Analysis.McGraw-Hill,NewYork1966 The firstcaseofFermat'slasttheorem.Invent.math.79(1985)409-416 for :-+ Bibliography 'a7 9,'C .-. 'L7 - valuations. r-' ,11 Cartan, H.,Eilenberg,S. Brumer, A. [21] [20] [19] Bruckner, H. 552 Faltings, G. Erdelyi, A.(Editor) Dwork, B. Dress, A. Dieudonne, I. [29] [28] Deuring, M. [27] [26] Deninger, C. [25] Chevalley, C. [24] Cassels, J.W.S.,Frohlich,A. [23] [22] [35] [34] [33] [32] [31] [30] ICJ ¢'O non COQ --. Algebraic NumberTheory.Thompson,Washington,D.C.1967 Homological Algebra.PrincetonUniversityPress,Princeton,N.J.1956 burg 1979 Hilbertsymbole zumExponentenp"andPfaffscheFormen.ManuscriptHam- bereich MathematikderUniversitatEssen,Heft2,1979 Explizites ReziprozitatsgesetzandAnwendungen.VorlesungenausdemFach- Inst. Oberwolfach1964.BibliographischesInstitut,Mannheim1966 Hasse, Roquette,AlgebraischeZahlentheorie.BerichteinerTagungdesMath. Eine expliziteFormelzumReziprozitatsgesetzfiirPrimzahlexponentenp.In: Endlichkeitssatze furabelscheVarietaten aberZahlkorpern.Invent.math.73 London 1953 Higher TranscendentalFunctions, vol.I.McGraw-Hill,NewYorkToronto 22 (1958)180-190 Norm residuesymbolinlocalnumberfields.Abh.Math.Sem.Univ. Hamburg, Mathematics, vol.342.Springer,BerlinHeidelbergNewYork 1973 Contributions tothetheoryofinducedrepresentations.Lecture Notesin Geschichte derMathematik1700-1900,Vieweg,Braunschweig Wiesbaden 415 Uber denTschebotareffschenDichtigkeitssatz.Math.Ann.110(1935)414- Heft 10,TeilII Die KlassenkorperderkomplexenMultiplikation.Enz.Math.Wiss.BandI2, Univ. Hamburg16(1949)32-47 Algebraische BegrindungderkomplexenMultiplikation.Abh.Math.Sem. Arithmetic Geometry,138-156,CambridgeUniv.Press1997 Motivic L-functionsandregularizeddeterminantsII.In:F.Catanese(ed.): 55 (1994),Part1,707-743 Serre (eds.):Seattleconferenceonmotives.Proc.Symp.PureMath.AMS Motivic L-functionsandregularizeddeterminants.In:Jannsen,Kleimann, Class FieldTheory.UniversitatNagoya1954 On theunitsofalgebraicnumberfields.Mathematika14(1967)121-124 (1983) 349-366 1985 Bibliography 55:t N F". `C) C,] tom. On Class Field Towers (in Russian). Izv. Akad. Nauk. SSSR 28 (1964) 261- On Class Field Towers (in Russian). Izv. Akad. (2) 48, 91-102] 272. [English translation in: AMS Translations Riemann-Roch. SGA 6, Lecture Th6orie des Intersections et Thooreme de Berlin Heidelberg New York 1971 Notes in Mathematics, vol. 225. Springer, Fields. Deutscher Verlag der Galois Cohomology of Algebraic Number Wissenschaften, Berlin 1978 Heidelberg Berlin 1977 Algebraic Geometry. Springer, New York Allgemeine Theorie der Gaul3schen Summen in algebraischen Zahlkorpern. Abh. d. Akad. Wiss. Math.-Naturwiss. Klasse 1 (1951) 4-23 Bericht fiber neuere Untersuchungen and Probleme aus der Theorie der algebraischen Zahlkorper. Physica, Wnrzburg Wien 1970 On class field theory of multidimensional local fields of positive characteristic local fields theory of multidimensional On class field pp. 103-127 in Soviet Math. 4 (1991) Advances the characteristic 0, with local fields of theory of multidimensional Class field Algebra i Analiz 3 issue (in Russian) of positive characteristic. residue field J. 3 (1992) [English trans]. in St. Petersburg Math. 3 (1991), pp. 165-196 pp. 649-678] 515-538 abelienne sur Z. Invent. math. 81 (1985) Il n'y a pas de variete 1977 Springer, Berlin Heidelberg New York Riemannsche Flachen. York 1983 Springer, Berlin Heidelberg New Siegelsche Modulfunktionen_ Academic (L-functions and Galois properties). Algebraic Number Fields San Francisco 1977 Press, London New York vol. 74. Springer, Berlin Formal Groups. Lecture Notes in Mathematics, Heidelberg New York 1968 eines beliebigen alge- Allgemeiner Existenzbeweis fiir den Klassenkbrper 1-37 braischen Zahlkorpers. Math. Ann. 63 (1907) algebraischer Zahlkorper. Beweis des Hauptidealsatzes fiir die Klassenkorper Hamb. Abh. 7 (1930) 14-36 82 (1921) 256-279 Punktgitter and Idealtheorie. Math. Ann. New Jersey 1971 Analytic Number Theory. Prentice-Hall Inc., Abelian extensions of complete discrete valuation fields. In: Number Theor1 discrete valuation fields. of complete Abelian extensions pp. 47-74 Press, Cambridge 1996, Cambridge Univ. Paris 1993194, Win [50] [51] [52] [53] [40] [43] [44] [45] [46] Goldstein, L.J. [47] [48] [49] Hartshorne, R. Hasse, H. [36] [37] [38] Fontaine, J.M. [39] Forster, O. Freitag, E. [41] Frohlich, A. (Editor) [42] Frohlich, A. Furtwangler, Ph. Golod, E.S., afarevic, I.R. Grothendieck, A. et at. Haberland, K. Bibliography I. Fesenko, ti" ,-r '.D [74] Hi bschke,E. [73] Holzer, L. Hilbert, D. Herrmann, 0. Hensel, K. [72] [71] [70] [69] Henniart, G. [68] [67] [66] Hecke, E. [65] [64] [63] [62] Hazewinkel, M. [61] [60] [59] [58] [57] [56] [55] [54] 554 matik derUniversitatRegensburg 1987 Arakelovtheorie furZahlkorper. Regensburger Trichter20,FakultatfdrMathe- Klassenkorpertheorie. Teubner, Leipzig1966 Verlag, Berlinetc.1998 son, withanintroductionbyF.LemmermeyerandN.Schappacher. Springer The TheoryofalgebraicNumberFields("Zahlbericht"),transl. byI.Adam- scher Produktentwicklung.Math.Ann.127(1954)357-400 Uber HilbertscheModulfunktionenanddieDirichletschenReihen mitEuler- Theorie deralgebraischenZahlen.Teubner,LeipzigBerlin1908 Lois der6ciprociteexplicites.SeminaireTheoriedesNombres, Paris1979- 80. Birkhauser,BostonBaselStuttgart1981,pp.135-149 Chelsea, NewYork1970 Vorlesungen nberdieTheoriederalgebraischenZahlen.Secondedition. Werke Nr.7,159-171.Vandenhoeck&Ruprecht,Gottingen1970 Uber dieZetafunktionbeliebigeralgebraischerZahlkorper.Mathematische Mathematische Werke.Vandenhoeck&Ruprecht,Gottingen1970 Vandenhoeck &Ruprecht,Gottingen1970 der Primzahlen.ZweiteMitteilung.MathematischeWerkeNr_14,249-289. Eine neueArtvonZetafunktionenandihreBeziehungenzurVerteilung Vandenhoeck &Ruprecht,Gottingen1970. der Primzahlen.ErsteMitteilung.MathematischeWerkeNr.12,215-234. Eine neueArtvonZetafunktionenandihreBeziehungenzurVerteilung Local classfieldtheoryiseasy.Adv.Math.18(1975)148-181 London 1978 Formal groupsandapplications.AcademicPress,NewYorkSanFrancisco Nachr. 5(1951)301-327 Zur ArbeitvonI.R.SafarevicfiberdasallgemeineReziprozitatsgesetz.Math. Zahlentheorie. Akademie-Verlag,Berlin1963 Vorlesungen fiberZahlentheorie.Springer,BerlinHeidelbergNewYork1964 Uber dieKlassenzahlabelscherZahIkorper.Akademie-Verlag,Berlin1952 Mathematische Abhandlungen.DeGruyter,BerlinNewYork1975 Theory. Thompson,Washington,D.C.1967 History; ofClassFieldTheory.In:Cassels-Frohlich,AlgebraicNumber Angew. Math.162(1930)169-184 Fi hrer,DiskriminanteandVerzweigungskorperabelscherZahlkorper.J.Reine braischen Zahlkorper.Math.Ann.107(1933)731-760 Die StrukturderR.BrauerschenAlgebrenklassengruppefibereinemalge- --O. 75" 'O'' .-+ CAD rig )V] ""+ c'" .5G t.. L.' 'O+ .U. ENO ^°- moo, .U1 °q' °H° Fem.,. coo ,s4 CI) !LIB Era Obi ."^ can coo can :'[ f-: FU. con Bibliography .a., C17 0 Landau, E. [92] [91] [90] Kunz, E. [89] Krull, W. [88] Kolcze, £P. [87] Koch, H.,Pieper,H. [86] Koch, H. Klingen, H. [85] [84] Kawada, Y. [83] Kato, K. [82] Kaplansky, I. [81] Janusz, G.J. Iwasawa, K. Ireland, K.,Rosen,M. [80] [79] [78] [77] [76] [75] Huppert, B. Bibliography n400 Zahlen andderIdeale.Chelsea, NewYork1949 Einfiihrung indieelementare andanalytischeTheoriederalgebraischen Wiesbaden 1986 Kahler Differentials.ViewegAdvanced LecturesinMath.,Braunschweig :RS Boston BaselStuttgart1985 Introduction toCommutativeAlgebraandAlgebraicGeometry. Birkhauser, 100 (1928)687-698 Galoissche TheoriederunendlichenalgebraischenErweiterungen. Math.Ann. Witt-Vektoren solcherFunktionen.Diplomarbeit,Regensburg1990 Wissenschaften, Berlin1976 Zahlentheorie (AusgewahlteMethodenandErgebnisse).Deutscher Verlagder Berlin 1970 Galoissche Theoriederp-Erweiterungen.DeutscherVerlagWissenschaften, 265-272 Uber dieWertederDedekindschenZetafunktion.Math.Ann.145(1962) Class formations.Proc.Symp.PureMath.20(1969)96-114 Univ. ofTokyo,Sec.IA26(1979)303-376 A generalizationoflocalclassfieldtheorybyusingK-groupsI.J.Fac.'!Sci. Commutative Rings.TheUniversityofChicagoPress1970 Algebraic NumberFields.AcademicPress,NewYorkLondon1973 (1968) On explicitformulasforthenormresiduesymbol.J.Math.Soc.Japan20 Press, Oxford1986 Local ClassFieldTheory.OxfordUniversityPress,NewYork;Clarendon Press 1972 Lectures onp-adicL-Functions.Ann.Math.Studies74,PrincetonUniversity Heidelberg Berlin1981 A classnumberformulaforcyclotomicfields.Ann.Math.76(1962)171-179 A ClassicalIntroductiontoModernNumberTheory.Springer,NewYork Endliche GruppenI.Springer,BerlinHeidelbergNewYork1967 z a P 'fl ; fax ti. ... ein AnalogonzumHilbertsymbolfuralgebraischeFunktionen and +-1 coo ,-, I,; °)o cam. )a, CAD 7G' cud 1 'C1 555 Lang, S. 556 Mumford, D. Milne, J.S. Meschkowski, H. Matsumura, H. [100] Lubin, J.,Tate,J. [99] Lichtenbaum, S. [98] [97] [96] [95] [94] [93] Neukirch, J. Narkiewicz, W. [104] 'J' [103] [102] [101] Patterson, S.J. O'Meara, O.T. Ogg, A. [111] Odlyzko, A.M. [110] [109] [108] [107] [106] [105] [114] [113] .-+[112] .r-. Real Analysis.Addison-Wesley1968 Introduction toModularForms.Springer,BerlinHeidelbergNewYork1976 Elliptic Functions(SecondEdition).Springer,NewYork1987 Cyclotomic Fields.Springer,BerlinHeidelbergNewYork1978 Algebraic NumberTheory.Addison-Wesley1970 Algebra. Addison-Wesley1971 The RedBookofVarietiesandSchemes.LectureNotesinMathematics,vol. Etale Cohomology.PrincetonUniversityPress,Princeton,NewJersey1980 Mathematiker-Lexikon. BibliographischesInstitut,Mannheim1968 Commutative ringtheory.CambridgeUniversityPress1980 387 Formal ComplexMultiplicationinLocalFields.Ann.Math.81(1965)380- matics, vol.1068.Springer,BerlinHeidelbergNewYork1984,pp.127-138 Values ofzeta-functionsatnon-negativeintegers.LectureNotesinMathe- Boston 1987 P. Schneider(Editors).PerspectivesinMathematics,vol.4.Academic Press, Conjectures onSpecialValuesofL-Functions.M.Rapoport,N. Schappacher, The BeilinsonConjectureforAlgebraicNumberFields.In: Beilinson's On SolvableNumberFields.Invent.math.53(1979)135-164 Klassenkorpertheorie. BibliographischesInstitut,Mannheim1969 Class FieldTheory.Springer,BerlinHeidelbergNewYorkTokyo1986 Jubilaum derDMV.Vieweg,Braunschweig1990 Algebraische Zahlentheorie.In:EinJahrhundertMathematik.Festschriftzurn Publishers, Warszawa1974 Elementary andAnalyticTheoryofAlgebraicNumbers.PolishScientific Jahrhundert Mathematik.Festschrift zumJubilaumderDMV.Vieweg,Braun- ERiCH HECKEanddieRolle derL-ReiheninZahlentheorie.In:Ein Introduction toquadraticforms. Springer,BerlinGottingenHeidelberg1963 Modular FormsandDirichletSeries.Benjamin,NewYorkAmsterdam 1969 Academic Press,LondonNewYorkSanFrancisco1977 On ConductorsandDiscriminants.In:A.Frohlich,AlgebraicNumber Fields. 1358. Springer,BerlinHeidelbergNewYork1988 schweig Wiesbaden1990,pp.629-655 (/] u04 ODD you Goo a.. O.: `/1 C/] F., '": Obi F'. Zoo 0-4 k.< Bibliography 'r1 -5, X00 Rapoport, M.,Schappacher,N.,Schneider,P.(Editors) Rapoport, M. Poitou, G. Bibliography Serre, J.-P. Schilling, O.F.G. Scharlau, W.,Opolka,H. [118] Ribenboim, P. [117] [116] [115] Takagi, T. [127] Aremarkonzetafunctionsofalgebraicnumberfields. In:Automorphic Shintani, T. [126] Areciprocitylawinnon-solvableextensions.J.ReineAngew.Math.221 Shimura, G. [125] [124] [123] [122] [121] [120] [119] [131] [130] [129] Siegel, C.L. [128] Mathematics, vol.4.AcademicPress,Boston1987 Boston 1987 Schneider (Editors)).PerspectivesinMathematics,vol.4.AcademicPress, Conjectures onSpecialValuesofL-Functions(Rapoport,Schappacher, Comparison oftheRegulatorsBeilinsonandBorel.In:Beilinson's Cohomologie GaloisiennedesModulesFinis.Dunod,Paris1967 '-' Representations lineairesdesgroupesfinis,2nded.Hermann,Paris1971 Groupes algebriquesetcorpsdeclasses.Hermann,Paris1959 Cours d'arithmetique.PressesuniversitairesdeFrance,Dunod,Paris1967 Corps locaux.Hermann,Paris1968 Berlin HeidelbergNewYork1964 Cohomologie Galoisienne.LectureNotesinMathematics,vol.5.Springer, The TheoryofValuations.Am.Math.Soc.,Providence,RhodeIsland1950 From FermattoMinkowski.Springer,BerlinHeidelbergNewYorkTokyo York 1979 13 LecturesonFermat'sLastTheorem.Springer,BerlinHeidelbergNew Beilinson's ConjecturesonSpecialValuesofL-Functions.Perspectivesin J. Coll.Sci.Univ.Tokyo44,5 (1922)1-50 Berechnung vonZetafunktionenanganzzahligenStellen.Nachr. Akad.Wiss. non-positive integers.J.ofFac.Sc.Univ.TokyoS.IA,vol. 23,393-417, On evaluationofzetafunctionstotallyrealalgebraicnumber fieldsat Springer, BerlinHeidelbergNewYork1981 Forms, RepresentationTheoryandArithmetic.BombayColloquium 1979. (1966) 209-220 1984 41, 9(1920)1-33 Uber eineTheoriedesrelativ-abelschen Zahlkorpers.J.Coll.Sci.Univ.Tokyo Uber dasReziprozitatsgesetzin einembeliebigenalgebraischenZahlkorper. Gottingen 1969,pp.87-102 1976 CDR `"" f3. L1. chi v°] 557 [145] Neukirch, J.,Schmidt,A.,Wingberg,K. [144] Cornell, G.,Silverman,J.H.,Stevens,G.(Editors) [143] Zariski, 0.,Samuel,P. Zagier, D. Witt, E. Weyl, H. [142] [141] [140] [139] Weiss, E. [137] Weil, A. Washington, L.C. Tate, J. ,7.p?? [138] [136] [135] Vostokov, S. [134] [133] [132] Tamme, G. 558 Cohomology ofNumberFields.SpringerVerlag,Berlinetc.1999 Modular FormsandFermat'sLastTheorem.SpringerVerlag,Berlin etc.1997 Commutative AlgebraI,H.VanNostrand,Princeton,NewJersey 1960 hauser, BaselBostonStuttgart1981 Die ersten50MillionenPrimzahlen.In:MathematischeMiniaturen 1.Birk- Amsterdam 1954,Ser.II,vol.2,71-73 Verlagerung vonGruppenandHauptidealsatz.Proc.Int.Congr.ofMath. Algebraische Zahlentheorie.BibliographischesInstitut,Mannheim1966 Algebraic NumberTheory.McGraw-Hill,NewYork1963 Heidelberg NewYork1979 fonctions algebriques.EuvresScientifiques,vol.I,1939a.Springer,Berlin Sur 1'analogieentrelescorpsdenombresalgebriqueset Basic NumberTheory.Springer,BerlinHeidelbergNewYork1967 Introduction toCyclotomicFields.Springer,BerlinHeidelbergNewYork (1978). [Englishtranslationin:Math.USSRIzvestija13(1979)] Explicit formofthereciprocitylaw.Izv.Akad.Nauk.SSSR.Ser.Math.42 ton 1950(reprintedinCassels,J.W.S.,Frohlich,A.[24]) Fourier analysisinnumberfieldsandHecke'szeta-functions.Thesis,Prince- Mathematics, vol.4.AcademicPress,Boston1988 Beilinson's ConjecturesonSpecialValuesofL-Functions.Perspectivesin The TheoremofRiemann-Roch.In:Rapoport,Schappacher,Schneider, Mathematik derUniversitatRegensburg1979.[Englishtranslation:Introduc- Einfiihrung indieetaleKohomologie.RegensburgerTrichter17,Fakultatfur 1982 tion toEtaleCohomology.Springer,BerlinHeidelbergNewYork1994] ``' Bibliography - ofagroupring...... 410 - oftheGrothendieckring...... 243 augmentation ideal associated ...... 3 Artin symbol...... 407 Artin-Schreier theory...... 281 Artin reciprocitylaw...... 390,407 -zeroes of...... 541 - functionalequation...... 540,541 - completedArtinL-series...... 537 - ArtinandHeckeL-series...... 539 Artin-Hasse, theoremof...... 339 Artin L-series...... 518 Artin, E. Artin conjecture...... 525 - localArtinconductor...... 532 Artin conductor...... 527, arithmetic progression...64,469,545 arithmetic algebraicgeometry archimedean valuation Arakelov divisor...... 189 Arakelov classgroup...... 190 - strongapproximationtheorem..193 approximation theorem...... 117 analytic classnumberformula algebraic numbers...... 5 algebraic numberfield...... 5 affine scheme...... 88 adele ...... 357 admissible mono/epimorphism abstract valuationtheory...... 284 - productformula.....108,109,185 - p-adic...... 107 - ofarepleteideal...... 186 abstract Galoistheory absolute value - ofaprime(place)...... 184 - ofanidele...... 361 - ofanideal...... 34 absolute norm absolute Galoisgroup...... 261 °w' ..tic ...... 406,413 v°. _°? ti. 7s' a°) °`9 ...... 275 ...... 118 527,533 .... 193 .... 468 ... 231 ^'! ,P, N .--, .-+ 'a' .-, f-+ .-+ v', t-+ Index - exponentpof...... 435 - Dirichletcharacter...... 434, 478 - conductorof...... 434,473, 478 - Cherncharacter...... 244,246 - charactergroup...... 273,280 character centrally symmetric...... 26 Cebotarev densitytheorem - ofanumberfield...... 219 - ofametrizednumberfield.....222 central function...... 519 canonical module canonical metric...... 29 - onR*...... 454 - oniR...... 446 - onMinkowskispace...... canonicai measure canonical divisor...... 209 Bruckner,theorem of...... 339 Bruckner, H...... 338,417 Brumer, A...... 394 Brauer, theoremof...... 522 Borel, A...... 432 Bloch,5 ...... 432 - generalized...... 441,515 Bernoulli numbers...... 38,427 big Hilbertclassfield...... 399 Bernoulli polynomial...433,443,'511 Beilinson conjecture...... 432,443 Bauer, M.,theoremof...... !548 battle ofHastings...... 44 -integral basis...... 12 - discriminantofabasis...... 1.11 - basisofalattice...... !.24 auflerwesentliche Diskriminanten- basis augmentation representation of aRiemannsurface...... 209 teiler ...... 207 ...... 545 ...... 520 N .-. V'1 N V'1 _.. y'1 N .:P I'D 129 - Euler-Minkowski....212,256,258 - trivialcharacter...... 434 - ofarepresentation...... 519 - principalcharacter...... 435,520 -primitive GroBencharacter.....472 - primitiveDirichletcharacter...434 - irreduciblecharacter...... 520 - inducedcharacter...... 521 - Heckecharacter...... 480 - GroBencharacter...... 436,470 560 - p-classfieldtheory..298,326,332 - global...... 357,390 - existencetheorem...... 322,396 - localclassfieldaxiom...... 317 - globalclassfieldaxiom...... 383 - rayclassfield...... 396 - problemofclassfieldtower....413 - localclassfield...... 322 - Hilbertclassfield...... 399,402 - globalclassfield...... 395 - bigHilbertclassfield...... 399 Chow theory...... 193 Chow group...... 83,190 Chinese remaindertheorem...... 21 Chevalley, C...... 357 Chern class...... 244 Chern character...... 244,246 characteristic polynomialofafieldele- - Euler-Poincare...... 209 characteristic - ofanorder...... 82 - ideaeclassgroup...... 359 - idealclassgroup...... 22 - divisorclassgroup...... 83 - connectedcomponentofideaeclass - Arakelovclassgroup...... 190 class group...... 22 class function...... 519 - tautological...... 306 - local...... 317,320 - higher-dimensional...... 310 class fieldtheory...... 300 class fieldaxiom...... 299 class field...... 304 ment ...... 9 group ...... 368 -06 CIO .x. «'S In. Pot l'1 s.. If) -z} N (.A t1) closedness relation...... 108,185 class numberformula...... 468 - finitenessof...... 36,81 class number...... 34,36,81 - repleteidealclassgroup...... 186 - repletedivisorclassgroup.....190 - rayclassgroup...... 363,365 - S-classgroup...... 71 0., - ofaGroBencharacter...... 47:3 - conductor-discriminantformula534 - Artinconductor.....527,532,53:3 complex prime...... 183 - ofavaluedfield...... 123 - profinitecompletion...... 274 -of aG-modulation...... 308 completion completely split...... 49 complete valuedfield...... 123, complete lattice...... 24 - functionalequation...... 425,466 completed zetafunction.....422,466 completed L-series..437,499,503,537 complementary module(Dedekind)195 compactified Grothendieckgroup.233 - Pic(O)°iscompact...... 193 -compact group...... 269 compact Coleman's normoperator...... 351 cohomology cohomological dimension...... 306 cocycle ...... 282 coboundary ...... 282 closure,integral ...... 7 - Safarevicconjecture...... 207 - Riemannhypothesis...... 432 - Mordellconjecture...... 207 - Lichtenbaumconjecture...... 516 - Leopoldtconjecture...... 394 - Fermatconjecture...... 37,38 - Beilinsonconjecture...... 432,443 - Artinconjecture...... 525 conjectures congruence subgroup...... 36:3 - ofaDirichletcharacter...434,4713 conductor ...... 47,79,323,397 complex multiplication...... 402 'C7 P'] (L) vii ()c Chi '.7 ...... .7. G]. .7G 000 N'0 N N 123,131 'd' N 'V' '-+ v-) V') Index .1. ... '.p :17 .-- 284 (.I) v°' -on Kc...... 226 -onC conjugation - primeideals...... 53 - embeddings...... 161 conjugate - Taniyama-Shimura-Weil Index - generalized...... 348 cyclotomic polynomial...... 59,166 - primefactorization...... 61 - generalizedcyclotomicfields..348 cyclotomic field.....58,158,273,398 - Zr-extension...... 326,386 - cyclotomic cycle ...... 193 crossed homomorphism...... 282 critical strip...... 432 covering transformation...... 93 - universal...... 93 - ramified...... 92 covering ...... 92,93 cotangent element...... 255 convex ...... 26 connected componentoftheideaeclass - ofprimenumbersinthecyclotomic - ofprimenumbers...... 61,409 - ofavaluation...... 167 - ofaprimeideal...... 54 - ofamicroprime...... 290 decomposition group - ofavaluation...... 171 - ofaprimeideal...... 54 - ofamicroprime...... 290 decomposition field cyclotomic units...... 44 Dedekind's complementarymodule 195 - functionalequation...... 467 Dedekind zetafunction...... 457 Dedekind domain...... 18 - ofprimenumbersinthering decomposition law...... 409 2-extension group ...... 368 gaussian integers...... 4 field ...... 61 `.y 'l7 'LS ...... 444 p.. p...... 385,386 ..... 38 opt 'd' 00: V'1 Opt t-' - ofarepletedivisor...... 190 - ofadivisor...... 96 degree Dedekind,R ...... 17 - ofDirichlet...... 543 - ofCebotarev...... 545 density theorem - naturaldensity...... 543 - Dirichletdensity...... 542 density Deninger, C...... 542 degree valuation...... 95 degree map...... 190 - ofarepresentation...... 519 - ofarepleteideal...... 213 -unit theorem...... 42,81,358 - densitytheorem...... 543 Dirichlet, G.P.Lejeune...... 42 Dirichlet density...... 542 Dirichlet character...... 434,478 - zeroesof...... 442 - specialvaluesof.....442,443,515 - functionalequation....:...... 440 - completed...... 437 Dirichlet L-series...... 435,496 direct (inductive)limit...... 266 diophantine equation...... 104 differentials ...... 200,341 differential module...... 200,254 - ofametrizednumberfield.....224 - ofanelement...... 197 different ...... 195,201,254 determinant ofametrizedo-module derivation ...... 200 - bound...... 204,223 - ofabasis...... 11 - auJ3erwesentlicheDiskri,ninanten- discriminant ...... 49,201, 251 - ofafunctionfield...... 95 discrete valuation...... 67,121 discrete subgroup...... 24 Dirichlet-character, conductorof..434, -prime numbertheorem64,469,543 231,245 478 teiler ...... 207 'r' 04N .O, p+' .,' 'L7 N N N (-n -j? Vii v') 561 Euler, L...... 64,431 Euler's identity...... 419, 435 Euler product...... 419,435 Euler factorsatinfinity..459,527,535, etale topology...... 90,93 equivalent valuations...... 116 equivalent representations...... 519 a4' equidistributed primeideals...... 545 elliptic curve...... 402 Dwork, theoremof...... 332 Dwork, B...... 298, duplication formula,Legendre's..421, - Tateduality...... 326,404 - Serreduality...... 209 - Pontryagindual...... 273 -dual ideal...... 194 duality Dress, A...... 307 double functor...... 307 double cosets...... 55,58 -replete principaldivisor...... 189 - repletedivisor...... 189 - principaldivisor - groupofrepletedivisorclasses.190 - groupofrepletedivisors...... 189 - divisorgroup...... 82,95 - divisorclassgroup...... 83 - degreeofadivisor...... 96 - Chowgroup...... 83 - canonicaldivisor...... 209 - Arakelovdivisorclassgroup...190 - Arakelovdivisor...... 189 divisor ...... 82 division pointsofaformalgroup..347 distribution ofprimenumbers.....432 discriminant anddifferent...... 201 - Stickelberger'srelation...... 15 - Minkowski'stheoremon...38,207 - ofanumberfield...... 15 - ofanideal...... 14 - ofanelement... - conductor-discriminantformula534 562 541 456 'L7 v,' 'L7 '-t ...... 11 ...... 83 298,332 .-. ..r .ti .ti N --i '-0 0110 .-a - HeckeL-series...... 502, 503 - DirichletL-series...... 440 - Dedekindzetafunction...... 467 - ArtinL-series...... 540,541 function field...... 94 - onWittvectors...... 134 - reciprocity...... 521 - correspondence...... 226 - automorphism...... 58,286,406 functional equation - abstract...... 285,287 Frobenius Frey, G...... 38 fractional ideal...... 21 Fourier transform...... 446 - divisionpointsof...... 347 - logarithmof...... 343,345 formal group...... 342 formal 0-module...... 343 Fontaine, J.M.,theoremof...... 207 finiteness ofclassnumber...... 36,81 finite prime(place)...... 183 Fibonacci numbers...... 53 Fesenko,I ...... 310 Fermat's lasttheorem...... 37,38 Faltings, theoremof(Mordellconjec- - ofacompletefield...... 131 - ofahenselianfield...... 144,147 factorial ring...... 7 extension ofavaluation...... 161, exponential valuation69,107,120,184 exponential function,p-adic...... 137 - ofaGrMencharacter...... 478 - ofanoperatorp...... 278 - exponentpofancharacter.....435 exponent - p-adicexpansion.....99,101,106 - row-columnexpansion...... 6 expansion existence theoremofclassfieldtheory Euler-Poincare characteristic.....209 Euler-Minkowski characteristic...212, ture) ...... 207 °O' 322,396 256,258 4'" CAD 161,163 Index global classfield...... 395 ghost componentsofWittvectors .134 - ofaRiemannsurface...... 209 - ofanumberfield...... 214,467 genus - abstract...... 275 generic point...... 86 generalized cyclotomictheory....346 general reciprocitylaw...... 300 gaussian units...... 3 gaussian primenumbers...... 3 gaussian integers...... 1 Gauss's reciprocitylaw....51,64,416 Gauss sum...... 51,438,473,488 - higher-dimensional...... 454 gamma function...... 421 - ofvaluations...... 166 - infinite...... 261 Galois theory Galois group,absolute...... 261 Galois descent...... 372 - induced...... 312,374 G-module ...... 276 G-modulation ...... 307 G°b (maximalabelianquotient) Furtwangler, P...... 406,413 fundamental units...... 42 - volumeofthefundamentalmesh - volumeofthefundamentalmesh - volumeofthefundamentalmesh - volumeofthefundamentalmesh26 - regulator...... 43,431,443 - ofalattice...... 24 fundamental mesh - ofvaluationtheory...150,155,165 - forprimeideals...... 46 - ofaG-set...... 307 - Riemann'szetafunction..425,426 - Mellintransform...... 422 Index fundamental identity fundamental group...... 93 274 a repleteideal...... 212 an ideal...... 31 the unitlattice...... 43 -°' .w- l., Fs: '7' }°a 't; - theoremof...... 180 - quotient...... 312,378 Herbrand henselization ...... 143 - P-valuation...... 298 - valuation...... 143,288,309,389 - localring...... 152 - field...... 143,147 henselian Hensel's lemma...... 129, Hensel, K...... 99 - transformationformula...... 490 - partial...... 489 Hecke thetaseries...... 489 - partial...... 496 - HeckeandArtinL-series...... 539 - functionalequation...... 502,503 - completed...... 499,503 Nab Hecke character...... 480 Hasse's Zahlbericht...... 363 Hasse-Arf, theoremof...... 355,530 - on118...... 446 - onap-adicnumberfield...... 142 - typeof...... 478 - primitive...... 472 - conductorof...... 473 Hecke L-series...... 493,496,497 Hasse-Minkowski, theoremof....385 Hasse normtheorem...... 384 cad Haar measure Gysin map...... 253 Grunwald, theoremof...... 405 group cohomology...... 284 Grothendieck-Riemann-Roch .....254 Grothendieck group,replete...... 233 Grothendieck, A...... 225,253 -exponent of...... 478 GrS]lencharacter ...... 436,470 Golod-Safarevic ...... 413 global Tateduality...... 404 global reciprocitylaw...... 390 global normresiduesymbol...... 391 global field...... 134 - theory...... 357,390 - axiom...... 383 !34 t.' wry G'. i., r`' .-, 129,148 N N .-+ 00 I'D 0000 O'. ..pp 563 fl. Hilbert, D...... 53 - gammafunction...... 454 - classfieldtheory...... 310 - ramificationgroups...... 176, higher hermitian scalarproduct..28,226,444 Hermite, theoremof...... 206 564 - normof...... 186 - invertible...... 74 - integral...... 21 - fractional...... 21 - dual...... 194 - discriminantof...... 14 - degreeofarepleteideal...... 213 - absolutenormof...... 34 Hurwitz formula...... 220 - tame...... 335 - productformula...... 414 - explicit...... 339 Hilbert symbol...... 305,333,414 Hilbert's ramificationtheory..53,166 Hilbert-Noether, theoremof...... 283 Hilbert classfield...... 399,400,402 Hilbert 90...... 278,281,283,284 - logarithm...... 445 higher-dimensional -unit group...... 122 - S-idele...... :...... 358 - principal...... 359 - normof...... 370 - idelegroup...... 357 - id6leclassgroup...... 359 - absolutenormof...... 361 ideae ...... 357 ideal number - definedmodm...... 408 ideal group...... 21,408 - replete...... 186 ideal classgroup...... 22,186 - volumeoffundamentalmesh....31, - repleteideal...... 185 - repleteprincipalideal...... 186 - principalidealtheorem...... 410 ideal ...... 16 212 _°: 'J' 't3 _°: ,7...... 16,496 176,352 N N '.D V'1 .-, 00.0 - ofametrizednumberfield.....224- - representation...... 52] - character...... 521 induced index ofspecialty...... 218 imaginary quadraticfield...... 402 - p-adic...... 100,104,111 - algebraic...... 5 infinite prime...... 183 - ofavaluation...... 168 - ofaprimeideal...... 57 - abstract...... 285 - ofavaluation...... 173 - ofaprimeideal...... 57 inertia field - ofavaluation...... 150,165 - ofaprimes(place)...... 184 - ofaprimeideal...... 46,49 - abstract...... 285,309 inertia degree...... 46,49,184,285 inductive system...... 265 inductive limit...... 266 - G-module...... 312,374,521 irreducible character...... 520 invertible 0-module...... 229,230 invertible ideal...... 74 inverse different...... 195 - ringextension...... 6 - integrallyclosed...... 7 - ideal...... 21 - closure...... 7 - basis...... 12 integral integer infinite primenumber...... 184 infinite Galoistheory...... 261 inertia group Jacobi's thetaseries...... 422, 424 Jacobi symbol...... 417 Iwasawa theory...... 63 Iwasawa, K. isometric ...... 229 irregular primenumber...... 38 irreducible representation...... 519 gaussian ...... 1 ate. ,may .`n 0,01 ...... 37 'GU v'1 Index '-O '.O ,-'-' ,-+ V'1 (-A r-, 00 Legendre's duplicationformula ...421, - Z-structure...... 24 - volumeoffundamentalmesh ....26 - unitlattice...... 40 - Minkowski'slatticepointtheorem - fundamentalmesh...... 24 - completelattice - basisof...... 24 lattice ...... 2,23 Langlands philosophy...... 549 -partial L-series...... 496 - p-adicL-series...... 516 - HeckeL-series...... 493,496 - functionalequation..440,502,540 - DirichletL-series...... 435,496 - completedHeckeL-series.499,503 - completedDirichletL-series...437 - completedArtinL-series...... 537 - ArtinandHeckeL-series...... 539 - ArtinL-series...... 518 L-series L-function ofaG(CIR)-set Kiirschak, J...... 107 Kummer, E...... 16,38 Kummer theory...... 277,279,340 Kummer extension...... 278,380 Krull valuation Krull topology...... 167,262 Krull dimension...... 73 Kronecker-Weber theorem Kronecker's programme Kronecker's Jugendtraum Krasner's lemma...... 152 Kato, K...... 310,432 Kahl er,E. Kahler differentials j-invariant ...... 402 Jannsen,U ...... 221 ti-ti Index K-theory ...... 193,310,431 Krull-Akizuki, theoremof K IK,maximalunramifiedextension 456 27 ,C, 154,285 CD" Oar ;;0 f07 CO) °`a ti) ...... 200 ...... 123 ...... 200 ...... 548 ...... 401 ...... 77 ... 324,398 ...... 455 Cpl N N N ...... 24 W'0 00\ vii I-- 000 N 000 I-- - Minkowskimeasure...... 221 - HaarmeasureonR...... 446 - Haarmeasureonap-adicnumber measure - unramifiedextensionofIF,,((t))176 - unramifiedextensionofQ....176 - unramifiedextension.....154,285 - tamelyramifiedextension.....157 - order...... 72 maximal Mackey functor...... 307 - series...... 328 - module...... 343 - extension...... 348 Lubin-Tate - p-adic...... 136,142 - higher-dimensional...... 445 - ofaformalgroup...... 343,345 logarithm local-to-global principle.....161,357, local ring...... 66 local reciprocitylaw...... 320 local normresiduesymbol - ofavaluedfield...... 160 localization ...... 65,71 - 2-localfield...... 310 - theory...... 317 - axiom...... 317 local classfield...... 322 line bundle...... 208,255 - projective...... 103,266 - inductive(direct)...... 266 limit Lichtenbaum conjecture...... 516 Leopoldt conjecture length ofamodule...... 82 - snakelemma...... 79 - Nakayama'slemma...... 72 - Krasner'slemma...... 152 - Hensel'slemma...... 129 lemma (renownedlemmas) Legendre symbol...... 50,336 J'0 local field...... 134 384, 385,391 field ...... 142 ...... 394 ...... 321 .-' .-+ r--. '-+ r-+ 565 nonarchimedean valuation Noether, E...... 282 Newton polygon...... 144 natural density...... 543 Nart, E...... 149 Nakayama's lemma...... 72 multiplicity ofarepresentation....519 Mordell conjecture...... 207 monogenous ringextension...... 178 - ofaHeckecharacter...... 480 module m...... 363 module ofdefinition...... 407 modulation ...... 307 modular function...... 402 modular form...... 434 Mobius inversionformula...... 484 Mobiusfunction ...... 474 Minkowski-Hasse, theoremof....385 - theory,multiplicativeversion...32 - theory...... 28 - theoremonlinearforms...... 28 - theoremondiscriminant...38,207 - space...... 29,444 - metric...... 31 - measure...... - latticepointtheorem...... 27 - bound...... 34 Minkowski, H...... 24 microprime ...... 290,299 - projectiveresolution...... 234 - 0-module...... 227 - numberfield...... 222 metrized - trivial...... 227,229 - standard...... 28,228 - Minkowski...... 31 - hermitian...... 226 - canonicalonMinkowskispace..29 metric Mellin transform...... 422 Mellin principle...... 423 - onR+...... 454 - onl[8...... 446 - onMinkowskispace...... 29 measure, canonical 566 .or Dice too U-. Goo Col ...... 118 N ..a -Pa '-+221 mot' -exponential valuation...... 107 -expansion ...... 99,101,114 - absolutevalue...... 107 Ostrowski, theoremof...... 124 - maximal...... 72 - p-adic...... 128,136 - p-adic...... 100,101,111 - ideal...... 16,486 - Fibonacci...... 53 - Bernoulli...... 38,427,441,515 - algebraic...... 5 p-adic order ofanumberfield...... 72 Odlyzko, A.M...... 223 numbers - quadratic...... 50 - p-adic...... 136 - metrized...... 222 - imaginaryquadratic...... 402 - genusof...... 214,467 - discriminantof...... 15 - algebraic...... 5 number field n-th ramificationgroup...... 176 norm topology...... 303 norm theorem,ofHasse...... 384 -product formula...... 393 - overQ...... 331 - local...... 321 - global...... 391 norm residuesymbol...... 302 norm residuegroup...... 277 norm-one surface...... 39,460 normalized valuation...... 121,184 normalization ...... 7,76,91 - universalnorms...... 304,308 - ofanid8le...... 370 - ofanideal...... 186 - Hassenormtheorem...... 384 - ofagaussianinteger...... 2 - Coleman'snormoperator...... 351 - onC...... 444 - absolute...... 34,184,186,361 norm ...... 8 0,' tar .ENO tad try f9' .-i N i-- .-+ .-- '-- Index prime numbertheorem,ofDirichlet 64, prime element...... 121,289 - ofgaussianintegers...... 1 - inthecyclotomicfield...... 61 prime decomposition...... 18,409 - real...... 183 - microprime...... 290,299 - infinite...... 183 - finite...... 183 - complex...... 183 - absolutenormof...... 184 prime (place) presheaf ...... 87 preparation theorem(Weierstrass)..116 power sum...... 433,443 3°° power seriesfield...... 127,136 power residuesymbol...... 336,415 Pontryagin dual...... 273 polyhedric cone...... 504 Poisson summationformula...... 447 Poincare homomorphism....234,237 place ...... 183 - replete...... 186,239 Picard group...... 75,185 p-function, ofWeierstrass...... 401 periodic p-adicexpansion...... 106 Pell's equation...... 43,84 p-class fieldtheory.....298,326,332 - zetafunction...... 458 - L-series...... 496 - Heckethetaseries...... 489 partial - numbers...... 128,136,271 - numberfield...... 136 - logarithm...... 136,142 - exponentialfunction...... 137 = zetafunction...... 516 - Weierstrasspreparationtheorem116 - valuation...... 69 - units...... 112 -unit rank...... 394 - periodicp-adicexpansion.....106 - numbers...... 100,101,111,271 - L-series...... 516 Index p-adic 469,543 'L7 R'. -4) ..y 2.4:1 .-a .-+ .-+ .-y N O\: 1.0 caw r-. 0'o 000 - ofvaluations...... 150, - ofprimes(places)...... 184 - ofprimeideals...... 45 - ofmetrizednumberfields..... 224 - abstract...... 285,309 ramification index - uppernumberingof...... 180 - higher...... 176,352 - Herbrand'stheorem...... 180 ramification group...... 168 ramification field...... 175 - 0-module...... 228 - limit...... 103,266 profinite group...... 264 profinite completion...... 274 - fortheHilbertsymbol...... 414 - forabsolutevalues...108,109,185 quadratic residuesymbol.....50,417 quadratic residue...... 50 quadratic numberfield...... 50 pythagorean triples...... 5 purely ramified...... 49,158,286 p $ylowsubgroup...... 274 'tiPri ferring7G...... 272 - system...... 266 - resolution...... 234 - line...... 97 projective projection formula...... 248 product, restricted...... 357 - forthenormresiduesymbol...393 product formula procyclic groups...... 273 principal units...... 122 principal idele...... 359 principal idealtheorem...... 410 principal ideal,replete...... 186 - replete...... 189 principal divisor...... 83 'c7 principal character...... 435,520 - rootofunity...... 59 - polynomial...... 129 - character...... 434,472 primitive '-t can CI. 150,165 .-. N V') 0000 .4- 567 - wildly...... 158 - totally...... 49,158,286 - tamely...... 154 - covering...... 92 ramified ...... 49 - ofvaluedfields...... 166 - higher...... 176,354 - Hilbert's...... 53 ramification theory ramification points...... 92 568 regular primenumber...... 38 regular primeideal...... 79 reciprocity map...... 291 - localreciprocitylaw...... 320 - globalreciprocitylaw...... 390 - general...... 300 - Gauss'sreciprocitylaw....51,416 - forpowerresidues...... 415 - Artinreciprocitylaw.....390,407 reciprocity law reciprocity homomorphism...... 294 reciprocity, Frobenius...... 521 real prime...... 183 ray classgroup...... 363,365 ray classfield...... 366,396,403 rationally equivalent...... 83 rank ofcoherent0-module...229,243 - degree...... 519 - augmentationrepresentation ...520 replete idealclassgroup...... 186 - volumeoffundamentalmesh..212 - repleteprincipalideal...... 186 - degreeof...... 213 - absolutenormof...... 186 replete ideal...... 185 - principalideal...... 186 - principaldivisor...... 189 - Picardgroup...... 186 - Grothendieckgroup...... 233 - divisorclassgroup...... 190 - divisor...... 189 replete regulator ...... 43,431,443 regular representation...... 520 - characterof...... 519 representation ofagroup...... 518 Iii: C's .-. N .-. d0' ,-- `-+ v'. - hypothesis...... 432 Ribet, K. residue classfield...... 121 - ofzetafunctions...... 425 - ofap-adicdifferential...... 341 -of L-series...... 423 residue - regular...... 520 - multiplicityof...... 519 - irreducible...... 519 - induced...... 521 - equivalentrepresentations.....519 - ofadeles...... 357 ring - theoremof...... 209 - Grothendieck-Riemann-Rochfor - fornumberfields....213,214,218 Riemann-Roch Riemann-Hurwitz formula...220,22 - trivialzeroes...... 432 - specialvaluesof.....427,431,432 - functionalequation...... 425,426 - Euler'sidentity...... 419,43.5 - completed...... 422, Riemann's zetafunction...... 419 - surface...... 208 Riemann, B. restricted product...... 357 row-column expansion...... 6 - ofWittvectors...... 134,283 - valuationring...... 121 - Priiferring - ofp-adicintegers....104,111, - localring...... 66 - henselianvaluationring...... 143 - henselian...... 152. - factorial...... 7 - Dedekinddomain...... 18 Schwartz function...... 446 Schmidt, F.K...... 58,152 scheme ...... 88,96 Safarevi6,I.R ...... 413 Safarevi6 conjecture 42° `.) number fields...... 254 224 .or coo y"" ,., two ...... 38 i ,-_. Q.. 'a...... 272 acs ...... 207 104,111,271 422,466 Index c'! N.,, ,.., r-. t, (11 - sectionof...... 87 - presheaf...... 87 section ofasheaf...... 87 Index solenoid ...... 368 snake lemma...... 79 - resolutionofsingularities...... 91 singularity ...... 73,91 simplicial cone...... 504 Siegel-Klingen, theoremof...... 515 S-ideae ...... 358 Shintani's unittheorem...... 507 - structuresheaf...... :...88 sheaf ...... 88 sesquilinear ...... 226 Serre duality...... 209,214 S-class group...... 71 - [x,a)...... 341 - tameHilbertsymbol...... 335 - quadraticresiduesymbol...... 417 - powerresiduesymbol....336,415 - normresiduesymbol.302,321,331, - Legendresymbol...... 50,336 - Jacobisymbol...... 417 - Hilbertsymbol...... 305,333 - Artinsymbol...... 407 symbols Sylow subgroup...... 274 supplementary theorems.....340,416 S-units ...... 71,358 structure sheaf...... 88 strong approximationtheorem....193 strict cohomologicaldimension...306 Stirling's formula...... 206 coo Stickelberger's discriminantrelation15 standard metric...... 28,228 stalk ofasheaf...... 88 spectrum ofaring...... 85 -maximal extensionoflFp((t)) ..176 - maximalextension...... 157 tamely ramified...... 154 tame Hilbertsymbol...... 335 Tamagawa measure...... 432 Takagi,T . stalk of...... 88 coo 391,393 ...... 406 '-' 'd' .-a -^+ (.A V'1 --l Tamme, G...... 225,240 - Dirichletdensitytheorem...... 543 - - Bruckner...... 339 - Brauer...... 522 - Artinreciprocitylaw...... 390 - Artin-Hasse...... 339 theorem (renownedtheorems) Taylor, R...... 38 tautological classfieldtheory.....306 Tate's thesis...... 503 Tate duality...... 326,404 Taniyama-Shimura-Weil conjecture38 - maximalextensionofOp...... 176 - Hermite...... 206 -Hasse normtheorem...... 384 - Hasse-Minkowski...... 385 - Hasse-Arf...... 355 - Grunwald...... 405 - Grothendieck-Riemann-Roch..254 - Gaussreciprocitylaw..51,64,416 - Fontaine...... 207 - F.K.Schmidt(primedecomposition) - F.K.Schmidt(henselianvaluation) - Faltings(Mordellconjecture)..207 - extensiontheorem(forvaluations) - Dwork...... 332 - Dirichletprimenumbertheorem - Dirichletunittheorem..42,81,358 - Riemann-Rock...... 209, - Ostrowski...... 124 - Minkowskitheoremonthediscrimi- - Minkowskilatticepointtheorem27 - Minkowski-Hasse...... 385 - Krull-Akizuki...... 77 - Kronecker-Weber...... 324,398 - Hilberttheorem90...... 281 - Hilbert-Noether...... 283 - Herbrand...... 180 - Riemann-Roch-Grothendieck ..254 - principalidealtheorem...... 410 58 469 Cebotarev 152 161,163 nant .....:...... 207 CIO !T7 sex ...... 545 ..077 209,218 569 r-' L/') ^`' - unitrank,p-adic...... 394 - unitlattice...... 40 - S-units...... 71,358 - Shintani'sunittheorem...... 507 - principal...... 122 - p-adic...... 112 - gaussian...... :...... 3 - fundamental...... 42 - Dirichlet'sunittheorem.42,81,358 - cyclotomic...... 44 units ...... 39 unique primedecomposition...... 18 trivial zeroesoftheRiemann'szeta trivial metric...... 227,229 55.5 transformation formulaforthetaseries - onWittvectors...... 134 type ofaGrSl3encharacter...... 478 trivial character...... 434 transfer ...... 296,410 trace-zero space...... 460 trace-zero hyperplane...... 39 trace form...... 194 trace ...... 8,444 totally split...... 49 totally ramified...... 49,158,286 totally disconnected...... 264 - Zariski...... 85 - norm...... 303 - Krull...... 167,262 - etale...... 90 topology Todd class...... 254 - transformationformula...425,437, - ofalattice...... 450 - Jacobi...... 422,424 - Hecke...... 489 - ofanalgebraicnumberfield...484 - Wilson...... 2 - Weierstrasspreparationtheorem116 - Siegel-Klingen...... 515 - Shintaniunittheorem...... 507 theta series...... 443 570 function ...... 432 425,437,438,439,452,490 438, 439,452,490 '"'. .n' 'L1 (D' 'J' dm' Nom... mo!!. Q)p.. ... .N-. 3333 I>>>a I>>II 0O, 000 000 0000 0000 (J1 wildly ramified...... 158 Weierstrass preparationtheorem ..116 Weierstrass f-function...... 401 Weber, H...... 366,405 Weber function...... 403 Ver ...... 296,410 vector bundle...... 193,255 Vandermonde matrix...... 11 - ofRiemann'szetafunction...427, - ofDirichletL-series..442,443,515 - abstract...... 284,309 - henselian...... 143 - p-adic...... 69 - normalized...... 121,184 - Krullvaluation...... 123 v p,foraprime...... 184 values (special) valuation theory...... 99 valuation ring...... 121 - nonarchimedean...... 118 -henselian P-valuation...... 298 - henselian...... 143,288,309,389 - extensionof...... 131,144 - exponential...... 69,107,120,184 - equivalentvaluations...... 116 - discretevaluationring...... 67 - discrete...... 67,95,121 - degreevaluation...... 95 - archimedean...... 118 - abstracthenselian....288,309,389 valuation ...... 67,116 upper numberingoframificationgroups upper half-space...... 445 upper half-plane...... 425 - maximalextension...... 154,285 - extensionofhenselianfields...153 - extensionofalgebraicnumberfields - lCIRisunramified...184,220,224 unramified .....49,184,202,286,309 universal norms...... 304,308 universal covering...... 93 431,432 r"4 "'' 180 202 w.. 'J' 414411 000 (.A N'0 0~0 .-r Index - ofArtinL-series...... 541 zeroes Zariski topology...... 85 Zahlbericht, Hasse's...... 363 Zagier, D...... 433 Z ...... 272 Witt vectors...... 134,283,340 Witt, E. Wilson's theorem...... 2 Wiles, A...... 38 Index ...... 410 .A. - Riemann...... 419 - p-adic...... 516 - partial...... 458 - ofRiemann'szetafunction....432 Z-structure ...... 24 Zr-extension, cyclotomic...... 326 - Dedekind...... 457 - completed...... 422,466 zeta function...... 419 - ofDirichletL-series...... 442 Vii 571