Lectures on Complete Discrete Valuation Fields

1: Discrete Valuation Fields

(1.1). Valuations. One can generalize the properties of the -adic valuation ¡£¢ and proceed to the concept of valuation. Let Γ be an additively written totally ordered abelian group. Add

Γ ¥§¦ ¤ ¤¨¦ ¤ ¥ ¤ ¤ to a formal element + ¤ with the properties + , + + , + (+ ) = + ,

Γ Γ Γ

¤ ¤ ¥ © ¤ (+ ¤ ) + (+ ) = + , for each ; denote = + .

A map ¡ : with the properties

¡ ( ) = + = 0

¡ ¡

¡ ( ) = ( ) + ( )

¡ ¡

¡ ( + ) min( ( ) ( ))

¡ is said to be a valuation on ; in this case is said to be a valuation ﬁeld. The map induces

Γ Γ ¡

a homomorphism of to and its value group ( ) is a totally ordered subgroup of .

¡ ¡

If ¡ ( ) = 0 , then is called the trivial valuation. It is easy to show that ( 1) = 0, and if

!¡

¡ ( ) ( ), then

¡ "¡ # ¡ "¡ ¡

¡ ( ) min( ( + ) ( )) min( ( ) ( )) = ( );

$ ¡ ¡ ¡ ¡

thus, if ¡ ( ) = ( ) then ( + ) = min( ( ) ( )).

%'& £(©) ¡ * £(©+ ¡ , * &

(1.2). Basic Objects. Let = : ( ) 0 , & = : ( ) 0 . Then %'&

coincides with the set of non-invertible elements of %-& . Therefore, is a local ring with the

* % ¡ &

unique maximal ideal & ; is called the ring of integers (with respect to ), and the ﬁeld

.& %-&0/1* 2©)%'&

= & is called the residue ﬁeld, or residue class ﬁeld. The image of an element

in .& is denoted by , it is called the residue of in . The set of invertible elements of

%'& 34& %-&5 6*

is a multiplicative group = & , it is called the group of units.

798;:-:-<=

$ , &

Assume that char( ) = char( & ). Then char( ) = 0 and char( ) = 0.

$ .&

Proof. Suppose that char( ) = = 0. Then = 0 in and therefore in . Hence

= char( .& ).

(1.3). >@?

¡ 1. A valuation ¡ on is said to be discrete if the totally ordered group ( ) is isomorphic

to the naturally ordered group X .

Z ¡£¢ Y

For a prime and a non-zero integer Y let = ( ) be the maximal integer such that

¡£¢ ¡£¢ Y)/E\ ¡£¢ Y ]¡£¢ \ ¡£¢ ¤

divides Y . Extend to rational numbers putting ( ) = ( ) ( ); (0) = + .

The -adic valuation ¡£¢ is a discrete valuation with the ring of integers

Kbdc

_ Y6\6©WX`a\

%'&^ = : is relatively prime to \

1 QiFkjRlN8POm8onp

2 e0fhg

r ^

The residue field & is a finite field of order .

s t t ¡ ¡

¢ ¢

2. Let = ( ). For an irreducible polynomial ( ) deﬁne ( u ) similarly to above.

¡ ¢ The map ( u ) is a discrete valuation with the ring of integers

( t )

y y

%'& w.x t t ©+s t t t z

^ = : ( ) ( ) [ ] ( ) is relatively prime to ( ) y

( v ) x

( t )

t / t s t s

and the residue ﬁeld is s [ ] ( ) [ ] which is a ﬁnite algebraic extension of .

s ¡P{

The ﬁeld has another discrete valuation trivial on : ( ) = deg( ), its residue ﬁeld

x x is s .

(1.4). Discrete valuations. A ﬁeld is said to be a discrete valuation ﬁeld if it admits a |}©~%'&

nontrivial discrete valuation ¡ (see Example 1 in (1.3)). An element is said to be

| ¡

a prime element (uniformizing element) if ¡ ( ) generates the group ( ). Without loss of X

generality we shall often assume that the homomorphism ¡ : is surjective.

798;:-:-<= |

Let be a discrete valuation ﬁeld, and be a prime element. Then the ring of

%'& |9N%'&

integers %'& is a principal ideal ring, and every proper ideal of can be written as for

\, * |p%'& %'& some 0. In particular, & = . The intersection of all proper ideals of is the zero

ideal.

|9; \©+X ©3 &

Each element 2©+ can be uniquely written as for some and .

% \ ¡ ©6m

Proof. Let be a proper ideal of & . Then there exists = min ( ) : and hence

|9N%'&§ |9N%'& |9N%-&

|9o©) for some unit . It follows that and = . If belongs to

%-& ¡ ¤

the intersection of all proper ideals |9A%'& in , then ( ) = + , i.e., = 0.

¡ |@©34& |9; (©34& |9 |95

Let \ = ( ). Then and = for . If 1 = 2, then

¡ Y ¡ ©3 & \ Y \ + ( 1) = + ( 2). As 1 2 , we deduce = , 1 = 2.

(1.5). Completion. Completion of a discrete valuation ﬁeld is an object which is easier to

work with than with the original ﬁeld.

¡ ¡ X

Let be a ﬁeld with a discrete valuation (as usual, ( ) = ). is a metric space &

( )

with respect to the norm = (1 2) . So one can introduce the notion of a fundamental

sequence: a sequence ( ) 0 of elements of is called a fundamental sequence if for every

\ ¡ Y6\6!\

real there is ( ) 0 such that ( ) for ( ).

If ( ) is a fundamental sequence then for every integer there is such that for all

ccc

¡ 2 \ ¦§\ ¦

\4Y\9 we have ( ) . We can assume 1 2 . If for every there

\9 \9 ¡ $ ¡ ¡ ¡ \!\9

is such that ( ) = ( +1), then ( ) and ( ) for , and

0 0

¤ ¡

hence lim ¡ ( ) = + . Otherwise lim ( ) is ﬁnite.

798;:-:-<=

The set of all fundamental sequences forms a ring with respect to componentwise

addition and multiplication. The set of all fundamental sequences ( ) 0 with 0 as

¤ /

\) + forms a maximal ideal of . The ﬁeld is a discrete valuation ﬁeld with its

¡ ¡ ¡ ¡

discrete valuation ¡ deﬁned by (( )) = lim ( ) for a fundamental sequence ( ) 0.

Proof. A sketch of the proof is as follows. It sufﬁces to show that is a maximal ideal of

\( ¤

. Let ( ) 0 be a fundamental sequence with 0 as + . Hence, there is an

$ \!~\ \! ~\ \!£\

\ 0 0 such that = 0 for 0. Put = 0 for 0 and = for 0.

Then ( ) 0 is a fundamental sequence and ( )( ) (1) + . Therefore, is maximal.

QSFkjRlm8DOm8on

(1.6). A discrete valuation ﬁeld is called a complete discrete valuation ﬁeld if every

©W

fundamental sequence ( ) 0 is convergent, i.e., there exists = lim with respect to

¡ ¡ ¡ ¡9 ¤ ¡

¡ . A ﬁeld with a discrete valuation is called a completion of if it is complete, = ,

¡ and is a dense subﬁeld in with respect to .

¥4lGAB;GLF¦Q§ORQSGAT= Every discrete valuation ﬁeld has a completion which is unique up to an iso-

morphism over .

Proof. We verify that the ﬁeld ¨/ with the valuation is a completion of . is

(

embedded in / by the formula ( ) mod . For a fundamental sequence ( ) 0

\ ¡ § Y"\\ ©

and real , let 0 0 be such that ( ) for all 0. Hence, for 0

( )

¡ © ¨/

we have ¡ (( 0 ) ( ) 0) , which shows that is dense in . Let (( ) )

ccc

¡ \ \ be a fundamental sequence in ¨/ with respect to . Let (0), (1), be an increasing

( ) ( ) ( )

Y \ \ \ Y sequence of integers such that ¡ ( ) for , ( ). Then ( ) is a

2 1 1 2 ( )

( )

¡ ¡ ¨/

fundamental sequence in and it is the limit of (( ) ) with respect to in . Thus,

¡ ¡

we obtain the existence of the completion / , .

¡ ¡

¡ ¡ ¡ ¡ ~©(

If there are two completions 1, 1 and 2, 2, then we put ( ) = for and

¡ ¡

extend this homomorphism by continuity from , as a dense subﬁeld in 1, to 1. It is easy

¡ ¡

¡ ¡ ¡

to verify that the extension : 1 2 is an isomorphism and 2 ª = 1.

x x

¡ ¡

We shall denote the completion of the ﬁeld with respect to by p& or simply .

r ¡

1.7. Examples of complete valuation ﬁelds. 1. The completion of with respect to ¢ of

(1.3) is denoted by ¢ and is called the ﬁeld of -adic numbers. Certainly, the completion of

r «1¬S«{ r r

with respect to the absolute value of (1.1) is . Embeddings of in ¢ for all prime

and in is a tool to solve various problems over . An example is the Minkowski–Hasse

¥°²±d©)r r

Theorem: an equation ®¯¥°²±t³°´td± = 0 for has a nontrivial solution in if and only

r ¢

if it admits a nontrivial solution in and in for all prime .

Xµ¢

The ring of integers of r-¢ is denoted by and is called the ring of -adic integers. The

¶¢

residue field of r ¢ is the finite field consisting of elements.

s t ¡ s t

2. The completion of ( ) with respect to u is the formal power series ﬁeld (( ))

+ {

t ©2s \

of all formal series { with and = 0 for almost all negative . The

+ {

¡ s t t

ring of integers with respect to u is [[ ]], that is, the set of all formal series 0 ,

©+s s

. Its residue ﬁeld may be identiﬁed with .

¡ 3 % &

(1.8). Representatives. For simplicity, we will often omit the index in notations & , ,

* | &

& , . We ﬁx a prime element of .

·¸¹% ©·

A set · is said to be a set of representatives for a valuation ﬁeld if , 0

%©º%/1* and · is mapped bijectively on under the canonical map = . Denote by

+ {

¼ ¼

rep: ©»· the inverse bijective map. For a set denote by ( ) the set of all sequences

{

+ { +

¥;° ° ¥°4©+¼ ¼ ¼ \)½ 5¤

( ) , . Let ( ) { denote the union of increasing sets ( ) where .

The additive group has a natural ﬁltration

° °

+1 ccc c

¬¬¬m¾!| %¾(| %¾ °

° +1

%µ/1| %¿ The factor ﬁltration of this ﬁltration is easy to calculate: | . QiFkjRlN8POm8onp

4 e0fhg

798;:-:-<=

¡ | ©

Let be a complete ﬁeld with respect to a discrete valuation . Let ° for each

À À

©+X ¡ | be an element of with ( ° ) = . Then the map

+ { Ä

+ {

¥° °´Á0ÂÃ ¥;° |A°

Rep: ( ) { ( ) rep( )

{

°´Á0Â$ °ÅÁ0Â ¡ ¥° ¥°µ$

is a bijection. Moreover, if ( ¥° ) = (0) then (Rep( )) = min : = 0 .

À ¥°

Proof. The map Rep is well deﬁned, because for almost all 0 we get rep( ) = 0 and the

|A° ¥° °´Á0Â$ Æk° °ÅÁ0Â

series rep( ¥° ) converges in . If ( ) = ( ) and

©WX ¥°µ$ Æk°D

\ = min : =

¡ ¥ | Æ | \ ¡ ¥ | ]Æ | ,Ç\ ,!\

° ° °

then ( ) = . Since ( ° ) for , we deduce that

¥° Æk° \ ¡ (Rep( ) Rep( )) =

Therefore Rep is injective.

¥° ¥°³$ È©§ |9;

In particular, ¡ (Rep( )) = min : = 0 . Further, let . Then = with

+©Ç3 | 0 0 '©!3 ¥ 0

\©2X , . We also get = for some . Let be the image of in ;

$ 6 ¥ | ©6| %

then ¥ = 0 and 1 = rep( ) . Continuing in this way for 1, we obtain a

® ¥ | °

convergent series = rep( ° ) . Therefore, Rep is surjective.

GAlGLCEC

|9 ]© We often take | = . Therefore, by the preceding Lemma, every element can be uniquely expanded as

+ {

°Å| °4©+· °

= and = 0 for almost all 0

{Ë Ë Ë

8;I1QSTmQiORQiGLT9=

) 6]©Ì|9L% 6Í |9

g If , we write mod . 3

(1.9). Units. The group 1 + |p% is called the group of principal units 1 and its elements are

3 | %

called principal units. Introduce also higher groups of units: ° = 1 + for 1.

ccc

¾Î3h¾Î3 ¾Ï3 ¾ The multiplicative group has a natural ﬁltration 1 2 . We

describe the factor ﬁltration of the introduced ﬁltration on . ¥4lGAB;GLF¦Q§ORQSGAT=

Let be a discrete valuation ﬁeld. Then

©X~Ð|~©! (1) The choice of a prime element | ( 1 ) induces an isomorphism

3}ÒWX .

(2) The canonical map %Ç%/E* = induces the surjective homomorphism

Ô`Ã

0: 3~ ;

0 maps 3H/P3 1 isomorphically onto .

(3) The map

34° o p| Ã

° : 1 +

2©)% 3 /P3

° ° for induces the isomorphism of ° +1 onto for 1.

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e0f 5

Ó Ó

Proof. (2) The kernel of 0 coincides with 3 1 and 0 is surjective. (3) The induced map

3 /P3 °

° +1 is a homomorphism, since

° ° ° °

2 c

| | | |

(1 + 1 )(1 + 2 ) = 1 + ( 1 + 2) + 1 2

This homomorphism is bijective, since ° (1 + rep( ) ) = .

GAlGLCEC

Let Õ be not divisible by char( ). Raising to the th power induces an auto-

À À

3 /P3 3

° ° morphism of ° +1 for 1. If is complete, then the group for 1 is uniquely

Õ -divisible.

° °

° +1

| ~©!% 1ÖµÍ Õ´p| |

Proof. If = 1 + with , then 1 + mod . Absence of nontrivial

34° Õ -torsion in the additive group implies the ﬁrst property. It also shows that has no

nontrivial Õ -torsion. °

° 1

× | ©]% × ÕØ 9| ÖÙ× × ©(3

For an element = 1 + with we have = (1 + ) 1 with 1 ° +1.

Õ × Applying the same argument to × 1 and so on, we get an th root of in in the case of

complete .

(1.10). Raising to th power. Let char( ) = 0. Lemma (1.2) shows that either

char( ) = or char( ) = 0. We shall study the operation of raising to the th power. Denote Û

this homomorphism by Ú

The ﬁrst and simplest case is char( ) = .

Ú Û

¥4lGAB;GLF¦Q§ORQSGAT=

, 3

Let char( ) = char( ) = 0. Then the homomorphism maps °

À À

injectively into 3Ü¢° for 1. For 1

¢ ¢ ¢° ¢°

° +1

Í | | Ý]©)%'¤

(1 + p| ) 1 + mod

° ¢ ¢ ¢°

| Proof. Since (1 + | ) = 1 + and there is no nontrivial -torsion in and , the

assertion follows.

GAlGLCEC

Ú Û

Let be a ﬁeld of characteristic 0 and let be perfect, i.e = . Then

3 °Þ/D3 3p¢°Ø/P3 ¢°

maps the quotient group ° +1 isomorphically onto the quotient group +1 for À

We now consider the case of char( ) = 0, char( ) = 0. As = 0 in the residue ﬁeld

©+* ¡

, we conclude that and, therefore, for the surjective discrete valuation of we get

ß.

¡ ( ) = 1.

8;I1QSTmQiORQiGLT9=

ß ß ¡

g The number = ( ) = ( ) is called the absolute ramiﬁcation index of .

· ©

Let | be a prime element in . Let be a set of representatives, and let 0 be the

+1 Ë

|à5©W|à %

element of uniquely determined by the relation 0 .

¥4lGAB;GLF¦Q§ORQSGAT=

Ú Û

Let be a discrete valuation ﬁeld of characteristic zero with residue ﬁeld of

¦!ß0/ 3 3 ¢°

positive characteristic . Then the homomorphism maps ° to for ( 1), and

!ß0/ 3 ° (©%

34° to + for ( 1). Moreover, for

° ¢ ¢ ¢° ¢°

!ß0/ Í | | (1 + p| ) 1 + mod if ( 1) (1) QiFkjRlN8POm8onp

6 e0fhg

° ¢ ¢ ¢° ¢°

Í | | ß0/ ©+X.

(1 + | ) 1 + ( + 0 ) mod if = ( 1) (2)

° ¢ ° °

+ + +1

à à

p| Í | |

(1 + ) 1 + 0 mod if ,Çß0/ ( 1) (3) Ë The induced homomorphisms on the quotient ﬁltration are injective in cases (1), (3) and

surjective in case (3).

¡ (á"¢ ß0/ If a primitive th root áâ¢ of unity is contained in , then (1 ) = ( 1) and the

kernel of the induced homomorphisms in case (2) is of order .

3 ° 3 ß0/ ©X ß0/

If is complete, then + = ° for ( 1). If is complete and ( 1) ,

then the homomorphism in (2) is injective iff there is no nontrivial -torsion in .

Proof. Let 1 + 2©)34° . We get

¢ ¢

¢ ( 1)

2 1

¬¬¬ (1 + ) = 1 + + + + +

À À À À ¢

( 1) ¢

ccc

2 ä 1

ß ¡Üã ß ¡ ß ¡ and ¡ ( ) = + , = + 2 , ( ) = + ( 1) , ( ) = , so

¢ ¢ ¢

¡ ¡ $ ¡

¡ ((1 + ) 1) = ( + ) if ( ) = ( )

¢ ¢

!¡

¡ ((1 + ) 1) ( + ) otherwise

¦(¡ ¦!ß0/

Note ¡ ( ) ( ) if and only if ( 1). For a unit we obtain the ﬁrst statement of Û the proposition. Ú

Further, the homomorphism is an isomorphism in case (3) and injective in case (1).

á ©£ ¡ á ß0/ ß1/ ©X ¢

Assume that ¢ . From the previous (1 ) = ( 1) and ( 1) .

æ ^å

¢ 1

Ã ¶

Therefore, the homomorphism + 0 is not injective. Its kernel 0 ¢ in this

case is of order . Ë

À ,

If is complete, then due to surjectivity of the homomorphisms in case (3) for

¢ ¢ ¢

ß0/ 34° 3 3 3 3 ¬¬¬ 3 ß0/

° °

( 1) we get = +1 ° = +2 = = . Now let ( 1) be an integer.

çà à çà

Assume that the horizontal homomorphism in case (2) is not injective. Let 0 satisfy the

¢ ¢

( 1)

|àè ©]3 é+, ß1/

equation 0 + 0 0 = 0. Then (1 + 0 ) ± for some ( 1). Therefore

¢ ¢ ¢

( 1)Ë ( 1) 1

|àè ©3 |àè ©3 ¢

(1 + 0) ) = 1 for some 1 (¢ 1)+1. Thus, (1 + 0) ) 1 ( 1) is a

àè àè

primitive th root of unity.

É GAlGLCEC

Let be a complete discrete valuation ﬁeld.

m \ ,

If char( ) = 0, then is an open subgroup in for 1. If char( ) = 0, then

R is an open subgroup in if and only if is relatively prime to .

contains ﬁnitely many roots of unity of order a power of .

R m

3 £ \!

Proof. If char( ) = 0, then we get 1 for 1. It means that is open. If

3 £R \4 m

char( ) = , then 1 for ( ) = 1 and is open. In this case, if char( ) = ,

° ¢ ¢

©/

then 1 + | for ( ) = 1. Then is not open. If char( ) = 0, then we obtain

¢ê

, ß0/ 3 Ç Yë ß R \ ° when ( 1) + ( 1) . Therefore is open for 1.

This corollary demonstrates that for complete discrete valuation ﬁelds of characteristic 0 with

residue ﬁeld of characteristic the topological properties are closely related with the algebraic

ones. The case char( ) = is very different.

QSFkjRlm8DOm8on

(1.11). Product representation. Now we deduce a multiplicative analog of the expansion in

the corollary of (1.8).

¥4lGAB;GLF¦Q§ORQSGATìîíH8;TNFk8;C1ïk= ·

Let be a complete discrete valuation ﬁeld. Let be a set of

!© \]©X ©·

representatives. Then for there exist uniquely determined , ° for 0,

0 ©W· , such that can be expanded in the convergent product

| | °

= 0 ð (1 + )

Ë Ë

° 1

K©)3

Proof. The existence and uniqueness of \ and 0 immediately follow. Assume that ,

1 Ë

then ﬁnd ©· with (1 + ) +1. Proceeding by induction, we obtain an

Ë Ë

expansion of in a convergent product. If there are two such expansions (1 + °î| ) =

ñ ° °

° °

(1 + ), then the residues , ° coincide in . Thus, = .

Ë Ë Ë Ë Ë

X (1.12). ¢ -Structure of The Group of Principal Units. Everywhere in this section is a

complete discrete valuation ﬁeld with residue ﬁeld of positive characteristic .

¢ò

\ ¤

If ©3 1 then 1 as + . This enables us to deﬁne

0ó Pó ¥ ¥©WX õ¥ ©WX

= lim if lim = ¢

{ {

Pô Dô

798;:-:-<=

ö ö ö

+ ö

W©(3 ¥©]X¢ ©(3

ó ó ó ó

Let 1, . Then ó 1 is well deﬁned and = , = ( ) ,

E× × R×]©3 ¥LâÆ ©(Xµ¢ 3 X¢

ó ó

( ) = ó for 1, . The multiplicative group 1 is a -module under 3

the operation of raising to a power. Moreover, the structure of the Xµ¢ -module 1 is compatible 3

with the topologies of Xµ¢ and 1.

ò ò

¥ Æ ¥ 2Æ \ ¤

Proof. Assume that lim = lim ; hence 0 as + and lim ó = 1.

X Ò)3 ÷3 ¥L X

¢ ¢ A map 1 1 ( ( ) ó ) is continuous with respect to the -adic topology on and the discrete valuation topology on 3 1. This argument can be applied to verify the other

assertions of the lemma.

¥4lGAB;GLF¦Q§ORQSGAT=

Let be of characteristic with perfect residue ﬁeld. Let be a set of

representatives, and let · 0 be a subset of it such that the residues of its elements in form a

¶ ø ·

basis of as a vector space over ¢ . Let an index-set numerate the elements of 0. Let

¡ ¢ be the -adic valuation.

Then every element 2©)3 1 can be uniquely represented as a convergent product

| "¥ ©X ©+·

± °þ± ¢ ±

ó üþý ð

= ð (1 + ) 0

Ë Ë

± ÁDû ¢ ( °´ù )=1

°´ú 0

ÿ ¦é³©ø ¡£¢ ¥°²± ¦ÇE ¨

and the sets ø;°Åù = : ( ) are ﬁnite for all 0 , ( ) = 1.

3 \

Proof. We ﬁrst show that the element can be written modulo for 1 in the desired

form with ¥°²±©!X . Proceeding by induction, it will sufﬁce to consider an element

Í |9 3 ©¹· \4

modulo 3 +1. Let 1 + mod +1, . If ( ) = 1, then one can ﬁnd

ccc ccc

ö¡

Ë Ë

©· Æ âÆ ©+X |9Í~ñ |9 3 [

1 0 and 1 such that 1 + [ =1(1 + ) mod +1 for

£¢ ccc

Ë Ë Ë Ë

\ \9 \9 \9 Å ©·

some Y . If = with an integer , ( ) = 1, then one can ﬁnd 1 0 and

Ë Ë QiFkjRlN8POm8onp

8 e0fhg

¢¥¤

ccc

ö¡

Æ âÆ ©)X 3 Y |9Í |9

ñ [

1 such that 1 + [ =1(1 + ) mod +1 for some . Now due

Ë Ë

to the continuity we get the desired expression for (©3 1 with the above conditions on the sets

ø ÿ

°´ù .

¥ é ©ø °þ±

Assume that there is a convergent product for 1 with ± , . Let ( 0 ) = 1 and 0 be

À À À

( ) ( ) Ë

ü ý

\ ¦ é+©(ø ·

0 0 ý

such that = 0 ókü for all ( ) = 1, . Then the choice of 0 imply

(1 + ) ókü +1, which concludes the proof.

GAlGLCEC

±| ±©£·

The group 3 1 has a free topological basis 1 + where where 0,

Ë Ë

( ) = 1.

¡ Ã

If ß = ( ) is divisible by 1, let : be the map + 0 . Then raising to

the th power in case (2) of the proposition in (1.10) is described by . Ë

¥4lGAB;GLF¦Q§ORQSGAT=

Let be of characteristic 0 with perfect residue ﬁeld of characteristic .

· ·

Let · be a set of representatives and let 0 (resp. 0 ) be a subset of it such that the residues

¶¢ ¶¢

of its elements in form a basis of as a vector space over (resp. are -generators of

ø øç · ·¨

d/ ( ) ). Let the index-set (resp. ) numerate the elements of 0 (resp. 0 ). Let

À À À À

©+X` ¦ ß0/ = : 1 ( 1) ( ) = 1

Then every element 2©)3 1 can be represented as a convergent product

¢ ¢

° ( 1)

àè

| × | ©+· "× ©+· "¥ "¥ ©WX

± ó ± ó ± ± °þ± ± ¢

üþý ý

ð ð

= (1 + ) ð (1 + ) 0 0

Ë Ë

°´Á¨§ ± ÁDû ± ÁDû

(the second product occurs when ß0/ ( 1) is an integer) and the sets

ø kéo©6ø ¡ ¥ ¦ED ø ¦é³©ø ¡ ¥ ¦ÇE

°´ù ÿ ¢ °þ± ¢ ±

= : ( ) ÿ = : ( )

À ©

are ﬁnite for all ¨ 0, . 3

Proof. We shall show how to obtain the required form for ©©3 modulo +1. Let

|9 3 ©+·

= 1 + mod +1, . There are four cases to consider:

ccc ccc

Ë Ë

©· Æ âÆ ©X

(1) \]© . One can ﬁnd 1 0 and 1 satisfying the congruence

ö¡

Ë Ë

|9 |9³Í 3 Y

ñ [

1 + [ =1(1 + ) mod +1 for some .

ccc ccc

Ë Ë

©W· ß1/ \ \ \ ©+ Æ ¦Æ ©

(2) \ ( 1), = with . Then there exist 1 0, 1

Ë Ë

X such that

¢©¤

| Í | 3 Y [

1 + ð (1 + ) mod +1 for some

Ë Ë

[ =1

©X \ ß0/ · \ \9

(3) ß0/ ( 1) , = ( 1). The deﬁnition of 0 imply that if = with

ccc ccc ccc ccc

©+· × "× ©+·¨ Æ âÆ â ©X

\9 ç© , then there exist 1 0, 1 0, 1 , 1 such

that Ë

¢¥¤ ÿ

ö¡

| Í | × | 3 Y6

[ ð

1 + (1 + ) ð (1 + ) mod +1 for some

Ë Ë

[ =1 =1

ß0/ \Ì Dßd¦ ß0/ \9 \Ì Dß

(4) \, ( 1). If = min : ( 1) and = , then

¢©

Í | 3 ©·

1 + | (1 + ) mod +1 for some

Ë Ë Ë

QSFkjRlm8DOm8on

e0f 9

|9 3

Now applying the arguments of the preceding cases to 1+ þ|9 , we can write 1+ mod +1

in the required form. Ë

GAlGLCEC

Let be of characteristic 0 with perfect residue ﬁeld of characteristic .

(1) If does not contain nontrivial th roots of unity then the representation in the proposition

3 ¤

is unique. Therefore the elements of the proposition form a topological basis of 1 ù .

(2) If contains a nontrivial th root of unity, let be the maximal integer such that

¥ â¥ ±

contains a primitive th root of unity. Then the numbers °²± of the proposition

are determined uniquely modulo . Therefore the elements of the proposition form a

/D3 3

ù ¤

topological basis of 1 1 ù .

3 X ß (3) If the residue ﬁeld of is ﬁnite then 1 is the direct sum of a free ¢ -module of rank

and the torsion part. x

á ©/ Proof. (1) If ¢ then all horizontal homomorphisms of the diagrams in the second

Proposition of (1.10) are injective.

á ¢

(2) Argue by induction on . Write a primitive th root in the form

° ¢ ¢ ÿ

ÿ ( 1)

àè

| á × |

± ¢ ±

ü ý ý

ð ð

= ð (1 + ) (1 + )

°ÅÁ¨§ ± ÁDû ± ÁDû

and raise the right hand side to the th power which demonstrates the non-uniqueness.

Now if

¢ ¢

° ( 1)

àè

±| ×1±|

ó ó

üþý ý

ð ð

1 = (1 + ) ð (1 + )

°ÅÁ¨§ ± ÁDû ± ÁDû

¥ Æ "¥ Æ Æ âÆ

°þ± ± ± °þ± ±

then we deduce that °þ± = = with -adic integers . Then

° ¢ ¢ ö

ö ( 1)

àè

| × |

± ±

üþý ý

ð ð

ð (1 + ) (1 + )

°ÅÁ¨§ ± ÁDû ± ÁDû is a th root of unity, and so is equal to

å 1

¢ ÿ

ÿ ¢ ¢ ÿ °

( 1) ä

àè

×E±| ±|

üþý ý

ð ð

ð (1 + ) (1 + )

± ÁDû °´Á¨§ ± ÁDû

1 1

Æ âÆ

°þ± ± ±

for some integer . Now by the induction assumption all °²± are

¥ "¥ ±

divisible by . Thus, all °þ± are divisible by . 3

(3) If the residue field of is finite then 1 is a module of finite type over the principal X

ideal domain ¢ , so by the structure theorem on such modules it is a direct sum of a free

module and a ﬁnite torsion module. If a primitive th root of unity is in , then the kernel of

¦ ¦

is of order . Hence : ( ) = , since is ﬁnite. The cardinality of is equal to

ß0/ ß0/ / ß = [ ( 1)] [[ ( 1)] ].

Om8DTAF¦QiGLTNF¨GLI GL:-B;CE8POm8oqpQi8;CEVAF f¸>@? 10

2: Extensions of Complete Fields t

(2.1). Henselian Property. Let be a commutative ring. For two polynomials ( ) =

ccc y

t ¥ t Æ t ¬¬¬ Æ

¥ + 0, ( ) = + + 0 their resultant is the determinant of the matrix

ccc

¥ ¥

1 ¥ 0

ccc

¥ ¥

¥ 1 0

. . . .

......

ccc

¥ ¥ ¥

1 0 c

ccc

¥ Æ

Æ 1 0

ccc

Æ Æ

Æ 1 0

5. . . .

......

ccc

Æ Æ

Æ 1 0

y y y y;y

This determinant · ( ) is zero iff and have a common root; in general ( ) = 1 + 1

x x x xçx

y y

©W t t ¥ t} W t Æ t}

ñ ° ñ ± ±

for some polynomials 1 1 [ ]. If ( ) = ° =1( ), ( ) = =1( ),

x x

y y y

· ¥ Æk ñ p°ç 6R± · t 6¥L t ¥ ±

then their resultant ( ) is °´ù ( ). In particular, ( ( )) = ( ).

y y

©Ç% t · ©Ç%

If [ ] then ( ) . We shall use the following properties of the resultant:

x x

y y ¢ ¢

y +1

Í t · Í · * · ©Ï* *

if mod * [ ] then ( ) ( 1 ) mod ; if ( ) then

x x x x x

¢ y

t % t % t

* [ ] [ ] + [ ].

x %

Let be a complete discrete valuation ﬁeld with the ring of integers , the maximal ideal

t ¥ t ¬¬¬ ¥ ©% t

* , and the residue ﬁeld . For a polynomial ( ) = + + 0 [ ] we will denote

t ¬¬¬ ¥ t © t

the polynomial ¥ + + 0 by ( ) [ ]. We will write

Í t *

( t ) ( ) mod

t ©Ì*! t

if ( t ) ( ) [ ].

¥4lGAB;GLF¦Q§ORQSGAT=

y y

Let 0 0 be polynomials over such that deg = deg 0 + deg 0 and the

x x y

leading coefﬁcient of coincides with that of 0 0. Let

¢ ¢ y

y +1 2 +1

©W*/ Í *

· ( 0 0) 0 0 mod

x .

for an integer 0.

Then there exist polynomials ( t ) ( ) such that

¢ ¢

y y y y c

y +1 +1

Í * Í *

= deg = deg 0 deg = deg 0 0 mod 0 mod

t A° t ©Ç% t

Proof. We ﬁrst construct polynomials ° ( ) ( ) [ ] with the following properties:

y y y

A°

deg( ° 0) deg 0, deg( 0) deg 0

° °

¢ ¢ y

y + +

° t Í t * N° t Í! t * °

( ) ° 1( ) mod ( ) 1( ) mod

¢ c

y +2 +1

Í ° t N° t *

( t ) ( ) ( ) mod

t R± t é]¦ Proceeding by induction, we can assume that the polynomials ± ( ) ( ), for 1,

have been constructed. For a prime element | put

° °

¢#" ¢#$ y

y + +

° t t | ° t N° t t | ° t °

( ) = ° 1( ) + ( ) ( ) = 1( ) + ( )

Om8;TNF¦QSGATAFUGLI GL:-B;CE8POm8³qpQS8DCEVLF Rfh>@?

" $ " $

t t ©% t t t t t

° ° °

with ° ( ) ( ) [ ], deg ( ) deg 0( ), deg ( ) deg 0( ). Then

° °

¢ ¢

$ "

y y y c

+ ä +2 +1

° t A° t t t Í| t ° t t ° t *

° ° °

( ) ( ) ° 1( ) 1( ) 1( ) ( ) + 1( ) ( ) mod

° ¢

y +2

t t t | t °

Since by the induction assumption ( ) ° 1( ) 1( ) = 1( ) for a suitable

x x " $

t ©% t t t °

1( ) [ ] of degree smaller than that of , we deduce that it sufﬁces for ° ( ), ( )

x x

$ "

y ¢ c

¢ +1

| t Í t t t t *

° ° °

to satisfy the congruence 1( ) ° 1( ) ( ) + 1( ) ( ) mod .

¢ y

y +1

· t t Í· t t Í $ * °

However, ( ° 1( ) 1( )) ( 0( ) 0( )) 0 mod . Then the properties

" $

¿ ¿ °

of the resultant imply the existence of polynomials ° , satisfying the congruence. Write

" " "

y y

° ° ° °

= 1 % + 1 with polynomial 1 of degree smaller than that of 1. Then it is easy

$ $

° °

to see that the degree of = + % 1 is smaller that the degree of 1. The polynomials

" $ °

°" are the required ones.

y y y

° t t A° t t t t

Now put ( t ) = lim ( ) ( ) = lim ( ) and get ( ) = ( ) ( ). x

The following statement is often called Hensel Lemma; it was proved by K. Hensel for -adic

numbers and by K. Rychl´ik for complete valuation ﬁelds.

GAlGLCEC

y y

e Let 0 0 be monic polynomials with coefﬁcients in . Let = 0 0 and

x x

y y

t suppose that 0 0 are relatively prime in [ ]. Then there exist monic polynomials with

coefﬁcients in % , such that

y y y c

t t t t t t

( t ) = ( ) ( ) ( ) = 0( ) ( ) = 0( )

t t ©+*/

Proof. We have · ( 0( ) 0( )) and we can apply the previous proposition for = 0.

t The polynomials ( t ) and ( ) may be assumed to be monic, as it follows from the proof of the proposition.

Valuation ﬁelds satisfying the assertion of Corollary 1 are said to be Henselian. Corollary 1

demonstrates that complete discrete valuation ﬁelds are Henselian.

GAlGLCEC

t %

Let ( ) be a monic polynomial with coefﬁcients in . Let

x ¢

2 ¢ +1 +1

©Ì* ©W*/

( 0) ( 0)

x x

¢ +1

( ) = 0.

t t t t & &]©£%

Proof. Put 0( t ) = 0 and write ( ) = 1( )( 0) + with . Then

x x

y ¢

2 ¢ +1 2 +1

2©* t t ©£% t t Í t t *

& . Put 0( ) = 1( ) [ ]. Hence ( ) 0( ) 0( ) mod and

x x

y ¢

¢ +1 +1

( 0) = 0( 0) . Hence ( 0( ) 0( )) , and the proposition implies the

y ¢

y +1

t ©% t t t¯ ]-âÇÍ£ *

existence of polynomials ( t ) ( ) [ ] such that ( ) = 0 mod ,

t t

and ( t ) = ( ) ( ).

>@? ¶(' ¥2©!¶Ü

If the residue ﬁeld of is ﬁnite , then for every ' the polynomial

' 1

t ¥

( t ) = 1 has a root such that = . So one has multiplicative representatives of

consisting of 0 and all roots of unity in of order dividing % 1 (or, equivalently, all roots

of unity in of order prime to ).

GAlGLCEC

\ *ÇÌ!m For every positive integer Y there is such that 1 + .

Om8DTAF¦QiGLTNF¨GLI GL:-B;CE8POm8oqpQi8;CEVAF f¸>@?

¢ ¢© ¢

¢© +1 +1

t 5¥ ¥© *Ç Y¸©+* * ©Ì* *

Proof. Put ( t ) = with 1+ . Let . Then (1) .

ó x

x 2 +1 ¤

¢ +1

* t Í *

Therefore for every ¥© 1 + the polynomial ( ) has a root 1 mod due to ó Corollary 2. x

798;:-:-<= 1

t t ¬¬¬

Let ( t ) = + 1 + + 0 be an irreducible polynomial with coefﬁcients

©% Ü°4©)% ¦ ¦Ç\W

in . Then the condition 0 implies for 0 1.

©% é ¡ 9± ¡ Ü°

° +

Proof. Assume that 0 and that is the maximal such that ( ) = min0 + 1 ( ).

©% If ç±/ , then put

t t

1( ) = ± ( )

x x

± ±

y 1 1 1

t t t ¬¬¬

± ±

0( ) = + ± 1 + + 0

1 ±

t t

0( ) = ± + 1

y y

t t t t

We have 1( t ) = 0( ) 0( ), and 0( ) 0( ) are relatively prime. Therefore, by the

x t

proposition 1( t ) and ( ) are not irreducible.

x x

Γ , - ,}

(2.2). ß and . Let be a ﬁeld and an extension of with a valuation : .

- -K Î ,µ/1 - Γ

Then induces the valuation 0 = ¤ : on . In this context is said to be

- ,

an extension of valuation ﬁelds. The group - 0( ) is a totally ordered subgroup of ( )

- , ß ,µ/0.-

and the index of - 0( ) in ( ) is called the ramiﬁcation index ( ). The ring of

%0/ %1/ * integers is a subring of the ring of integers and the maximal ideal / coincides with

0 0

/ /

* %

/32 0 . Hence, the residue field 0 can be considered as a subfield of the residue field

%0/ 4/

,0/ . Therefore, if is an element of 0 , then its residue in the ﬁeld 0 can be identiﬁed

%1/ ,5/

with the image of as an element of in the ﬁeld . We shall denote this image of

,5/ / 6/ ,/1.- by . The degree of the extension 0 is called the intertia degree ( ). An

immediate consequence is the following lemma. x

798;:-:-<=

- , ,¾ ¾

Let , be an extension of and let be a valuation on . Let and

let - 0 be the induced valuation on . Then

,µ/17- ß ,/ ~7- ß §/1.-

ß ( ) = ( ) ( 0)

µ/17- ,µ/ ~7- §/17-

( , ) = ( ) ( 0)

x x x

µ/1 -

(2.3). Extension of Discrete Valuation. Assume that , is a ﬁnite extension and 0 is

ccc

â ©8, ß¦ß ,µ/17-

a discrete valuation. Let elements 1 for natural ( ) be such that

ccc

- - .- - - , /9- ° p°

( 1) + ( ) ( ) + ( ) are distinct in ( ) ( ). If ° =1 = 0 holds

- °µ© - ° p° -ã ° Ü° ° p°

° +

with , then, as ( ) are all distinct, we get =1 = min1 + ( ) and

ccc

¦ ¦Èß â

so ° = 0 for 1 . This shows that 1 are linearly independent over and

,µ/0.- | -

hence ß ( ) is ﬁnite. Let be a prime element with respect to 0. Then we deduce that

, ¦:- |

there are only a ﬁnite number of positive elements in - ( ) which are ( ). Consider the

, - ,¨ - smallest positive element in - ( ). It generates the group ( ), and we conclude that is

a discrete valuation. Thus, we have proved the following result.

798;:-:-<=

µ/0 - - , - Let , be a ﬁnite extension and 0 discrete for a valuation on . Then is discrete.

Om8;TNF¦QSGATAFUGLI GL:-B;CE8POm8³qpQS8DCEVLF Rfh>@?

, ¡ -

(2.4). ß; for Complete Fields. Let and be ﬁelds with discrete valuations and

;, - ¡

respectively and h . The valuation is said to be an extension of the valuation , if

k¡ -K ¡ ß -K ¡ -K ¡

- 0 = for a positive . We shall write and use the notations ( ) ( ) instead of

,/17- ,/17- ]© - ß -K ¡ ¡

ß ( ) ( ). If then ( ) = ( ) ( ).

798;:-:-<=

Let , be a ﬁnite extension of of degree ; then

- ¡ -K ¡ ¦!\

ß ( ) ( )

ccc

ß ß -K ¡ -K ¡

Proof. Let = ( ) and let be a positive integer such that ¦ ( ). Let 1

x x x

Ë Ë=<

%1/ ,1/

be elements of such that their residues in are linearly independent over & . It sufﬁces

| ¦ ¦ ¦(é¦§ßU °

to show that / are linearly independent over for 1 0 1. Assume x

that Ë

°²± °

/ = 0

°´ù ± °þ±

for °þ±.©+ and not all = 0.

|L& °þ±o©2%'&

Multiplying the coefﬁcients °þ± by a suitable power of , we may assume that

°²±)©* ® °²± °`©!* ® °²± °þ±)©* °

& & °

and not all . Note that if ° , then = 0 and so .

Ë Ë

é ® °þ± °U/©W* é /

Therefore, there exists an index such that ° . Let 0 be the minimal such index.

°þ± °´| é - °²± ß ¦\

Then 0 = ( / ), which is impossible. We conclude that all = 0. Hence,

-K ¡ -K ¡ ¦Ç\

and ß ( ) ( ) .

¡ ¡ ß ¡¡ ¡

>@? Let be the completion of with the discrete valuation . Then ( ) =

¡¡ ¡ L$ ß ¡ ¡ ¡ ¡ ¡L ¡

1 ( ) = 1. Note that if is not complete, then : = ( ) ( ). On the contrary, in x

thex case of complete discrete valuation ﬁelds we have

¥4lGAB;GLF¦Q§ORQSGAT=

Let , be an extension of and let be complete with respect to discrete

7- - ¡L -K ¡ ß ß -K ¡ ¤ |>/©?,

valuations ¡A . Let = ( ) and = ( ) . Let be a prime element

x x

ccc

with respect to - and 1 elements of such that their residues form a basis of

Ë Ë=<

| 4/ , % %1/

° & &

over . Then / is a basis of the -space and of the -module , with

¦ ¦é ¦Çß5 ¤ ,µ/0 \ ß

1 ¦ 0 1. If , then is a ﬁnite extension of degree = .

x x x

Proof. Let · be a set of representatives for . Then the set

@ ¥° ° ¥;° ©W· ¥° A

· = : and almost all = 0 Ë

° =1

[

, | ¡ | | |

& /

is the set of representatives for . For a prime element with respect to put = & ,

ß1Z éD ¦(éÌ §ß !©B, where Y = + 0 . Using Corollary (1.8) we obtain that an element can be

expressed as a convergent series

× | × ©·

= with

Writing

× × × ©W·

° ° ù °

= ù with

Ë ° =1

Om8DTAF¦QiGLTNF¨GLI GL:-B;CE8POm8oqpQi8;CEVAF f¸>@? 14

we get

Ä Ä

[

| | ×

±âù ° ° ã

[ /

= + &

°Åù ±

[

®DC

°Åù ± °î|

Thus, can be expressed as / with

[

°Åù ± × ±"ù °î| ©W ¦ ¦ ¦6é ¦!ßU [

= + & 1 0 1

[

@ | A °

By the proof of the previous lemma this expression for is unique. We conclude that /

, %1/ % form a basis of over and of over & .

(2.5). Uniqueness of Extension of Discrete Valuation From a Complete Field. Further we

X ¡ ß -K ¡ X -

shall assume that ¡ ( ) = for a discrete valuation . Then ( ) = : ( ) for an ¡

extension - of .

EGFN8;GAlN8D:U=

¡ ,

Let be a complete ﬁeld with respect to a discrete valuation and a ﬁnite

- , ¡ - extension of . Then there is precisely one extension on of the valuation and =

¡ -K ¡ , -

¤ IHKJ

ª with = ( ). The ﬁeld is complete with respect to .

x x

-U ¡ -U ,

¤ LHMJ

Proof. Let = ª . First we verify that is a valuation on . It is clear that

U ¤ -U p -U -U -U !-U

- ( ) = + if and only if = 0 and ( ) = ( ) + ( ). Assume that ( ) ( )

for '© , then

- - ã

- ( + ) = ( ) + 1 +

5 N -U N and it sufﬁces to show that if - ( ) 0, then (1 + ) 0. Let

t ¥ t ¬¬¬ ¥

( t ) = + 1 + + 0

N H¥ N

¤ ¤ H

be the monic irreducible polynomial of over . Then we get ( 1) 0 = ( O ) ( ) and if

, N H¥ N ¡ ¥ ¤

= : ( ) , then (( 1) 0) = HMJ ( ). We deduce that ( 0) 0, and making use of

¥° ¦ ¦(Y (2.1), we get ¡ ( ) 0 for 0 1. Now

N ¥ ¬¬¬ ¥

¤ ¤ H

( 1) ( O ) (1 + ) = ( 1) = ( 1) + 1( 1) + + 0

è x

hence

ä ä

¡ã N ¡Kã N

¤ ¤ ¤

H HMJ

( O ) (1 + ) 0 and (1 + ) 0

è è

U N -U ,

i.e., - (1 + ) 0. Thus, we have shown that is a valuation on .

, -U \ç¡ ~©( /E\ -U

Let \ = : ; then ( ) = ( ) for . Hence, the valuation (1 ) is an

, /1\ -U , $ X ß ß ,µ/1 /E\ -U ß

extension of ¡ to (note that (1 ) ( ) = in general). Let = ( (1 ) ); is

/ /

ß1/1\ -5 ,ÇÐr - , - | X X |

ﬁnite. Put - = ( ) : , hence ( ) = ( ) = with a prime element

- - ß1/1\ ¡ ,

¤ IHKJ

with respect to . Therefore, = ( ) ª is at once a discrete valuation on and an è

extension of ¡ .

ccc

.N , ¦£W¦£\ Let N 1 be a basis of the -vector space . By induction on , 1 , we

shall show that

Ä À

( ) ( ) ccc

N ¥ Y ¤QPSR¥ Y ¤ "

° ° ° 0 0 for = 1

° =1

Om8;TNF¦QSGATAFUGLI GL:-B;CE8POm8³qpQS8DCEVLF Rfh>@? 15

( )

¥ © where ° .

The left arrow and the case = 1 are clear. For the induction step we can assume that À

( ) ccc ( )

¥ $ ¡ ¥ °

° 0 for each = 1 . Therefore we can assume that ( ) is bounded for

ccc

= 1 . Hence

Ä Ä 1

( ) ( ) ä ( )

N Æ N ã ¥ ¥ N

° ° °

1 + ° = 1 0 ° ° =2 =1 1

( ) ( ) ä ( )

Æ ¥ ¥ °

where ° = ( 1 . Then

Ä Ä

Ä ( ) ( +1) ( ) ( +1)

Æ ]Æ Æ N Æ N

° °

° ° °

( ° ) = 0

° °

° =2 =2 =2 À

( ) ( +1) ccc

Æ Æ ° and the induction hypothesis shows that ° 0 for = 2 . Thus, each

( ) ( )

Æ Æk°µ©6 N Æ N°

° °

( ) converges to, say, . Finally, the sequence 1 + ° =2 converges both to

® ®

N Æk°TNR° N Æ °TNR° N° ° 0 and to 1 + ° =2 , so 0 = 1 + =2 which contradicts the choice of .

( ) ( )

¥ NR° ¥

° °

Similarly one shows that a sequence ° =1 is fundamental if and only if is

ccc "

fundamental for each = 1 . , Thus, the completeness of implies the completeness of its ﬁnite extension with respect

to any extension of ¡ . We also have the uniqueness of the extension.

GAlGLCEC

, - - VU

Let be a ﬁnite Galois extension of . Then ª = for the discrete valuation

- , © ,/1 | , | U

on and U Gal( ). If is a prime element in , then is a prime element and

%1/ * %1/ *

/ / U

U = , = .

798;:-:-<=

/1 ©2% t

Let , be a ﬁnite extension. Let and let ( ) be the monic irreducible

t ©6%'& t t

polynomial of over . Then ( ) [ ]. Conversely, let ( ) be a monic polynomial

x x

]©W, t ]©%

with coefﬁcients in %'& . If is a root of ( ), then .

¢ ê

Y½

Proof. It is well known that = is separable over for some 0. Let be a

£©! £©%

ﬁnite Galois extension of with . Then, in fact, and the monic irreducible

polynomial ( t ) of over can be written as

y c

t t °î °4© §/1

U U ( ) = ( U ) Gal( ) 1 = 1

° =1

t ©% t t }©%

& °

By the previous corollary we get U . Hence we obtain ( ) [ ] and ( ) =

¢ê y

ä 1

ãît ©)% t ]©3, t t ¥ t ¬¬¬ ¥ ©

& [ ]. If is a root of the polynomial ( ) = + 1 + + 0

1 x

% t /©)%1/ H¥ ¬¬¬î `¥ -LW©Ì* 2©)%1/ &

[ ] and , then 1 = 1 0 / , contradiction. Thus, .

(2.6). Types of Ramiﬁed Extensions. From now on is a complete discrete valuation ﬁeld.

,/1 ¡ ,

Let be an algebraic extension. If J is the unique discrete valuation on which

¡ ¡0¤ ß ,¨ ,5 ß ¡ ¡0¤

extends the valuation = on , then we shall write ( ) ( ) instead of ( J ),

¤ ¤ ¤ ¤ ¤

¡ ¡ % % "* * ¦3 3 "| |

( ). We shall write or or ¤ or or for the ring of integers x

Om8DTAF¦QiGLTNF¨GLI GL:-B;CE8POm8oqpQi8;CEVAF f¸>@?

% * 3 | ¡

& & &

, the maximal ideal & , the group of units , a prime element with respect to , and

the residue ﬁeld .& , respectively.

,'/

A finite extension , of a complete discrete valuation field is called unramified if

µ/1 ,/1

is a separable extension of the same degree as , . A ﬁnite extension is called totally

5 ,µ/0 ,'/

ramified if ( , ) = 1. A finite extension is called tamely ramified if is a separable

,¨ , ,/1

extension and ß ( ) where = char( ) 0. A ﬁnite extension is called totally

,µ/1 ,

ramified if , = . A finite extension is called wildly totally ramified if = and the

µ/0

degree of , is a power of = char( ). µ/1

(2.7). Unramiﬁed Extensions. If , is unramiﬁed then we deduce From Lemma (2.4) that

,¨ ,5 ,

ß ( ) = 1, ( ) = : .

¥4lGAB;GLF¦Q§ORQSGAT=

,/1 , © , ]©%

(1) Let be an unramiﬁed extension, and = ( ) for some . Let J be such

Ë Ë

, , % %

that = . Then = ( ), and is separable over , J = [ ]; is a simple

Ë Ë

root of the polynomial ( t ) irreducible over , where ( ) is the monic irreducible

x x

polynomial of over .

¤ % (2) Let ( t ) be a monic polynomial over , such that its residue is a monic separable

x alg

t ,

polynomial over . Let be a root of ( ) in , and let = ( ). Then the

µ/0 ,

extension , is unramiﬁed and = ( ) for = .

Ë Ë

©~% t

Proof. (1) By the preceding lemma ( t ) [ ]. We have ( ) = 0 and ( ) = 0,

x x x t

deg ( t ) = deg ( ). Furthermore,

x x

, t t ,

: ( ) : = deg ( ) = deg ( ) ( ) : = :

x x

It follows that , = ( ) and is a simple root of the irreducible polynomial ( ). Therefore,

$ $ % %'¤

( ) = 0 and ( ) = 0, i.e., is separable over . Thus, J = [ ].

x x

° t t ñ

(2) Since the residue of is separable, it is separable too. Let ( ) = ° =1 ( ) be the

x x x

t t

decomposition of ( t ) into irreducible monic factors in [ ]. Then every root of ( ) is in

x x

the ring of integers of any ﬁnite Galois extension of which contains all of them; and therefore

t ©% t t t t

° ( ) [ ]. Suppose that is a root of 1( ). Then 1( ) = 1( ) is a monic separable

x x x

polynomial over . The Henselian property of implies that 1( ) is irreducible over .

2©)% ]© , ,!¾

We get J . Since = , we obtain ( ) and

Ë Ë

y c

, R£ , ³R£ t t

deg 1( t ) = : : ( ) : = deg 1( ) = deg 1( )

x x

,µ/0

Thus, , = ( ), and is unramiﬁed.

GAlGLCEC

µ/1¦§/9, §/0

(1) If , are unramiﬁed, then is unramiﬁed.

(2) If , is unramiﬁed, is an algebraic extension of and is the discrete valuation

Y,/

ﬁeld with respect to the extension of the valuation of , then is unramiﬁed.

/1, /0 , , /1 (3) If , 1 2 are unramiﬁed, then 1 2 is unramiﬁed.

Proof. (1) follows from the multiplicativity of the ramiﬁcation index. To verify (2) let

, Ç©% t ©% t ~/©+*

= ( ) with J , ( ) [ ] as in the ﬁrst part of the proposition. Then

Y, because , = ( ). Observing that = ( ), we denote the irreducible monic polynomial

Om8;TNF¦QSGATAFUGLI GL:-B;CE8POm8³qpQS8DCEVLF Rfh>@?

t t

of over by 1( ). By the Henselian property of we obtain that 1( ) is a power of

x x

t t t

an irreducible polynomial over .However, 1( ) divides ( ), hence 1( ) is irreducible

x x x

Y,µ/ separable over . Applying the second part of the proposition, we conclude that is unramiﬁed.

(3) follows from (1) and (2).

,µ/0 ,'/

An algebraic extension , of is called unramiﬁed if are separable extensions

-K ¡ ¡ - ¡ and ß ( ) = 1, where is the discrete valuation on , and is the unique extension of

on , . The third assertion of the Corollary shows that the compositum of all ﬁnite unramiﬁed

alg

extensions of in a fixed algebraic closure of is unramified. This extension is not a complete field in general, but a Henselian discrete valuation field. It is called the maximal

ur ur ur

unramiﬁed extension of . Its maximality implies U = for any automorphism of

sep ur

the separable closure over . Thus, is Galois.

¥4lGAB;GLF¦Q§ORQSGAT=

µ/0 ,'/ ,µ/0 (1) Let , be an unramiﬁed extension and let be a Galois extension. Then is

Galois.

/1 ,/

(2) Let , be an unramiﬁed Galois extension. Then is Galois. For an automorphism

U U U

U Gal( ) let be the automorphism in Gal( ) satisfying the relation ¯ ¯ =

©©£% ,/1

J U

for every . Then the map U induces an isomorphism of Gal( ) onto

Gal( , ). µ/1

Proof. (1) It suffices to verify the first assertion for a finite unramified extension , . Let

, = ( ) and let ( ) be the irreducible monic polynomial of over . Then

Ë Ë

t t °

( ) = ð ( ) Ë

° =1

with ° 1 = . Let ( ) be a monic polynomial over of the same degree as ( )

Ë Ë Ë t

and ( t ) = ( ). The Henselian property implies

t t °

( ) = ð ( ) x

° =1

©% ,

° ° °

with J = . The ﬁrst proposition above shows that = ( 1), and we deduce that

Ë µ/0

, is Galois.

©% J

(2) Note that the automorphism U is well deﬁned. Indeed, if with = , then

6 ©+*

U U

U ( ) and = . It suffices to verify the second assertion for a finite unramified

/1 - t

Galois extension , . Let ( ) be as in the ﬁrst part of the ﬁrst proposition. Since all

, t , ,µ/

roots of ( t ) belong to , we obtain that all roots of ( ) belong to and is Galois. The

x x

,µ/1 ,'/ U

homomorphism Gal( ) Gal( ) deﬁned by U is surjective because the condition

° Ü° p° t Ü° ° ,µ/0 ,µ/ U

U = implies = for the root of ( ) with = . Since Gal( ) Gal( )

Ë Ë Ë

µ/1 ,'/

are of the same order, we conclude that Gal( , ) is isomorphic to Gal( ). É

GAlGLCEC

The residue ﬁeld of coincides with the separable closure of and

ur sep

/1 / Gal( ) Gal( ).

Om8DTAF¦QiGLTNF¨GLI GL:-B;CE8POm8oqpQi8;CEVAF f¸>@? 18

sep

t t

Proof. Let © , let ( ) be the monic irreducible polynomial of over , and ( ) as

Ë Ë

E t , £ °

in the second part of the ﬁrst proposition. Let ° be all the roots of ( ) and = ( ). À

ur ur ur sep x

!! Ü°4©

Then , and = for a suitable . Hence, = .

, (2.8). Maximal Unramiﬁed Field. Let , be an algebraic extension of , and let be a

alg alg ,

discrete valuation ﬁeld with perfect residue ﬁeld. We will assume that = .

¥4lGAB;GLF¦Q§ORQSGAT=

, Let , be an algebraic extension of and let be a discrete valuation ﬁeld.

ur ur ur

, ,- , ,

Then = , 0 = 2 is the maximal unramiﬁed subextension of which is

,/9, contained in , , and 0 is totally ramiﬁed.

ur ur ur

¾:,- ,- Proof. We have , . Since the residue ﬁeld of coincides with the residue ﬁeld

ur ur ur

, ,- , ,

of , we deduce = . An unramiﬁed subextension of in is contained in 0, /0

and , 0 is unramiﬁed.

¥4lGAB;GLF¦Q§ORQSGAT=

, /0

Let , be a ﬁnite Galois extension of and let 0 be the maximal unramiﬁed

,/1 , /1 , / U subextension in . Then 0 and 0 are Galois, and the map U deﬁned in the

second proposition of (2.7) induces the surjective homomorphism

µ/1 , /0 , / ,'/ Gal( , ) Gal( 0 ) Gal( 0 ) = Gal( )

ur ur ur ur ur

/0 , /9, , /=, Ò , /1 The extension , is Galois and Gal( 0) Gal( ) Gal( ),

ur ur ur ur

Ñ Ñ

/1 ,µ/9, , /9, /9,

Gal( , ) Gal( 0), Gal( ) Gal( 0).

U U

Proof. Let U Gal( ). Then 0 is unramiﬁed over , hence 0 = 0 and 0

µ/0 , / is Galois. The surjectivity of the homomorphism Gal( , ) Gal( 0 ) follows from the

µ/0 /1 second proposition of (2.7). Since , and are Galois extensions, we obtain that

ur ur ur

- /1 , ,' , is a Galois extension. Then = by the previous proposition. The remaining assertions are easily deduced by Galois theory.

(2.9). Tamely Ramiﬁed Extensions.

¥4lGAB;GLF¦Q§ORQSGAT=

, /1

(1) Let , be a finite tamely ramified extension of , and let 0 be the maximal unramified

,µ/1 , , | % % | | , J

subextension in . Then = 0( ) and J = 0 [ ] with a prime element in

| , ß ß ,5

satisfying the equation tà | 0 = 0 for some prime element 0 in 0, where = ( ).

/0 , , @à £©Z, ß

(2) Let , 0 be a ﬁnite unramiﬁed extension, = 0( ) with = 0. Let if

, ,µ/1

= char( ) 0. Then is separable tamely ramiﬁed.

µ/=, | ,

Proof. (1) (2.8) shows that , 0 is totally ramiﬁed. Let 1 be a prime element in 0, then

| |à | , ©3 , , ×W©%

J J = for a prime element J in and . Since = , there exists such 1 J 0 0

1 1

C C

× | ×L |à 1×L ©(% that = . Hence = for the principal unit = J . For the polynomial

1 J

tà4 ©W* ß

( t ) = we have (1) , (1) = . Now Corollary 2 of (2.1) shows the existence

x x

x 1

[W©% [ [ | | × | | [ J of an element J with = , = 1. Therefore, = 1 , 0 = are the elements

desired for the ﬁrst part of the Proposition.

| | , ©§3 ó

(2) Let = 1 for a prime element 1 in 0 and a unit J 0 . The polynomial

tàD , t t tàD ¨

( t ) = is separable in 0[ ] and we can apply the Henselian property to ( ) = x

Om8;TNF¦QSGATAFUGLI GL:-B;CE8POm8³qpQS8DCEVLF Rfh>@? 19

sep

t , × /=,

and a root ×©2 of ( ). We deduce that 0( ) 0 is unramiﬁed and hence it sufﬁces

, × â , × to verify that §/ 0 for = ( ) 0 = 0( ), is tamely ramiﬁed. We get = 0( 1)

àè\ è\

p×L @à | ß;"¥ ¨È ¦á |

with 1 = , 1 = 1ó . Put = g.c.d.( ). Then 0( 2 ) with 2 = 1

á á /

and a primitive ß th root of unity. Since the extension 0 ( ) 0 is unramiﬁed (this can

á ¡

be veriﬁed by the same arguments as above), | 1 is a prime element in 0 ( ). Let be the

âá ¥R/= ¡ | © ß0/9 X ¡ | © ß0/9 X

discrete valuation on 0 ( 2 ). Then ( ) ( 1) ( ) and ( 1) ( ) , because

¥m/9 ß0/9 ß ¦á A á =

and are relatively prime. This shows that 0( 2 ) 0( ) ß1/ . However,

áR" á U¦ ß0/= ám" / á

0( 2) : 0( ) , and we conclude that 0( 2) 0( ) is tamely and totally

ám" / §/

ramiﬁed. Thus, 0 ( 2) 0 and 0 are tamely ramiﬁed extensions.

GAlGLCEC

µ/1¦§/9, §/0

(1) If , are tamely ramiﬁed, then is separable tamely ramiﬁed.

/1 §/0

(2) If , is tamely ramiﬁed, is an algebraic extension, and is discrete, then

Y,µ/

is tamely ramiﬁed.

/1, /0 , , /0 (3) If , 1 2 are tamely ramiﬁed, then 1 2 is tamely ramiﬁed.

Proof. It is carried out similarly to (2.8). To verify (2) one can ﬁnd the maximal unramiﬁed

/1 ,µ/0 Y,/ Y,

subextension , 0 in . Then it remains to show that 0 is tamely ramiﬁed. Put

, | |à | Y, Y, | , = 0( ) with = 0. Then we get = 0( ), and the second part of the proposition yields the required assertion.

(2.10). Totally Ramiﬁed Extensions. A polynomial

t ¥ t ¬¬¬ ¥ %

( t ) = + 1 + + 0 over

is called an Eisenstein polynomial if c

ccc 2

"¥ ©Ì*õ¥ ©W*/

¥ 0 1 0

¥4lGAB;GLF¦Q§ORQSGAT=

(1) The Eisenstein polynomial ( t ) is irreducible over . If is a root of ( ), then

x x

/1 \ % ¤

( ) is a totally ramiﬁed extension of degree , is a prime element in ( ), ( )

= % [ ].

µ/1 \ |

(2) Let , be a separable totally ramiﬁed extension of degree , and let be a prime

| \

element in , . Then is a root of an Eisenstein polynomial over of degree .

t , ß ß ,5

Proof. (1) Let be a root of ( ), = ( ), = ( ). Then x

\ç¡ ¡ ã ¥° ß¡ ¥° ¡

J J

J ( ) = min ( ( ) + ( ))

° +

0 + 1 R

° =0

¡ ¡ , ¡ , J

where and J are the discrete valuations on and . It follows that ( ) 0. Since

À À

¤ ¤ ¤

ß£¡ ¥ ~ß£¡ ¥° ¡ , \ç¡ ß£¡ ¥ ß ¡ "\

J J

( 0) ( ) + J ( ) for 0, one has ( ) = ( 0) = .Then ( ) = 1 =

ß; % %

= 1, and J = [ ].

, , | (2) Let | be a prime element in . Then = ( ). Let

t ¥ t ¬¬¬ ¥

( t ) = + 1 + + 0

| \ ß \ç¡ | ¥ \ç¡0¤

° ã

J +

be the irreducible polynomial of over . Then = and ( ) = min0 + 1 ( ) +

¤ ¤ ¤

¥° , \ \ç¡ ¥ ¡ ¥ , hence ¡ ( ) 0, and = ( 0), ( 0) = 1.

Om8DTAF¦QiGLTNF¨GLI GL:-B;CE8POm8oqpQi8;CEVAF f¸>@?

,µ/0

(2.11). Ramiﬁcation Groups. Let , be a ﬁnite Galois extension of , = Gal( ). Put

" "

+1 c

@ A

J U

= : U for all 1

" " " " °

Then 1 = by Lemma (2.11) and ° +1 is a subset of .

¡ , ]

Let J be the discrete valuation of . For a real number deﬁne

"4^ "

A @

J J U

= U : ( ) + 1 for all

" "

Certainly each of is equal to ° with the least integer .

798;:-:-<=

" "

° are normal subgroups of .

° ° °

" +1 1 1 +1 +1

U U U

Proof. Let U . Then . Hence ( ) ( ) =

J J J "

1 "

° +1

` 6 ` 6 ]©+*

U U

U ( ) = ( ( ) ) + ( )

" " "

`©© ° © °.`©© ` ©Ï% ¸©Ï%

J J U

i.e., U . Furthermore, let . Then ( ) for and

° "

° +1 1 +1 1

`m ?`m(©Ì* ` ` ]©W* ` `+© °

U U

U ( ) , ( ) , .

J J

" "

^ µ/1

The groups are called (lower) ramiﬁcation groups of = Gal( , ).

¥4lGAB;GLF¦Q§ORQSGAT=

Let , be a ﬁnite Galois extension of , and let be a separable extension of

À À

" " "

,/9, . Then 0 = Gal( 0) and the th ramiﬁcation groups of 0 and coincide for 0.

Moreover,

" "

+1 b

° _ © |W 6|6©W* U = U :

0 J

, °

for a prime element | in , and = 1 for sufﬁciently large .

" "

Proof. Note that U 0 if and only if Gal( ) is trivial. Then 0 coincides with

µ/0 ,'/

the kernel of the homomorphism Gal( , ) Gal( ). The ﬁrst proposition of (2.11) and

µ/9, °

the second proposition of (2.7) imply that this kernel is equal to Gal( , 0). Since is a

À À

" "

subgroup of 0 for 0, we deduce the assertion about the th ramiﬁcation group of 0. We

% % | ¥°´| ©]% %

J J J J

get = 0 [ ]. Let = ° =0 be an expansion of with coefﬁcients in 0 .

¥ ¥ ©

° ° U

As U = for 0 it follows that

° °

6 ¥ ã | 6|

° U U = ( )

° =0

" ° ° " "

° | ¨| © ° ¡ |' ¨| ©

U U

Now we deduce the description of , since U ( ) . If max ( ) : ,

then = 1 .

" "

The group 0 is called the inertia group of , and the ﬁeld , 0 is called the inertia subﬁeld µ/1

of , .

¥4lGAB;GLF¦Q§ORQSGAT=

, | Let , be a ﬁnite Galois extension of , a separable extension of , and

a prime element in , . Introduce the maps

¦ ¦

" "

A , L , , °

0: 0 ° : ( 0)

¦ Ó ¦

° ° ° |p/E|

° U

by the formulas ( U ) = ( ). Then is a homomorphism with the kernel +1 for À

Om8;TNF¦QSGATAFUGLI GL:-B;CE8POm8³qpQS8DCEVLF Rfh>@? 21

Proof. The proof follows from the congruence

` | `R| | `R| |

U U

( ) U

3 ¬ Í ¬ °

= U3a mod +1

| |Zb | | |

" "

7`Î© ©

° ° ° U

for U . The kernel of consists of those automorphisms , for which °

° +1 +2

|p/E|© * |+ 6|©+* U

U 1 + , i.e., .

J J

GAlGLCEC

, ,

e Let be a ﬁnite Galois extension of , and a separable extension of .

" " " "

, /

If char( ) = 0, then 1 = 1 and 0 is cyclic. If char( ) = 0, then the group 0 1

" " "

°Þ/

is cyclic of order relatively prime to , ° +1 are abelian -groups, and 1 is the maximal

-subgroup of 0.

Proof. The previous proposition permits us to transform the assertions of this corollary into

the following: a ﬁnite subgroup in , is cyclic (of order relatively prime to char( ) when

$ , ,

char( , ) = 0 ); there are no nontrivial ﬁnite subgroups in the additive group of if char( ) = 0;

, ,

if char( , ) = 0 then a ﬁnite subgroup in is a -group.

GAlGLCEC

, ,

Let be a ﬁnite Galois extension of and a separable extension of .

µ/=, , /0

Then the group 1 coincides with Gal( , 1), where 1 is the maximal tamely ramiﬁed µ/1

subextension in , .

/9,

Proof. The extension , 1 0 is totally ramiﬁed by the ﬁrst proposition of (2.11) and is the

µ/=,

maximal subextension in , 0 of degree relatively prime with char( ). Now Corollary 1

/9,

implies 1 = Gal( , 1).

GAlGLCEC

Let , be a ﬁnite Galois extension of and a separable extension of .

/ ,µ/1 Then 0 is a solvable group. If, in addition, , is a solvable extension, then is solvable.

Proof. It follows from Corollary 1. , (2.12). The Norm Map for Cyclic Extensions. Let be a local ﬁeld and its Galois

extension of prime degree \ . Then there are four possible cases: /1

, is unramiﬁed; /1

, is tamely and totally ramiﬁed;

/1 ,

, is totally ramiﬁed of degree = char( ) 0;

798;:-:-<=

/1 \ N©W*

Let , be a separable extension of prime degree , . Then

N N N &

¤ ¤ ¤ ¤

HKJ J J

HMJ (1 + ) = 1 + ( ) + Tr ( ) + Tr ( )

è è è è

&©]% ¡ & ¡ N

¤ ¤

J J J J

with some such that ( ) 2 ( ) ( HMJ and Tr are the norm and the trace è

maps, respectively). è

, °

Proof. Recall that for distinct embeddings U of over into the algebraic closure of ,

À ¦Ç\

1 ¦ , one has

2©W,

° °

¤ ¤

U J U

HMJ = ( ) Tr = ( )

è è ° ° =1 =1

Om8DTAF¦QiGLTNF¨GLI GL:-B;CE8POm8oqpQi8;CEVAF f¸>@? 22

Hence

Ä Ä Ä

N N N c c N ¬¬¬ N N

° ° °ed d ° ±

ð ð

HMJ U U U U

(1 + ) = (1 + ( )) = 1 + ( ) + U ( ) + + ( )

° ° ° ° ± +

=1 =1 =1 1 + =1

& N ± N ¬¬¬ ¡ & ¡ N

J J ±

For = U ( ) + we get ( ) 2 ( ). +

1 +

Our nearest purpose is to describe the action of the norm map HMJ with respect to the

ﬁltration on , and .

¥4lGAB;GLF¦Q§ORQSGAT=

/1 \ |

Let , be an unramiﬁed extension of degree . Then a prime element

° °

¤ ¤

, 3 °Åù | % 34°´ù | %

¤ ¤ J in is a prime element in . Let J = 1 + , = 1 + . Then the following

diagrams are commutative:

l Û

0 Û

Û Û Û Û

3 , m m k

°´ù

,¨ m m kaX 3 m m k ,

Û J k Û k

Û Û Û Û

g g

f=h i

f=h i f=h i

f=h i i

f9h Tr

j j

j j j j

i l

0 l

m m kaX

3 ¤ m m k

°Åù

3 m m k

k k

Proof. The commutativity of the ﬁrst two diagrams is easy. The preceding Lemma shows that

¢ °

° °

| | | &

¤ ¤ ¤ ¤

¤ ¤

J HMJ J

HMJ (1 + ) = 1 + (Tr ) + ( ) + Tr ( )

è è è è

Ë Ë Ë

À À

¡ & ¡0¤ &

¤ J

with ( ) 2 and, consequently, Tr J ( ) 2 . Thus, we get

° °

° +1

Í | | |

¤ ¤

¤ ¤ ¤ J

HMJ (1 + ) 1 + (Tr ) mod

è è

Ë Ë

and the commutativity of the third diagram.

GAlGLCEC

3 3

¤ ù ù J

In the case under consideration HMJ 1 = 1 .

¥4lGAB;GLF¦Q§ORQSGAT=

µ/0 \

Let , be a totally and tamely ramiﬁed cyclic extension of degree . Then

| , |L¤ |9 ,

for some prime element J in , the element = is prime in and = . Let

° °

3 | % 3 ¤ | %'¤

°´ù °´ù

¤ J

J = 1 + , = 1 + . Then the following diagrams

ò Û

0 Û

Û Û Û Û

3 °´ù m m ¦ ,

, m A â X

3 £ m m â ,

k k

Û Û =

Û Û Û Û

Ú Û

g g

f=h i

f=h i f=hmi

j j

j j j

j id

i l

0 l

m A â X

m m ¦ 3 °Åù

3 £ m m â

Ú Û

\ Ò \

are commutative, where id is the identity map, \ takes an element to its th power, is

À À

3 3 \© \

°Åù

¤ ¤ ° ù

J HMJ

the multiplication by , 1. Moreover, HKJ = +1 if .

è è

| ,µ/0 ,µ/0 \

Proof. Since | = and is Galois, then Gal( ) is cyclic of order and

| á0| ,µ/0 á \ áo©+

J J U U ( ) = for a generator of Gal( ), where is a primitive th root of unity, .

It is easy to see that the ﬁrst diagram is commutative.

,µ/0 ©% ©

J U Corollary in (2.5) shows that U ( ) = for Gal( ), , and we get the commutativity of the second diagram.

Om8;TNF¦QSGATAFUGLI GL:-B;CE8POm8³qpQS8DCEVLF Rfh>@?

é \ ©+ ©%'¤

If = , then 1 + | for , and

Ë Ë

° ° °

+1 c

| | Í \ | |

¤ ¤ ¤

HMJ (1 + ) = (1 + ) 1 + mod

Ë Ë Ë

3 ¤ 3. 3

°´ù °´ù

°Åù J

We deduce = = HMJ

À ±

1 X

ñ ¤

t t á \ ©)%

Finally, 1 = ± =0 ( ), therefore for and for one has Ë

° ± ° °

| á | |

¤ ð

HMJ (1 + ) = (1 + ) = 1 ( )

J J

Ë Ë Ë

± =0

3 °Åù 3

¤ ¤ ° ù

J HMJ

Thus HMJ = +1 .

è è

GAlGLCEC

3 3

¤ ù ù J

In the case under consideration HMJ 1 = 1 . If is algebraically closed

,¨ ¤

then HKJ = .

è /1

Now we treat the most complicated case where , is a totally ramiﬁed Galois extension

, % % | , | |

J J J

of degree = char( ) 0. In this case J = [ ], = ( ) for a prime element

, , ,/1 | /1| ©3

J J J U

in , and = . Let U be a generator of Gal( ), then ( ) . One can write

| /E| * ©3 ¤©

J J

U ( ) = with 1 + . Then

Ë Ë

2 2

| /1| ¬ 5¬

J J

U U

U ( ) = ( ) = ( )

Ë Ë Ë

and

¢ ¢ ¢

1 c

| /E| 5¬ ¬ ¬¬¬4¬

J J

U U

1 = U ( ) = ( ) ( )

This shows that 1 + and 1 + ¤ , because raising to the th power is an injective

Ë Ë

| /1| © *

J J

homomorphism of . Thus, we obtain U ( ) 1 + . Put

J ¢

U ( ) c

×;| ×©)3 n ,5 o

= 1 + with J = ( ) 1 ( )

| J

Note that does not depend on the choice of the prime element and of the generator U of

" /1

= Gal( , ). Indeed, we have

U U

¢ ¢

( ) ¢ +1 ( ) +1

×| | Í |

Í 1 + mod and 1 mod

J J J

C ©)3

for an element J . We also deduce that

U ( )

©)3 ù

" " "

]©3,¨ ¢

for every element . This means that = , ¢ +1 = 1 .

798;:-:-<= ¢

¢ 1

t t ¥ t ¬¬¬ ¥ | ¢

Let ( ) = + 1 + + 0 be the irreducible polynomial of J over .

Then x

¦]é³¦

| 0 if 0 2

¤qp

Tr J =

| r

( J ) 1 if = 1 x

Om8DTAF¦QiGLTNF¨GLI GL:-B;CE8POm8oqpQi8;CEVAF f¸>@?

| ¦ ¦ t J

Proof. Since U ( ) for 0 1 are all the roots of the polynomial ( ), we get x

¢ 1 Ä 1 1

= c

° °

ä ä

ã ã

| t |

( ) J U

( ) U ( ) °

x =0 x

t] s

Putting s = and performing the calculations in the ﬁeld (( )), we consequently deduce

¢ ¢

t s ¥ s ¬¬¬ ¥ s

( ) = (1 + ¢ 1 + + 0 )

s ¢

1 ¢ +1

!s s

= Í mod

t ¥ s ¬¬¬ ¥ s

( ) 1 + ¢ 1 + + 0

± °

1 ± +1

| s

= = U ( )

° °

t | | s

J J U U ( ) 1 ( )

± 0

/ s s s

(because 1 (1 ) = ° 0 in (( )) ). Hence

± ±

1 °

Ä Ä

| s

¢ ¢

U ( )

J +1

qs s

Í mod

U ( ) °

± 0 =0

± ±

1 °

¦6é ¦ | | 0 if 0 2

U ( )

J J

¤qp

Tr J = =

J J

( ) U ( ) 1 if = 1 °

x =0 x

as desired.

À À

¥4lGAB;GLF¦Q§ORQSGAT=

¦ ¥ é

Let [ ¥ ] denote the maximal integer . For an integer 0 put ( ) =

t u /

+ 1 + ( 1 ) . Then

± ° °

( ) c

| % | %

J ¤

Tr J ( ) =

° °

¢ ¢

1 ä +1

ñ ã

| | | /E| Í ×| | |

J J J J J

° U

Proof. One has ( ) = U ( ) and ( ) 1 + mod . J =1 J

Then x

¢ ¢

1 ( 1)( +1)

| #× |

( J ) = ( 1)!( )

x ¢

( 1)( ¢ +1)+1

* , |L¤

with some )© 1 + . Since = , for a prime element in one has the

|L¤ | 1 0 ç©)3

representation = with J . The previous lemma implies

¦é ±

+ ¢ +1 0 if 0 1

¤ ±

¢ J

Tr a =

J ¢

+ +1

è +1

| é

¤ if = 1

¢ ° °

+1 1 ä

0 /çã #× | |

¤ ¤ ±

¤ ¤ J

for + +1 = ( ) ( 1)!( ) . Since Tr J ( ) = Tr ( ) we can choose

è è

¢ xX

+ +1

é | é

± ±

¢ ¢

the units , for every integer , such that Tr J ( ) = 0 if ( + 1) and

+ +1 J + +1

± ¢

¢ °

+( +1)

| é % | % | é £± J

= ¤ if ( + 1). Thus, since the -module is generated by , , we

° ±

° ( )

| % | %

J ¤

conclude that Tr J ( ) = .

J è

Om8;TNF¦QSGATAFUGLI GL:-B;CE8POm8³qpQS8DCEVLF Rfh>@?

¥4lGAB;GLF¦Q§ORQSGAT=

µ/0 ,

Let , be a totally ramiﬁed Galois extension of degree = char( ) 0.

| , |L¤ 3 |

°Åù

J J J

Let be a prime element in . Then = HMJ is a prime element in . Let =

° °

¤ ¤

| % k34°´ù | % ¤

1 + J = 1 + . Then the following diagrams are commutative:

0 l

Û Û

, N m ¦ X , 3 m A â

J k

Û Û Û Û

g g

f=hmi f=h i

j j y j

j id

i i

0 l

N m ¦ X

3 ¤ m A â

f f f

l l

+ l

¤ ¤

Û Û Û Û Û Û

ü ü

3 °Åù N m ¦ , 3 ù m A â , 3 ¢°´ù , m N m ;

¢ ¢

J J J

k k k

Û Û Û Û Û

Û = = + =

g g g

^ å ^ å

f=h i f=h i f=h i

¢ 1 1

j£y j j z7{ z z j j

j ( )

ô | |

i i i

l l

l +

¤ ¤

ü ü

¤ ¤ ¤

34°´ù 3 ù m A â 3 °´ù N m ¦ N m ¦

¢ ¢

k k

k +

À À

: ,

where 1 ¦ in the third diagram and 0 is the last diagram.

À À

, 3 °´ù 3

¤ ¤ ° ù

J HMJ

Moreover, HMJ ( + ) = ( + +1 ) for 0 .

è è

Proof. The commutativity of the ﬁrst and the second diagrams is easy.

| ©3

To treat the remaining diagrams, put = 1 + with J . Then, by the ﬁrst lemma,

J Ë

we get Ë

° °

| | &

¤ ¤ ¤ ¤

HMJ J J

HMJ = 1 + ( ) + Tr ( ) + Tr ( )

è è è è

Ë Ë Ë

¡ &

with J ( ) 2 . The previous proposition implies that

À À

° 1 2 1

ä ä

¡0¤6ã | ~ & ~ : m¡0¤ã :

¤ ¤ J

Tr J ( ) + 1 + Tr ( ) + 1 +

è è

À Z

and for

À À

ä ä

¤ ¤

¡ ã | ¡ ã &

¤ ¤ J

Tr J ( ) + 1 Tr ( ) + 1

è è Therefore, the third diagram is commutative.

Further, using ( o ), write

U ¢ ¢ ( ) ¢

+1 c

Í c × | ×| |

¤ ¤ ¤

¤ ¤

HMJ J

1 = HMJ 1 + ( ) + Tr ( ) mod

è è è

| J

We deduce that

¢ ¢ ¢

+1 c

×| Í£ × | |

¤ ¤

¤ ¤ HKJ

Tr J ( ) ( ) mod

è è

× Í× | 3 §3 3

¤ ù

J J J

Since HMJ ( ) mod in view of 1 , we conclude that

¢ ¢ ¢

¢ +1

×| 6× | ©Ì| %

¤ J

HMJ (1 + ) 1 ( )

Ë Ë Ë Ë ©)%'¤

for ©%'¤ . This implies the commutativity of the fourth (putting ) and the ﬁfth (when

Ë Ë

©W| %'¤

¤ ) diagrams.

Ë ©)%'¤

Finally, if , then

| À

° °

¢ ¢

U (1 + ) c

J + + +1

×| Í |

° 1 + mod

J J Ë

1 + |

Ë Ë

) 79GLj

26 f

À °

+ ¢

¤ ¤ ° ù

J HMJ This means that HKJ (1 + ) and

J + +1

è è

3 3

°´ù

¤ ¤ ° ù

J HMJ

HMJ ( + ) = ( + +1 ).

è è

#8;:

power in Proposition (1.10).

GAlGLCEC

3 3

ù ¤ ù

¢ ¢ J

+1 = HKJ +1 .

, ¤

If is algebraically closed then HKJ = . è Proof. It follows immediately from the last diagram of the proposition, since the multiplication

¢ 1

× by ( ) is an isomorphism of the additive group .

3: Local Class Field Theory

This section focuses on complete discrete valuation fields with finite residue field.

3.1. Useful Results on Local Fields with Finite Residue Field

(3.1.1). Structures. Let be a local field with finite residue field = , % = elements.

Since char( ¶ ) = , is of characteristic 0 or of characteristic .

, ¡ ¡ r

In the ﬁrst case ¡ ( ) 0 for the discrete valuation in , hence the restriction of on

is equivalent to the -adic valuation (by Ostrowski’s Theorem). Then we can view the ﬁeld r-¢

ß ¡ ß

of -adic numbers as a subﬁeld of . Let = ( ) = ( ) be the absolute ramiﬁcation index

r \ ß of . Then is a finite extension of ¢ of degree = . Such a field was called a local number field. x

In the second case is isomorphic (with respect to the ﬁeld structure and the discrete

>' t t valuation topology) to the ﬁeld of formal power series ¶ (( )) with prime element , since

the multiplicative representatives of the residue field form a finite subfield of . Such a field

was called a local functional ﬁeld. 798;:-:-<=

is a locally compact topological space with respect to the discrete valuation *

topology. The ring of integers % and the maximal ideal are compact. The multiplicative 3

group is locally compact, and the group of units is compact.

A° °´Á¨§ %

Proof. Assume that % is not compact. Let ( ) be a covering by open subsets in , i.e.,

A° % A° | %

% = , such that isn’t covered by a ﬁnite union of . Let be a prime element of .

©)% |p%

Since %/1|p% is ﬁnite, there exists an element 0 such that the set 0 + is not contained

ccc

Ë Ë

A° ©%

in the union of a ﬁnite number of . Similarly, there exist elements 1 such that

Ë +1 Ë

¬¬¬ |9 |9 % A°

0 + 1 | + + + is not contained in the union of a ﬁnite number of . However,

Ë Ë Ë

{®

|9 A° %

the element = lim + =0 belongs to some , a contradiction. Hence, is

Pô

|p% $

compact and 3 , as the union of + with = 0, is compact.

Ë Ë

)R= = HFk8DIPMNCx#8;FkMNCâOmFUGLT79GLj

e 27

(3.1.2). Galois Extensions. 798;:-:-<=

The Galois group of every ﬁnite extension of is solvable.

Proof. Follows from (2.11).

¥4lGAB;GLF¦Q§ORQSGAT=

, For every \ 1 there exists a unique unramiﬁed extension of of degree

\ , ,µ/0 '

: = ( ò 1). The extension is cyclic and the maximal unramiﬁed extension of

ur ¡

/0 X

is a Galois extension. Gal( ) is isomorphic to and topologically generated by an ¤

automorphism such that

¤ ¤

Í§ * (©% ur

( ) mod ¤ ur for

@¤

The automorphism is called the Frobenius automorphism of .

Proof. First we note that, by (2.1) contains the group 1 of ( % 1) th roots of unity

which coincides with the set of nonzero multiplicative representatives of in . Moreover,

3 Ò)3

¤ '

the unit group is isomorphic to 1 1 ù .

' ' '

¶ ¶ \ ¶

The ﬁeld has the unique extension ò of degree , which is cyclic over . (2.7) shows

, \ , '

that there is a unique unramiﬁed extension of degree over and hence = ( ò 1).

]©3 /0 Now let be an unramiﬁed extension of and . Then ( ) is of ﬁnite degree.

Therefore, is contained in the union of all ﬁnite unramiﬁed extensions of . We have c

ur ¡

Ñ Ñ

' '

/£¶ /0 ¶ X

Gal( ) lim Gal( ò )

sep

¶ /£¶ '

It is well known that Gal( ) is topologically generated by the automorphism U such

' sep ur

¥ ¥ ¥Ì©+¶ /1 '

that U ( ) = for . Hence, Gal( ) is topologically generated by the Frobenius ¤

automorphism .

#8;:

If ò 1, then

@¤ Í *

( ) mod

Ë Ë

and ( ) ò 1. The uniqueness of the multiplicative representative for implies

Ë Ë ¤

now that ( ) = .

Ë Ë

( )

>@? á"¢ r r ¢ á"¢

¢ ê

(3.1.3). Let ê be a primitive th root of unity. Put = ( ). Then

¡9 á ¢

( ) ê

ê ( ) = 0 ^

( )

á r

¢ ¢

and ê belongs to the ring of integers of . Let

¢ê

å å

t 1 1

¢ê ¢ ¢ê 1 ¢

( 1) ( 2) c

t t t ¬¬¬ ( ) = å = + + + 1

¢ 1

t 1

( ) 1

á"¢ t r r4¢ÜÜ¦ á

ê ¢ ¢

Then is a root of ( ), and hence : ( 1) . The elements ê ,

À À

d t

0 , are roots of ( ). Hence

° °

t t ]á ]á

¢ ¢

ð ð

( ) = ( ê ) and = (1) = (1 )

x x

¢§° ¢§°

° ¢ê ° ¢ê

0 0

) 79GLj

28 f

However, °

° 1 1

]á á á ¬¬¬ á

¢ ¢

ê ê ê

(1 ê )(1 ) = 1 + + +

( ) ° 1

r á á

¢ ¢ ê

belongs to the ring of integers of . For the same reason, (1 )(1 ê ) belongs to

å 1

° ¢ê ¢

( ) 1 ( 1)

r +á Wá +á

¢ ¢

¢ ¢ ê ê

the ring of integers of . Thus, (1 ê )(1 ) is a unit and = (1 )

( ) 1 ( )

ß r r-¢ U r ¢

for some unit . Therefore, ( ¢ ) ( 1) , and is a cyclic totally ramiﬁed

6á"¢ U r ¢

extension with the prime element 1 ê , and of degree ( 1) over . In particular,

% % ^ ]á"¢ % ^ á"¢ ê

( ) = [1 ê ] = [ ]

(3.1.4). The Group of Principal Units. If char( ) = , then Proposition (6.2) Ch. I shows

that every element ]©3 1 can be uniquely expressed as the convergent product

± °

óküþý ð

= ð (1 + )

± ÁDû

¢§°

°´ú 0

¤ %

where the index-set ø numerates elements in , such that their residues form a basis of

¶>' ¶ | %'¤

± °

over ¢ , and the elements belong to this set of elements; are elements of with

| ¡ | ¥ ©+X 3

± ° °þ± ¢

( ° ) = , and . Thus, 1 has the inﬁnite topological basis 1 + .

Ë ]©3

Now let char( ) = 0. Every element 1 can be expressed as a convergent product

| ó ó

± ° üþý ð

= ð (1 + )

°´Á¨§ ± ÁDû

À À

¦ ß0/ ß ß ø

where = 1 ( 1), , = ( ); the index-set numerates elements in

¤ '

% ¶ ¶;¢

, such that their residues form a basis of over , and the elements ± belong to this set

% ¡ |A° ¥°þ±©+X¢

of elements; |A° are elements of with ( ) = , and .

"¥

If a primitive th root of unity does not belong to , then = 1 = 0 and the above

3 Xµ¢ \ ß r-¢p

expression for is unique; 1 is a free -module of rank = = : .

If a primitive th root of unity belongs to , then = 1 + ( ¢ 1) is a principal unit

àè

©W/ ¥©WXµ¢ 3

such that , and . In this case the above expression for is not unique. 1 is

\ Xµ¢ L¢ K

isomorphic to the product of copies of and the -torsion group , where 1 is the

! ¢

maximal integer such that .

798;:-:-<=

\Ç

If char( ) = 0, then is an open subgroup of ﬁnite index in for 1. If

char( ) = , then is an open subgroup of ﬁnite index in for . If char( ) =

\ and , then is not open and is not of ﬁnite index in .

Proof. It follows (1.9) and (1.10) and the previous considerations.

, ¤

(3.1.5). The Norm Groups. Now we have a look at the norm group HKJ ( ) for a ﬁnite

extension , of . Recall that the norm map

¶ A ¶

H4© :

' ' ¾Ç¶

is surjective when ¶ .

3 ,µ/0 3

¤ J

Let be a ﬁnite unramiﬁed extension. Then (2.12) implies that HKJ = in the

,µ/0 , Å|9w Ò 3 | ¤

case of an unramiﬁed extension and HKJ = , where is a prime element

, \ , 9/ \ ¤

in , = : . This means, in particular, that HKJ is a cyclic group of order è

)R= = HFk8DIPMNCx#8;FkMNCâOmFUGLT79GLj

e 29

¤ 3

in this case. Conversely, every subgroup of ﬁnite index in that contains coincides with

,µ/0 , ¤

the norm group HKJ for a suitable unramiﬁed extension .

µ/1 \

Let , be a totally and tamely ramiﬁed Galois of degree . (2.12) shows that

|© 3 3 ,

¤ ù ù ¤

J HMJ

HKJ 1 = 1

è è

| , #| © , ¤

for a suitable prime element in (e.g. such that = ( ), and HKJ for

if and only if . Since is Galois, we get and ( % 1).

Ë Ë

¶R \ ¶ ¶ /E¶R

' ' ' Hence, the subgroup ' is of index in , and the quotient group is cyclic. We

conclude that

, Å|Ò #Ò)3

¤ ù

HKJ = 1

©§3 ¶Ü /£¶m 9/ ,¨

' '

with an element , such that its residue generates . So HKJ is

Ë Ë

cyclic of order \ . Conversely, every subgroup of index relatively prime to char( ) coincides

,¨ ,µ/0 ¤

with the norm group HKJ for a suitable cyclic extension .

/1 ,/1

Let , be a totally ramiﬁed Galois extension of degree . The right vertical ¢

¢ 1

homomorphism of the fourth diagram in the last proposition of (2.12) has

Ë Ë Ë

kernel of order ; therefore its cokernel is also of order . Let ©~3 be such that

Ë Ë

does not belong to the image of this homomorphism. Since is perfect, we deduce that

3 | © / ,¨ 9/

¤ ¤ ù

J HKJ

1 + 1 . The other diagrams imply that HMJ is a cyclic group of order

è è

| , Ì¦ ß1/

¤ ¤

and generated by 1 + mod HMJ . If char( ) = 0, then ( 1), where

ß ß0/ á"¢

ß = ( ), and if then = ( 1) and a primitive th root of unity belongs to ,

, , | | / ¤

and = ( ) for a suitable prime element in . In this case HMJ is generated

,¨

by mod HKJ . è

ur ur (3.1.6). Completion of . The ﬁeld is a Henselian discrete valuation ﬁeld with

algebraically closed residue ﬁeld and its completion is a local ﬁeld with algebraically closed

sep ¶

residue ﬁeld ' .

Let · be the set of multiplicative representatives of the residue ﬁeld of if its characteristic

· \6 '

is . is the union of all sets ò 1 1 (which coincides with the set of all roots of unity

of order relatively prime to ) and of 0.

Let be a ﬁnite separable extension of . Since the residue ﬁeld of is algebraically

U/¨ closed, is totally ramiﬁed.

798;:-:-<= The norm maps

3 ÷3

HK : :

è è

are surjective.

H/¨

Proof. Since the Galois group of is solvable, it sufﬁces to consider the case of a Galois extension of prime degree Õ . Certainly, the norm of a prime element of is a prime element of

. The surjectivity of the norm maps follows from (2.12).

U/=

(3.1.7). Augmentation Group 3 ( ) .

8;I1QSTmQiORQiGLT9=

g H/¨ 3 U/= 3 ù For a ﬁnite Galois extension denote by ( ) the subgroup of 1

; 3

generated by where runs through all elements of 1 and U runs through all elements

U/= of Gal( ).

) 79GLj

30 f

3 ©¹3 ©}· ;

Every unit in can be factorized as with , 1 Since = 1

Ë Ë

Ë 1

3 U/= 3 £; ©£3

we deduce that ( ) coincides with the subgroup of generated by , ,

U Gal( ).

¥4lGAB;GLF¦Q§ORQSGAT=

Let be a ﬁnite Galois extension of . For a prime element of deﬁne

1 c

;

U/= 3 /P3 U/= | 3 U/=

: Gal( ) ( ) ( U ) = mod ( )

The map is a homomorphism which does not depend on the choice of | . It induces a

ab " ab

U/= 3 /P3 U/= monomorphism : Gal( ) ( ) where for a group the notation stands for the maximal abelian quotient of " .

The sequence

g¢¡ h7£

N m ¦¨3 U/= m N â 3 /P3 U/=

1 Gal( ) ( ) 1 is exact.

1 1 1

|¥¤0 3 |¥¤P ; ©3 U/=

Proof. Since belongs to , we deduce that ( ) ( ) and

1 1 1 c

9¤0 ¤P ;

Í| | 3 U/¨ | mod ( )

| ; Í Thus, the map is a homomorphism. It does not depend on the choice of | , since ( )

¥; 3 H/¨ | mod ( ).

Surjectivity of the norm map has already been proved.

H/¨ HK

Suppose ﬁrst that Gal( ) is cyclic with generator U . The kernel of coincides

1 ê 1 1

d(; |¥ Í |¥; 3 U/= with . Since is a homomorphism, we have ( ) mod ( ). So

dV; 3 U/= we deduce that is equal to the product of ( ) and the image of . This shows the exactness in the middle term.

å 1

ê ê ê

1 1+ + ¦§¦§¦ + 1 1 1

;

3 U/= 3 | ©]3 U/=

Note that = ( ) , so ( ) = . If ( ), then

1 1 1

|¥; £; ©3 |9Lp ,

( ) = for some . Hence belongs to and therefore : divides

and U = 1. This shows the injectivity of .

U/=

Now in the general case we use the solvability of Gal( ) and argue by induction. Let

*/¨ H/¨ $ *¨$ | ¨ |

be a Galois cyclic subextension of such that = = . Put = HK .

è©

3 3 3 U/= 3 */=

Since HK : is surjective, we deduce that ( ) = ( ).

è© è©

Let HK = 1 for . Then by the induction hypothesis there is Gal( )

è 1

¤P

| × ×©63 */= × H/¨ ©63

¨ HK

such that Hª = with ( ). Write = with ( ). Then

è© è©¬«

1 1 «

p |¥¤P

belongs to the kernel of Hª and therefore by the induction hypothesis can be

è©

« 1 1

C C

|¥; © U/1* ©3 H/1* WÍ§|¥9¤0 3 H/¨

written as with U Gal( ), ( ). Altogether, mod ( ) which shows the exactness in the middle term.

1 1

;

To show the injectivity of assume that ( ). Then ¨ ( ) and

* U

by the previous considerations of the cyclic case U acts trivially on . So belongs to

H/1*

Gal( ). Now the maximal abelian extension of in is the compositum of all cyclic

* *

extensions of in . Since U acts trivially on each , we conclude that is injective.

¨ \ 2©B (3.1.8). 1 Acting on . For every every element can be uniquely expanded

°@©·

= °î|

Ë Ë

+ ó

)R= = HFk8DIPMNCx#8;FkMNCâOmFUGLT79GLj

e 31 where | is a prime element in .

ur ur

Since : is continuous, it has exactly one extension : which acts as

° °

® ®

| Ã |

° °

° .

ó ó

Ë Ë

1 m We shall study the action of ® on the multiplicative group.

798;:-:-<= 1

2¯¨põÃ ¢®N The kernel of the homomorphism is equal to and the

®N

3 3 3 ù \

image is ; ù = for every 1.

°-©W· ° ° °´| ©3

Proof. For = ° with the condition ( ) = implies that ( ) =

Ë Ë Ë Ë

° ©

for ¥ . Hence, belongs to the residue ﬁeld of and . Similarly one shows the

3 /P3 ù

exactness of the sequence in the central term ù + .

Let ©§3 . We shall show the existence of a sequence such that

3 4 ©3 ù

mod +1 ù and +1 +1 .

' 1 1

C C C

Let = 0 with , 0 1 ù . Let be such that = . Then = ;

Ë Ë Ë Ë C

put 0 = .

ccc

" " ©)3

Now assume that the elements 0 1 have already been constructed. Deﬁne

the element +1 ©· from the congruence

1 1 +1 +2 c

®N

Í | |

1 + +1 mod

Ë ©+·

There is an element × +1 such that

× × × 6× Í |

( +1) +1 + +1 = +1 +1 + +1 0 mod

Ë Ë

+1 1 1 1

®N

× |9 ; ©)3 4 ©)3 ù

Now put +1 = (1 + +1 ). Then +1 +2 ù and +1 +1 .

There exists = lim , and = . When ù the element can be

chosen in ù as well.

ur ur

µ/0 , /1 Let , be a ﬁnite Galois totally ramiﬁed extension. The extension is Galois with

µ/0 the group isomorphic to that of , . We may assume that the completion of is a subﬁeld ur

of the completion of , .

H/¨ ,/1

The extension is totally ramiﬁed of the same degree as . From (2.7) and (2.8) we

U/¨ ,µ/0 deduce that the extension is Galois with the group isomorphic to that of .

¥4lGAB;GLF¦Q§ORQSGAT= 1

®N

N N© N ©)3 H/¨

Let be such that ( ). Then HK belongs to the group

, ¤

HMJ . è

1 1

N±®N ©£3 ` © U/=

ñ ± ± ± Proof. We have = for some and Gal( ). By the pre-

1 1 1 1

®N

× × ©~3 NÜ × ®N

± ± ñ

± vious proposition we have = ± for some . So ( ) = 1 and

1 1 1

N ¥N , NÜ ñ× ¥ ¥ ©W, ©

¤ ¤

HKJ HKJ

= with . Then Hª = .

è è è

) 79GLj

32 f

3.2. The Neukirch Map

ur ur ur

, ,- /1

(3.2.1). Let , be a ﬁnite Galois extension of . Then = . Recall that Gal( )

¤ consists of X -powers of .

8;I1QSTmQiORQiGLT9= ur

¿ ¿

¤ ¤

,µ/0 © , /0

g ur U

Put Frob( ) = U Gal( ) : is a positive integer power of .

¥4lGAB;GLF¦Q§ORQSGAT= µ/0

The set Frob( , ) is closed with respect to multiplication; it is not closed with

ur ur

¿ ¿

Σ © ,/1 Σ , U The fixed field of U Frob( ) is of finite degree over , = , and is the

Frobenius automorphism of Σ.

µ/0 Thus, the set Frob( , ) consists of the Frobenius automorphisms Σ of ﬁnite extensions

ur ur ¡

, , / Σ X

Σ of in with Gal( ) .

¿ ¿

,/1 L ,µ/0 Ã U

The map Frob( ) Gal( ) U is surjective. J Proof. The ﬁrst assertion is obvious.

ur ur ur ur ur

Σ ², Σ ³, , / Σ

Since we deduce that . The Galois group of is

¿ X

topologically generated by U and isomorphic to , therefore it does not have nontrivial closed

ur ur / subgroups of ﬁnite order. So the group Gal( , Σ ) being a subgroup of the ﬁnite group

ur ur ur ur

/0 , Gal( , ) should be trivial. So = Σ .

¤ @ Y

Σ Σ ur Σ

2 ¤ Put 0 = . This field is the fixed field of U = , therefore 0 : = is finite. We deduce that

Σ Σ Σur ur ur ur

, , , : 0 = : = : = : 0

is ﬁnite. Thus, Σ /1 is a ﬁnite extension.

Σ0: ¤

¿ ¿ ¿

¤ ¤ ¤ ¤

G´ ´ @

ur ur ur ur

U U Now U is a power of Σ and Σ = = = . Therefore, =

, Σ. Certainly, the Frobenius automorphism Σ of a ﬁnite extension Σ of in with

ur ¡

/ X ,/1 Gal( , Σ) belongs to Frob( ).

, /0 @¤ © ,µ/0

J U Denote by an extension in Gal( ) of . Let U Gal( ), then 0 is equal

¿ ¿

@ \ @ , ` -

J J U to for some positive integer . Hence U acts trivially on 0, and so =

ur ur

¿ ¿ ¿ ¿ ¿

,µ/=, `]© , /0 ` ` ` Ü J

belongs to Gal( 0). Let Gal( ) be such that = . Then for U = we

¿ ¿ ¿ ¿

@ ` @ © ,µ/0

J J

U U U

deduce that U = and = = . Then the element Frob( ) is mapped

to U Gal( ).

8;I1QSTmQiORQiGLT9=

,µ/0

(3.2.2). g Let be a ﬁnite Galois extension. Introduce

, Ã µ/0 / | ,

ϒ ,

¤ ¤ ¤ ¤

U H HKJ HMJ

J : Frob( ) Σ Σ mod

è è è è

Σ ,µ/1 | Σ ©

where is the ﬁxed ﬁeld of U Frob( ) and Σ is any prime element of .

798;:-:-<=

¿ ¿ ¿

ϒ ¿ ϒ

¤ ¤

J J

U J U

The map J is well deﬁned. If = id then ( ) = 1.

è è

| | | ³©3 Proof. Let | 1 2 be prime elements in Σ. Then 1 = 2 for a unit Σ. Let be the

:È /

compositum of Σ and , . Since Σ Σ , the extension Σ is unramiﬁed. From (3.1.5)

× ×Ì©)3 µ

we know that = HKµ Σ for some . Hence

| | | ¬ × | ¬ ×

¤ ¤ ¤ ¤ ¤

H HMJ H H H HMµ H H

Σ 1 = Σ ( 2 ) = Σ 2 Σ ( Σ ) = Σ 2 ( J )

è è

è è è è è è

)R= =}EGFN8¶H8;MwAQSl;jVF3·Ì

| Í | ,¨

¤ ¤

H HKJ

We obtain that H Σ 1 Σ 2 mod .

è è

¿ Σ

,Ç | © ,¨

¤ ¤

J J

U H

If = id then and therefore Σ Σ HKJ .

è è

(3.2.3). The deﬁnition of the Neukirch map is very natural from the point of view of the well known principle that a prime element in an unramiﬁed extension should correspond to the Frobenius automorphism (see Theorem (3.2.4) below) and the functorial property of the

ϒ ¤

reciprocity map (see (3.2.5) and (3.3.4)) which forces the reciprocity map J to be deﬁned

as it is. è

,µ/1 L 9/ ¨ ϒ ,

¤ ¤ HMJ

Already at this stage one can prove that the map J : Frob( ) è

induces the Neukirch homomorphism è

,/1 L /

ϒ ,

¤ ¤ J

: Gal( ) HMJ

è è

¿ ¿

U U

In other words, J ( ) does not depend on the choice of Frob( ) which extends

¿ ¿ ¿

¿ ¿

,µ/0 ¸ © ϒ ϒ ϒ

¤ ¤ ¤

J U J U J U U

U Gal( ), and moreover, ( 1) ( 2) = ( 1 2).

è è We will choose a different route,è which is a little longer but perhaps is more satisfying.

ϒ ¤

The plan is the following: ﬁrst we easily show the existence of J for unramiﬁed extensions and even prove that it is an isomorphism. Then we deduce some functorialè properties

ϒ¿ ¤

of J . To treat the case of totally ramiﬁed extensions in the next section, we introduce the

è Ψ ϒ

¤ ¤ J

Hazewinkel homomorphism J which acts in the opposite direction to . Calculating è

Ψ è ϒ

¤ ¤ J

composites of the latter with J we shall deduce the existence of which is expressed è

by the commutative diagram è ¿

ϒ f=h i

Û Û

, ,/1 N m â /

HMJ

Frob( Û )

è j j id

ϒ f=h i

,¨ ,µ/0 N m â 9/ ¤

Gal( ) HMJ è

Ψ ϒ

¤ ¤ J

Then using J we prove that is a homomorphism and that its abelian part

è è

ab ab

, ϒ ,/1 /

¤ ¤

: Gal( ) HMJ

è è is an isomorphism. Then we treat the general case of abelian extensions and then Galois extensions reducing it

ϒ¿ ¤

to the two cases described above and using functorial properties of J . This route not only

ϒ è ¤

establishes the existence of J , but also implies its isomorphism properties.

EGFN8;GAlm8;:H=

(3.2.4). Let , be an unramiﬁed extension of of ﬁnite degree.

¿ ¿

,µ/0

ϒ ©

U U U

Then J ( ) does not depend on the choice of for Gal( ). It induces an

,¨ µ/0 L9/ ϒ ,

¤ ¤ HMJ

isomorphism J : Gal( ) and

è è

¤ ¤

, ϒ Í§|

¤ ¤ HMJ

J mod

è è

for a prime element |A¤ in .

,µ/1 \6 Y , ¤

Proof. Since is unramiﬁed, U is equal to sor some 1. Let = : . Then ¿

¿ Σ \ \ Õ´Y¸, Õ

¤ U U must be in the form with = + 0 for some integer . The ﬁxed ﬁeld of is

) 79GLj

34 f

|A¤

the unramiﬁed extension of of degree . We can take as a prime element of Σ. Then

ϒ | | Í| ,

¤ ¤ ¤

¤ ¤

U H HMJ

J ( ) = Σ = mod

è è è

¿ ¿

¤ ϒ |9 |

¤ ¤

J U U

since = HKJ . Thus, ( ) does not depend on the choice of .

è è

ϒ ¤ | ,

¤ ¤ HMJ

It is now clear that J is a homomorphism and it sends to mod .

è è

| / , ,

¤ ¤ HKJ

Results of (3.1.5) show that mod HMJ generates the group which is è

ϒ è

, ¤

cyclic of order : . Hence, J is an isomorphism. è

ϒ¿ ¤

(3.2.5). Now we describe ﬁrst functorial properties of J .

798;:-:-<=

,/ Let §/1 be a ﬁnite separable extension and let be a ﬁnite Galois extension,

sep

U Gal( ). Then the diagram of maps ¿

ϒ f=hº¹

Û Û

,¨ ,/ m m ¦ ë9/

HKJ

Frob( Û )

j j

¨»

¿ ¹

ϒ f=h

¼ ¼

,µ/ E m m N / ,

U U H J U

Frob( U ) ( ) ( )

è7

¿ ¿ ¿

` `+© ,/

` ur

U U

is commutative; here U ( ) = for Frob( ).

1 ¿

Σ ¿ Σ

` ` |

U U Proof. If is the fixed field of , then U is the fixed field of . For a prime element

Σ Σ

| | |

¨ ¨

U H U U±H

in , the element U is prime in . Since Σ ( ) = Σ , the proof is completed.

è è

¥4lGAB;GLF¦Q§ORQSGAT=

K/9, ,µ/0 K/ Let §/1 and be ﬁnite separable extensions, and let and be

ﬁnite Galois extensions. Then the diagram of maps ¿

ϒ he¹

Û Û

Ì K/ m m k¨ëç/

H Û

Frob( Û )

¹5h i

j j

» ¿

ϒ f=h i

, ,/1 m m k 9/ ¤

Frob( ) HMJ

¿ ¿

is commutative. Here the left vertical homomorphism is the restriction U ur of Frob( )

¨ ¤

and the right vertical homomorphism is induced by the norm map H .

è The left vertical map is surjective if = .

¿ ¿ ¿

Proof. Indeed, if U Frob( ) then for = ur Gal( ) we deduce that

¿ ¿ ¿ ¿

¤ ¤

¤ Σ p `6© ,/1

` ur ur U = U is a positive power of , i.e., Frob( ). Let be the ﬁxed ﬁeld of .

ur ur ur

, / , Σ ` Σ Then T = 2 is the fixed field of . The extension T is totally ramified, since = T

| |

Σ | Σ 2

and so T = T . Hence for a prime element Σ in the element T = H Σ T Σ is prime in

| | |

¤ ¤ ¨ ¤ ¨

H H H

T and we get H T T = Σ Σ = ( Σ Σ).

è è è è

If = , then the left vertical map is surjective, since every extension of U Frob( )

/1 K/1 to Gal( ) belongs to Frob( ).

R=§)R= EGFN8³í#< ¾8¨¨QSTwL8;C`íHGL:-GL:-GLlmB¿FNQiFk:

) 35

GAlGLCEC

,µ/1 Let §/1 be a Galois subextension in a ﬁnite Galois extension . Then the

diagram of maps

Û Û Û

,µ/ N N â ,/1 m m k §/1

Û Û

Frob( Û ) Frob( ) Frob( )

¿ ¿ ¿

f=hmi ¹5h i

ϒ f=he¹ ϒ ϒ

j j j

¹5h i

ë9/ N N â 9/ ,¨ ,¨ë m m k 9/ ë© A m â

¨ ¤ ¨ ¤

HMJ HKJ

H 1

è è è is commutative; here the central homomorphism of the lower exact sequence is induced by the

identity map of . Proof. An easy consequence of the preceding Proposition.

3.3. The Hazewinkel Homomorphism

(3.3.1). Let , be a ﬁnite Galois totally ramiﬁed extension of . The Galois group of the

U/¨ ,µ/0

extension is isomorphic to Gal( ).

8;I1QSTmQiORQiGLT9=

Let be the continuous extension on of the Frobenius automorphism J . Let

, 2©W | be a prime element of . Let be the maximal abelian extension of in . For

1 1

2©3d °®N ¢®N HK

by Lemma (3.1.6) there is such that = HK . Then = = 1 and

è by Proposition (3.1.7) è

1 1

®N

Í§| 3 U/=

mod ( )

for some U Gal( ) which is uniquely determined as an element of Gal( ) where M À = . Deﬁne the Hazewinkel (reciprocity) homomorphism

ab 1 1

®N £

, L / ,µ/1 õ6Ã õ õ Í§|

¤ ¤

HKJ U Hª

J : Gal( ) =

è è è

798;:-:-<= Ψ ¤

The map J is well deﬁned and is a homomorphism. è

Proof. First, independence on the choice of | follows from Proposition (3.1.7). So we can W, assume that |6© .

N NL Hª

If = Hª then belongs to the kernel of . Therefore by Proposition (3.1.7)

1 1 è 1 1 1 1

4 |¥¤P ©§3 U/= N¥®N °®N ®m ÍÏ°®N 3 U/=

N = with ( ). Then = mod ( ) which

« «

proves correctness« of the deﬁnition.

If Hª ( 1) = 1 and ( 2) = 2, then we can choose 1 2 for 1 2 and then from

è Ψ ¤

Proposition (3.1.7) we deduce that J is a homomorphism.

#8;:

/1 ,

1. Since , is totally ramiﬁed, the norm of a prime element of is a prime element of

, 3 / 3 / ,µ/1

¤ ¤

J HMJ

. So HMJ = . Moreover, if is a totally ramiﬁed -extension

è è

,¨ 9/ 3 3 /

¤ ù ¤ ù

J HKJ

(i.e. its degree is a power of ), then = 1 HMJ 1 , since all multiplicative

è è representatives are th powers. 2. The Hazewinkel homomorphism can be defined for every finite Galois extension, but it has the simplest form for totally ramified extensions.

) 79GLj

36 f

Ψ ϒab

¤ J

(3.3.2). Now we prove that is inverse to ¤ .

EGFN8;GAlN8D:U=

µ/1 K/1

Let , be a ﬁnite Galois totally ramiﬁed extension. Let be the maximal µ/0

abelian subextension of , . Then

(1) For every U Frob( )

¿ ¿

Ψ ϒ ¿

¤ ¤

J U U

J ( ) =

è è

¿ ¿

U J

(2) Let and let U Frob( ) be such that = ( ). Then

ϒ ¿ Í ,

¤ ¤

U HKJ

J ( ) mod

è è

¿ ¿

Ψ ϒ ¿

¤ ¤

J U U

Therefore, J is an isomorphism, ( ) does not depend on the choice of for

è è

U Gal( ) and induces the Neukirch homomorphism

,/1 L / ϒ ,

¤ ¤ J

: Gal( ) HMJ

è è

ϒab ab ,µ/1 /1

The latter induces an isomorphism ¤ between the groups Gal( ) = Gal( ) and

,¨ 9/ Ψ

¤ ¤ J

HMJ , which is inverse to .

è è

ur ur ur ur

/0 , /9, Ò , /0 Proof. To show (1) note at ﬁrst that Gal( , ) is isomorphic to Gal( ) Gal( )

ur ur

@ Y © , /1 |

U U and so U is equal to for some positive integer and Gal( ). Let Σ be a ur ur

Σ ¿ Σ | ,

prime element of the ﬁxed ﬁeld of U . Since Σ is a prime element of = we have

1 ê 1

| K©)3 | , | 9=® | ur Σ = for some J , where is a prime element of . Therefore = .

Σ Σ ur Σ Y

¤ ¤

ªnH H H

Let 0 = , then 0 : = . Then H Σ = Σ0 Σ Σ0 and Σ Σ0 acts as

è è è

è 1

¬¬¬ ¤

ur ur ur ¤ ur

HMJ HK H

H Σ Σ = = , Σ0 acts as 1 + + + . We have

è è è è 0

å 1

ê c

1+ + ¦§¦§¦ +

® ®

| |

¤ ¤

¤ ur ur ur ur

HKJ HKJ

H Σ Σ = 1 where 1 =

è è è

,¨ | Í

¤ ¤ ¤

¤ ur ur

HMJ HKJ J

So = H Σ Σ 1 mod and ( ) can be calculated by looking at

è è è

ê ê

1 1 1 1 1 1 ¿

®N ®N

¨® Í!==® | £ | 3 U/= 1 . We deduce 1 = = = mod ( ). This proves (1).

1 1

°®N Í| 3 U/= © ,µ/0 U

To show (2) let = Hª and mod ( ) with Gal( ). Then

¿ ¿

¿ ϒ ,¨ @ | Í

¤ ¤ ¤

¤ ur ur

U J U H HKJ HMJ

again U = and similarly to the previous ( ) = Σ Σ 1 mod

è è è

1 1 1 è

®N

®m

Í| Í 3 U/= and 1 mod ( ). From the second proposition of (3.1.9) applied to

, N 4 N Í

HKJ HK Hª

= 1 we deduce that HK belongs to and therefore 1 =

è è è è

, ¤

mod HKJ which proves (2).

è Ψ ¤

Now from (1) we deduce the surjectivity of J . From (2) and Lemma in (3.2.2) by taking

¿ ¿

, Ψ © Ψ

¤ ¤ ¤

µ µ

U J HKJ J

U = , so that = id = ( ), we deduce that , i.e. is injective.

è è è

Ψ ϒ¿

¤ ¤ J

Hence J is an isomorphism. Now from (1) we conclude that does not depend on the

è è

,µ/0 © ϒ

U J

choice of a lifting U of Gal( ) and therefore determines the map .

¿ ¿

¸ ϒ

U U U J Since we can take U 1 2 = 1 2, from (1) we deduce that is a homomorphism.

Proposition (3.2.1) and (2) show that this homomorphism is surjectiè ve. From (1) we deduce

/9 that its kernel is contained in Gal( , ). The latter coincides with the kernel, since the image

ϒ ¤

of J is abelian.

è É

GAlGLCEC

,/1 ×©Ád ×(®N | ©

For U Gal( ) there exists such that = . Then =

ϒ × ×

J U Hª

HK belongs to and ( ) = .

è è è

R=§)R= EGFN8³í#< ¾8¨¨QSTwL8;C`íHGL:-GL:-GLlmB¿FNQiFk:

) 37

×Ì©W` Conversely, for every ©W there exists such that

1 1 c

®m

HKJ U

HK mod = for some Gal( )

è è

µ U

Then J ( ) = . è

× |®N

Proof. We can assume that is a unit, since = 1. Denote by the same notation U the

H/¨ Σ

U U

element of Gal( ) which corresponds to U . Let be the ﬁxed ﬁeld of = . Applying

the lemma of (3.1.8) to the continuous extension to of the Frobenius automorphism U we

1 1

C C

1 1 1

=® =®N

| = =

so belongs to the ﬁxed ﬁeld of U in which by the lemma of (3.1.8) equals to the

Σ , | | Σ

ﬁxed ﬁeld of U in . The element Σ = is a prime element of . Note that

1 1 1 1 1

C C C

®N ®N |¥; ®¹©Ï3 H/¨

( ×L ) = = ( ) ( ); hence from the proposition of (3.1.8) we

Í × ,

Hª HKJ

deduce that HK mod . Finally,

è è è

| Í × , Í

¤ ¤

Hª HK HMJ

H Σ Σ mod

è è è è To prove the second assertion use the ﬁrst assertion and the congruence supplied by the

ϒ Ψ ,¨ .Í © ,/1

¤ ¤ ¤

U HMJ U U J

Theorem: J ( ) mod where Gal( ) is such that = ( ).

è è è /1 The Theorem demonstrates that for a ﬁnite Galois totally ramiﬁed extension , in the

definition of the Neukirch map one can fix the choice of Σ as the field invariant under the action

of U .

(3.3.3). The following Lemma will be useful in the proof of the main theorem.

798;:-:-<= /1

Let , be a finite abelian extension. Then there is a finite unramified extension

§/ such that is an abelian extension of , is the compositum of an unramiﬁed

extension 0 of and an abelian totally ramiﬁed extension of . For every such we

s! ë

¤ ¨ ¨ ¤ ¤

H HKÂ

have H = 0 0 .

è è è

/1 ,' Proof. Since , is abelian, the extension is an abelian extension of . Let

ur ur

¿ ¿

Ç© ,- /0 @¤ s s

Gal( ) be an extension of . Let be the ﬁxed ﬁeld of . Then 2 = ,

s , so s is an abelian totally ramiﬁed extension of . The compositum of and is an

ur ur

s ,

unramiﬁed extension of , , since = . The ﬁeld is an abelian extension of and

§/1s Ò §/ Gal( §/1 ) Gal( ) Gal( 0).

Now the left hand side of the formula of the Lemma is contained in the right hand side Ã . We

35¨ 3 Ã 34¤§ 3 30¨ |£¨

¨ ¤ ¤ ¨

Â Â

H H Â

have HMÂ , since . If is a prime element

è è è

| ¨ ©3Ã |£¨ ¡0¤ X ¡0¤ !©BÃ

¨ ¤ ¨ ¤ ¨ ¤

H H

of , then H . Then ( ) = ( 0 0 ). So every can

è è è

|9 6©ÁÃ 3 Y Ã

¨ ¤

2 ¨

be written as = H with and some . Thence is contained in

ë ë Ã

¨ ¤ ¨ ¤ H

H and we have = .

è è

Now we state and prove the ﬁrst main theorem of local class ﬁeld theory.

EGFN8;GAlN8D:U=

µ/0 K/1

Let , be a ﬁnite Galois extension. Let be the maximal abelian subexten- µ/1 sion of , .

) 79GLj

38 f

¿ ¿

Ψ ϒ ©

¤ ¤

J U U U

Then J is an isomorphism, ( ) does not depend on the choice of for

è è µ/1

Gal( , ) and induces the Neukirch (reciprocity) homomorphism

ϒ ,/1 L / ,

¤ ¤ J

: Gal( ) HMJ

è è

ab ab

,µ/0 K/1 9/ ¨

ϒ ,

¤ HMJ

The latter induces an isomorphism ¤ between Gal( ) = Gal( ) and

J è

Ψ è ¤

(which is inverse to J for totally ramiﬁed extensions).

µ/0 ,

Proof. First, we consider the case of an abelian extension , such that is the com-

positum of the maximal unramiﬁed extension , 0 of in and an abelian totally ramiﬁed

s , , s!

¤ ¤ ¤

HMJ HMÂ

extension of . Then by the previous lemma = HMJ 0 0 . From

è è

Proposition (3.2.5) applied to surjective maps è

/1 , /1 ,µ/1 s+/1 Frob( , ) Frob( 0 ) and Frob( ) Frob( )

ϒ¿ ¤

and from Theorem (3.2.4) and Theorem (3.3.2) we deduce that J does not depend on the

s , ,

¤ ¤ ¤

HKJ HMÂ HKJ

choice of U modulo 0 0 , therefore, modulo . So we get the map

è è

ϒ è ¤

J .

è ϒ ¤

Now from Proposition (3.2.5) and Theorem (3.2.4), Theorem (3.3.2) we deduce that J is

s , ,

¤ ¤ ¤

HMJ HMJ

a homomorphism modulo HMÂ 0 0 , so it is a homomorphism modulo .

è è è

,¨ , s , ϒ ©

¤ ¤

HMJ U U

It is injective, since if J ( ) , then acts trivially on 0 and , and so on . è

Its surjectivity follows fromè the commutative diagram of Corollary in (3.2.5). /1

Second, we consider the case of an arbitrary ﬁnite abelian extension , . By the previous 9, Lemma and the preceding arguments there is an unramiﬁed extension §/ such that the map

ϒ¿ ϒ

§/1 ,µ/1

¤ ¨ ¤

¨ induces the isomorphism . The map Frob( ) Frob( ) is surjective and

è è

ϒ¿ ϒ

¤ ¤ J

we deduce using Proposition (3.2.5) that J induces the well deﬁned map , which è

è ϒ

J U is a surjective homomorphism. If U Gal( ) is such that ( ) = 1, then from the commutative diagram of Corollary in (3.2.5) and surjectivity of ϒ è for every ﬁnite abelian

ϒ ϒ

` `]© §/9,

¨ ¤ ¨ ¤

extension we deduce that ( U ) = ( ) for some Gal( ). The injectivity of

è è

ϒ `

¨ ¤

now implies that U = acts trivially on . è

Finally, we consider the general case of a ﬁnite Galois extension where we argue by induction

/1 ,µ/0

on the degree of , . We can assume that is not an abelian extension.

Every U Gal( ) belongs to the cyclic subgroup of Gal( ) generated by it, and by ¿

ϒ ¿

¤ U

what has already been proved and by Proposition in (3.2.5) J ( ) does not depend on the è

¿ ϒ

¤ J

choice of U and therefore determines the map .

è µ/1 Since Gal( , ) is solvable, we conclude similarly to the second case above using the

ϒ ¤

induction hypothesis that J is surjective. In the next several paragraphs we shall show that è

ϒ ϒ ,/9

J H

(Gal( )) = 1. Due to surjectivity of this implies that the map ¤ in the diagram

ϒ ¤ of Corollary (3.2.5) (where we put = ) is zero. Since µ is an isomorphism we see from

ϒ è ,µ/= ¤

the diagram of the Corollary that J is a surjective homomorphism with kernel Gal( ).

/9 ϒ , ¤

So it remains to prove that J maps every element of the derived group Gal( ) to

,µ/1 $ ϒ ¤

1. Since Gal( ) is solvable, we have = . Proposition (3.2.5) shows that J ( ) =

C C

µ µ

H J J

¤ ( ( )) for every Gal( ). Since by the induction assumption is a

è è homomorphism,è it sufﬁces to show that

1 1 1 1

ϒ ϒ ` ` ` `

J U U H J U U

( ) = ¤ ( ( )) = 1

è è è

R=§)R= EGFN8³í#< ¾8¨¨QSTwL8;C`íHGL:-GL:-GLlmB¿FNQiFk:

) 39

.`© ,/1

for every U Gal( ). To achieve that we use Lemma (3.2.5) and the induction hypothesis.

µ/1s ,µ/0 ,µ/9 ` Suppose that the subgroup Gal( , ) of = Gal( ) generated by Gal( ) and is not equal to " . Then from the induction hypothesis and Lemma (3.2.5)

1 1 1 1 1

` ` ` `

ϒ ` ϒ ϒ ϒ

Â U U J Â J Â U U J Â

J ( ) = ( ) ( ) = ( )

è è è è

and so c

1 1 1 ä

ϒ ϒ

` ` ` ã

J U U H J Â

( ) = ¤ ( ) = 1

è è

` K/0 `m

In the remaining case the image of generates Gal( ). Hence U = for some

1 1 1 1

C C C

the preceding c

1 1 1 1 1 ä

L ¤0

C C C C C

` ` ` ` `

ϒ ` ϒ ϒ

¤ ¤

J J H J

( ( ) ) = ( ) = ¤ ( ) = 1

è è è

GAlGLCEC

/1 /1

(1) Let , be a ﬁnite Galois extension and let be the maximal abelian subextension in

, ,µ/0

¤ ¤

µ H

. Then HKJ = .

è è

(2) Let , be a ﬁnite abelian extension, and a subextension in . Then

,¨ , ]©

¨ ¤ ¨ ¤

H HMJ

HMJ if and only if .

è è è Proof. The ﬁrst assertion follows immediately from the theorem. The second assertion follows

the diagram of the corollary in (3.2.5) (with Frob being replaced with Gal ) in which the

H ¤

homomorphism ¨ is injective due to the theorem. è

ϒ ¤

(3.3.4). We now list functorial properties of the homomorphism J . Immediately from the

previous theorem and (3.2.5) we deduce the following è

¥4lGAB;GLF¦Q§ORQSGAT=

,/ (1) Let §/1 be a ﬁnite separable extension and let be a ﬁnite Galois extension,

sep

U Gal( ). Then the diagram

ϒ f=hº¹

Û Û

,¨ ,/ m m k ë9/

HMJ

Û Gal( Û )

j j

¨»

f=h ¹

¼ ϒ ¼

,µ/ E m N m k ç/ ,

U U H J U

Gal( U ) ( ) ( )

è7

is commutative.

K/9, ,µ/0 / (2) Let §/1 be ﬁnite separable extensions, and let and be ﬁnite Galois extensions. Then the diagram

ϒ hº¹

Û Û

/ m N ¦ /

H Û

Gal( Û )

g4Ä

¹5h i

j j

ϒ f=h i

, ,/1 m N ¦ / ¤

Gal( ) HMJ è is commutative.

) 79GLj

40 f

3.4. The Reciprocity Map

In this section we deﬁne and describe properties of the reciprocity map

L /1 Ψ : Gal( )

ϒ ¤

using the Neukirch map J studied in the previous sections. è

ϒ ¤

(3.4.1). The homomorphism inverse to J induces the surjective homomorphism è

ab c

7,/1 L ,/1 ( ¬ ): Gal( )

Ψ ¤

It coincides with J for totally ramiﬁed extensions.

è sep ab

Denote the maximal abelian extension of in by .

¥4lGAB;GLF¦Q§ORQSGAT= $

$ ab

/1 Let be a subgroup in Gal( , ) , and let be the ﬁxed ﬁeld of in

ab 1 $

, ¬§,µ/0 ë

¨ ¤ 2

. Then ( ) ( ) = H .

, , , , , , , , Let 1 2 be abelian extensions of ﬁnite degree over , and let 3 = 1 2, 4 = 1 2 2.

Then

, , , , , ,

¤ ¤ ¤ ¤ ¤ ¤

HMJ HKJ HMJ HMJ HMJ

HKJ 3 3 = 1 1 2 2 4 4 = 1 1 2 2

è è è è è è

, , , ,

¤ ¤ HKJ

The field 1 is a subfield of the field 2 if and only if HMJ 2 2 1 1 ; in particular,

è è

, , , ,

¤ ¤ HMJ

1 = 2 if and only if HMJ 1 1 = 2 2 .

è è

¤ HKJ

If a subgroup H in contains a norm subgroup for some ﬁnite Galois extension

,µ/0

, then H itself is a norm subgroup.

, /=,-°

Proof. The ﬁrst assertion follows immediately from (3.3.3) and (3.3.4). Put ° = Gal( 3 ), À

= 1 2. Then $

1 1 $

, ¬i7, /1 ¬i7, /1

¤ 2

HMJ 3 3 = ( 3 ) (1) = ( 3 ) ( 1 2)

è $

1 $ 1

, , ¬i7, /1 ¬i7, /0

¤ ¤

2 2 HKJ

= ( 3 ) ( 1) ( 3 ) ( 2) = HKJ 1 1 2 2

è è

$ $ $

1 $ 1 1

, ¬i7, /1 ¬i7, /0 ¬i7, /0 ¤

HMJ 4 4 = ( 3 ) ( 1 2) = ( 3 ) ( 1)( 3 ) ( 2)

, ,

¤ ¤ HMJ

= HMJ 1 1 2 2

è è

, Å, , , , ,

¤ ¤ ¤ ¤

HMJ HMJ HKJ

If 1 2, then HMJ 2 2 1 1 . Conversely, if 2 2 1 1 , then

è è è è

, , ,

¤ ¤

J HMJ

HMJ 1 2 ( 1 2) coincides with 2 2 , and Theorem (3.3.3) shows that the extension

è è

, /1 , /0 , Z,

, 1 2 is of the same degree as 2 , hence 1 2.

, ë ¾ ,µ/0

¤ ¨ ¤

HMJ H H H

Finally, if H , then = , where is the ﬁxed ﬁeld of ( ).

è è

Passing to the projective limit, we get

ab ab

Ψ ¤

, L L / ,/1 /1 ¤

: lim HKJ lim Gal( ) = Gal( )

where , runs through all ﬁnite Galois (or all ﬁnite abelian) extensions of . The homomorphism

Ψ ¤ is called the reciprocity map.

)R=ÇÆ=}EGFN8 QSF¦Om8;TNjR8EGFN8;GAlN8D:

>@? 41 N8;GAlm8;:H= (3.4.2). EGF The reciprocity map is well deﬁned.

ab /0

Its image is dense in Gal( ), and its kernel coincides with the intersection of all norm

,¨ ,µ/1 ¤

subgroups HKJ in for ﬁnite Galois (or ﬁnite abelian) extensions .

, 2© ,

¤ 2

trivially on if and only if HMJ .

ur ( )

Ψ ¤ ]©W

The restriction of ( ) on coincides with ¤ for .

sep

, /1

Let be a ﬁnite separable extension of , and let U be an automorphism of Gal( ). Then the diagrams

Ψ f Û

Û ab

,¨ N m â , /9, Û

Û Gal( )

j j

¨» f

Ψ ¼

ab ä

, , N m â / ,

U U ( ) Gal ( ) U

Ψ f Û

Û ab

, N m ¦ , /=, Û

Û Gal( )

f=h i

j j

Ψ i

ab ä

/1 N m ¦ Gal ã

` `

U U are commutative, where U ( ) = , the right vertical homomorphism of the second

diagram is the restriction.

/1, /0 , , Proof. Let , 1 2 be ﬁnite extensions and 1 2. Then Proposition (3.3.4) shows that the restriction of the automorphism

is well deﬁned.

, £© -,µ/0 ¤

The condition HMJ is equivalent ( ) = 1 and the last relation means that

è ab

, Ψ ¤

( ) acts trivially on 2 .

, È , Ψ ¤ ¤

Hence, the kernel of is equal to HMJ , where runs through all ﬁnite Galois

Ψ ¤

,µ/1 ,/1

extensions of . Since ( ) = Gal( ) for a ﬁnite abelian extension , we deduce J

Ψ ab /1

that ( ) is dense in Gal( ).

|L¤ @¤ |L¤ Ψ ¤

Theorem (3.2.4) shows that ( ) ¤ ur = for a prime element in . Hence,

( )

¤ ¤

Ψ ¤ Ψ

¤ ¤ ( ) ¤ ur = and ( ) ur = 1. The commutativity of the diagrams follow from Propositions (3.3.4) and (3.3.5).

3.5. The Existence Theorem

¥4lGAB;GAF¦QiORQSGAT=

(3.5.1). Let , be a ﬁnite separable extension of . Then the norm map

, ¤

HMJ :

,¨ ¤

is continuous and HKJ is an open subgroup of ﬁnite index in . è

) 79GLj

42 f

K/1 ,¯É

Proof. Let be a ﬁnite Galois extension with . Then, by Theorem (3.4.2),

Ì K/1

¤ µ

H is of ﬁnite index in . The Galois group of the extension is solvable.

,¨ ¤

Therefore, in order to show that HKJ is open, it sufﬁces to verify that the norm map for a cyclic extension of prime degree transformsè open subgroups to open subgroups. This follows from the description of the behavior of the norm map in (2.13). The same description of the

H ¤ norm map implies that the pre-image ¨ of an open subgroup is an open subgroup for a

è 1

§/1

H H

cyclic extension . Therefore, the pre-image ¤ of an open subgroup in is

µ è

1 1 ä 1

W ¾

J H H H H H H H

¤ ¤

an open subgroup in . Since ¤ ( ) ( ) , we obtain that ( ) is

J J

è è è

, ¤

open in and HKJ is continuous.

EGFN8;GAlm8;:§ì¥Ê QSF¦Om8;TNjR8ªEIFA8DGLlm8;:0ËPï =

(3.5.2). >@? There is a one-to-one correspondence between

open subgroups of ﬁnite index in and the norm subgroups of ﬁnite abelian extensions:

, ¤

HÉÌÍHMJ . This correspondence is an order reversing bijection between the lattice of open

2 H

subgroups of ﬁnite index in (with respect to the intersection H 1 2 and the product

, ,

2 H

H 1 2 ) and the lattice of ﬁnite abelian extensions of (with respect to the intersection 1 2 ,

and the compositum , 1 2 ).

Proof. We verify that an open subgroup H of ﬁnite index in coincides with the norm

,¨ ,µ/0

¤ H

subgroup HKJ of some ﬁnite abelian extension . It sufﬁces to verify that contains

, ,/1 ¤

the norm subgroup HMJ of some ﬁnite separable extension . Indeed, in this case

Ì K/1 ¾;,

µ H

H contains , where is a ﬁnite Galois extension, . Then by Proposi-

ë 7K/0

¨ ¤

H H

tion (3.4.1) we deduce that H = , where is the ﬁxed ﬁeld of ( ) and è

§/1 is abelian.

\ \

Assume char( ) , where is the index of H in . If , then the Kummer

,¨ ,/1 , ¤

theory implies that = HMJ for the abelian extension , where = ( ).

, ,µ/0

¤ H

Since is of ﬁnite index in , the extension is ﬁnite. Then HMJ . If

,¨ ¤

is not contained in , then put 1 = ( ). By the same arguments, 1 = HKJ 1 for

,¨)! ,/1

¤ H

some ﬁnite abelian extension 1. Then HKJ .

Yº

Assume now that char( ) = . We will show by induction on 1 that any open

subgroup H of index in contains a norm subgroup. This is true for = 1, as YÝ, follows from the theory of Artin–Schreier extensions. Now let 1, and let H 1 be an

U H

open subgroup of index in such that H 1. By the induction assumption,

, , , ¾

¤ ¤ ¤

HKJ H HKJ HMJ

H 1 1 1 . The subgroup 1 1 is of index 1 or in 1 1 . In the ﬁrst

è è è

, ¾ ,µ/9,

¤ HMJ

case H 1 1 , and in the second case let 1 be a ﬁnite separable extension with è

1 ä

, , ¾ ,¨ ¾

¤ ¤

HKJ HKJ J HMJ H H H

¤ ; then .

J 1 1 1

è è

1 è

\ \

For an open subgroup H of index in with we now take open subgroups

H H H H

H 1 and 2 of indices and , respectively, in such that 1 2. Then

¾ , , ¾ , ,

¤ ¤ ¤

2 2

H H HMJ HKJ HMJ J

H = 1 2 1 1 2 2 1 2 ( 1 2) and we have proved the desired

è è è

assertion for H .

Finally, Proposition (3.4.1) implies all remaining assertions.

É GAlGLCEC

The reciprocity map Ψ ¤ is injective and continuous. " 1 "

Proof. By Theorem (3.4.2) the preimage ¤ ( ) of an open subgroup of the group " ab " 1

/0 , ,

¤ ¤

Gal( ) coincides with HMJ , where is the ﬁxed ﬁeld of . Hence, ( ) is è

)R=ÇÆ=}EGFN8 QSF¦Om8;TNjR8EGFN8;GAlN8D:

>@? 43

open and Ψ ¤ is continuous. Since the intersection of all norm subgroups coincides with the ¤

intersection of all open subgroups of ﬁnite index in , we conclude that Ψ is injective.

#8;:

One may omit the word “open” in the Theorem if char( ) = 0.

8;I1QSTmQiORQiGLT9=

g The ﬁeld , which is an abelian extension of ﬁnite degree over , with the

H H

property HKJ = is called the class ﬁeld of the subgroup . è

(3.5.3). Now we will generalize Theorem (3.5.2) for abelian (not necessarily ﬁnite) extensions

,µ/0

of . For an abelian extension , we put

Î ,

¨ ¤ ¤ H

HKJ =

è è

, ,¨ ¤

where runs through all ﬁnite subextensions of in . Then the norm subgroup HKJ , as

, ¤

the intersection of closed subgroups, is closed in . Theorem (3.4.2) implies that HMJ =

è ä

1 ab ä 1 ab

/ /=,

È Ψ Ψ

ã ã

¨ ¤ ¤

Gal( ) = Gal( ) . Moreover, for a closed subgroup H in

" " g

ab g

Ψ ¤

denote the topological closure of ( H ) in Gal( ) by ( ). In other words, ( ) $

$ ab

/0 ¾ Ψ

coincides with the intersection of all open subgroups in Gal( ) with ( H ). If "

1 g

Ψ Ψ ¤

~©(

an element belongs to ¤ ( ( )), then the automorphism ( ) acts trivially on

$ $

¤ ¾ Ψ

the ﬁxed ﬁeld of an open subgroup with ( H ). From Theorem (3.4.2) we deduce

,¨ ]© ë

¤ ¨ ¤

H HKJ

that H , where corresponds to . We conclude that = for the

è è

Ψ ¤ ,

ﬁxed ﬁeld of ( ) (or of ( H ) ).

EGFN8;GAlN8D:U=

,¨ ,} ¤

The correspondence HKJ is an order reversing bijection between the

lattice of abelian extensions of and the lattice of closed subgroups in . The quotient

,¨ 9/ ,µ/1 ¤

group HKJ is isomorphic to a dense subgroup in Gal( ). è

Proof. It remains to use the injectivity of Ψ ¤ and the arguments in the proof of Proposi-

tion (3.4.2) and Theorem (3.5.2) (replacing the word “open” by “closed”).

µ/0 ,

(3.5.4). Let , be a ﬁnite abelian extension, and 0 be the maximal unramiﬁed subextension

¤ ¤

, Ψ 3 ,µ/=, ©

of in . Theorem (3.4.2) shows that ( ) Gal( ). Conversely, if U

J 0

Gal( ) and U = ( ) for , then Theorem (3.4.2) implies that ( ) = 0, i.e.,

0 J

¤ ¤

Ψ ¤

J 0

ur ur

¤ ¤ ¤ ¤

, /1 3 Ψ

deduce that Ψ ( 3 ) ur = Gal( ). Since is compact and is continuous, the J

ab ur

3 ¤ /0

group Ψ ¤ ( ) is closed and equal to Gal( ).

¤ ¤

| Ψ | #

Let be a prime element in and ( ) = . Then ¤ ur = , and for the ﬁxed ﬁeld

ÐÏ of we get

ur ur ab

ÐÏ Ý°ÏR

2 = =

(the second equality can be deduced by the same arguments as in the proof of Proposition (3.2.1).

| ,µ/0 /1 The prime element belongs to the norm group of every ﬁnite subextension of Ï .

ab ur

/0°Ï /0 ÐÏN/1 The group Gal( ) is mapped isomorphically onto Gal( ) and the group Gal( )

ab ur

/1 ¤ is isomorphic Gal( ). The latter group is often denoted by and called the inertia

" ab ab

/1 subgroup of ¤ = Gal( ).

) 79GLj

44 f

We have

ab ur ur ¡

Ñ Ñ Ñ

/0 /0 Ò /0 /1 /1 34¤ X Ï Gal( ) Gal( Ï ) Gal( ) Gal( ) Gal( ) and

ab ur

Ψ ¤

eÒ /1

( ) = Gal( )

where e is the cyclic group generated by . We observe that the distinction between and ¡

ab ¡

/0 X X 9/P3

Gal( ) is the same as that between and . So if we deﬁne the group as lim

3 Ò X where 3 runs over all open subgroups of ﬁnite index in , then = and the

reciprocity map Ψ ¤ extends to the isomorphism (and homeomorphism of topological spaces)

" c

¡ ab ab

Ψ ¤ L /0 : Gal( ) = ¤

Deﬁne c

¡ 1 ab

ϒ Ψ /1 L

= ¤ : Gal( )

¤ 3 ¤

Then ϒ ¤ maps homeomorphically onto . The ﬁeld Ï can be explicitly generated by roots of iterated powers of the isogeny of a

formal Lubin–Tate group associated to | .

3.6. Comments on other approaches to the local reciprocity map

(3.6.1). The approaches of Hazewinkel and Neukirch for local fields with finite residue field can be developed without using each other; but each of them has to go through some “unpleasant” lemmas. In characteristic there is a very elegant elementary approach by Y. Kawada and I. Satake

which employs Artin–Schreier–Witt theory.

r-¢ L¢©Ñ

(3.6.2). The maximal abelian totally ramiﬁed extension of r-¢ coincides with ( ) where

L¢Ñ

is the group of all roots of order a power of . By using formal Lubin–Tate groups

ÐÏ associated to a prime element | one can similarly construct the field of (3.5.5). Due to explicit results on the extensions generated by roots of iterated powers of the isogeny of the formal group, one can develop an explicit class field theory for local fields with finite residue field. Disadvantage of this approach is that it is not apparently generalizable to local fields with infinite residue field.

(3.6.3). All other approaches prove and use the fact (or its equivalent) that for the Brauer group

of a local ﬁeld there is a (canonical) isomorphism

¤ Lr#/1X inv : Br( ) Historically this is the ﬁrst approach due to H. Hasse.

Recall that the Brauer group of a ﬁeld s is the group of equivalence classes of central

simple algebras over s . A ﬁnite dimensional algebra over is called central simple if

/1s ,

there exists a ﬁnite Galois extension , such that the algebra viewed over isomorphic to

a matrix algebra over , (in this case is said to split over ). A central simple algebra

Ò Ò s Y¸

over s is isomorphic to ( ) where is a division algebra with centre , 1. Two

R=§ÓR= GA:':-8;TmOmFHGLTGNO(FA8DlW

central simple algebras `âU are said to be equivalent if the associated division algebras are s isomorphic over s . The group structure of Br( ) is given by the class of the tensor product of representatives.

A standard way to prove the assertion about Br( ) is the show that every central simple

/0 algebra over splits over some ﬁnite unramiﬁed extension of , and then using Gal( )

sep

¶ /E¶

Gal( ) reduce the calculation to the fact that the group of continuous characters X of

r#/1X

is canonically (due to the canonical Frobenius automorphism) isomorphic with .

Now let a character X = Hom ÿ ( ) correspond to a cyclic extension of

1 °

U called cyclic algebra deﬁned as ° =0 where = , = ( ) for every .

We have a pairing

Ò ¤Ç LrH/0X` '.Ô Ã ¤

ù Õ

X ( ) inv ([ ]) This pairing induces then a homomorphism

ab c

/1 Gal( ) = Hom(X âr/0X ) Then one proves that this homomorphism possesses all nice properties, i.e. establishes local class ﬁeld theory. The just described approach does not require cohomological tools and was known before the invention of those.

(3.6.4). Using cohomology groups one can perhaps simplify the proofs in the approach described in (3.6.3). From our point of view the exposition of class field theory for local fields with finite residue field given in this chapter is the most appropriate for a beginner; at a later stage the cohomological approach can be mastered. The real disadvantage of the cohomological approach is its unexplicitness whereas the approach in this chapter in addition to quite an explicit nature

can be easily extended to many other situations. If , is a ﬁnite Galois extension of then one has an exact sequence

$ 2

,µ/0 , ,

1 (Gal( ) ) Br( ) Br( ) 1 , and Br( ) is the union of classes of algebras which split over (i.e. the image of all

$ 2

/1 7, ,/1 (Gal( , ) ) for all ﬁnite Galois extensions ). So inv induces a canonical isomorphism

$ 1

2 c

,µ/0 , X¨/EX

¿ ¤

inv J : (Gal( ) )

/m , ¤

Denote the element which is mapped to 1 : by J . If stands for the modiﬁed

Tate’s cohomology group, then the cup product with J induces an isomorphism

$ $

¡ +2

,/1 "X ,µ/0 , (Gal( ) ) ¿ (Gal( ) )

For = 0 we have

$ $

¡ ab 0 ¡ 2

, ,/1 ,/1 "X ¿ ,/1 7, / ¤

Gal( ) = (Gal( ) ) (Gal( ) ) = HMJ è which leads to the analog of Theorems (3.3.3) and (3.4.2). Certainly the last isomorphism in ϒab

much more explicit form is ¤ .

J è

) 79GLj

46 f ¤! Ar#/1X Using cohomology groups one can interpret the pairing WÒ X of the previous

subsection as arising from the cup product

$ $

$ 0 2 2

µ/1 7, Ò ,µ/1 "X ,µ/0 ,

(Gal( , ) ) (Gal( ) ) (Gal( ) ) $

$ 1 2

/1 "r'/0X ,/1 "X and the border homomorphism (Gal( , ) ) (Gal( ) ) associated to the

exact sequence c

0 Xrr#/1X§ 0

For a ﬁeld s one can try to axiomatize those properties of its cohomology groups which are

" ab

sufﬁcient to get a reciprocity map from s! to , as it is well known this leads to the notion Â

of class formation.

(3.6.5). Assume that is of characteristic zero with ﬁnite residue ﬁeld of characteristic .

)Ò LrH/1X

For \6 1 the -component of the pairing X deﬁned in (3.6.3) is a pairing

$ " $ " $ " c

1 1 2

¤ ¤ ¤

7L¢ Ò âX¨/ X .L¢ ò ( ò ) ( ) ( )

If for every \ one knows that this pairing is a perfect pairing, and the right hand side is a cyclic

group of order , then one deduces the -part of class ﬁeld theory of the ﬁeld .

More generally, for a ﬁnitely generated Xµ¢ -module equipped with the action of and

ë ~.L¢ annihilated by deﬁne (1) = Hom( ò ). The previous pairing can be generalized to

the pairing given by the cup product

° °

$ " $ " $ "

2 2 c

¤â Ò ¤â ¤. ¢

( ) ( (1)) ( ò ) By Tate local duality it is a perfect pairing of ﬁnite groups. So, if one can establish Tate local duality independently of local class ﬁeld theory, then one obtains another approach to the -part

of local class ﬁeld theory in characteristic zero.

$)° "

¤ Γ â J.-M. Fontaine’s theory of Φ -modules was used by L. Herr to relate ( ) with Φ Γ cohomology groups of a simple complex of -modules. Then Tate local duality can be established by working with the complex above and this provides another approach to the -part of local class ﬁeld theory. Exercises

1. Go through the proof of (1.6).

2. Go through the proofs of (1.12).

3. Multiplicative Representatives.

, ¥© 2©)%

Assume that char( ) = 0 Let . An element is said to be a multiplicative

¢ê

justiﬁed by the following Proposition. U

¥4lGAB;GLF¦Q§ORQSGAT=

¥}©

¢ê

¥ ¥ Æ

2 . A multiplicative representative for such is unique. If and have the multiplicative

¥RÆ representatives and , then is the multiplicative representative of .

Proof. We need the following Lemma.

798;:-:-<=

¢ ò ¢ò

¡ 6 !Y Y¯, ¡ 6 (\ Y

¢ ¢ ¢ ¢ ¢ ¢

1 1

|9×N |9×N ¬¬¬ |9nN | N

Proof. Put = + ; then = + + + ( ) + , and as

¢ ¢ ¢

ccc

¡ |9×N Y ¡ | N !Y

¡ ( ) 1 (recall char( ) = ), we have ( ) + 1 ( ) + 1, and

¢ ¢

+1 c

6 ©+|9 %

Now the required assertion follows by induction.

¢ê

¥]©

To prove the ﬁrst assertion of the Proposition, suppose that 2 . Since has

¢ ê

no nontrivial -torsion, there exist unique elements ¥ satisfying the equations = .

¢ ¢

©)% ¥ ¡

Let be such that = . Then +1 = and ( +1 ) 1. The previous

+1 å

¢ò

¢ò ¢ê ò

6 \

lemma implies ¡ ( +1 ) + 1 Hence, the sequence ( ) is fundamental. It

¢ê ò ¢ò

©% \ ¥

has the limit = lim . We see that = 0 for 0 and 0 = , i.e., 0 is a

¥Ì©

multiplicative representative of ¥ . Conversely, if has a multiplicative representative ,

¢ ê

¥Ç©

Furthermore, if and are multiplicative representatives of , then writing =

¢ê

¢ ê ¢ê

Ú Û

" ©§%

= for some , we have = and = because of the

¡ (Y

injectivity of in . Now the previous lemma implies ( ) + 1, hence = .

¥ Æ p ¥RÆ

Finally, if and are the multiplicative representatives of and , then = and

¢ ê

p]© p ¥RÆ

2 . Therefore, is the multiplicative representative of .

5 Ø Denote the set of multiplicative representatives in % by .

47 8DljQSFk8DF

48 >@?

GAlGLCEC

e If is perfect (i.e. is a local ﬁeld) then every element of has its

Ø ÙØ ÚØ

multiplicative representative in . The map : induces an isomorphism ¿ 0 .

ÙØ

The correspondence : is called the Teichmuller¨ map.

Ø E

If is ﬁnite then 0 is a cyclic group of order equal to 1.

GAlGLCEC

' ¥LâÆ³©

Let char( ) = . If are the multiplicative representatives of ,

¥ Æ

then + is the multiplicative representative of + .

¢ ê

¢ ê ¢ê ¢ ê

" !©

Proof. Let = = . Then + = ( + ) hence + 2 and

¥ Æ + = + .

4. Go through the proof of (2.9) and (2.10). 5. Go through (2.11). 6. Go through the proofs of (2.12).

7. Structure theorems for complete discrete valuation ﬁelds. Let be a complete discrete

valuation ﬁeld with perfect residue ﬁeld.

There are three possible cases: two equal-characteristic cases, when char( ) = char( ) = 0

or char( ) = char( ) = 0, and one mixed-characteristic case, when char( ) = 0 char( ) =

, 0.

798;:-:-< =

% e The ring of integers contains a nontrivial ﬁeld if and only if char( ) =

char( ).

* õ 2

Proof. Since ¤ = (0), is mapped isomorphically onto the ﬁeld , therefore

%'¤

char( ) = char( ). Conversely, let be the subring in generated by 1. Then is a ﬁeld

* 2 if char( ) = , and ¤ = (0) if char( ) = 0. Hence, the quotient ﬁeld of is the desired

one.

A ﬁeld , that is mapped isomorphically onto the residue ﬁeld = is called a

%'¤

coefficient field in %'¤ . Such a field, if it exists, is a set of representatives of in . (1.8)

implies that in this case is isomorphic (algebraically and topologically) to the ﬁeld (( )):

t a prime element | in corresponds to . Note that this isomorphism depends on the choice of a coefﬁcient ﬁeld (which is sometimes unique, see Proposition 2 below) and the choice of a

prime element of .

We shall show below that a coefﬁcient ﬁeld exists in an equal-characteristic case.

The simplest case is that of char( ) = char( ) = 0.

¥4lGAB;GLF¦Q§ORQSGAT =

e Let char( ) = 0. Then there exists a coefficient field in . A coefficient field r

can be selected in inﬁnitely many ways if and only if is not algebraic over .

Proof. Let be a maximal subﬁeld in , in other words, be not contained in any other

larger subfield of % . We assert that = , i.e., is a coefficient field. Indeed, if is

algebraic over , then is separable over and we can apply the Henselian property to ﬁnd

2©~ © ©

maximality of , we get . Furthermore, let be transcendental over .

Ë Ë 8;l;jQSFk8;F

>@? 49

¤ ®

!©6% ¥°´

Let be such that = . Then is not algebraic over , because if ° =0 = 0

¥ © ® ¥ ¥ ¥

° ° ° °

with , then ° =0 = 0. Hence, = 0 and = 0 ( is mapped isomorphically

onto ). By the same reason [ ] 2 = (0). Hence, the quotient ﬁeld ( ) is contained

$ in % and = ( ), contradiction. Thus, we have been convinced ourselves in the existence

of a coefﬁcient ﬁeld.

r (©%

If is not algebraic over , let be an element transcendental over the prime subﬁeld

¤ ¤

in . Then the maximal subﬁeld in , which contains ( + ) with ¤ ,

r r is a coefﬁcient ﬁeld. If is algebraic over , then is algebraic over and is uniquely

determined by our previous constructions.

¥4lGAB;GLF¦Q§ORQSGAT =

Let char( ) = . Then a coefﬁcient ﬁeld exists and is unique; it coincides

¤ % with the set of multiplicative representatives of in .

Proof. Convince yourself.

We conclude with the case of mixed-characteristic: char( ) = 0, char( ) = . with the s

residue ﬁeld isomorphic to . Here is an analog:

¥4lGAB;GLF¦Q§ORQSGATÛ)R=

Let be a complete discrete valuation ﬁeld of characteristic 0 with residue

s s

ﬁeld s of characteristic . Let 1 be any extension of . Then there exists a complete

ß s

discrete valuation ﬁeld 1 which is an extension of , such that ( 1 ) = 1 and 1 = 1.

s ¥ s

Proof. It is sufﬁces to consider two cases: s 1 = ( ) is an algebraic extension over

s Ü s s /1s

and s 1 = ( ) is a transcendental extension over . If, in addition, in the ﬁrst case 1

¥ s t

is separable, then let ( t ) be the monic irreducible polynomial of over , and let ( )

t t

be a monic polynomial over the ring of integers of s such that ( ) = ( ). By the

t ¥

Hensel Lemma (1.2) there exists a root of ( ) such that = . Then 1 = ( ) is the

¥ Æ)©s

desired extension of . Next, if = and is an element in the ring of integers

of such that = , then 1 = ( ) is the desired extension of for = . Finally,

- - Ü ° ¡ p°

[ +

in the case of transcendental extension let be deﬁned as ( ( )) = min + ( ) for

[

( ) = ° = [ ] with = 0, . Extending to ( ) we obtain the discrete

x s

valuation - with residue ﬁeld 1. É GAlGLCEC

residue ﬁeld of characteristic and the absolute index of ramiﬁcation is equal to 1. r ¢ Proof. One can set = and apply the Proposition.

8. Hasse–Herbrand function. Let be a complete discrete valuation ﬁeld with perfect

residue ﬁeld.

¥4lGAB;GLF¦Q§ORQSGAT=

,µ/0

Let the residue field be infinite. Let be a finite Galois extension,

¤ HMJ

H = . Then there exists a unique function

Ý!QÝ ¤

= J : è

such that (0) = 0 and

35Þ 34°´ù 30Þ §3$ 30Þ §3

¤ ¤

° ù ° ù ù ° ù ù ° °

J J J

H H H ( ) ( ) +1 ( )+1 +1 8DljQSFk8DF

50 >@?

é6,Ú 3 ¹3

±"ù

° ù J

Proof. The uniqueness of follows immediately. Indeed, for ( ) H +1 ,

À À À À

¿ ¿ ¿

¦ ¦

hence if is another function with the required properties, then ( ) ( ) ( ) ( ), i.e.,

= .

,/1

As for the existence of , we ﬁrst consider the case of an unramiﬁed extension . (2.12)

À À

, $ ,

H ¤

shows that in this case ( ) = (because ¤ ( ) = 1 and Tr = ).

J J

è è µ/0

The next case to consider is a totally ramiﬁed cyclic extension , of prime degree. In

, , ¤

this case the map HKJ is determined by some nonzero polynomials over . The image of è

under the action of such a polynomial is not zero since , is inﬁnite. Hence, we obtain

À À

( ) = : µ/0

if , is totally tamely ramiﬁed, and

À À

¦ÁP

w À

( ) = À

ÁP

(1 ) +

µ/0 ,

if , is totally ramiﬁed of degree = char( ) 0.

¨ ¨ ¤

Now we consider the general case. Note that if we have the functions J and for the

è è

,µ/~â§/0 ,/1

¤ ¨ ¨ ¤

ª J

Galois extensions , then for the extension one can put J = .

è è

Indeed, è

Þ ¤ Þ

3 §34°´ù 3

° ù ° ù

¤ ¨ ¤

f=hmi ¹5h i

J H

HMJ ( ) ( )

è è ¤

Furthermore, the behavior of HKJ is determined by some nonzero polynomials (the compo-

¨ ¨ ¤ H

sition of the polynomials for HKJ and , the existence of which can be assumed by è

induction). Hence è

35Þ 3$

¤ ° ù ° ù

f=h i J

HKJ ( ) +1 è

Since

30Þ 35Þ 3

¨ ¨

¤ ° ù ¨ ¤ ° ù ° ù

f=hmi ¹5h i

J H

HMJ ( )+1 ( )+1 +1

è è

we deduce that = J is the desired function.

è ur

, ,

J J J J

In the general case we put J = 0 for 0 = and determine 0 by

è è è

/9, induction using solvability of , 0.

To treat the case of ﬁnite residue ﬁelds we need

798;:-:-<=

we get

Hªß ß HKJ ( ) = ( )

ur ¤ ur

è è

ur ur ß ur ur

, ,5à

where à is the completion of , = .

= °²± with 0 1 = :

J J

± =0 8;l;jQSFk8;F

>@? 51

ur ur

, |

°²±

¤ J

Then HMJ ( ) = det( ). Since = ( ) and è

ur ur ur ur ur

, ß , ß , ß ,5 , : = ( ) = ( ) = ( ) = :

we get

°²± ¤

ur ur ¤ HKJ

HMJ ( ) = det( ) = ( )

è è

ur ur "

K/1 ¾Ú,

Finally, let be a ﬁnite totally ramiﬁed Galois extension with . Let =

" $ "

ur $ ur

ur ur

: . Then

ur ¤ ur

U HSß ß HKJ ( ) = ( ) = ( )

ur ¤ ur

" $

ß ¡

¡ ur ur

K/ K/ ,

because and are isomorphic to Gal( à ) and Gal( ). µ/0

This Lemma shows that for a ﬁnite totally ramiﬁed Galois extension , the functions

,/1

ß ß J and coincide. Now, if is a ﬁnite Galois extension, we put

ur ¤ ur

J J ß ß J = =

ur ¤ ur

0 J

è è è

ß ß In particular, if is finite we put J = (the residue field of is infinite as the

ur ¤ ur

J è separable closure of a ﬁnite ﬁeld). è

It is useful to extend this function to real numbers. For unramiﬁed extension, or tamely totally

ramiﬁed extension of prime degree, or totally ramiﬁed extension of degree = char( ) 0 put

]p ])¦ÁP

] ] ] , ]á ]

¤ ¤ ¤

J J

J ( ) = ( ) = : ( ) =

è è è

]pÉ])Á

(1 ) +

) ,µ/=,

for real ] 0 respectively. Using the solvability of 0 (Corollary 3 of (4.4) Ch. II) and the

]

¤ ¨ ¨ ¤ ¤

ª J J

equality J = deﬁne now ( ) as the composite of the functions for a

è è è è

µ/9,

tower of cyclic subextensions in , 0.

EGFN8;GAlN8D:U=

¤ ¤ ¤

Thus deﬁned function J : [0 + ) [0 + ) is independent on the choice of

,/1 ¤

a tower of subﬁelds. The function J is called the Hasse–Herbrand function of . It is

piecewise linear, continuous and increasingè .

/ /

Proof. It sufﬁces to show that if 1 , 2 are cyclic extensions of prime degree, then

¨ ¨ ¨ ¨ ¨ ¨

µ µ ª

(*) 1 ª 1 = 2 2

è è è è

where = 1 2.

] ]

¨ ¨ ¨

Note that each of ¨ 1 ( ), 2 ( ) has at most one point at which its derivate is not è continuous. Therefore thereè are at most two points at which the function of the left (resp. right)

hand side of ( o ) has discontinuous derivative. By looking at graphs of the functions it is obvious that at such points the derivative strictly increases and there is at most one such noninteger point

for at most one of the composed functions of the left hand side and the right hand side of ( o ).

At this point (if it exists) the derivative jumps from to 2. From the uniqueness in the preceding proposition we deduce that the left and right hand sides

of ( o ) are equal at all nonnegative integers. Thus, elementary calculus shows that the left and

right hand sides of ( o ) are equal at all nonnegative real numbers. 8DljQSFk8DF

52 >@?

9. Selfduality of local fields with finite residue field. Let be a local field with finite residue

¦ ¦

ÍâH

ﬁeld. Prove that there is a nontrivial character : and that every character of is

ÌÃ ¥ã] ¥©W of the form ] ( ) for a uniquely deﬁned . This means that the additive group of is selfdual. It is one of the ﬁrst observations which lead to the theory of J. Tate and K. Iwasawa

on harmonic analysis interpretation of the zeta function.

t Y ¶;¢¹ X/ â#

[Hint: For a local functional ﬁeld ¶ (( )) let : be induced by

X] | ¥m/

¥ + exp(2 ), deﬁne

¥ t Yã ¥

: Tr 1

For ¢ let ( ) be such that ( ) mod where = ( ). Deﬁne

| / ^ : exp(2 ( ) )

For a local number ﬁeld deﬁne

¦ ¦

¤ ^

= ª Tr

| ¥ °

To show the second part for a charater ﬁnd as ° constructing step by step. ]

Ë Ë 10. Go through details of (3.3.3). 11. Go through details of (3.4). 12. Go through details of (3.5).

13. The reciprocity map and Hasse–Arf theorem.

EGFN8;GAlN8D:U=

µ/0 ,µ/0

Let , be a ﬁnite abelian extension, = Gal( ). Denote by the Hasse–

3 3 3 ¤

¤ ¤

¤ ù ù

Herbrand function J . Put 1 = , 0 = , and ( 1) = 1. Then for every

,¨ ,¨9/ 3 ¤

\] Ψ

¤ ¤ ¤

HMJ HMJ

integer 1 the reciprocity map J maps the quotient group

è è è

" "

,¨ ,¨ç/D3 \ Þ 3 ¤

¤ ¤ ù HMJ

isomorphically onto the ramiﬁcation group ( ) = ( ) and HMJ +1

è è

" "

/ Þ

isomorphically onto Þ ( ) ( )+1.

" "

Þ \

Thus Þ ( )+1 = ( +1) for integer ; this is called the Hasse–Arf theorem.

Proof. Let , 0 be the maximal unramiﬁed extension of in . We know from Exercise 8 that

3 ù 3 ù \!

¤ ¤

J J HMJ

J = 0 , and from section 1 that the norm 0 maps 0 onto for 0.

è è è

,Uâ 7, ,

Using the second commutative diagram of (3.4.2) (for = = = 0 ) we can therefore

µ/0 \

assume that , is totally ramiﬁed and 0.

We use the notations of Corollary (3.3.2), so and are complete ﬁelds. Let U ( ),

1 1 1

then belongs to ( ) J . Let be such that = . The ﬁrst proposition

× 35Þ ù

of (3.1.8) shows that can be chosen in ( ) . Now from Corollary (3.3.2) and the previous

ϒ ¤ ϒ × 3 ù ,¨ Þ

¤ ¤

U HK HMJ

Exercise we deduce that J ( ) = = ( ) belongs to . So ( ( ))

è è è

¤ ϒ 3 ù ,¨ ,¨ Þ 3

¤ ù ¤ HKJ

HMJ . Similarly, we establish that ( ( )+1) +1 .

è è

3 ù , ,µ/1 ¤

Conversely, let belong to HKJ . For the abelian extension we will prove below a stronger assertion than that in Corollaryè (3.3.2):

1 1

®N

Hª

there exists such that ( ) mod HKJ and = for some

1 è

U Gal( ). For every such we have ( ) .

Ψ ¤ Ψ 3 ù ,¨ Þ 3 ù ,

¤ ¤ HKJ

From this assertion we deduce that ( HKJ ) ( ). Hence ( ) =

è è

" " " "

" ϒ ϒ Þ Þ Þ Þ Þ

¤ ¤ J

( ) and J ( ( )+1) = ( ( +1)), so ( )+1 = ( +1).

è è 8;l;jQSFk8;F

>@? 53

µ/1 \

It remains to prove the assertion by induction on the degree of , . If = 0, the assertion

,/1 `

is obvious, so we assume that \6, 0. If Gal( ) is of prime order with generator , then from

×©Ç3 ÌÍ × ,

HMJ

Corollary (3.3.2) we know that there is such that Hª ( ) mod and

è è ê

1 1 ê 1 1

×(®N | ¤ Y é ¡ ×V®N ¡ | ¤ `m

= for some integer . So = ( 1) = ( 1). If = 1 then

`më$ 3 ,

ù ± ¤

the assertion is obvious, so assume that = 1. From (2.12) we know that +1 HMJ . è

× 3 × `

ù ±

J HK

If Hª ( ) belongs to +1 , then ( ( )) is 1, not , a contradiction. Therefore,

è è è

¡0¤ × é é

¤ J

( HK ( ) 1) = = ( ).

è è

,µ/0 ,µ/

For the induction step let §/0 be a subextension of such that Gal( ) is of

1 1 1

× ×V®N | ,¨ ×V®N

HMJ HK

HK ( ) mod and = . By the induction hypothesis belongs

è è©

è 1

35Þ ®N ù

C C

ù ¹5hmi

from the same proposition we deduce that HK = with ( ) and .

è©

©)30Þ

ù f=h i

According to the deﬁnition of the Hasse–Herbrand function in Exercise 8 there is ( )

1 1 «

®N £

| Hª

such that Hª ( ) = . Then = for some in the kernel of .

è© è© « «

1 ê

,µ/ 2Í§| ¤ Let ` be a generator of Gal( ). Using Proposition (3.1.7) we deduce that

1 1 ê 1

mod ( ) and so = for an appropriate and some integer . «

1 1 1 ê 1

By the ﬁrst proposition of (3.1.8) there is such that = . Then =

1 1

× N±¤0 35Þ &; ù

where = . All we need to show is that belongs to ( ) . If it does not, then

1 1 ê

é ¡ N£¤P ¡ | 9¤ , ,5 N J

= ( 1) = ( 1) 0. Let = ( ) as deﬁned in (2.12). Since is a

unit, from (2.12) we deduce that é` is prime to . On the other hand, the following Lemma

implies that é is congruent to modulo , a contradiction.

798;:-:-<= µ/1

Let , be a ﬁnite Galois extension with separable residue ﬁeld extension. Let

À À

" " " " "

1 1

± °

± ° ± °

U U U +1 and +1 with 1. Then + +1 and

mod .

| ,

Proof. Let J be a prime element of . Then

| `R|

J J

9| -"©)%

= 1 + p| = 1 + with

J J

| |

J J Therefore

± +1

`R| | |

J J J

U U U

U = + ( )( )

± ± ± ° ° °

+1 +1 + +1 + +2 c

Í| p| 9| é p9| *

J + + + ( + 1) mod

J J J

± ° ±

° + +1 + +2

`W 8` | Í é³ p9| * |

J J U

Hence ( U ) ( ) mod . Substituting instead of the other

J J

1 1

` | , J prime element U of we deduce that

1 1

` | `

° ± ° ±

U U

+ + +1 c

é. p9| *

Í 1 + ( ) mod

J J

é ,µ/0 Now if is the maximal ramiﬁcation number of , then ± +1 = 1 . Therefore

the last formula in the previous paragraph shows that every positive ramiﬁcation number À

/1 é

of , is congruent to modulo . Therefore every two positive ramiﬁcation number of

µ/0 , are congruent to each other modulo . Finally, from the same formula we deduce that

1 1 "

` © `

° ± U U + +1. 8DljQSFk8DF

54 >@?

14. The Hilbert symbol. Let be a local field with finite residue field . Let the group X

sep

\ \

of all th roots of unity in the separable closure be contained in and let if

char( ) = .

6Ò)ä

The norm residue symbol or Hilbert symbol or Hilbert pairing ( ¬§¬ ) : is deﬁned by the formula

1 sep c

Ψ ¤

N N N mN©W

( -" ) = ( )( ) where =

sep 1

9©W Nç NÜ N ç©

If N is another element with = , then and

1 1 c

Ψ ¤ Ψ

N N N N ( )( ) = ( )( )

Hence the Hilbert symbol is well deﬁned.

¥4lGAB;GLF¦Q§ORQSGAT =

e The norm residue symbol possesses the following properties:

(1) ( ¬§¬ ) is bilinear;

2©W â$

(2) (1 6'" ) = 1 for = 1 (Steinberg property);

(4) ( -" ) = ( ) ;

-" ]©

(5) ( ) = 1 if and only if H ( ) and if and only if

¤ ¤ ò

( )

¤ ¤

òæ

H ( ) ;

( )

]©+ 2©+

(6) ( -" ) = 1 for all if and only if ,

2©+ ]©+

( -" ) = 1 for all if and only if ;

-" Y¸ 7

(7) ( -" ) = ( ) for 1 ;

J J

(8) ( ) = ( HMJ ) for , where ( ) is the Hilbert symbol

¬i¬ ù , in , , ( ) is the Hilbert symbol in , and is a ﬁnite separable extension of ;

sep

J J

U U U

(9) ( U ) = ( ) , where is ﬁnite separable, Gal( ), and

:,¨ Ç

but not necessarily .

sep

)© .N

Proof. (1): For N = we get ä

1 1 ä 1

¤ ¤ ¤

Ψ ¤ Ψ Ψ Ψ

N ãeN N ¬NãçN N

N ( 1 2)( ) = ( 1) ( 2)( ) ( 1)( ) ä

1 ä 1

Ψ ¤ Ψ

ãçN N N çN

= ( 2)( ) ã ( 1)( )

©è

since Ψ ¤ ( 1) acts trivially on ( 2 ) . We also obtain

ä ä

1 1 ä 1 1

¤ ¤

Ψ ¤ Ψ Ψ

ãeN N N N ãçN N ãçN N

( ' 1 2) = 1 2 ( )( 1 2) = 1 ( )( 1) 2 ( )( 2)

c -"

= ( ' 1) ( 2)

sep

1 2

Ψ ¤

(5),(2),(3),(4): ( ' ) = 1 if and only if ( ) acts trivially on ( ) and if and only if

ã 2©

by Theorem (3.4.2) H ( )) .

¤ ¤

( ò )

Ü © á \

\çY . Let = 1 with 1 and let be a primitive th root of unity. Then for 8;l;jQSFk8;F

>@? 55

sep

å 1

° °

6 ]á & ]á á &

ð ð ð

1 = (1 ) = a 1 1

° °

=1 =1 ± =1

ã ]á & d£©

¤ ¤ ¤ ¤

ò æ ò æ

ð H

= H 1 ( )

( ) ( )

è è

° =1

1 1

H ) )- $ "$ Hence, (1 '" ) = 1. Further, = (1 )(1 ) for = 0 = 1. This means that

1 1 1

-â â

( H-â ) = (1 ) (1 ) = 1. Moreover,

H'" -" â # ' Ækß;"

1 = ( Hp "p ) = ( ) ( ) ( ) ( ) = ( ) ( )

i.e., ( -" ) = ( ) .

Finally, if ( -" ) = 1, then ( ) = 1, which is equivalent to

6©

¤ ¤

òæ H ( )

( )

,¨ ,µ/0 ¤

and is a nontrivial abelian extension. By Theorem (3.4.2) the subgroup HMJ does not

,¨ !© /© ¤

coincide with . If we take an element such that HKJ then, by property

è $ (5), we get ( ' ) = 1.

sep

(7): For N = , one has ä

1 ä

Ψ ¤ Ψ

ãºN N ãeN N '

( -" ) = ( )( ) = ( )( ) = ( )

ç because ( N ) = .

(8): Theorem (3.4.2) shows that ä

1 1 ä

Ψ Ψ ¤

-" -" ù N N N ã N ã

¤ ¤

J J HMJ

( ) = ( )( ) = HKJ ( ) ( ) =

è è

sep

)© .N

(9): Theorem (3.4.2) shows that for N = , c

1 ä

N N -" ' Ψ

ã ù ù

J J J

U U U

( U ) = ( )( ) = ( )

GAlGLCEC

LÍ Ò+ /0 /1

( ¬§¬ ) :

/1 \ £Ü \

The Kummer theory asserts that abelian extensions , of exponent (

¾ J

if char( ) = ) are in one-to-one correspondence with subgroups B J , such that B

J J

, = ( B J ) = ( : B ) and the group B is dual to Gal( ).

EGFN8;GAlN8D:U=

¯Ü \

Let , if char( ) = . Let A be a subgroup in such that

¬§¬

A. Denote its orthogonal complement with respect to the Hilbert symbol ( ) by

B = A é , i.e.,

' ]©

,¨ , êKé ¤

Then A = HMJ , where = ( B) and = . è 8DljQSFk8DF

56 >@?

Proof. Recall that is of ﬁnite index in .

Y é

Let B be a subgroup in with B and B : = . Let A = B . Then

Ψ ¤ ( ), for A, acts trivially on ( ) for B. This means that Ψ ( ) acts trivially

, 2© ,¨ ¤

on = ( B) and, by Theorem (3.4.2), HKJ . Hence

, ¤

A HKJ

]© ¤ Z,

, Ψ ¤

Conversely, if HKJ , then ( ) acts trivially on ( ) and

(©

H ( )

¤ ¤

( ò )

è -"

, ,¨

¤ ¤ HKJ

HMJ A. Thus, A = .

è è

Furthermore, to complete the proof it sufﬁces to verify that a subgroup A in with

é é Ò)

R A coincides with (A ) . Restricting the Hilbert symbol on A we obtain that

é~ë

it induces the nondegenerate pairing A /0R)Ò9/ A . The theory of ﬁnite abelian

/ é

groups implies that the order of A /1m coincides with the order of A . Similarly, one

-/0 9/ é é

can verify that the order of A é is the same as that of (A ) , and hence the order of

é é é é é é é

/ A equals the order of (A ) /1 . From A (A ) we deduce that A = (A ) .

The problem to ﬁnd explicit formulas for the norm residue symbol originates from Hilbert.

In the case under consideration the challenge is to ﬁnd a formula for the Hilbert symbol ( -" )

in terms of the elements ' of the ﬁeld . This problem is very complicated when .

×X

Nevertheless, there is a simple answer when \ .

¥4lGAB;GLF¦Q§ORQSGAT=

Let \ be relatively prime with and . Then

( ' 1)

è"

ì '

( ' ) = ( )

where % is the cardinality of the residue ﬁeld and

ì Ò+ Ù

: ' 1

is the tame symbol: ì ( ) is the multiplicative representative of the residue of

å å

& & & &

i i i i

( ) ( ) ( ) ( )

( 1)

IX \

Proof. Note that the elements of the group , for , are isomorphically mapped onto the

¶Ü \- ' subgroup in the multiplicative group . Hence, ( % 1). Note also that the prime elements

| ]©3 ¤ | | ó

generate . Indeed, if = ó with , then = 1 for the prime element

| ¥~$ | | | ¥

| 1 = , when = 1, and = 2 for the prime element 2 = , when = 1. Using

ì |4" |4"

properties (1) and (7) of the Hilbert symbol it sufﬁces to verify that ( ) = ( ) ' 1 for

]©+ .

' ¤ ù

Let = ( ) ó with = ( ) 1 1 . Then, as ( ) = 1 ( ) = 1,

Ë Ë

ì |4 ì |4 |4 #|

we obtain ( ) = ( ) = . Property (3) of the Hilbert symbol shows that ( ) ' 1 = 1.

Ë Ë

3 |4 /0

¤ ' ù

Since the group 1 is ( % 1)-divisible, ( ) 1 = 1. Finally, since the extension ( )

sep 1 Ë

Ë '

1 1 1 c

× ¤ | × × @¤ × × |4 Ψ

( ) ' 1 = ( )( ) = ( ) = =

Ë Ë

|4 ì |4

We conclude that ( ) ' 1 = = ( ).

Ë 8;l;jQSFk8;F

>@? 57

15. Local Kroneker–Weber theorem.

Read (3.1.3).

Let á 1 be a primitive ( 1) th root of unity, let 2 be a primitive th root of unity, and

, r á , r á ¢

1 = ¢ ( 1) 2 = ( 2). Show that

, Ò)3 ^ , Ò)3 ù ^

^ ^ HMJ

1 = HKJ 2 =

1 2

è è

Prove the local Kronecker–Weber Theorem: every ﬁnite abelian extension of r4¢ is contained

á \

in a cyclotomic ﬁeld r-¢ ( ) for a suitable primitive th root of unity.