Downloaded By: [Behn, Antonio] At: 19:03 4 March 2008 o all for the ril eipoeo hsrsl ysoigta vr eirm ylcrn is ring cyclic semiprime every 2 that that showing assume not by do result We commutative. this and on associative improve we article x ih-irns epoeta vr ih-irn sslal,adta farn is ring a if that and solvable, is right-nilring every that prove We right-nilrings. .INTRODUCTION 1. 10.1080/00927870701665248 LLC online DOI: Group, print/1532-4125 Francis 0092-7872 & ISSN: Taylor © Copyright Algebra in Communications nvria eCie ail 5,Snig,Cie a:+627-82 -al [email protected] E-mail: +56-2271-3882; Fax: Chile; Santiago, 653, Casilla Chile, de Universidad 00MteaisSbetClassification: Subject Mathematics 2000 Words: Key = ylcidentity cyclic ,i omttv n soitv ig nScin n ftepresent the of 3 and 2 Sections In ring. associative and commutative a is 0, lifl 19)poe hteeypiecci igwee2 where ring cyclic prime every that proves (1995) Kleinfeld nScin n esuytenloec n ovblt fcyclic of solvability and nilpotency the study we 5 and 4 Sections In edfiea define We eevdSpebr8 06 eie aur 2 07 omnctdb .P Shestakov. Ciencias, P. de I. Facultad by Matemáticas, Communicated de Departamento 2007. Behn, 22, Antonio January to Revised correspondence Address 2006; 8, September Received y z y x oascaiesmpienloetiett algebra. identity nilpotent semiprime Nonassociative in atao Chile Santiago, atao Chile Santiago, eut ooti ufiin odtosfrtenloec fcci right-nilalgebras. these cyclic apply of and nilpotency right-nilrings the cyclic for Moreover, of conditions solvable. nilpotency sufficient is the obtain right-nilring for to cyclic conditions results every sufficient that find identity and we polynomial commutative the and satisfying associative rings nonassociative study y zx we article this In 3 2 1 Behn Antonio SATISFYING NONASSOCIATIVERINGS OF NILPOTENCY AND SEMIPRIMALITY eateto ahmtc,Iw tt nvriy ms oa USA Iowa, Ames, University, State Iowa Mathematics, of Department Chile, of University Science, of Faculty Mathematics, of Department eateto ahmtc,Mtooia nvriyC.Educación, Cs. University Metropolitan Mathematics, of Department R . hc ecl cci ig. epoeta vr eirm ylcrn is ring cyclic semiprime every that prove We rings.” “cyclic call we which , ylcring cyclic ® 6 3–4,2008 132–141, 36: , 1 vnCorrea Iván , ob o eesrl soitv ring associative necessarily not a be to x yz 73;16N60. 17A30; x yz 132 = 2 n ri o Hentzel Roy Irvin and , y zx = y zx x = implies 0 3 R x x htsatisfies that = = 0. implies 0 x yz (1) = Downloaded By: [Behn, Antonio] At: 19:03 4 March 2008 ideals nilpotent. not are which n right-nilindex and vr ideal every a that prove to results our use we Finally, nilpotent. right-nilalgebra is cyclic it then right-nilpotent n of subring the h olwnspwr of powers followings the omttrby commutator o some for 3. 2. 1. 1. Lemma RINGS CYCLIC IN IDENTITIES 2. L and ring the in elements 4. Proof .Uigtecci dniy()adpr eobtain we 1 part and (1) identity cyclic the Using 2. x uhthat such x s uv st xy xy st uv xy st x st uv y x and ercl oedfiiin.Aring A definitions. some recall We o o eesrl soitv o omttv ring commutative nor associative necessarily not a For ntrso hs prtr,tecci dniy()i qiaetto equivalent is (1) identity cyclic the operators, these of terms In R suul ewl eoeteascao by associator the denote will we usual, As .Drc acltosuigtecci dniy()give (1) identity cyclic the using calculations Direct 1. . I R ssi obe to said is R R R x and 1 x n 1 1 epciey where respectively, , 1 R 1 , = xy st uv = I x R = = = = J Let of tx sy R n = n R x R of R R OASCAIERNSSATISFYING RINGS NONASSOCIATIVE = = R = 0 x y x R eeae by generated = ; , R 0 .Argtnligi adt have to said is right-nilring A 0. 0( eacci ig hntefloigiette odin hold identities following the Then ring. cyclic a be n I , . tx sy uv 2 and and and and and IJ + ; nilpotent = xy st n = A .W rsn xmlso ih-iagba fdmnin5 dimension of right-nilalgebras of examples present We 1. = R implies 0 xy n snloetwe dim when nilpotent is = R = mle that implies 0 stesals oiieitgrfrwihti strue. is this which for integer positive smallest the is − n of and uv xy st R R R x 0), n n n n yx = x n R ffrsome for if + aL = solvable 1 L snilpotent, is = s t xy = = n h etadrgtmlilcto prtr by operators multiplication right and left the and ; y i x I = + L R R R x j = = x = x ∈ n n = xa .I scerta rm mle semiprime. implies prime that clear is It 0. n n R n − − − R 1 i R x 1 1 ffrsome for if R = I : R and x R R j = 2 = L n s y tx scle a called is , y right-nil uv tx sy 0or R = aR for for for for for n R x = A xy n> n> n> n> n> J = ≤ 0, = right-nilindex n = ax X YZ , ffrevery for if rwhen or 4 ih-iptn (left-nilpotent) right-nilpotent tx sy rm ring prime 1( 1( 1( 1( 1( R a c b a 0. . n R rnia powers principal lnr powers plenary ih powers right ih powers right etpowers left = = = scalled is Y ZX R 0, tx sy uv edfierecursively define we , = nil A x ffreeypi of pair every for if n ab c ∈ ffrevery for if if a dimension has semiprime ) R x ) ) R hr ssome is there − n : ) = a bc ) o all for 0 x ffor if the , ∈ 133 (2) R if n Downloaded By: [Behn, Antonio] At: 19:03 4 March 2008 3 ENE AL. ET BEHN 134 .W s at n n h ylciett oget to identity cyclic the and 2 and 1 parts use We 3. oolr 2. Corollary Proof 2. 1. 3. Lemma in elements of products of sum hsproves This .W s at – n h ylciett oobtain: to identity cyclic the and 1–3 parts use We 4. oget to 1. part proves This Proof hspoe at2. part proves This If and If R R .Since 2. hsflosfo at3o em ,sneeeyeeetof element every since 1, Lemma of 3 part from follows This . .Uigtehptei n h ylciett 1 eget we (1) identity cyclic the and hypothesis the Using 1. . x z st y x aifisteidentity the satisfies sascaie then associative, is xy st uv xy st uv x st xy Let Let R xy z st R sascaie tstse h identity the satisfies it associative, is R = eacci ring. cyclic a be = eacci ig Then ring. cyclic a be 0 xy st .Uigi,w obtain we it, Using 0. . ======R 2 1 c c 1 3 c x st y x xy st = = yt sx uv u x yt vs vs tx uy xy vs ut x y st uv ys xt uv xy st uv aifisteidentity the satisfies x st yz xy z st = = R st xy y x st . = = x y st 0 then , ======R = = 2 1 1 c 3 1 x yz st z st xy y st x yx st 2 xy st uv u v s x yt vs yt ux sx vy ut ys xt uv xy vs ut R sassociative. is = lostse h identities the satisfies also x y st x st xy = y s tx = = x st y x x yz st x y z st ======0 1 c c 2 1 . s x yt uv vs u x yt sx tv uy st xy uv ys vx ut = .W s at1 part use We 0. R 2 x st xy safinite a is = 0 Downloaded By: [Behn, Antonio] At: 19:03 4 March 2008 eiy o eso h nutv tpfrpr .W suethat assume We 1. part for step inductive the show we Now verify. Proof 3. em 4. Lemma RINGS CYCLIC SEMIPRIME 3. 4. of 2. most em at4, part 1 Lemma Proof h nutv yohssadtecci dniyw obtain we identity cyclic the and hypothesis inductive the R of Proof if em 7. Lemma 1. iptn n therefore and nilpotent ial,w s em gi oso that show to again 3 Lemma use we Finally, identity order 6. Theorem RINGS CYCLIC IN NILPOTENCY 4. h ylciett eget we identity cyclic the and hoe 5. Theorem 4. Lemma proves This Proof k x R R R RR R = 2 ∈ . n Now, . 2 n n RR n epoedb nuto on induction by proceed We . yCrlay2 Corollary By . ic ti la that clear is it Since . Let . R ≤ n = I ⊆ − yteIao–unvNgt–imnterm(omnk 1990), (Formanek, theorem Ivanov–Dubnov–Nagata–Higman the By . n and 1 = RR R x st y x k Then . R − = n R 1 + n n + N 1 − y Let n I Let 2 . 1 = 2 R ∈ R Let Let n ea nLma4adlet and 4 Lemma in as be + ⊆ OASCAIERNSSATISFYING RINGS NONASSOCIATIVE R ; ; k k R R − R = n 2 R = x st y x 1 NR R 1 R eacci ig hn o every for Then, ring. cyclic a be R then snloetand nilpotent is n eacci igadlet and ring cyclic a be R − .Lma3sosthat shows 3 Lemma 0. easmpiecci ig Then ring. cyclic semiprime a be k o 2 for 1 eacci ih-irn findex of right-nilring cyclic a be 2 R R = n ; R xy + R yx st n ie that given and 0 xy st 1 2 − 2 ≤ ∈ and sslal.Thus solvable. is k sascaie oi sa soitv irn fidxat index of nilring associative an is it so associative, is N k = I ≤ RR o n hieof choice any for scoe ne diinw nyne oso that show to need only we addition under closed is yx n = = ewn oso that show to want We . n R r in are s t yx + s t xy ssolvable. is R n For . n R I nfc,let fact, In . R N + I = = R ssmpie eko that know we semiprime, is = n k = x z y x R n = − s x ty s x yt sas solvable. also is 2 1 x I = scommutative. is R ∩ y t s y x k ,alfu ttmnsaees to easy are statements four all 2, n ∈ RR R R ≥ R X YZ 2 yLma4, Lemma By . ∈ sascaieadcommutative. and associative is n = 2 = n x xR − : N ihu lmnso additive of elements without ∈ k 0 0 ∈ R = 2 + othat so y t s y I n = R Y ZX + R Hence . 1 n 0 − = k Then . R RR ∈ R n n R R sassociative. is + N hn using Then, . ≥ I aifisthe satisfies R sa ideal an is n that and 2 sa ideal an is N n R = Using . .By 0. R 2 135 is Downloaded By: [Behn, Antonio] At: 19:03 4 March 2008 3 ENE AL. ET BEHN 136 92,te o every for then 1982), obtain that Proof hspoe u li.I atclrw banthat obtain we particular In claim. our proves This em 9. Lemma Proof 8. Theorem lemma. the of proof the completes This that k and rm() o eassume we Now (2). from R 2 n ≥ n + + 2 .For 2. 1 R R RR R o at4 eassume we 4, part For o at3 easm that assume we 3, part For opoetenloec of nilpotency the prove To o at2 easm that step. inductive assume the we proves 2, this so part direct, For is inclusion reverse The sn o at – n hnteidciehptei eget we hypothesis inductive the then and 1–3 parts now Using . 2 x ewl rtso htfrevery for that show first will We . hsflosdrcl rmLma7pr 4. part 7 Lemma from directly follows This . 2 n = − n sascaie eget we associative, is 1 = R = k R 2 0 n RR If = 2 + 2 . = = n n n Let 1 + + + ,w have we 2, R 2 1 n R RR RR 2 sn h nutv yohss h ylciett n recalling and identity cyclic the hypothesis, inductive the Using . = ⊆ = sn at1 h ylciett n h nutv yohsswe hypothesis inductive the and identity cyclic the 1, part Using . R R satisfies R n n RR R = R RR eacci ig If ring. cyclic a be x x 2 + + L n R R in + 2 R R y n 1 + = n n R R R n 1 L + − L , x R + x k + + 1 L x k 2 + R x n n + n x R 1 R = R n k and , R n = − = = 2 x ≥ R n n 2 k 1 L R R n + = − n R R 0 L y x and 2 1 − 2 1 ⊆ ≥ = R n R R k x k 1 = n t rtlnaiain(e hvao tal., et Zhevlakov (see linearization first its and R R L sn 2 gi eget we again (2) Using . x ≥ = n that and 2 R R R R efis hwthat show first we , n n x x R − − = r iptn prtr.Mr precisely More operators. nilpotent are n that and 2 = R x n k k + L 2 R R R n 1 srgtnloet then right-nilpotent, is R R RR y + k k 2 L k n 2 x + + 2 x ≥ k ⊆ R n = R R L + 2, R k k x n = L R R − 2 = R n n 1 L y n n L R − − 2 ewn oso that show to want We . R x k RR n R k k = x + L = L x 2 = 1 x n ⊆ R n k R x + n R − = R 1 R = 1 ⊆ x RR n = k R 2 2 L R R R For . x n n y x k = R n ewn oso that show to want We . n n + L + = x R = y 1 2 R R R ewn oshow to want We . = n + k n 0. 2 snilpotent. is + R = L 1 n R + + y R 2 R 1 hsi direct is this 2 n RR R x k n o every for R L 2 n x n + = 2 ⊆ 0 Downloaded By: [Behn, Antonio] At: 19:03 4 March 2008 rdc novd ecnie rttecs where case the first consider We involved. product n ewl rv hsb nuto on induction by this prove will we and assume we Now utpyn hsls dniyby identity last this Multiplying Proof o,when Now, hsfiihstepofo h lemma. the of proof the finishes This ti la that clear is it hspoe u li.Nww e httefis ierzto of linearization first the that see we Now claim. our proves This rte sasingle a as written 10. Lemma lemma. the of proof the completes This n a eal owietesm lmn nmr hnoewy.Ltu carefully us Let way). one than more in element expressions, same these the of write two to multiply able be may one rte ntrso prtr as operators of terms in written ecamta ntermiigcss(when cases remaining the in that claim We ubro ie by times of number o ml xoet hsi la since clear is this exponents small For . m Let = v OASCAIERNSSATISFYING RINGS NONASSOCIATIVE R = uv R ,the 0, x x k R x + x k x ntergt nohrwords, other In right. the on uv L = = 1 eacci ih-irn n let and right-nilring cyclic a be apdsm ubro ie by times of number some tapped L n that and y y xL x xL 0 = = = n = = = L x xL xL = j R y R R R j R + x k and 0 k 2 2 x x R j k x n n R xL R + − − j x k + x n n k 1 1 L − Then . + k x xL + + + y 1 R m uv n + = u x n n + n n k k uv R = = − − = m ntelf iew get: we side left the on = n k = R 1 1 1 1 m = − L R = xL x k − 1 1 n = xL xL L 1 − x x n L x j L xL 1 y k j xL x j L + j + x R R + k k x j k k + x R k j = = R + k + = , R n n v n x n + xL k k v + − n x + R + + x xL x xL x m a erea es ) then 2), least at degree has L − 1 m 1 n = + n x − 1 n n + y = + − − 1 R k R m 1 1 xL + k x − + = m x xL m k 1 n j x k m ∈ r l h oesof powers the all are X YZ n R − 0 = L 0 R 1 ntelf n hnsm other some then and left the on = k R R xL x ro h oa ereo the of degree total the on or m xL R 0 hnaypwrof power any Then . ntecs that case the In . = 0 , 2 x x n = − L 2 j n 1 Y ZX R y R = R k m x xL − k 1 + 1 = x xR n x = ntc that (notice emyget. may we m a be can 0 x + a be can n = 137 0 Downloaded By: [Behn, Antonio] At: 19:03 4 March 2008 3 ENE AL. ET BEHN 138 either iagba.W ilpoeta fdim if that prove will We nilalgebras. ALGEBRAS CYCLIC IN NILPOTENCY 5. rirr lmn in element arbitrary so 11. Corollary but ylcsnei aifistesrne identity stronger the satisfies it since cyclic ae:()when (1) cases: ozr rdcso ai elements basis of products nonzero ozr rdcso ai elements basis of products nonzero o iptn.I at oiethat notice fact, In nilpotent. not h diiegopof group additive the a o eipoe nayovoswya ssonb h olwn examples. following the by shown is 1. as Example way obvious any in improved be not can rte salna obnto felements of combination linear a as written ih-iagbao ieso n ih-iidx5 u ti o iptn since nilpotent not is it but 5, right-nilindex and x 5 dimension of right-nilalgebra eeae by generated Proof oneeapei bs osbe ntesneta,eeycmuaiepower commutative every that that, proved been sense also Suttle’s has the that It 1975). proved in been possible” nilalgebra has associative it “best conjecture, is Albert’s nilalgebras counterexample Regarding power-associative 1948). commutative of Albert, nilpotency (see the on Albert of conjecture ba / to xml 2. Example shows. example following rv hti h eann fiiedmninl ae,nllersaesolvable. is are right-nilindex nilalgebras the cases, dimensional) (finite remaining the in that prove n A x ,i smrhct nw oneeapegvni ute 17)t a to (1972) Suttles in given counterexample known a to isomorphic is 2, n h iidxis nilindex the and nti eto eoti oerslso h iptnyo ylcright- cyclic of nilpotency the on results some obtain we section this In nitrsigrmr nEape1i httealgebra the that is 1 Example on remark interesting An A A norcs,cci ih-iagba r iptn hntedmninis dimension the when nilpotent are right-nilalgebras cyclic case, our In sntrgtnloet tagtowr aclto ftefut oe fan of power fourth the of calculation Straightforward right-nilpotent. not is ewl assume will We . 1 j x 2 ≥ 1 sacci ih-iagbao ieso n ih-iidx4 hc is which 4, right-nilindex and 5 dimension of right-nilalgebra cyclic a is aifisteidentity the satisfies x x x 1 n x a x 2 1 x 1 or x 1 x x 1 Let osdrtealgebra the Consider 1 = snloet nohrwords, other In nilpotent. is 1 k = n Let a = x ≥ ≤ x 3 A 2 0. R 2 n,()we h ih-iidxof right-nilindex the when (2) and, 4 n A n ea ler ihbasis with algebra an be eacci ih-irn n let and right-nilring cyclic a be − A + x hw that shows = x 4 A ≥ .B em eget we 9 Lemma By 1. 2 x ,btntncsaiywe h ih-iidxis right-nilindex the when necessarily not but 1, fdimension of .I hscase this In 0. x n 2 x 1 ihtenwmultiplication new the with = seCre n uz,19) ti noe rbe to problem open an is It 1999). Suazo, and Correa (see n = x = x yz x 3 3 .B em 0 n element any 10, Lemma By 0. A A A 2 x A = x a ih-iidx4. right-nilindex has 5 = snloetwe h ieso sfiieequal finite is dimension the when nilpotent is 3 x x ,teeoei scci,moreover, cyclic, is it therefore 0, ihbasis with 1 A ≤ 1 x =− snloet(esehbradMyung, and (Gerstenhaber nilpotent is 4 A = = 3 R x yz x + x n snil. is x 4 sntpwrascaiesince power-associative not is 4 then , x 3 u u = 5 x = = x 1 x a .W a losethat see also can We 0. x x and 3 xL a n therefore and 0 3 x A x 2 = 1 j ∈ x 1 R x snloeti h following the in nilpotent is AA = A R k x 3 2 x ∗ 2 hntesubring the Then . x x where is 2 4 4 ie by given = 3 n x x x + othat so 0 5 4 4 x a .Teeconditions These 1. n h following the and 3 j n h following the and x =− u + 2 A ∈ = x k + a x endon defined , ≥ x x 3 ∗ snilpotent. is 5 A 3 b 3 A n n sclearly is n = 2 − x A a be can A x sthe as 1 4 n ab ⊆ ,so 1, = = sa is and A x + R 4 2 Downloaded By: [Behn, Antonio] At: 19:03 4 March 2008 oieso of codimension em 13. Lemma conclude can we part first Ib a R dim em 12. Lemma P Proof n 15. Theorem nilpotent. fwrsof words of olw that follows xml 3. Example nilpotent. not hoe 14. Theorem ,te yLma13, Lemma by then 3, nec od esyta word a that say We word. each in in Proof a Ix x Proof particular, Proof em 2for 12 Lemma hrfr odhslength has word a Therefore x x x rprycnandin contained properly that of ideal ,weget x + h e ftewrsi h letters the in words the of set the I snloet Hence, nilpotent. is . 1 I 2 I Then . x fdim If nodrt banornx eut ene h olwn w ehia lemmas: technical two following the need we result, next our obtain to order In o epeetacci ih-iagbawihi nnt-iesoa n is and infinite-dimensional is which right-nilalgebra cyclic a present we Now suenwdim now Assume + = ≤ Let . . Since . Assume . = b o every for 0 2 A A Ia ups rtthat first Suppose . Ia then = A = + A P I A n = snilpotent. is I a I x Ib = snilpotent. is Let Let Let n ihcefiin in coefficient with and A + IA A snilpotent. is I Ib a I OASCAIERNSSATISFYING RINGS NONASSOCIATIVE ic o every for Since . Let Let ea lmn in element an be ,then 1, r ieryidpnets that so independent linearly are + I s1and 1 is xF = sslal,w nwthat know we solvable, is N A A = b x A A A Similarly . = eacci ih-iagbaadlet and right-nilalgebra cyclic a be eacci ih-iagbaadlet and right-nilalgebra cyclic a be o some for .Fo em eko htfrevery for that know we 9 Lemma From 0. 2 = in = A eacci ih-iagbawt dim with right-nilalgebra cyclic a be I/I eacci ih-iagbao ieso n right-nilindex and n dimension of right-nilalgebra cyclic a be I x Ix 0 Ix A I A = eget we . hc sacnrdcin hrfr dim Therefore contradiction. a is which , 1 2 hrfr,i emultiply we if Therefore, . I 2 x spoel otie in contained properly is = IA A A 2 and .Since 2. 2 spoel otie in contained properly is ≥ o every for 0 x sargtielof ideal right a is IA .Let 1. IA x Ib 3 A x = = ∈ in = 5 spoel otie in contained properly is IA A I .Ti rvstelemma. the proves This 0. A = hnteems xs elements exist must there Then . A Ia b K u Since . K x , ⊆ onal nnt e fidtriae and indeterminates of set infinite countably a edfieannomttv multiplication noncommutative a define We . uhthat such A a length has i R eafil and field a be x I uhta ahlte cusa otonce most at occurs letter each that such x eas have also we , hrfr 0 Therefore . 2 ∈ snloetaddim and nilpotent is A A A A A IA 2 n therefore and = sapoe da of ideal proper a is A x = I k I A fdmnina ot2 Using 2. most at dimension of 2 . n I X YZ I fi sfre by formed is it if .Term8sosthat shows 8 Theorem 0. + = then , A = = Ix = h e ffiiefra sums formal finite of set the IA A Ia .I ses osethat see to easy is It 0. x I I eget we , = I a + A rmCrlay1 it 11 Corollary From . eargtielof right-ideal a be I ea da of ideal an be IA Y ZX . IA ≤ Ib + ⊆ spoel contained properly is I 4 x ntergtsd by side right the on = = Then . b a IA in IA 0. IA/I ,i olw that follows it 2, so A A ≤ h operator the , b a + IA fdim If . 2 k ,uigthe using 1, A b = saright- a is letters 5 ∈ A = Ix/I fthe If . = A Ia A 0 such A A 2 .In .If 139 x + = is is i . Downloaded By: [Behn, Antonio] At: 19:03 4 March 2008 4 ENE AL. ET BEHN 140 ewl hwb nuto on induction by show will We hne sn h ylciett,w get we identity, cyclic the using whence, 2. ora .(06.Feil ih-iagba satisfying right-nilalgebras Flexible (2006). I. Correa, 1. in let .A 14) oe-soitv rings. Power-associative (1948). A. A. Albert, iptn n hrfr iptn yTerm8 rmCrlay2, Corollary From 8. Theorem by nilpotent therefore and nilpotent Proof than greater or 0 characteristic of field a ow a s h aaaHga hoe oso hti snilpotent. is it that show to Theorem Nagata–Higman the use can we so upre yFNEY 002.IvnRyHnzli upre by supported is Hentzel REFERENCES Roy Irvin 1060229. 10-05. FIBAS and FONDECYT 7060096 FONDECYT by supported true. be to expect could we results what ACKNOWLEDGMENT of idea better formal a a get attempting to before general conjectures check in to and 1993) proof al., et (Jacobs Albert program (3), every hoe 16. index. the Theorem bounded replacing of by nil result with this dimension improves finite theorem of following hypothesis The nilpotent. is nilalgebra x y x rvsteassertion. the proves o easm that assume we Now A ux uv A by: noi eni upre yFNEY 004.Iá orais Correa Iván 1070243. FONDECYT by supported is Behn Antonio computer the of use the mention to like would we comment, final a As A ercl hta ler scalled is algebra an that recall We o eol edt hwthat show to need only we Now n t Applications Its and rmteflxbeiett n h iert fteascao ehave we associator the of linearity the and identity flexible the From . = n stewr bandadn h letter the adding obtained word the is 2 , n sacci ih-iagbao ih-iidx2wihi o iptn ic for since nilpotent not is which 2 right-nilindex of right-nilalgebra cyclic a is + = 0if 2 ··· .Cre 20)poe htacci eil nt-iesoa right- finite-dimensional flexible cyclic a that proves (2006) Correa 0. 0 = v x = = A a length has z x y z z Let 1 2 x + n 2 A A x A y z y x 3 eacci n eil ih-iagbao right-nilindex of right-nilalgebra flexible and cyclic a be x etr oe nPr n ple ahmtc.246:103–106. Mathematics. Applied and Pure in Notes Lecture . + A ⊆ 4 > x z y x 2 + A ··· zy x n 1, x 2 ⊆ n u x A = = A n n = 2 = z z y z v 2 + that z yx or 0. n A 2 n ewl prove will we and v A A A n − salte hti ntecmoiinof composition the in is that letter a is 2 2 Then . + 2 rn.Ae.Mt.Soc. Math. Amer. Trans. n flexible n xy z snloett ocuethat conclude to nilpotent is z x y z hc,uigteidciehypothesis, inductive the using which, x ⊆ tteedo h word the of end the at A A + x yz 2 fi aifisteflxbeidentity flexible the satisfies it if snilpotent. is z xy n + For . = x z y x y zx n A = . 2 n o-soitv Algebra Non-Associative + + ti rval true. trivially is it 1 64:552–593. 2 x x y x ⊆ A A 2 u sassociative is 2 . A n + 1 sright- is From . n over u (3) ; Downloaded By: [Behn, Antonio] At: 19:03 4 March 2008 hvao,K . lnk,A . hsao,I . hrhv .I (1982). I. A. Shirshov, P., I. Shestakov, M., A. Slin’ko, A., K. Zhevlakov, ora . uz,A 19) nacaso omttv oe-soitv nilalgebras. power-associative commutative of class a On (1999). A. Suazo, I., Correa, ute,D 17) oneeapet ojcueo Albert. of conjecture a to counterexample A (1972). D. Suttles, aos .P,Le . udn,S . fu,A . rbu . htly .(1993). T. Whiteley, K., Prabhu, J., A. Offut, V., S. of Muddana, nilalgebras D., power-associative Lee, P., commutative D. On Jacobs, (1975). C. H. Myung, theorem. M., Nagata–Higman Gerstenhaber, The (1990). E. Formanek, cae,R .(1966). D. R. with Schafer, Rings (1995). M. Kleinfeld, eryAssociative Nearly 72T-A169. Abstract 19:A–556. .Algebra J. cdmcPress. Academic o dimension. low srGuide User S.Aalbea tp/wwc.lmo.d/dpj/albertstuff/albert.html. http://www.cs.clemson.edu/ at Available USA. 215:412–417. OASCAIERNSSATISFYING RINGS NONASSOCIATIVE eateto optr cec,CesnUiest,Cesn SC, Clemson, University, Clemson Science, Computers of Department . rc mr ah Soc. Math. Amer. Proc. e okLno:Aaei Press. Academic York/London: New . nItouto oNnsoitv Algebras Nonassociative to Introduction An x yz = y zx 48:29–32. . om Algebra Comm. caAp.Math. Appl. 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