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When Is a Graded PI Algebra a PI Algebra?K. I. Beidar a c & M. A. Chebotar ba Department of Mathematics, National Cheng-Kung University, Tainan, Taiwanb Department of Mechanics and Mathematics, Tula State University, Tula, Russiac Department of Mathematics, National Cheng-Kung University, Tainan, TaiwanPublished online: 20 Oct 2011.

To cite this article: K. I. Beidar & M. A. Chebotar (2003): When Is a Graded PI Algebra a PI Algebra?, Communications inAlgebra, 31:6, 2951-2964

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When Is a Graded PI Algebra a PI Algebra?

K. I. Beidar1,* and M. A. Chebotar2

1Department of Mathematics, National Cheng-Kung University,Tainan, Taiwan

2Department of Mechanics and Mathematics, Tula State University,Tula, Russia

ABSTRACT

Let G be a monoid with cancellation and let R be a strongly G-graded algebra with finite G-grading satisfying a graded polynomialidentity. We show that R is a PI algebra.

Key Words: Graded algebras; PI-algebras; Graded PI-algebras.

1. INTRODUCTION

We refer the reader to Beidar et al. (1996), Rowen (1980) for the basicconcepts and results on rings with (generalized) polynomial identities. Inwhat follows G is a monoid with unity e and cancellation, and F is a

*Correspondence: K. I. Beidar, Department of Mathematics, National Cheng-Kung University, Tainan, Taiwan; E-mail: beidar@mail.ncku.edu.tw.

COMMUNICATIONS IN ALGEBRA�

Vol. 31, No. 6, pp. 2951–2964, 2003

2951

DOI: 10.1081/AGB-120021901 0092-7872 (Print); 1532-4125 (Online)

Copyright # 2003 by Marcel Dekker, Inc. www.dekker.com

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commutative ring with 1. Let R be an F-algebra. Recall that the algebraR is called almost G-graded if there are F-submodules Rg�R, g2G,such that R¼P

g2GRg and RgRh�Rgh for all g, h2G. IfP

g2GRg isdirect (i.e.,

Pg2GRg¼

Lg2GRg), then the algebra R is called G-graded.

Further, set supp(R)¼fg2G jRg 6¼ 0g. We say that the G-grading isfinite, if jsupp(R)j<1. A G-graded algebra R is said to be strongly G-graded if R has an identity 12Re, supp(R) consists of invertible elementsand RgRg�1¼Re for all g2 supp(R). Throughout the rest of the paper Ris an almost G-graded F-algebra with finite G-grading.

Let X¼Sg2GXg be the union of disjoint infinite sets Xg, g2G,

and let A¼FhX i be the free algebra on X. Given x2X, we set d(x)¼ gprovided that x2Xg, g2G. Clearly d :X!G is a well-defined map.Next, given a monomial M¼ x1x2 . . . xn2A and an element g2G, we setd(M)¼ d(x1)d(x2) . . . d(xn) andAg¼

Pd(M)¼gFM. ClearlyA¼L

g2GAg

is a G-graded F-algebra. Let R be an almost G-graded F-algebra. Anelement f (x1, x2, . . . , xn)2A is said to be a G-graded polynomial identityon R provided that f( f )¼ 0 for all algebra homomorphisms f :A!Rsuch that f(Xg)�Rg for all g2G. The algebra R is said to be an almostG-graded PI algebra if it satisfies a G-graded polynomial identity at leastone of whose coefficients is equal to 1. If R is both a G-graded algebraand an almost G-graded PI algebra, we shall simply say that R is aG-graded PI algebra. It is easy to see that if R is an almost G-graded PIalgebra, then it satisfies a multilinear G-graded polynomial identityf (x1, x2, . . . , xn) at least one of whose coefficients is equal to 1. Let R bean almost G-graded PI algebra with multilinear G-graded polynomialidentity f (x1, x2, . . . , xn). We may assume without loss of generality thatthe monomial x1x2 . . . xn is involved in f with coefficient 1. We set

Gf ¼ fdðx1Þ; dðx2Þ; . . . ; dðxnÞg�G:

If in additionR is aG-graded algebra, then we may assume without loss ofgenerality that there exists g2G such that d(N)¼ g for all monomials Ninvolved in f.

Bergen and Cohen (1986) proved thatR is PI provided that jGj < 1,F is a field and Re is a PI algebra. This result was extended to algebrasover arbitrary commutative rings by Kelarev (1993). Bahturin and Zaicev(1998) obtained an analogous result for algebras over a field with finite G-grading, where G is a monoid with cancellation. Recently, Sehgal andZaicev proved that if H is a normal subgroup of a group G with finiteindex and the group algebra F[G], considered as G=H-graded algebra,is G=H-graded PI, then F[G] is a PI algebra (Sehgal and Zaicev, 2002).Note that in this case F[G] is a strongly G=H-graded algebra.

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The main result of the present paper is the following generalization ofthese results.

Theorem 1.1. Let G be a monoid with unity e and cancellation, let F be acommutative ring with 1 and let R be an almost G-graded F-algebra withfinite G-grading satisfying a G-graded multilinear polynomial identityf (x1, x2, . . . , xn). Suppose that the monomial x1x2 . . . xn is involved in f withcoefficient 1, d(N)¼ d(x1x2 . . . xn) for all monomials N involved in f andGf� supp(R). Then:

(i) If R is a prime algebra and jsupp(R)j ¼ 2, then the algebra Re

contains a nonzero ideal satisfying the standard identity St2n�2

of degree 2n� 2 and the algebra R satisfies a nontrivial general-ized polynomial identity. If in addition R is a simple algebra with1, then R is a PI algebra.

(ii) If both R and Re are prime algebras, then Re satisfies St2n�2 andR is a PI algebra.

(iii) If R has an identity 12Re, and Gf consists of invertible elementsand RgRg�1¼Re for all g2Gf, then R is a PI algebra.

Corollary 1.2. Let G be a monoid with cancellation and let R be a G-graded PI algebra with identity. Suppose that R is strongly G-graded withfinite G-grading. Then R is a PI algebra.

The following examples illustrate the necessity of conditions inTheorem 1.1.

Example 1. Let D be a skew field which is not a PI ring. Let F¼Z(D)be the center of D, let R¼M2(D) be theF-algebra of 2� 2 matrices overD and let feij j 1� i, j� 2g be a system of matrix units of R. Further, letG¼hai be a cyclic group of order 3. Set u¼ e11 and v¼ e22. Next, set

Re ¼ uRuþ vRv; Ra ¼ uRv and Ra2 ¼ vRu

and note that R is a G-graded algebra satisfying a G-graded polynomialidentity f (x, y)¼ xy, x, y2Ra. Further, Re is a direct sum of two skewfields, Gf� supp(R), R is a simple algebra with 1 and R is not a PI alge-bra. Therefore the first statement of the theorem does not hold in generalif jsupp(R)j ¼ 3. Next, the second statement does not hold in general ifRe

is not prime even if R is a simple Artinian algebra and Re is a direct sumof two skew fields.

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Example 2. Let F be a field, let R be the F-algebra of infinite matriceswith finite number of nonzero entries and let u be the matrix whose (1, 1)entry is equal to 1 and all the other ones are equal to 0. Further, letG¼fe, gg be a cyclic group of order 2. Set Re¼ uRuþ (1� u)R(1� u)and Rg¼ uR(1� u)þ (1� u)Ru. Clearly R is a G-graded algebra satisfy-ing a G-graded polynomial identity

f ðx1; . . . ; x5Þ ¼ ½x1; x2�x3½x4; x5�; x1; x2; x4; x5 2 Re; x3 2 Rg:

Moreover, R2g ¼Re, R is a simple algebra and is not a PI algebra. We see

that the first statement of the theorem does not hold in general if R con-tains no an identity. Moreover, a G-graded PI algebra need not to be PIalgebra even if it is a simple algebra satisfying a nonzero generalized poly-nomial identity and jsupp(R)j ¼ 2.

Example 3. Let F be a field, let n be a positive integer, let R¼Mn(F),let u¼ e11 and v¼ 1� u. Let G¼fe, gg be a cyclic group of order 2. Asabove set Re¼ uRuþ (1� u)R(1� u) and Rg¼ uR(1� u)þ (1� u)Ru.We already know that R is a strongly G-graded algebra and

f ðx1; . . . ; x5Þ ¼ ½x1; x2�x3½x4; x5�; x1; x2; x4; x5 2 Re; x3 2 Rg

is a G-graded polynomial identity on R. On the other hand the minimaldegree of a polynomial identity on R is 2n (Rowen, 1980, Lemma 1.4.3).Therefore there exists no function m¼m(deg( f )) such that a simple alge-bra with 1 satisfying the G-graded polynomial identity f satisfies a poly-nomial identity of degree m.

2. PROOF OF MAIN THEOREM

We first set some further notation in place. Let A be an algebra.Given right A-modules U and V and a module map h :U!V, wedenote by hx the image of x2U under h. If ; 6¼S�A and W�U aresubsets, we set

‘ðW;SÞ ¼ fx 2 W j xS ¼ 0g:

Let n be a positive integer and let M1,M2, . . . ,Mn, M be right A-modules. Given n-tuple (m1, m2, . . . ,mn)2

Qnj¼1Mj, we shall write

mn ¼ ðm1;m2; . . . ;mnÞ. Let 1� i� n and let G :Q

j6¼iMj!HomA(Mi, M)

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be a map. We define a map Gi :Qn

j¼1Mj!HomA(Mi,M) by the rule

GiðmnÞ ¼ Gðm1; . . . ;mi�1;miþ1; . . . ;mnÞ for all mn 2Ynj¼1

Mj:

Let t2A, let k, t be nonnegative integers with t� n and let

Giib :

Yn�t

j¼1

Mj ! HomAðMi;MÞ; 1 � i � n� t; 0 � b � tþ k;

be maps such that

Xn�t

i¼1

Xtþk

b¼0

Giibðmn�tÞmit

b ¼ 0 for all mn�t 2Yn�t

j¼1

Mj: ð1Þ

We are now in a position to prove the following generalization of Beidar(1998, Theorem 2.5).

Lemma 2.1. Suppose that the following conditions are satisfied:

(i) For any 0� g� nþ k� 1 there exist a positive integer k¼ k(g) andelements aga, bga2A, a¼ 1, 2, . . . , k, such that

dg ¼Xka¼1

agatgbga 6¼ 0 and M

Xka¼1

agatsbga ¼ 0

for all s¼ 0, 1, . . . , nþ k� 1, s 6¼ g.(ii) ‘(M;Adg)¼ 0 for all g¼ 0, 1, . . . , nþ k� 1.

Then each Gib¼ 0.

Proof. We proceed by induction on n� t. First assume that n� t¼ 1,that is t¼ n� 1. Then (1) reads

Xnþk�1

b¼0

G1bm1tb ¼ 0 for all m1 2 M1

where each G1b2HomR(M1,M). We have to show that each G1b¼ 0.Pick any 0� g� nþ k� 1 and let aga, bga, dg2A be as in (i). Substituting

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m1xaga for m1, multiplying the resulting expression by bga from the rightand summing up, we obtain that

G1gm1xdg ¼ 0 for all m1 2 M1 and x 2 A.

Therefore (ii) yields that G1gm1¼ 0 for all m12M1 and so G1g¼ 0 for allg¼ 0, 1, . . . , nþ k� 1.

In the inductive case n� t > 1 we substitute mn�tt for mn�t in (1) andobtain

Xn�t�1

i¼1

Xtþk

b¼0

Giibðmn�t�1;mn�ttÞmit

b þXtþk

b¼0

Gn�t;bðmn�t�1Þmn�ttbþ1 ¼ 0

ð2Þ

for all mn�t 2Qn�t

j¼1 Mj. Fix mn�t2Mn�t. Multiplying (1) by t from theright and subtracting the resulting expression from (2), we see that

Xn�t�1

i¼1

Xtþ1þk

b¼0

Hiibðmn�t�1Þmit

b ¼ 0 for all mn�t 2Yn�t�1

j¼1

Mj; ð3Þ

where

Hii0ðmn�t�1Þ ¼ Gi

i0ðmn�t�1;mn�ttÞ; 1 � i � n� t� 1;

Hiibðmn�t�1Þ ¼ Gi

ibðmn�t�1;mn�ttÞ � Gii;b�1ðmn�tÞ;

1 � i � n� t� 1; 1 � b � tþ k;

ð4Þ

Hii;tþ1þkðmn�t�1Þ ¼ �Gi

i;tþkðmn�tÞ; 1 � i � n� t� 1: ð5Þ

It follows from both (3) and the induction assumption that each Hib¼ 0.Therefore (5) implies that Gi,tþk¼ 0, 1� i� n� t� 1. It follows from (4)that Gib¼ 0 for all 1� i� n� t� 1 and 0� b� tþ k. Now (1) reads

Xtþk

b¼0

Gn�t;bðmn�t�1Þmn�ttb ¼ 0 for all mn�t 2

Yn�t

j¼1

Mj:

Fixingmn�t�1 2Qn�t�1

j¼1 Mj , we conclude from the case n� t¼ 1 that eachGn�t, b¼ 0. Therefore each Gib¼ 0 and the proof is now complete.

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Corollary 2.2. Let Fi :Q

j 6¼iMj!HomA(Mi,M), i¼ 1, 2, . . . , n, be mapssuch that

Xni¼1

Fii ðmnÞmi ¼ 0 for all mn 2

Ynj¼1

Mj :

Suppose that there exists an element t2A such that the following condi-tions are satisfied:

(i) There exist a positive integer k and elements aga, bga2A,0� g� n� 1, 1� a� k, such that

dg ¼Xka¼1

agatgbga 6¼ 0 and M

Xka¼1

agatsbga ¼ 0

for all g, s¼ 0, 1, . . . , n� 1, s 6¼ g.(ii) ‘(M;Adg)¼ 0 for all g¼ 0, 1, . . . , n� 1.

Then each Fi¼ 0.

The proof of Proposition 2.3 is inspired by that of Cohen and Rowen(1983, Proposition 1.2). We note that Cohen and Rowen (1983, Lemma1.1) is proved for groups, but the same proof works for monoids withcancellation.

Proposition 2.3. Let G be a monoid with cancellation and let R be analmost G-graded algebra with finite G-grading. Let n¼ jsupp(R)j, let dbe a positive integer, let L¼P

g2GLg be a G-graded subalgebra of R(i.e., Lg�Rg is an additive subgroup and LgLh�Lgh for all g, h2G)and let K be a right ideal of Re. Further, let

H ¼ fg 2 suppðRÞ j g is not invertible in G or RgRg�1 ¼ 0g

and let I be the ideal of R generated byP

h2HRh. Then:

(i) If Lde ¼ 0, then Lnd¼ 0.

(ii) If Kd¼ 0, then (KR)nd¼ 0.(iii) I is a nilpotent ideal of R.

Proof. (i) Assume that Lde ¼ 0. Clearly Lnd is the sum of elements of

the form a1 . . . and where each ai2Lgiand gi2 supp(R). Therefore it is

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enough to show that each such product is equal to 0. Set hi¼ g1g2 . . . gi,i¼ 1, 2, . . . , nd. If some hi 62 supp(R), then a1a2 . . . ai2Rhi

¼ 0 and weare done. Therefore we may assume without loss of generality that eachhi2 supp(R). Since jsupp(R)j ¼ n, we conclude from Cohen and Rowen(1983, Lemma 1.1) that there exist integers 1� j0 < j1 < . . . < jd� ndsuch that

gj0þ1gj0þ2 . . . gj1 ¼ gj1þ1gj1þ2 . . . gj2 ¼ � � � ¼ gjd�1þ1gjd�1þ2 . . . gjd ¼ e

and so

ðaj0þ1 . . . aj1Þðaj1þ1 . . . aj2Þ . . . ðajd�1þ1 . . . ajd Þ 2 Lde ¼ 0

forcing a1a2� � �and¼ 0. Therefore Lnd¼ 0.

(ii) Suppose thatKd¼ 0. SetLg¼KRg, g2G, andL¼Pg2GLg.

Since Le¼KRe�K, we conclude that Lde ¼ 0 and so Lnd¼ 0 by (i).

(iii) Let H and I be as in the proposition. Since G is a monoid withcancellation, e is the only idempotent of G. Therefore every one-sidedinvertible element in G is invertible. Given g2G, we set Ig¼Rg ifg2H; otherwise

Ig ¼X

h2H; p2G;ph¼g

RpRh þX

h2H;q2G;hq¼g

RhRq þX

h2H;p;q2G;phq¼g

RpRhRq:

Clearly I¼Pg2GIg. In view of (i) it is enough to show that Ie is a nil-

potent ideal of the algebra Re. If e2H, then R2e ¼ 0 and we are done.

Assume that e 62H. Clearly

Ie ¼X

h2H;p2G;ph¼e

RpRh þX

h2H;q2G;hq¼e

RhRq þX

h2H;p;q2G;phq¼e

RpRhRq:

Fix h2H. If f2G and either fh¼ e or hf¼ e, then f¼ h�1. ThereforeRhRh�1¼ 0 and so (Rh�1Rh)

2¼Rh�1(RhRh�1)Rh¼ 0. We see that Rh�1Rh

is a nilpotent ideal of the algebra Re. Let p, q2G with phq¼ e. Thenhqp¼ e and so qp¼ h�1 forcing RqRp�Rh�1. Therefore

ðRpRhRqÞ2 ¼ RpRhRqRpRhRq � RpRhRh�1RhRq ¼ 0

and hence Ie, being the sum of a finite number of nilpotent ideals, is anilpotent ideal of Re. The proof is thereby complete.

Proposition 2.4. Let G be a monoid with cancellation, R be an almost G-graded algebra and n¼ jsupp(R)j < 1. Suppose that R is a prime algebra.

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Then Re is a semiprime algebra containing nonzero ideals I1, . . . ,Im

such that:

(a) Ik\I‘¼ 0 for all k 6¼ ‘;(b) I¼L

mj¼1Ij is an essential ideal of Re;

(c) each Ij is a prime algebra;(d) m� n;(e) if 0 6¼ dj2Ij and d¼Pm

j¼1dj, then ‘(Rg, Red)¼ 0 for all g2G.

Proof. It follows at once from Proposition 2.3(i) that Re is a semi-prime algebra and e2 supp(R). We now prove (a), (b) and (c) together.If Re is prime, then there is nothing to prove. Assume that Re is notprime. Then it contains a nonzero ideal I1 such that J1¼ ‘(Re;I1) 6¼ 0.It is easy to see that I1\J1¼ 0 and I1�J1 is an essential ideal of Re.Clearly both I1 and J1 are semiprime algebras. If both algebras I1

and J1 contain ideals satisfying (a), (b) and (c), then it follows easilyfrom Andrunakievich’s lemma that the same is true for Re. Thereforewe may assume without loss of generality that, say J1, does not containdesired ideals. From Andrunakievich’s lemma we conclude that J1

contains two nonzero ideals I2 and J2 such that I2\J2¼ 0 and thealgebra J2 contains no the desired ideals. Continuing in this fashion,we get nonzero ideals I1,I2, . . . ,Inþ1 of Re such that Ik\I‘¼ 0for all k 6¼ ‘. Clearly IkI‘¼ 0 for all k 6¼ ‘. Set Kj¼RIjR.Obviously Kj 6¼ 0 and so K¼K1K2 . . .Knþ1 6¼ 0. On the other handevery element of K is a sum of elements of the forma0b1a1b2 . . . anbnþ1anþ1 where each aj2Rgj

and each bj2Ij. Thereforea0b1a1b2 . . . anbnþ1anþ1 6¼ 0 for some aj2Rgj

and bj2Ij. Sethj¼ g1g2 . . . gj, j¼ 1, . . . , n. Since jsupp(R)j ¼ n, either hi¼ e (andso b1a1 . . . aibi2I1ReIi¼ 0) for some 1� i� n or hi¼ hj for some1� i < j� n (and so giþ1giþ2 . . . gj¼ e forcing biþ1aiþ1 . . . ajbjþ1¼ 0).In both cases a0b1a1b2 . . . anbnþ1anþ1¼ 0 forcing K¼ 0, a contradiction.Therefore (a), (b) and (c) are satisfied and moreover, m� n.

Finally, let the dj’s and d be as in (e). Suppose thatJ¼ ‘(Re;Red) 6¼ 0.Clearly J is an ideal of Re and JI 6¼ 0 by (b) because Re is semiprime.Therefore JIk 6¼ 0 for some 1� k�m. Take 0 6¼ x2JIk. As x2J,xRed¼ 0. Since xRedj2Ij, 1� j� n, and the sum

Pmj¼1Ij is direct by

(b), we have that xRedk¼ 0. In particular, xIkdk¼ 0 contradicting (c).Therefore ‘(Re;Red)¼ 0. Now let

L ¼Xg2G

rg j each rg 2 Rg and rgRed ¼ 0

( ):

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Clearly L is a left ideal of R, L¼Pg2GL\Rg and L\Re�

‘(Re; Red)¼ 0. Therefore Proposition 2.3(i) implies that Ln¼ 0. SinceR is prime, L¼ 0 and the proof is now complete.

The following result is a generalization of Cohen and Montgomery(1984, Theorem 7.3(1)) proved under the additional assumption that Gis a finite group.

Corollary 2.5. Let G be a monoid with cancellation, R be an almost G-graded algebra and n¼ jsupp(R)j < 1. Let P be a prime ideal of the alge-bra R. Then there exist prime ideals P1, P2, . . . , Pm, m� n, of Re such thatT

mi¼1Pi¼P\Re and each Pi is a minimal prime ideal with respect to the

property Pi�P\Re. The set fP1, P2, . . . , Pmg is uniquely determined byP.

Proof. Replacing R by R=P we may assume without loss of generalitythat P¼ 0. Let the Ijs and I be as in Lemma 2.4. It is now easy tosee that the only minimal prime ideals of Re are that of the formPj¼ ‘(Re;Ij). Since (

Tmj¼1Pj)I¼ 0, we conclude that

Tmj¼1Pj¼ 0.

We need the following result which is a special case of Bahturin(1998, Theorem 3).

Thoerem 2.6. Let G be a monoid with cancellation, let F be a field and letR be an almost G-graded algebra with finite G-grading. If Re is a PI alge-bra, then so is R.

Theorem 2.7 (Kepczyk and Puczyłowski, 2001, Theorem 5). Let H be aclass of rings which is closed under arbitrary direct powers and homo-morphic images. If every prime ring in H satisfies a generalized polynomialidentity, then H consists of PI rings.

Proof of Theorem 1.1. Let R be a prime F-algebra with G-graded poly-nomial identity f (x1, . . . , xn) satisfying the conditions of the theorem. Wefirst make some general observations.

Replacing F by F=‘(F;R) we may assume that F is an integraldomain. Setting S¼F n f0g and considering S�1F-algebra S�1R, wereduce the proof to the case when F is a field.

Assume that the algebra Re contains a nonzero ideal I which is a PIalgebra. We claim that

R satisfies a nonzero generalized polynomial identity. ð6ÞIndeed, setLp¼IRp, p2G, andL¼P

p2GLp. SinceLe is PI, Theorem2.6 implies that L is a PI algebra. It now follows from Jain (1971) that R

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satisfies a nonzero generalized polynomial identity (see also Beidar et al.,1996, Theorem 6.3.20).

It follows from Proposition 2.3(iii) that supp(R) consists of invertibleelements, e2 supp(R) and

RgRg�1 6¼ 0 for all g 2 suppðRÞ: ð7Þ

Next, Proposition 2.4 implies that Re is a semiprime algebra. Write

f ðx1; . . . ; xnÞ ¼Xni¼1

f ii ðxnÞxi

where each f ii ðxnÞ ¼ f ðx1; . . . ; xi�1; xiþ1; . . . ; xnÞ is a suitable multilinearpolynomial which does not depend on xi. Assume that Re has no nonzeroideals satisfying St2n�2. We claim that

f ii ðanÞai ¼ 0 for all i¼ 1;2; . . . ;n, and aj 2RdðxjÞ, j ¼ 1;2; . . . ;n: ð8Þ

Indeed, letI1,I2, . . . ,Im be as in Proposition 2.4. Then eachIj does notsatisfy St2n�2. Fix 1� j�m and recall thatIj is a prime algebra. Let Cj bethe Martindale (extended) centroid of Ij. If every element of Ij is alge-braic of degree �n� 1 over Cj, then Ij is a subalgebra of the algebra of(n� 1)� (n� 1) matrices over the algebraic closure of Cj (see Beidar andChebotar, 2000, p. 3928) and so Amitsur-Levitzki theorem (Rowen,1980) implies that Ij satisfies St2n�2, a contradiction. Therefore Ij con-tains an element tj which is not algebraic of degree n� 1 over Cj. That isto say 1, tj, t

2j , . . . , t

n�1j are linearly independent over Cj and so Beidar

et al. (1996, Theorem 2.3.3) yields that for any 0� g� n� 1 there exista positive integer k¼ k(j, g) and elements ajga, bjga2Ij, a¼ 1, 2, . . . , k,such that djg¼

Pka¼1ajgat

gj bjga 6¼ 0 and

Pka¼1ajgat

sj bjga¼ 0 for all s¼ 0,

1, . . . , n� 1, s 6¼ g. We may assume without loss of generality that k(j, g)does not depend on both j and g. Setting t¼Pm

j¼1tj, aga¼Pm

j¼1ajga andbga¼

Pmj¼1bjga, we see that

dg ¼Xka¼1

agatgbga ¼

Xmj¼1

djg andXka¼1

agatsbga ¼ 0 ð9Þ

for all g, s¼ 1, 2, . . . , n� 1, s 6¼ g. It follows from Proposition 2.4(e) that

‘ðRg;RedgÞ ¼ 0 for all g2G: ð10Þ

Consider the left multiplication by each f ii ðanÞ, an 2Rn, as an elementof HomRe

(Rd(xi), Rh) where h¼ d(x1x2 . . .xn). It now follows from

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Corollary 2.2 that each f ii ðxnÞxi is a G-graded polynomial identity on Rand our claim is proved.

(i) Assume that R is prime and jsupp(R)j ¼ 2. Then (7) implies thatsupp(R)¼fe, gg with g2¼ e. We claim that ‘(Re; Rg)¼ 0. Indeed, leta2 ‘(Re; Rg). Then

aRRg ¼ aðReRgÞ þ ðaRgÞRg � aRg ¼ 0

and so a¼ 0 because R is prime and Rg 6¼ 0.We now set h¼ d(xn)2Gf� supp(R)¼fe, gg. Set Ke¼ ‘(Re;Rh),

Kg¼fa2Rg j aRh¼ 0g and K¼KeþKg. It follows from the aboveresult together with semiprimeness of Re that Ke¼ ‘(Re;Rh)¼ 0. ClearlyK is an almost G-graded algebra and a left ideal of R and so Proposition2.3(i) implies that K2¼ 0. As R is prime, K¼ 0.

Assume that Re has no nonzero ideals satisfying St2n�2. Then (8)implies that fnðan�1Þ 2 K ¼ 0, and so we conclude that thatfn(x1, . . . , xn�1) is a G-graded polynomial identity on R. Making use ofinduction on n¼ deg( f ), we get that Re contains a nonzero ideal satisfy-ing St2n�4 and so St2n�2, a contradiction. Therefore Re contains a non-zero ideal I satisfying St2n�2. It now follows that (6) holds true.

Suppose that in addition R is a simple algebra with 1. Then the cen-tral closure of R is equal to R. It now follows from Martindale’s theoremon prime rings with generalized polynomial identity (Martindale, 1969)that R has a nonzero socle and the associated skew field is finite dimen-sional over its center (see also Beidar et al., 1996, Theorem 6.1.6). Since Ris simple, it coincides with its socle. In particular, 1 is an idempotent ofthe finite rank and so Litoff’s theorem (Beidar et al., 1996, Theorem4.3.11) yields that R is a matrix ring over a skew field finite dimensionalover its center. Therefore R is a PI algebra and the first statement of thetheorem is proved.

(ii) Now assume that R and Re both are prime algebras. If Re con-tains a nonzero ideal satisfying St2n�2, then Re satisfies St2n�2 and soTheorem 2.6 implies that R is PI. Assume that Re has no nonzero idealssatisfying St2n�2. Then (8) implies that fnðan�1Þan ¼ 0 for all aj2Rd(xj)

,j¼ 1,2, . . . , n. Setting I¼Rd(xn)

Rd(xn)�1, Kg¼fa2Rg j aI¼ 0g and

K¼Pg2GKg, we see that K is a left ideal of R. Since I is a nonzero

ideal of Re by (7), we conclude that Ke¼ 0. Therefore Proposition2.3(i) yields that K is a nilpotent ideal of R, forcing K¼ 0. Asfnðan�1Þ 2 K, we see that fnðxn�1Þ is a G-graded polynomial identity onR. Making use of induction on deg( f ), we complete the proof of thestatement.

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(iii) Suppose that 12Re, and that Gf consists of invertible elementsand RgRg�1¼Re for all g2Gf.

Let r be a positive integer and let Hr be the class of all almost G-graded algebras B with finite G-grading satisfying multilinear G-gradedpolynomial identity f and such that for any g2Gf there exist x1,x2, . . . , xr2Bg and y1, y2, . . . , yr2Bg�1 with

Pri¼1xiyi¼ 1. Clearly the

class Hr is homomorphically closed and is closed under arbitrary directpowers. Further, R2Hr for some integer r. In view of Theorem 2.7 it isenough to show that every prime algebra B in Hr satisfies a nonzerogeneralized polynomial identity.

If Be contains a nonzero ideal satisfying St2n�2, then (6) implies thatB satisfies a nonzero generalized polynomial identity. Therefore we mayassume without loss of generality that Be has no nonzero ideals satisfyingSt2n�2. Set g¼ d(xn). It follows from (8) that fn(a1, . . . , an�1)Bg¼ 0 forall ai2Bd(xi)

. Since 12BgBg�1, we conclude that fn(x1, . . . , xn�1) is aG-graded polynomial identity of B. Proceeding inductively on deg( f ),we get that B satisfies a nonzero generalized polynomial identity. Theproof is thereby complete.

ACKNOWLEDGMENTS

We express our gratitude to Prof. M.V. Zaicev for bringing to ourattention his joint paper Bahturin and Zaicev (1998) with Prof. Yu. A.Bahturin. We are also grateful to the referee for his=her useful comments.

REFERENCES

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Jain, S. K. (1971). Prime rings having one-sided ideal with polynomialidentity coincide with special Johnson rings. J. Algebra 19:125–130.

Kelarev, A. V. (1993). On semigroup graded PI algebras. SemigroupForum 47:294–298.

Kepczyk, M., Puczyłowski, E. (2001). Rings which are sums of two sub-rings satisfying polynomial identities. Comm. Algebra 29:2059–2065.

Martindale 3rd, W. S. (1969). Prime rings satisfying a generalized polyno-mial identity. J. Algebra 12:576–584.

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Received February 2002Revised April 2002

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