Canad. Math. Bull. Vol. 41 (1), 1998 pp. 81±85

THE CARDINALITY OF THE CENTER OF A PI RING

CHARLES LANSKI

ABSTRACT. The main result shows that if R is a semiprime ring satisfying a poly- nomial identity, and if Z(R) is the center of R, then card R Ä 2cardZ(R). Examples show that this bound can be achieved, and that the inequality fails to hold for rings which are not semiprime.

The purpose of this note is to compare the cardinality of a ring satisfying a polynomial identity (a PI ring) with the cardinality of its center. Before proceeding, we recall the def- inition of a central identity, a notion crucial for us, and a basic result about polynomial f g≥ f g identities. Let C be a commutative ring with 1, F X Cn x1,...,xn the free algebra f g ≥ 2 f gj over C in noncommuting indeterminateso xi ,andsetG f(x1,...,xn) F X some coef®cient of f is a unit in C .IfRis an algebra over C,thenf(x1,...,xn)2Gis a polyno- 2 ≥ mial identity (PI) for RPif for all ri R, f (r1,...,rn) 0. The standard identity of degree sgõ n is Sn(x1,...,xn) ≥ õ(1) xõ(1) ÐÐÐxõ(n) where õ ranges over the symmetric group on n letters. The Amitsur-Levitzki theorem is an important result about Sn and shows that Mk(C) satis®es Sn exactly for n ½ 2k [5; Lemma 2, p. 18 and Theorem, p. 21]. Call f 2 G a central identity for R if f (r1,...,rn) 2Z(R), the center of R,forallri 2R, but f is not a polynomial identity. One can obtain a trivial example of a central identity by adding a polynomial identity to a ®xed element from Z(R). A result with major consequences for the theory of PI rings was the proof of the existence of nonconstant central identities for

matrix rings Mn(F) by E. Formanek [1]. One among the many important applications of this work was a result of L. Rowen [6] showing that any nonzero ideal in a semiprime PI ring must intersect the center of the ring nontrivially. Thus, for a semiprime PI ring, there is an important and interesting relationship between the ring and its center. In particular, the center cannot be too small. A natural and intriguing question which arises is how small the center can be relative to the size of the ring? When R is a prime PI ring with center Z(R), then R and Z(R) have the same cardinality unless R is ®nite. This follows from a theorem of E. Formanek [2; Theorem 1, p. 79], which uses Rowen's result and shows that when R is a prime PI ring, the Z(R) module R embeds in a free Z(R) module of ®nite rank. Another approach is to observe that if S is the central localization of R at Z(R) (0), then S is a ®nite dimensional (simple) algebra over the quotient ®eld of Z(R) [5; Theorem 2, p. 57]. For future reference we record this observation as a theorem. Denote the cardinality of S by jSj.

Received by the editors June 26, 1996. AMS subject classi®cation: Primary: 16R20; secondary: 16N60, 16R99, 16U50. c Canadian Mathematical Society 1998. 81

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THEOREM A. If R is a prime PI ring then either R is ®nite or jRj≥jZ(R)j. We shall need to refer to the theorem of L. Rowen [6; Theorem 2, p. 221] mentioned above, so we state it for convenience.

THEOREM B. If R is a semiprime PI ring and I Â≥ 0 is an ideal of R then I\Z(R) Â≥ 0. We begin with some examples, the ®rst of which is easy and shows that when R is not semiprime, no particular relation exists between jRj and jZ(R)j, except for jZ(R)jÄjRj.

EXAMPLE 1. Let 1 ÚãÚåbe cardinal numbers with å in®nite, F a ®eld with j"FjÄã,and# Xa set of commuting indeterminates over F with jXj≥å.ThenR≥ FF[X] ÂM(F[X]) satis®es the standard identity S , Z(R) ≥ F Ð I ,sojZ(R)jÄ 0 F 2 4 2 ãÚå≥jRj. If the ring R in Example 1 satis®es a central identity, then by linearization it satis®es one which is additive in each variable. Substituting elements from fFe11, F[X]e12, Fe22g into this central identity shows that it must be a PI for R. Thus, R satis®es no central iden- tity. We present another less obvious, but still easy example which satis®es a nonconstant central identity.

EXAMPLE 2. Again let 2 ÚãÚåbe cardinal numbers with å in®nite, F a®eld with char F Â≥ 2, jFjÄã,andV≥fvi ji2Wg,Y≥fyi ji2Wg,andfzgdisjoint sets of noncommutingn indeterminateso over F with jWj≥å.LetHbe the ideal of the free algebra F V [ Y [fzg generated by viyi z and yivi + z,foralli2W, and all other products of twon elements fromo V [ Y [fzgexcept for fviyi and yivi j i 2 Wg.IfRis [ [f g Û the quotient F V Y P z HP, then by identifying indeterminatesP P with their images, consider R ≥ F + Fz + W Fvi + W Fyi. Note that, Fz + W Fvi + W Fyi is an ideal of R whose cube is zero because all its products are zero except that viyi ≥ z and yivi ≥z, for all i 2 W.NowjRj≥å,Z(R)≥F+Fz is ®nite or jZ(R)jÄãwhen ã is in®nite, 2 2 ≥ Â≥ uv vu Z(R)forallu,v R,andviyi yivih 2z 0.i Therefore, R satis®es the central identity [x1, x2] ≥ x1x2 x2x1,andthePI[x1,x2], x3 . In view of these examples and Theorem A, only semiprime PI rings which are not prime are left for consideration. Here the situation is not as clear as for prime rings since jZ(R)jÚjRjcan hold when Z(R) is in®nite, as our next example shows. å EXAMPLE 3. Let be an in®nite cardinal, C a commutativeQ semiprime ring with I jCj≥å,Ia set with jIj≥å,andkÙ1 an integer. Set H ≥ I Mk(C) ≥ Mk(C) ,the å ² complete directÁ product of copies of Mk(C). Fix a nonzero subring S Mk(C)sothat S\Z Mk(C) ≥ (0) and let R ≥fh:I!Mk(C)jh(i)2Sfor all but ®nitely many 2 g iQ I with pointwise addition and multiplication; that is, R consists of all elements in I Mk(C) having ®nitely many coordinates arbitrary in Mk(C) and all other coordinates in S. To see that R is a semiprime ring let h 2 R with h(i) Â≥ 0 and observe that (hRh)(i) ≥ Â≥ \ h(i)Mk (CÁ)h(i) 0, since C a semiprime ring forces Mk(C)tobesemiprime.UsingS ¾ L Z Mk(C) ≥(0), it is easy to see that Z(R) ≥ I C,andsojZ(R)j≥å. Finally, jRjÄ I å å å å I å å jMk(C)j≥å ≥2, and in fact jRj≥2, because 2 ÄjSjÄå ≥2 and there is an

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I I obvious inclusion of S into R ≥ Mk(C) . Note that although jZ(R)j≥åand ideals of R intersect Z(R), R has 2å different ideals de®ned by the subsets A of I as T(A) ≥fr2Rj r(i)≥0foralli2Ag. For a speci®c example one could let C ≥ F,a®eld,orC≥Fp[X], and let S ≥ Ce11,orS≥Ce12. The same construction for C ®nite and I countable yields R uncountable with Z(R) countable. Our ®rst result for ®nite centers is presumably well known, but we could not locate it speci®cally in the literature. Its proof is easy and it will be convenient to have the result, so we present it.

THEOREM 1. If R is a semiprime PI ring with ®nite center, then R is ®nite.

PROOF.SinceRis a semiprime ring, Z(R) is a ®nite commutative ring with no nonzero nilpotent elements, so Z(R) is a direct sum of ®nite ®elds. Let Z(R) ≥ Z ≥ Ze1 ýÐÐÐýZek ≥ Ze,wheree≥1z and feig are minimal orthogonal idempotents in Z whose sum is e. Therefore, R ≥ Re ý R(1 e), where R(1 e) ≥fÁrre j r 2 Rg,and Z(R)\R(1 e) ≥ 0. But R(1 e) is an ideal of R and Z R(1 e) ≥ Z(R) \ R(1 e), so by Theorem B, R(1 e) ≥ 0 forcing e ≥ 1R, and it follows that R ≥ Re1 ýÐÐÐýRek. Hence, Rei is a semiprime PI ring with Z(Rei) ≥ Zei, a ®nite ®eld, so Theorem B forces each Re1 to be simple, and so ®nite by Theorem A, proving that R is ®nite. We come to our main result, which shows that Example 3 illustrates the largest differ- ence which can occur between jZ(R)j and jRj,forRa semiprime PI ring. We shall need to know that there is a central identity gn(x1,...,xk) for the matrix ring Mn(C) which has no constant term and is linear in x1 ([1] or [5: p. 45]). The construction of gn in[1]or[5] shows that for any commutative ring K, gn is not a PI for Mn(K). j j THEOREM 2. If R is a semiprime PI ring and Z(R) is in®nite, then jRjÄ2Z(R).

PROOF. There is a natural embedding of R into the direct product of its prime images, each satisfying the same PI as R, so a well known result of S. A. Amitsur [5; Lemma 2, p. 55] forces R to satisfy a standard identity S2n for some n ½ 1. Let n be minimal so that R satis®es S2n.IfRsatis®es S2 ≥ x1x2 x2x1,thenRis commutative and R ≥ Z(R), so we may assume that n Ù 1 and proceed by induction on n;thatis,ifAis a semiprime PI jZ(A)j ring satisfying S2m for m Ú n,andifZ(A) is in®nite, then jAjÄ2 . Since no nonzero z 2 Z(R) is nilpotent, by using Zorn's Lemma one produces an ideal \f i j ½ g≥; Pz of R maximal with respect to Pz z i 1 , and it is straightforwardT to see that \ ≥ Pz is a prime ideal ofTR. It follows from the de®nition of Pz that Z(R) T( Z(R) Pz) 0, ≥ ≥ where P0 R.But Z(R)PzTis an ideal in the semiprime ring R,soZ( Z(R)Pz) 0[3; ≥ Lemma 1.1.5, p.Q 6], forcing Z(R) Pz 0byTheoremB,andRembeds naturally in the direct product Z(R) RÛPz. Now each RÛPz satis®es S2n and some of these quotients do not satisfy S2(n1) since R does not. Let gn(x1,...,xk) be a central identity for Mn(F), F a ®eld, where gn has integer coef®cients, one of which is 1, no constant term, is linear in x1, and is not a PI for any Mn(D)whereDis a commutative ring ([1] or [5; p. 45]). We argue that gn is a central identity for R. A result of C. Procesi [5; Proposition, p. 43] shows that gn is a polynomial identity for Mn1(F), so for Mk(F) with k Ä n 1, but not

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aPIforMn(F) by choice of gn.IfTis any prime ring satisfying a PI p(x1,...,xk), then 1 W ≥ TZ , the localization of T at Z(T) (0), also satis®es p(x1,...,xk) and is a simple algebra, ®nite dimensional over its center K [5; Theorem 2, p. 57]. Either W ≥ Mt(K)or W KL≥Mt(L), for L an algebraic closure of K,andW K Lalso satis®es p(x1,...,xk) [5; Lemma 1, p. 89], so 2t Ä deg p by the Amitsur-Levitzki theorem. In our case, if RÛPz satis®es S2(n1) then it embeds in some Mn1(F)andgn is an identity for RÛPz.IfRÛPz does not satisfy S2(n1) but gn is an identity for it, then we conclude ®rst that RÛPz does not embed in any Mk(F)forkÚnby the Amitsur-Levitzki theorem. Secondly, since S2n and gn are identities for RÛPz, as above, RÛPz embeds in Mn(F), which satis®es the PI gn, contradicting the choice of gn. Therefore, gn is a central identity or a PI for each RÛPz, so a central identity for R; it is not a PI for R since it is not a PI for any quotient RÛPz which fails to satisfy S2(n1). Choose z 2 Z(R)sothatgn is not a PI on RÛPz. Writing r + Pz ≥ rÅ 2 RÛPz, the fact that gn is a central identity for RÛPz means that there are rÅi 2 RÛPz,sothatgn(Åri)≥ cÅ2Z(RÛPz)(0). For any yÅ 2 Z(RÛPz), yÅÅc ≥ gn(yÅÅri,...,Årk), since gn is linear in its ®rst variable. But gn( yÅÅr1,...,Årk) ≥ gn(yr1,...,rk)+Pz with gn( yr1,...,rk) 2 Z(R), so cZÅ (RÛPz)  Z(R) and since Z(RÛPz) is a domain, jZ(RÛPz)jÄjZ(R)jresults. Applying j Û jÄj j ≥f 2 j Û g ≥ Theoremn A giveso R PTz Z(R).IfTI z Z(R) gTnis not a PI of R Pz , J [f g ≥ ≥ \  ≥ Z(R) I 0 ,A I Pz,andB QJPz,thenA B Z(R)Pz 0, so R embeds in Û ý Û Û Û 2 R A R B.NowR Aitself embedsQ in I R Pz, and as we have just observed, for z I, jZ(R)j jZ(R)j jRÛPzjÄjZ(R)j. Therefore, j I RÛPzjÄjZ(R)j ≥2 since Z(R) is in®nite. Hence jRÛAjÄ2jZ(R)jand the proof is complete when A ≥ 0. Â≥ Û \ ≥ Assuming thatQ A 0, it follows that A embeds in R B since A B 0, and of course Û Û 2 Û R B embeds in J R Pz. For each z J, gn is a PI of R Pz, so byQ our observations above and Procesi's theorem, RÛPz satis®es S2(n1) which means that J RÛPz and RÛB satisfy S2(n1).SinceAembeds in RÛB, A is a semiprime ring satisfying S2(n1). By Theorem 1 and the induction assumption, either A is ®nite or jAjÄ2jZ(A)j Ä2jZ(R)jsince Z(A)  Z(R) [3; Lemma 1.1.5, p. 6]. Thus jRj≥jRÛAjjAjÄ2jZ(R)j2jZ(R)j ≥2jZ(R)jcompleting the proof of the theorem. We record a simple consequence of Theorem 2 for algebraic algebras.

THEOREM 3. Let R be a semiprime ring and algebra over the integral domain C. If R is integral over C of bounded degree, then either R is ®nite or jRjÄ2jZ(R)j.

PROOF.IfRis integral over C of bounded degree n, then every r 2 R satis®es some n+1 relation r + cnrn + ÐÐÐ+c1r≥0. It is well known and straightforward to show that this relation implies that f[xi, y], j 1 Ä i Ä n +1gis C-dependent for any x, y 2 R,sothat n+1 2 Rsatis®es the polynomial identity Sn+1([x , y],...,[x ,y], [x, y]) [4; p. 230]. Applying Theorem 2 ®nishes the proof. We note that Example 1 and Example 2 show that if R is not semiprime, then R alge- braic of bounded degree over a ®eld does not imply any particular relationship between jZ(R)j and jRj. A similar example for semiprime rings is provided by Example 3 when C ≥ F,a®eld,andS≥Fe12. Our last example shows that the assumption of bounded degree in Theorem 3 is essential, even for prime or simple algebras.

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EXAMPLE 4. Let 1 ÚãÚåbe cardinal numbers with å in®nite, F a ®eld with jFjÄã,andVF an F-vector space with dimF V ≥å.Iffvi ji2Igis an F-basis of V, for I a well ordered set with jIj≥å, then one can represent the elements of HomF(V, V) as column (or row) ®nite åðåmatrices, say Må(F) with matrix units feij j i, j 2 Ig. Set M0 ≥fA2Må(F)jAhas only ®nitely many nonzero entriesg, or equivalently, M0 ≥fT2HomF(V, V) j vj 2 ker T for all but ®nitely many j 2 Ig. It is easy to see that M0 is a simple algebraic F-algebra with jM0j≥å,andthatZ(M0)≥0. By taking R ≥ M0 + F Ð IV , so adding scalar matrices to M0, it follows that R is a prime algebraic algebra over F, jRj≥å,andjZ(R)j≥jFjÄã.

REFERENCES 1. E. Formanek, Central polynomials for matrix rings.J.Algebra23(1972), 129±132. 2. , Noetherian PI-rings. Comm. Algebra 1(1974), 79±86. 3. I. N. Herstein, Rings with involution. Chicago Lectures in Mathematics, University of Chicago Press, Chicago, 1976. 4. N. Jacobson, Structure of rings. Amer. Math. Soc. Colloq. Publ. 37, Amer. Math. Soc., Providence, RI, 1964. 5. ,PI-algebras. Lecture Notes in Math. 441, Springer-Verlag, New York, 1975. 6. L. Rowen, Some results on the center of a ring with polynomial identity. Bull. Amer. Math. Soc. 79(1973), 219±223.

Department of Mathematics University of Southern California Los Angeles, CA 90089-1113 USA

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