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Linear and Multilinear Algebra, 1989, Vol. 24, pp. 127-134 Reprints available directly from the publisher Photocopying permitted by license only 0 1989 Gordon and Breach Science Publishers, S.A. Printed in the United States of America Nill *-polynomials in Rings A. GIAMBRUNO AND A. VALENTI Dipartimento di Maternatica, Universita di Palerrno. Via Archirafi 34, 90123 Palerrno, ltalv (Received February 25, 1988; in final form June 8, 1988) Let R be an algebra with involution * over a field F of characteristic different from 2 and let f(x,,..., x, x:,..., x,*) be a multilinear *-polynomial over F which, when evaluated in R, takes nilpotent values. We prove that if R has no nil right ideals and IF1 > 5, then f is a *-polynomial identity for R. 1. INTRODUCTION Let F be a field of characteristic different from two and R an F-algebra with involution *. In this paper we study multilinear *-polynomials that take nilpotent values in R. We shall prove that, under suitable hypotheses on R, such polynomials are *-polynomial identities for R. Let X = {x,,..., x,,...) be a countable set of unknowns and F{X, *) = Fix,, x:,..., x,, x:,...) the free algebra with involution in the xi's and xf's. The elements of FIX, *) are called *-polynomials. A *-polynomial f (x,,..., x,, xt,..., x,*)e F{X, *) is multilinear if, for each i (1 < i < n) either x, or x:, but not both, appears in each monomial off. If R is an F-algebra with involution * we say that a *-polynomial f(x,,..., x,, xt,..., x:) is nil valued in R if for all r,,..., r, R ~ there exists an integer m = m(r,,..., r,,) 2 1 such that f (r,,..., r,, rt,..., r,*)m = 0. We shall prove that if F has more than five elements and R is a ring with no nil right ideals then every multilinear *-polynomial nil valued in R is a *-polynomial identity (*-PI) for R. One could ask if this result is still valid if one replaces the hypothesis "no nil right ideals" with its two-sided version "no nil ideals". Unfortunately the following two special cases are still open: on one hand it is a long-standing question of Herstein whether, in a ring with no nil ideals, xlx2 - x2x, nil valued implies x,x, - x2x1 a polynomial identity for R (i.e., R is commutative) (see [7, p. 8001); on the other hand if x, + xt is nil valued in R (or more generally if all the symmetric elements are nilpotent) it is an open question of McCrimmon whether this forces R to have some non-zero nil ideal (see [3, p. 831). However, both these questions have a positive answer if R has no nil right ideals. We remark that if R is a ring and f is a multilinear polynomial (not a *-polynomial) an analogous theorem was proved in [I]; also, if we set [a,, a,] = a,a2 - a2a, and, The research of the first author was partly supported by a grant from MPI, Italy. 127
128 A. GIAMBRUNO AND A. VALENTI more generally, [a,,..., a,] = [[[a,, a,], aj,..., a,], then in [2] and [6] it was proved that if R has no nil right ideals and [x, + x:,..., x, + x,*] is nil valued in R then [x, + xt, x, + x:] is a *-polynomial identity and R satisfies the standard identity of degree four. Throughout this paper F will be a field with more than five elements, char F # 2 and R will be an F-algebra with no nil right ideals. S = { x R ~ 1 x = x*) and K = (x E R 1 x = -x*) will be the sets of symmetric and skew elements of R respectively. Also, f(x,,..., x,, x:,..., x,*) will be a multilinear *-polynomial nil valued in R. Let Z, = {I, *) be the group with two elements, S, the symmetric group on n symbols and H, = Z, - S, the wreath product of Z, and S,. It is clear that we may assume (as we will) that f is of the following form xi, if g, = 1 (g,o)=(gl,...,gn; 0) and xs1= x*, if g, = *. 2. - SIMPLE ARTlNlAN RINGS In this section R will be a simple artinian ring. By Wedderburn's theorem R M,(D). the ring of k x k matrices over a division ring D. In R are defined two different types of involutions [4, p. 0.101: (1) the symplectic type involution: in this case D = F is a field, k is even and * is given by (Aij)* = (A;) where the Aij's are 2 x 2 matrices over F with involution given by (2) The transpose type involution: let -: D + D be an involution in D and X = diag(c,,..., c,) G M,(D) such that c, = ci, for all i. If A = (a,,) E MJD), then * is given by A* = (alj)* = X(ii,,)X - l. Let eij (i, j = 1,..., k ) be the usual matrix units of M,(D). Given a sequence u = (A,,..., A,) of matrices in M,(D), the value of u is the product lul= AIA,...A,. Also, if (g, a) H, we write u(~-")= (A:(,,,..., A;;,,) where the symbols have the meaning mentioned above. If a,,..., a, E D, the sequence u = (a, e,, j,,..., a, eircj,,) is called simple. A simple sequence u is even if there exists (g, o ) H, ~ such that lu'g,")i = be,, # 0, for some b E D; u is odd if ~ u'~~"'~ = beij f 0, for some (g, a) H,, b~ D and i # j. Now, given a simple sequence u and an integer t, 1 < t < k, we define o(u, t) to be the total number of occurrences of t as a left or right index of the unit matrices eimjm appearing in u. It is clear that if u is simple, then (ul = be,, # 0 only if o(u, t) is even for all t, 16 t f k ; also, (u( = beij # 0, i # j, only if o(u, t) is even for all t # i, j and o(u, i), o(u, j) are odd.
*-POLYNOMIALS IN RINGS The basic fact about simple sequences is the following LEMMA 1 Let u E M,(D) be a simple sequence and (g, o) E H,. Then (1 ) if I U I = aeii, then lu(g,")l = bejj for some b E D, 1 6 j < k. (2) If IuI = aeij, i # j, then for some b, CE D u ( q, ~ ) l = {;'ij + ceji$ if * = transpose type e,,, t # m, if * = symplectic type. Suppose * = transpose type. Since for all t, ~(u'~~"', t) = o(u, t), by the above remark the conclusion of the lemma is clear in this case. Suppose now * = symplectic type. Recall that the involution * acts in the following way on the matrix units: e l l if i=odd,j=even +l i+t, if i, j odd \ej-i i-13 if i, jeven. Now, suppose that eij occurs once in the simple sequence u and assume for instance that i is even and j is odd. If u' is the sequence obtained from u by replacing eij with e;, then in u' four indices, namely i, i - 1, j, j + 1 have changed parity. From this fact it is easy to see that in u'~."' the number of indices that change parity is always a multiple of four. The conclusion of the lemma now follows easily. The previous lemma allows us to compute the value of a *-polynomial on a simple sequence. In fact we have LEMMA 2 Let u E Mk(D) be a simple sequence. Then k (1) if u is even, f (u, u*) = ccieii; i= 1 (2) if u is odd, for some a, b, aij D Write aeij + beji, if * = transpose type f&, u*, = { ' aijeij, if * = symplectic type. i#j and an application of the previous lemma completes the proof. For the proof of the next lemma it is crucial that f is a nil valued *-polynomial.
130 A. GIAMBRUNO AND A. VALENTI LEMMA 3 If f (u, u*) = 0 for all even simple sequences u, then,f is a *-polynomial identity for Mk(D). have Since f vanishes on even simple sequences, for all A,,..., A, MJD) we f(a1,...,a,, AT,..., A,*)=zaif(ui,u:) where the ui's are odd simple sequences. Therefore in order to prove the lemma it is enough to prove that f (u, u*) = 0 for all odd simple sequences. We distinguish two cases : * = transpose type. If u is odd, by the previous lemma f (u, u*) = aeij + beji, Case 1 for some ifj and a,b~d. Now, if a# 0, b# 0 take m> 1 such that 0 = f *"(u, u*) = (ab),eii + (ba),ejj. It follows ah = ba = 0 and this is a contradiction. Suppose then that f (u, u*) = aeij, a # 0. Let y ED be such that ycj!c;' # 0, i 1; such an element exists since I F / > 5. If d = cjjcl1 and I = unit matnx, the element is a unitary element (that is v* = v-') of Mk(D). Let be the inner automorphism of Mk(D) induced by v. By acting diagonally, q, induces an automorphism on M,(D) X-x Mk(D); let uq be the image of u under this tl automorphism. Then, by the hypotheses placed on f, f(uq, u*~) is a sum of odd simple sequences and, so, its diagonal entries are all zero. We have: f ( u~,!i*") = f (u, u*)"' = vj-(u, u*)v* and, equating to zero the coefficient of eii we get hence a = 0 and the result is proved in this case. Case 2 * = symplectic type. By the previous lemma, if u is an odd simple sequence, We have to prove that aij = 0, for all i, j. Now, if t # m are even, the matrix f (u, u*) = C ai.ieii. i f j is a unitary element of Mk(D). As above, computing explicitly the following expression f (uq, u*") = f (u, u*)'+' = ( 1 + e,, - ern_,,-,) C aijeij(l - elm + ern-,,-,) and equating to zero the coefficients of e,, and em_,,-, we get a,, = a,-,,-, = 0. i# j
i Now, similar calculations using the unitary matrices and when t is even, m is odd and t is odd, m is even respectively, yields the desired conclusion. We can now prove the final result for simple artinian rings. THEOREM 1 Let R be a simple artinian ring. If f is a multilinear *-polynomial nil valued in R then f is a *-polynomial identity for R. By the previous lemma it is enough to prove that f (u, u*) = 0 for all even simple sequences u. Let u be such a sequence. Then f (u, u*) = 1 aieii and, since f (u, u*) is nilpotent, this forces ai = 0, for all i. Thus f (u, u*) = 0. 3. GENERAL RINGS Recall that R is a ring with no nil right ideals and f is a multilinear *-polynomial nil valued in R. Our approach to this general situation will be that of looking at prime images of R satisfying a generalized polynomial identity (briefly GPI). In fact, in the presence of a GPI it is possible to prove the result for prime rings as the following lemmas show. Recall that by Posner's theorem [3], if R is a prime PI-ring then its ring of central quotients Q is a central simple algebra of dimension m2 over its center and one defines PI-deg(R) = m. LEMMA 4 If R is prime and satisfies a generalized polynomial identity then f is a *-polynomial identity for R. By [3, Corollary to Theorem 2.5.11 either R satisfies a PI or for every integer m > 1, there exists a prime PI*-subring R'"' of R such that PI-deg(R(")) 3 m. If R is PI then R is an order in a finite dimensional central simple algebra Q and the involution * can be extended to Q in a natural way so that f will still be nil valued in Q (see for instance [2]). But then by Theorem 1 f is a *-polynomial identity for Q and so for R. Suppose now that for every m 3 1 there exists R'"' c R, prime PI*-subring and PI-deg(R'"') 3 m. In this case, since f is nil valued in R'"', by the first part of the proof f is a *-PI for R'"', for all m. But then by 13, Lemma 5.1.51 R'") satisfies a PI of degree 2 deg( f ); so m d PI-deg(R'"') d def( f ), a contradiction. LEMMA 5 Let P be a prime ideal of R such that RIP satisfies a generalized polynomial Identity. Then for all r,,..., rn E R, f (rl,..., rn) E P and f (r,,..., rn)* E P. Suppose first that P* c P. Then RIP has induced involution and f is nil valued in RIP. By Lemma 4, f is a *-PI for RIP and the lemma is proved in this case. If P* Q P, (P + P*)/P is a two-sided ideal of RIP and, since RIP satisfies a GPI,
132 A. GIAMBRUNO AND A. VALENTI RIP, and so (P + P*)/P, has no nil right ideals (see [6, Lemma 23). Now write is the sum of all the monomials of f in which the variables xf',..., xin appear. It is clear that if g = (g,,..., g,) E B; then, for all ry',..., r,9"~ P*, Hence f, is a polynomial (not a *-polynomial) nil valued in (P+ P*)/P. Since (P + P*)/P has no nil right ideals, by [I, Theorem 31 f, is a PI for (P + P")/P. But then, since by [3, Theorem 1.4.21 0 # Z(P + P*/P) c Z(R/P), it can be easily proved that f, is also a PI for RIP. It follows that f (rl,..., r,) = 2 fg(ry1,..., r,"") E P. 9 ~ 2 3 This same argument applies to the polynomial f (x,,..., x,)* leading to the desired conclusion. The following result which holds for general rings will be used later. LEMMA 6 Let A be a ring with no nil right ideals and a A such that a2 #O. Let B = uaa and N = {b E B / aba = 0). Then BIN has no nil right ideals. Let p be a nil right ideal of B = BIN. In order to prove that p = 0 we may clearly assume p = 3B. For b E B let m 3 1 be such that (xb)" E N. Then a(xb)"a = 0. If x = aza and b = ara we get a(aza2ra)"a = 0; hence (a2za2r)"+' = 0, i.e., a2za2a is a nil right ideal of A. By the hypotheses on A, a2za2 = 0; thus x = aza E Nand 5 = 0. LEMMA 7 Suppose f is not a *-polynomial identity for R. If y,,..., y, E R are such that = f (yl,..., y,)"- l # 0 and f (yl,..., y,)" = 0 then yy* is not nilpotent. Suppose that yy* is nilpotent. Since y2 = 0, y + y* and y - y* are both nilpotent. Now if y + y* # 0 take t > 1 such that s = (y + y*)'-' # 0 and (y + y*)' = 0; if y - y* # 0 we let t > 1 such that s = (y- y*)'-' # 0 and (y- y*)' = 0. In any case ~ Eor S K,s#O and s2=0. Let r,,..., r, E R and let m 2 1 such that f (sr,,..., sr,)" = 0. Since s2 = 0 we get = g (sr,,..., sr,)"s where g(xl,..., x,) is a polynomial (not a *-polynomial). So, if lr(s) is the left annihilator of s in R, the above equalitfays that g is nil valued in the ring Since R, has no nil right ideals, by [I, Theorem 31 g is a PI for R,; hence g(sx,,..., sx,)s is a GPI for R.
Let now P be a prime ideal of R such that s$p. Then s= s + P # 0 and, so, g(sx,,..., Sx,)S is a GPI for l? = RIP. By Lemma 5 it follows that f(y,,..., ~,)EP and f (y,,..., yn)* E P which imply y, y* E P. Hence s = (y k y*)'-' E P, and this is a contradiction. We are now in a position to prove our result in its full generality. THEOREM 2 Let R be a ring with no nil right ideals. If f is a multilinear *-polynomial nil ualued in R, then f is a *-polynomial identity for R. We will prove the theorem by induction on n = def( f ). Suppose first deg( f ) = 1 ; then f = f (x, x*) = x + ax*. If a # - 1, then for all r E R, r E R}, and f (r + r*) = (1 + a)(r + r*); hence, since f is nil valued we get that {r + r* / so S is a nil set. Suppose that f is not a *-PI for R. Take O # SES such that s2 = 0 and let r E R. Then, if m 3 1 is such that (sr + r*sjm = 0, we get 0 = (sr + r*s)"s = (sr)"~. This says that Rs is a nil right ideal of R; hence s = 0 and this is a contradiction. On the other hand, if cc = - 1, f (x) = x - x*; thus (r- r* 1 r E R), and so K is a nil set. As above this leads to a contradiction also in this case. We assume now that deg( f ) > 1. Suppose that f is not a *-PI for R; let y,,..., ~,ER be such that Y = f(y,,..., y,)"-' # 0 and f (y,,..., y,)" = 0, for a suitable m > 1. Then y2 = 0 and by Lemma 7, yy* is not nilpotent. Let r,,..., re-, E R; since y2 = 0 then for a suitable m 3 1 we have where h(x,,..., x,-,) is a non-zero *-polynomial. Set a = yy*; let B = ara and N = {be B 1 aba = 0). Clearly B has induced involution and N = N*. Also, by Lemma 6, BIN has no nil right ideals. By what we proved above we have that f ~ all r r,,..., rn-, E R, m = m(r,,..., rn-,) 3 1. This says that h(x,,..., x, -,) is a "-polynomial nil valued in BIN. Since deg(h) < deg( f ), by the inductive hypothesis we have that h is a *-polynomial identity for BIN. Thus ah(ax,a,..., ax,_,a)a i$ a *-GPI for R. Let now P be a prime ideal of R such that a2 $P. If P* c P, then RIP has the induced involution and it satisfies the same *-GPI of R. By [5, Theorem 71 then RIP satisfies a GPI; but then by Lemma 5, f (y,,..., y,) E P and, so, a2 = ( ~y*)~ E P, a contradiction. If P* @ P, then as in the proof of Lemma 5, we can obtain a GPI for (P + P*)/P and, so, for R. By Lemma 5, f (y,,..., ~,)EP and, so, a2 E P, a contradiction. References [I] B. Felzenszwalb and A. Giambruno, Periodic and nil polynomials in rings, Can. Math. Bull. 23 (1980), 473-476. [PI A. Giambruno, Algebraic conditions on rings with involution, J. Algebra 50 (1978), 190-212.
134 A. GIAMBRUNO AND A. VALENTI 131 I. N. Herstein, Rings with Inr~olutron, University of Chicago Press, Chicago, 1976. 141 N. Jacobson, Lectures on Quadratic Jordan Algebras, Tata Institute, Bombay, 1969. [5] C. Lanski, Differential identities in prime rings with involution, Trans. Amer. Math. Soc. 291 (1985), 765-787. [6] A. Valenti, A note on rmgs with involution, Boll. UMI (7) 1A (1987), 271-274. [7] F. VanOystaeyen (Editor), Ring Theory, Proceedings of the 1978 Antwerp Conference, Marcel Dekker, New York, 1978.