Symmetric identities in graded algebras By Y. A. BAHTURIN, A. GIAMBRUNO and M. V. ZAICEV Abstract. Let P k be the symmetric polynomial of degree k i.e., the full linearization of the polynomial x k : Let G be a cancellation semigroup with 1 and R a G-graded ring with finite support of order n. We prove that if R 1 satisfies P k 0 then R satisfies P kn 0. If x 1 ; ... ; x n are noncommuting indeterminates, the polynomial P n x 1 ; ... ; x n P s2Sn x s1 x sn is called the symmetric polynomial of degree n. Being the full linearization of x n , P n determines the variety of nil algebras of exponent n in characteristic zero. This is no more true in positive characteristic; in fact Kemer in [3] has recently shown that every PI-algebra over a field of positive characteristic satisfies P n 0 for some n. An explicit result about P n was first obtained by Zalesskii in [5] by proving that if F is a field of characteristic p > 0 then M k F ; the algebra of k k matrices over F, satisfies the identity P kp 0 and no other symmetric polynomial of lower degree vanishes in M k F . By exploiting methods of graph theory, Domokos in [2] extended this result by proving that if a ring R satisfies the identity P n 0 then M k R satisfies P nk 0. Recall that if G is a (semi)group, a ring R is G-graded if R P g2G R g and R g R h 7 R gh for all g; h 2 G. Since M k F is a G-graded algebra for any finite group of order k, it is natural to wonder if the above results are peculiar of matrix rings or if they still hold in the setting of graded rings. In this direction Smirnov and Zalesskii in [4], by establishing a conjecture of [6], recently proved that if R P g2G R g is an algebra over a field of characteristic p > 0 graded by a finite abelian group G of order n and R 1 is commutative, then R satisfies P np 0. In this note we shall show that all the above results can be deduced from the combinatorial properties of the symmetric polynomial and the grading of a ring by proving the following: let G be a cancellation semigroup with 1 and R P g2G R g a G-graded ring such that the set fg 2 GjR g j 0g is of finite order n. If R 1 satisfies P k 0, then R satisfies P nk 0. The proof of this theorem is purely combinatorial and, as a consequence, we shall derive the results of [2] and [4]. Arch. Math. 69 (1997) 461 ± 464 0003-889X/97/060461-04 $ 2.30/0 Birkhäuser Verlag, Basel, 1997 Archiv der Mathematik Mathematics Subject Classification (1991): Primary 16R10; Secondary 16W50.