Prime rings satisfying a polynomial identity

Abstract

PROOF. Sufficiency is easy and omitted. We recall that B is a twosided quotient ring of its subring A if every element of B can be written ab-1, a, bCA, and also c-ld, c, dCA, and if every element of B not a left (right) zero divisor has a right (left) inverse in B. B is unique given A if it exists. To prove necessity we invoke Goldie's Theorem (2): Let R be a prime ring satisfying (11), (1r), (21), (2r). Then R has an mXm matrix ring over a division ring as its full ring of quotients. Here (11) is: every direct sum of nonzero left ideals of R has a finite number of terms. (21) is: the ascending chain condition holds for the annihilator left ideals of R. (1r) and (2r) are analogous. We will prove that if R is a prime ring satisfying a polynomial identity, then R satisfies (il) and (21). The conditions for right ideals will follow similarly. Let Zarx,(l) * X7r(n)= 0, aE C, the centroid of R, ir a permutation of 1, 2, * *. , n, be a homogeneous multilinear identity for R. We will show that the length of a direct sum of nonzero left ideals is at most n 1. First let Ij, 1 _j _ n, be left ideals invariant under the centroid of R, and let 1E . .* . In be direct. Let xjEIj. Then all terms in the identity whose rightmost factor is x1, say, must add up to zero by directness and the fact that the I; are invariant under the centroid. Here x1 was so numbered that at least one nonzero coefficient occurs. Since R is prime, I1 has no left annihilator, that is, we can now cancel xi from this identity. Continuing in this fashion, renumbering if necessary, we find In= 0. If Ij are not invariant under the centroid C, the Clj are, and C1 0 .* .. ED CIn is still direct. For if Vn 1 cjij =O, C eC, ijCIj, 1 _j_n, then Yr CR, Zn=1 (rcj)ij -O. But rcj1ijEIj, 1 <j_n, so each rcjij=O. Since R has no absolute right divisors of zero, cjij = 0, 1 <j _ n. To prove that R satisfies the ascending chain condition for left annihilator ideals, suppose I1gI2_ * * TIn are left annihilator