Wedderburn'sTheorem, asserting that a finite division ring is necessarily commutative, has been generalized in several directions [2]. Our present object is to establish the following additional generalization.Theorem 1. Suppose R is any associative ring with Jacobson radical J such that R/J is a finite ring with exactly q elements. Suppose also that (i) R/J is a division ring, (ii) J2 = (0), and (iii) the number of distinct qth powers of all elements of R is at most q. Then R is commutative.
Proof.For any x in R, let x -x + J. First, observe that xq =yq (mod J) if and only if &9 = yq which, in turn, is equivalent.to x = y, or x=y (mod J). Hence there are at least q distinct qth powers of elements of P. Thus, by (iii), there are exactly q distinct qth powers of elements of the ground ring R. It further follows that (1) x = y(mod /) implies xq = yq.For, otherwise, there would be more than q distinct 3th powers of elements of P, since x^y (mod J) implies xq^yq (mod /). Now, let aEJ and let bER. Then hq = b and hence bq -bEJ.Therefore, by (ii), we have(2) a(b" -b) = 0, (bq -b)a = 0 (aEJ,b arbitrary).Again, let aEJ. Since J2 = (0), by (ii), therefore(3) (ab + b)q = ab" + babq~l + b2ab"~2 + • • • + bq~xab + b", (4) (ba + b)q = bqa+ bab^1 + Pab^2 + • • • + bq~lab + bq.We have thus shown that (5) (ab + b)q -(ba + b)q = ab" -bqa (aEJ,b arbitrary).Moreover, aEJ implies ab+b = ba+b (mod J), and hence, by (1), (6) (ab + b)q = (ba + b)q (aEJ,b arbitrary). Now, an easy combination of (2), (5), (6), shows thatReceived by the editors April 22, 1966.