A generalization of Wedderburn’s theorem

Abstract

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.

Completing Segre's proof of Wedderburn's little theorem

“…There have been further simplifications, generalizations, and novel approaches taken in reproving Wedderburn's theorem and we refer to the preprint [1] for a brief synopsis of some of the known proofs. (Their synopsis does not include the notable proofs of Henderson [16], Outcalt and Yaqub [26], McCrimmon [25] and Tecklenburg [31].) To the authors' knowledge, with the possible exception of the proof of Tecklenburg [31], every known proof of Wedderburn's theorem uses only techniques that could be typified as number theoretic or stemming from abstract algebra, and in particular, the theory of rings and groups.…”

A generalization of a theorem of Wedderburn

“…We simply assume the ring has a left identity and the set of nilpotent elements i s an ideal. But f i r s t we state another generalization of Wedderburn's Theorem given by Herstein [ 7 ] . Proof.…”

“…Wedderburn 1 s Theorem, asserting that a finite division ring is necessarily commutative, has been generalized in several directions [1, 6,7,8]. A survey on a few papers concerning commutativity theorems for rings revealed that the two non-commutative rings defined on the Klein group {G, +) have been cited quite often as counterexamples to show that certain hypotheses cannot be deleted.…”

“…This class of noncommutative rings serves as counterexamples to show that certain hypotheses of various commutativity theorems cannot be omitted. For example, see [3,4,6,7 …”

Commutativity theorems for nonassociative rings with a finite division ring homomorphic image

“…Moreover, R is not associative. Other models for our theorems appear in [2,3]. Also, examples are exhibited in those papers which show that the theorems fail in case R fails to satisfy all of the hypotheses of these theorems.…”