A Commutativity Theorem

TLDR: Yen as mentioned in this paper showed that if R is a semiprime ring with 1, then it is not commutative or anticommutative, and he also showed that R need not be a ring with l. He also conjectured in this paper that R can be replaced by $'R^2=R'$ in the above conjecture.

Abstract: Let R be a ring, not necessarily with 1. Call potent (strongly potent) if for some natural (even natural number) n>1. The well known Jacobson's Commutavity Theorem [3] states : If every element x of a ring is potent, then R is commutative. Herstein [2, Theorem 3.13, pages 74] improved this: If, for every x, y in R, xy - yx is potent then R is commutative. Recently Machale [4] showed : If, for every x, y in R, xy+yx is strongly potent, then R is anticommutative. Yen [5] showed : If, for every x, y in R, either xy-yx is potent or xy+yx is strongly potent, then R is either commutative or anticommutative. Yen [6], also see [1], showed : If R is a semiprime ring and, for every x, y in R, xy+yx is potent, then R is commutative. The first author [1] showed : If R has 1 and, for every x, y in R, xy+yx is potent, then R is commutative. He also gave an example to show that in general if, for every x, y in R, xy+yx is potent, then R need not be commutative. Yen [7], proved : If R is a semiprime ring such that for every x, y in R, either xy+yx or xy-yx is potent then R is commutative. He conjectured in this paper : If R is a ring with 1 such that for every x, y in R either xy-yx is potent or xy+yx is potent, then R is commutative. Here we prove this conjecture. He also asked the following stronger question in a personal communication : Can we replace 'R with l' by $'R^2=R'$ in the above conjecture? Here we answer this question also as an application of the above result.