English

Herstein's question about simple rings with involution

Rings and Algebras 2012-10-12 v1

Abstract

The aim of this paper is to try to answer Herstein's question concerning simple rings with involution, namely: If RR is a simple ring with an involution of the first kind, with dimZ(R)R>4dim_{Z(R)}R > 4 and \Char(Z(R))2\Char(Z(R))\neq 2, is it true that S2=RS^2=R? We shall see that in such a ring RR, R=S3R=S^3. We shall bring two possible criteria, each shows when R=S2R=S^2. The first criterion: There exist x,ySx,y \in S such that xyyx0xy-yx \neq 0 and xSyS2xSy \subseteq S^2 \Leftrightarrow S2=RS^2=R. The second criterion: There exist x,ySx,y \in S such that xy+yx0xy+yx \neq 0 and xKyS2xKy \subseteq S^2 \Leftrightarrow S2=RS^2=R. Actually, those results are true without any restriction on the dimension of RR over Z(R)Z(R). In the special case of matrices (with the transpose involution and with the symplectic involution) over a field of characteristic not equal to 2, it is not difficult to find, for example, x,ySx,y \in S such that xyyx0xy-yx \neq 0 and for every sSs \in S, xsyS2xsy \in S^2. Therefore, proving Herstein's remark that for matrices the answer is known to be positive. Similar results for K6K^6, K4K^4, K+KSKK+KSK, KS+K2KS+K^2, SKSSKS and S2KS^2K can also be found.

Keywords

Cite

@article{arxiv.1210.3353,
  title  = {Herstein's question about simple rings with involution},
  author = {Vered Moskowicz},
  journal= {arXiv preprint arXiv:1210.3353},
  year   = {2012}
}
R2 v1 2026-06-21T22:20:16.357Z