Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $R$ is a ring such that $x^5=x$ for all $x\in R$, is $R$ commutative?

If the answer to the above question is yes, then what is the least positive integer $k \ge 6$, such that there exists a noncommutative ring $R$ with $x^k=x$ for all $x\in R$?

share|cite|improve this question
Sounds like a hard homework (: – curious Jan 6 '11 at 5:27
No. I came across with proofs that rings with $x^k=x$ for $k=2,3,4$ are commutative; hence the question. – TCL Jan 6 '11 at 5:34
@Jasper. Not necessary, from the link by curious. – TCL Jan 6 '11 at 5:39

The answer is that if $R$ is a ring such that for all $x \in R$ there is an integer $n(x) > 1$ such that $x^n(x) = x$ then $R$ is commutative. See Herstein's "Non Commutative Rings".

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.