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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 $x\in R$?

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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

2 Answers 2

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".

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