Let $R$ be a ring such that $x^4=x$ for every $x\in R$. Is this ring commutative?
Ring such that $x^4=x$ for all $x$ is commutative
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abstract-algebra
ring-theory
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8Have you tried anything so far which you could share? – 2011-10-28
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6There is a theorem of Jacobson that you may find useful, see here http://mathoverflow.net/questions/32032/on-a-theorem-of-jacobson. Short of that I feel like the easiest path is to just mess around with expansions until you get the desired result. – 2011-10-28
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2The analogous problem with $3$ in place of $4$ is a classic (perhaps due to its appearance in a book by Herstein). It turns out that if there is a positive integer $n > 1$ with the property that $x^n = x$ holds for all $x$ in $R$, then $R$ must be commutative. I do not know if the $n=4$ case is easier to prove than the general case (a wild guess: maybe try to show that if it holds for $n=4$, then it holds for $n=2$). The link http://www.math.niu.edu/~rusin/known-math/99/commut_ring of some sci.math discussion of the $n=3$ problem may be of interest. – 2011-10-28
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0The n=4 case appears in Herstein as well, actually; IIRC, it appears earlier. From what I recall of the solution, it's slightly easier than the n=3 case. – 2011-10-29