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Ring such that $x^4=x$ for all $x$

Let $R$ be a ring such that $a^4=a$ $ ,\forall a \in R$. How do I show that $R$ is commutative?

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marked as duplicate by lhf, t.b., Martin Sleziak, Asaf Karagila, Gerry Myerson Nov 16 '11 at 12:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I answered this here:… – user641 Nov 16 '11 at 9:43

1 Answer 1

If 2 is invertible, it may work this way:

For arbitrary elements $a,b$ calculate $ab-ba = (ab-ba)^4 = (ba-ab)^4 = ba-ab$. So $2ab=2ba$ and hence $ab=ba$.

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2 cannot be invertible, right? Otherwise, for any $a\in R$, $a=a^4=(-a)^4=-a$. So $2a=0$, which implies $a=0$. – Paul Nov 16 '11 at 9:19
you are totally right, thanks. So the above is useless. – MaoK Nov 16 '11 at 9:22
no, it's a good try. I think it's not easy to prove it, maybe I am not good at algebra. – Paul Nov 16 '11 at 9:28

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