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Let $A$ be a ring. Suppose that there exists $n\in\mathbb{N},n\geq 2$ such that $\forall x\in A,x^n=x$. Does that imply that $A$ is commutative ?

I have read all the links suggested in the comments, but found no real proof that can be extended to $n\geq 2$. Is there any way to show the property without having to prove the whole Jacobson's theorem ?

Is is obvious for $n=2$, and I can prove it for $n=3$ and $n=4$. However, the demonstrations uses identities which I cannot extend to higher degrees to obtain the result I want.

user26857
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Hippalectryon
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    You're assuming that every element in the ring satisfies the equation $x^n=x$ for a fixed $n$? – Pedro Jun 11 '14 at 22:44
  • @user26857 You should post that as an answer! – Bruno Joyal Jun 11 '14 at 22:45
  • @PedroTamaroff yes – Hippalectryon Jun 11 '14 at 22:46
  • @user26857 I do not know anything about it, could you give me some more details ? – Hippalectryon Jun 11 '14 at 22:47
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    Dear Hyppalectryon, Jacobson's result has been discussed a bit already on MO (see e.g. this question and the links there) and on MSE (e.g. this question deals with possible extensions, and this question is the case $n = 3$). I don't think the various links answer your question, but they may help to provide some background and context. Regards, – Matt E Jun 11 '14 at 22:56
  • This should be closed as a dupe of the suggested dupe. There is already a nice answer that covers this with Jacobson's theorem. We can't just keep reporting these links and referring to the special cases forever, as it seems we have been doing until now. – rschwieb Jun 12 '14 at 00:44
  • @rschwieb But there does not seem to be an answer that actually states and proves this for general $n$. –  Jun 12 '14 at 01:20
  • @wordsthatendingry I thought the point of http://math.stackexchange.com/a/360980/29335 is that it can be adapted to any $n$(?) Of course if you felt up to proving Jacobson's theorem as a solution, I'd be more than happy to use that instead. – rschwieb Jun 12 '14 at 01:54

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