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here the user proved that if in a ring we had $x^4=x$ then the idempotent element of the ring is central.

can we make a generalization and say in any ring that satisfies this condition for any $n$ the idempotent element is central?

"if $x^n=x$ then $e$ the idempotent element is central."

what can we say about the nilpotent element?

reiji sakamaki
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  • You can show that $Nil(R)={0}$. ($R$ is the ring). Also I think this wouldn't work for any $n$, but $n>1$ – Hoda Bibo Sep 27 '20 at 16:25
  • how did you get this? – reiji sakamaki Sep 27 '20 at 16:27
  • "can we make a generalization and say in any ring that satisfies this condition for any the idempotent element is central". The phrase "any ring that satisfies this condition for any n" is doubly ambiguous. When you say "the ring satisfies" that condition, you mean every element? When you say "any n" do you mean there exists an $n$ such that all elements satisfy the condition for that n? Or you mean elements can have different $n$'s? – rschwieb Sep 28 '20 at 12:58
  • At any rate, if you mean that each element satisfies $x^n=x$ for some $n>1$, possibly not the same $n$ for all elements, then the ring is commutative, so everything is central. – rschwieb Sep 28 '20 at 12:58

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