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Let $R$ be a Jacobson ring, namely for each element $a \in R$ there exists an integer $n$, $n＞1$ such that $a ^ n = a$. How can we prove that an idempotent element of $R$ must be a central element of $R$?

KUOW
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  • sorry, I left out $n＞1$. – KUOW Apr 05 '19 at 03:37
  • see https://math.stackexchange.com/questions/67148/if-a3-a-for-all-a-in-a-ring-r-then-r-is-commutative Math Gems's answer. – Idele Apr 05 '19 at 04:35
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    For future reference, "Jacobson ring" is more commonly used to refer to a ring whose prime ideals are intersections of maximal ideals. I thought the rings you are describing were called "periodic", but I have found that actually that may be used for a broader class of rings where $x^n=x^m$ for some $n<m$ depending on $x$. – rschwieb Apr 05 '19 at 13:38

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You can show that $R$ has no nonzero nilpotents. Indeed, if $a^n=0$ and $a^m=a$, then $a=a^{m^k}$ for any $k$, so taking $k$ large enough so that $m^k>n$ we get $a=a^{m^k}=0$.

Now, in a ring with no nonzero nilpotents, every idempotent is central. To see this, let $x\in R$ and $e=e^2$; we want to show the commutator $[e,x]$ is zero. But $[e,ex]^2=0$ (compute it!), so since there are no nilpotents, this shows $ex=exe$. Similarly $[e,xe]^2=0$, so $xe=exe$. Therefore $ex=exe=xe$ as required.

Ehsaan
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