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Assume that $R$ is a ring and $r-r^2$ is nilpotent for an element $r\in R$. If $r$ is not nilpotent, then $R$ has a nonzero idempotent.

user108209
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  • Take any nilpotent element $s$ in a nontrivial ring. Then $r:=1-s$ is not nilpotent. Then $r-r^2=r(1-r)=rs$ is nilpotent, but $r$ is not. Therefore the claim seems to be false. We need that $r$ and $1-r$ are not nilpotent. – Martin Brandenburg Dec 21 '13 at 14:41
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    Probably you also want the idempotent $\neq 1$. Otherwise... – Julien Dec 21 '13 at 14:41
  • @julien I think he or she forgot $\neq 1$ – Bobby Dec 21 '13 at 14:54
  • Dear @user108209 Can you confirm whether or not you want to assume a multiplicative identity, and whether or not your question says "idempotent other than zero and 1"? And if you are assuming commutativity without telling us? – rschwieb Dec 21 '13 at 22:00

2 Answers2

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Let $(r-r^2)^n=0$. Then we can conclude that $r^n=r^{n+1}f(r)$. It is not hard to see that $(r^{n}f(r)^n)^2=r^{2n}f(r)^{2n}$. Now, we compute $r^n$. Note that $r$ and $f(r)$ commuate

$$r^n=r^{n+1}f(r)=rf(r)r^n=rf(r)r^{n+1}f(r)=r^2f^2(r)r^n=r^{n+2}f^2(r)$$

We can repeat procedure and we can find $r^n=r^{2n}f^n(r)$. Thus we have $r^{2n}f^{2n}(r)=r^{2n}f^n(r)f^n(r)=r^nf^n(r)$ and the idempotent element is $r^nf^n(r)$.

Bobby
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By considering the subring generated by $r$, we may assume that $R$ is commutative. But then we can give a "geometric" solution as follows.

If $\mathfrak{p} \subseteq R$ is a prime ideal, then we have $r(1-r) \in \mathfrak{p}$, hence $r \in \mathfrak{p}$ or $1-r \in \mathfrak{p}$. Both $r$ and $1-r$ cannot be contained in $\mathfrak{p}$. This shows that $\mathrm{Spec}(R)$ is the disjoint union of the closed subsets $V(r)$ and $V(1-r)$. Since $r$ and $1-r$ are not nilpotent (see my comment), these closed subsets are proper subsets. It follows that $\mathrm{Spec}(R)$ is disconnected and hence has a nontrivial idempotent element.

Community
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Martin Brandenburg
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  • What if $R$ has no unit as the question strongly suggests? –  Dec 21 '13 at 15:46
  • I assumed that $R$ has a unit. Why the downvote? – Martin Brandenburg Dec 21 '13 at 18:17
  • I just interpreted the question a little bit differently. I thought $R$ is a ring in the usual sense (hence with unit) and that one wants to have a nontrivial idempotent (i.e. $\neq 0,1$). Therefore I also added the stronger assumption that $r$ and $1-r$ are not nilpotent. – Martin Brandenburg Dec 21 '13 at 18:48
  • @MartinBrandenburg But it does seem strange to choose an interpretation that requires two extra assumptions! (Although I agree that the downvote is excessive.) – Derek Holt Dec 21 '13 at 19:04