0
$\begingroup$

Let $R $ be a commutative ring with 1 and $s $ a nonnilpotent element of $R $ such that for each idempotent $e $ of $R $ there exists a natural number $n_e $ such that either $s^{n_e}e=0$ or $s^{n_e}(1-e)=0$. Can we deduce that for each idempotent $e$ of $R$ either $s^me=0$ or $s^m(1-e)=0$ for a fix natural number $m $?

$\endgroup$
6
$\begingroup$

No. Consider the subring $R\subseteq\prod_{n\geq 1}\mathbb R[x]/(x^n)$ consisting of sequences $(a_n)$ whose constant terms converge to some limit. Let $s=(x,x,x,\dots)$. Then $s$ is nonnilpotent, since $s^n$ has nonzero coefficient in $\mathbb R[x]/(x^{n+1})$.

Now every idempotent $e$ in $R$ is equal to $0$ or $1$ in each coordinate, and is eventually constant. Replacing $e$ with $1-e$ if necessary, assume $e=(e_n)$ satisfies $e_n=0$ for $n>N$. Then $s^Ne=0$.

However, for $e=(1,\dots,1,0,0,\dots)$ with $n$ ones we have $s^{n-1}e\neq 0,s^{n-1}(1-e)\neq 0$, so we cannot pick one exponent for all idempotents.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.