# Annihilator of idempotent elements

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 $$m$$?

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.