An example of a commutative ring with an annihilator condition I am looking for an example of  a non-zero commutative ring $R $ with identity such that $R $ has no idempotent elements, and there exists a non-nilpotent element $s\in R $ such that for each $n\in\mathbb{ N }$ there exists  $r\in R$ with $r (r-1)s=0$,  $rs^n\not=0 $ and $(r-1)s^n\not=0$.
 A: For this type of problem, I find that free algebras modulo the relations you want often suffice.  That is the case here.  Take $R=\mathbb{Z}\langle r,s : sr=rs,r^2s=rs\rangle$.  We check that
$$
r^2(sr)=r^3s=r^2s=rs \text{ and } (r^2s)r=rsr=r^2s=rs
$$
as well as
$$
s(r^2s)=srs=rs^2 \text{ and } (sr)rs=rsrs=r^2s^2=rs^2.
$$
So the two overlaps between the two relations both resolve to the same expression, and so Bergman's Diamond Lemma tells us that we can write elements of $R$ in a unique reduced form as
$$
t=f(r) + sg(s) + rsh(s)
$$
where $f(x),g(x),h(x)\in \mathbb{Z}[x]$ are polynomials.  We achieve this normal form by repeatedly applying the two rules $sr\mapsto rs$ and $r^2s\mapsto rs$ to simplify.
Let us now show that there are no nontrivial idempotents.  If we were to have $t^2=t$, then $f(r)^2=f(r)$ (work modulo the ideal generated by $s$), and so $f(r)\in \{0,1\}$ (since those are the only idempotents in $\mathbb{Z}[r]$).  After replacing $t$ by $1-t$ as necessary, we may assume $f(r)=0$.  Now, $t^2=s^2(g(s)+rh(s))^2$ has degree (in $s$) larger than $t$, unless $g(s)=h(s)=0$.   Thus the only idempotents are $0,1$.
It is also clear that $rs^n\neq 0$ and $rs^n\neq s^n$, from the normal form, as desired.
