Let $k$ be a field of characteristic zero. Let $R$ be a (commutative) $k$-algebra, and $f$ a $k$-algebra endomorphism of $R$.
What can be said about $I:=\cap_{i \in \mathbb{N}} f^i(R)$? Is it possible to determine when $I=k$?
Remarks:
(i) Of course, if $f$ happens to be surjective, then $I=R$.
(ii) We will require that the dimension of $R$ as a $k$-vector space is infinite. (If $\dim_k(R) = n < \infty$, then from $\dim_k(R) \geq \dim_k(f(R)) \geq \dim_k(f^2(R)) \ldots$ and pigeonhole principle, there exist $i<j$ such that $\dim_k(f^i(R))=\dim_k(f^j(R)) \in \{1,\ldots,n\}$. Then, since $f^j(R) \subseteq f^i(R)$, we get that $f^j(R) = f^i(R)$, etc.).
Special cases: I guess that my question is too general, so I do not mind to concentrate on $R=k[t]$ or $R=k[x,y]$, though I prefer a more general $R$, for example, a Noetherian ring.
A somewhat relevant question is this question.
Thank you very much!