Does $\text{End}_R(I)=R$ always hold when $R$ is an integrally closed domain?

Question

Let $R$ be a commutative ring with identity, and $I\neq 0$ an ideal of $R$, I'm thinking how to calculate $\text{End}_R(I)$. I have proved that when $R$ is a integral domain, $\text{End}_R(I)=\{r\in\text{Frac}(R)\mid rI\subset I\}$. So I'm curious about if $\text{End}_R(I)=R$ always holds when $R$ is also integrally closed. Can anyone help me with this? Thanks in advance.

Answer 1

For an integral domain $R$ we have $\operatorname{End}_R(I)=I:I$, so the question is when $I:I=R$.

It holds for completely integrally closed domains, or for integrally closed domains and finitely generated ideals.

Set $R=K[X, XY, XY^2, XY^3,...]$, and $I=(X, XY, XY^2, XY^3,...)$. ($R$ is integrally closed, but not completely integrally closed.) Now note that $I:I\ne R$.

Answer 2

This answer is based on user26857, I just make some point clearer. By Nakayama lemma, $I:I=R$ holds for all finitely generated ideals $I$ iff $R$ is integrally closed. And it's also straightforward to check that $I:I=R$ holds for all ideals iff $R$ is completely integrally closed.