One can show (see here) that for a Noetherian integral domain, the following are equivalent:
- $R$ is integrally closed; and
- $\text{End}_R(I) = R$ for all nonzero ideals $I$.
Is the following condition also equivalent?
- $\text{End}_R(\mathfrak{p}) = R$ for all nonzero prime ideals $\mathfrak{p}$.
The proof of (2) $\Rightarrow$ (1) that I know (and hopefully this is correct!) is as follows:
We show that $R$ is completely integrally closed, which is equivalent to integrally closed when $R$ is Noetherian. Let $Q$ be the fraction field of $R$: suppose that $q \in Q$ is almost integral, so there exists $0 \neq a \in R$ such that $a q^n \in R$ for all $n \geqslant 0$. Set $T = R[q]$, and $I = \{ x \in R \mid xT \subseteq R \}$, which is nonzero since $a \in I$. Then we have a chain of inclusions $$ R \subseteq T \subseteq \text{End}_R(I) = R $$ so that $R = T$, forcing $q \in R$, which means that $R$ is completely integrally closed.
Assuming this isn't flawed, if I can show that the ideal $I$ is prime, then I'm done. However, I can't see why this should be the case.