$\newcommand{\alg}{\overline{\mathbb{Z}}}$ Consider the integral closure $\alg$ of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$. My question is:
Is there a prime ideal $P \subset \overline{\mathbb{Z}}$ such that $\bigcap\limits_{n \geq 1} P^n \neq (0)$ ?
The property $\bigcap\limits_{n \geq 1} P^n = (0)$ holds in any noetherian domain (see Zariski–Samuel, Commutative algebra, volume 1, Chap. IV, §7, corollary 1, p. 216) but $\alg$ is known to be a non-noetherian ring. However, it is a Bézout domain ; in particular it is Prüfer, so apparently such an intersection is always a prime ideal.
Thank you!
Thanks to this helpful comment, I can provide an answer.
Proposition. Let $K$ be any algebraically closed field and $R \subset K$ be any subring. Let $A$ be the integral closure of $R$ in $K$ (we could just take an integrally closed subring $A \subset K$). Then every prime ideal $P$ of $A$ satisfies $P^n = P$ for all $n \geq 1$. In particular, $\bigcap_{n \geq 1} P^n \neq (0)$ iff $P\neq(0)$.
Proof. Clearly, $P^n \subseteq P$ holds. For the reverse inclusion, pick $x \in P$. The polynomial $T^n - x$ has a root $y$ in $A$. Then $y^n = x \in P$ implies $y \in P$ since $P$ is a prime ideal. We get $x = y^n \in P^n$, as desired. $\blacksquare$
We see that in our case $A = \overline{\Bbb Z}$ is a Prüfer domain but not a Dedekind domain. Typically, the localizations $A_P$ are valuation rings, but not DVR (since $P^n=P \leq A$, one cannot just have an integer-valued valuation, as in the case of Dedekind domains...).