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Let $R$ be an integral domain with every prime ideal having finite height. Then is $\bigcap_{n>1} P^n$ a prime ideal of $R$ for every prime ideal $P$ of $R$ ?

The Noetherian case obvious from Krull Intersection theorem, so any possible counterexample would have to be non-Noetherian.

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David Speyer's example of $$R = \bigcup_{n=1}^{\infty} k\left[x,\ y,\ x^{1/n!} y^{1/n!} \right]$$ for any field $k$ also works for this question. The ideal $P=(x, y, x y, x^{1/2} y^{1/2}, x^{1/3} y^{1/3}, x^{1/4} y^{1/4}, \cdots )$ is prime but $\bigcap P^n$ is not prime (see the linked answer for details).

Moreover, I claim every prime ideal in $R$ has height at most $2$. Indeed, suppose $Q_0\subset Q_1\subset Q_2\subset Q_3$ is a chain of prime ideals in $R$. There is then some $n$ such that this chain of inclusions remains strict when restricted to $R_n=k[x,y,x^{1/n!}y^{1/n!}]$ (since $R$ is the direct limit of these subrings $R_n$). But $R_n$ has dimension $2$ (it is an integral extension of $k[x,y]$), so this is impossible.

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