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Let $\iota \colon A \subset B$ be a finite integral extension between domains. Suppose that $A$ is UFD, so $A$ is an integrally closed domain. $A$ and $B$ may not be noetherian ring.

Choose a prime ideal ${\frak p}$ of $A$ such that ${\mathrm{ht}}(\frak p) < \infty$. By "lying-over" theorem, there are finitely many prime ideals ${\frak P}_1,\ldots,{\frak P}_n$ of $B$ such that ${\frak P}_i \cap A = {\frak p}$ for $1 \leq i \leq n$.

Q. Does the following equality holds$\colon$ $\sqrt{{\frak p}B} = {\frak P}_1 \cap {\frak P}_2 \cap \ldots \cap {\frak P}_n$?

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    $\begingroup$ Doesn't this follow from the usual going-down theorem and the characterization that $\sqrt{\mathfrak p B}$ is the intersection of all prime ideals of $B$ containing $\mathfrak p$? i.e. the intersection of all primes of $B$ lying over a prime containing $\mathfrak p$, i.e. by going-down the intersection of all primes of $B$ containing a prime lying over $A$, QED. $\endgroup$ – Will Sawin Oct 29 '16 at 20:41
  • $\begingroup$ Dear Will Sawin, very many thanks!! Pierre $\endgroup$ – Pierre MATSUMI Oct 30 '16 at 4:53

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