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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 $R$ ? If that is not true in general, then what if we also assume $R$ has finite dimension, or say very low dimension (say 1) ?

The Noetherian case obvious from Krull Intersection theorem, and so is the case when $R$ is s Prufer domain , due to Corollary 2.4 (a) of https://cms.math.ca/openaccess/cjm/v18/cjm1966v18.1024-1030.pdf . So any possible counterexample would have to be non-Noetherian and non-Prufer.

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    $\begingroup$ @Fred Rohrer : you claim to have a finite dimensional Valuation ring counterexample ..? That would be great ... it would say that we can't do any much better without Noetherian hypothesis ... $\endgroup$
    – user111492
    Jun 26, 2018 at 8:57
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    $\begingroup$ Dear @misao, the claim in the question you mention is not true. See the remark after the proof of Theorem 2 in D. D. Anderson, The Krull intersection theorem, Pacific J. Math. 57 (1975), 11-14. $\endgroup$ Jun 26, 2018 at 12:27
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    $\begingroup$ Check "The Krull intersection theorem II" by D. D. Anderson, you'll find a couple of positive results. In particular if $P$ is principal, or invertible, or locally principal. Theorem 3.1 tells you that the intersection is a prime ideal whenever $R$ is a valuation domain. $\endgroup$
    – Luc Guyot
    Jun 26, 2018 at 15:25
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    $\begingroup$ @FredRohrer: in the remark to Theorem 2 of Anderson's first paper on Krull Intersection, the ideal $I$ itself is not necessarily taken to be prime ... but here I start with a prime ideal $P$ to begin with ... $\endgroup$
    – user111492
    Jun 26, 2018 at 16:55
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    $\begingroup$ @FredRohrer: As it appears, the claim is indeed true for Prufer domains ... Corollary 2.4 (a) in cms.math.ca/openaccess/cjm/v18/cjm1966v18.1024-1030.pdf $\endgroup$
    – user111492
    Jun 26, 2018 at 18:26

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