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Let $R$ be the localization of a ring at a height $1$ prime ideal. Is it always a valuation ring, even if it is no discrete valuation ring?

Edit: Another related question: Let $X/k$ be a smooth variety and $x \in X$ a codimension $1$ point. Is $\mathcal{O}_{X,x}$ a DVR?

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Valuation rings are integrally closed, but there is no reason to think that the local rings of dimension $1$ are so. –  user26857 Jan 21 '13 at 15:30
    
@user5262: yes to the edited question. See Hagen's answer. –  user18119 Jan 29 '13 at 15:32
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-1 for editing your question one week later. The right thing would have been to post it as another question. (Of course, during this time you could have said thanks to those who bothered to answer you, but this is another story.) –  user26857 Jan 29 '13 at 17:05

2 Answers 2

A concrete counterexample is the following: $R=K[X^2,X^3]$. Then $\dim R=1$ and $R$ is not integrally closed (why?). If all the localizations of $R$ are valuation rings, in particular integrally closed, then $R=\cap_{\mathfrak p\in\operatorname{Spec(R)-\{(0)\}}} R_{\mathfrak p}$ is integrally closed, a contradiction.

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This is a good counter example. –  Bombyx mori Jan 21 '13 at 17:41

In general the answer is no.

The localization of a noetherian ring $R$ at a height-1-prime is a (discrete) valuation ring, if $R$ is normal.

H

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