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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. – YACP Jan 21 at 15:30
@user5262: yes to the edited question. See Hagen's answer. – QiL'8 Jan 29 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.) – YACP Jan 29 at 17:05

2 Answers

up vote 3 down vote

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. – user32240 Jan 21 at 17:41
up vote 2 down vote

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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