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Take an integral, proper variety $X$ over $k$ with function field $k(X)$. Let $A$ be a DVR containing $k$ having field of fractions $k(X)$. Take $P \in X$. Does there always exist an injection $\mathcal{O}_{X,P} \to A$?

EDIT: I changed the question a very little bit (replaced $\subseteq$ by the existence of an injection).

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I would say the answer is no. Suppose we take a smooth irreducible curve over $k$. Then the local rings are DVR's containing $k$, but they don't contain each other.

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It sounds right, but what's wrong if I directly apply valuative criterion in the original question? – Soarer Oct 5 '11 at 1:26
Dear Soarer, it is a nice idea to apply the valuative criterion for properness. That criterion will tell you that there is indeed a point $Q\in X$ and a morphism $\mathcal O_{X,Q} \to A$, but there is no reason why $Q$ should equal $P$. – Georges Elencwajg Oct 5 '11 at 9:39
In the valuative criterion you can select the point of $X$ to which the generic point of $\mathrm{Spec}(A)$ is mapped to, but not the point $P$ to which the closed point of $\mathrm{Spec}(A)$ is mapped to. The valuative criterion only ensures the existence of such a point $P$ (that makes the relevant diagram commutative). – Hagen Knaf Oct 5 '11 at 9:41
So is it true or false? I'm terribly confused... I'm reading a text which seems to suggest that it's true, but I don't see why. – Evariste Oct 5 '11 at 10:41
@Georges, Thanks! Now I see where my confusion was: I was somehow thinking interchangeably $Spec O_{X,P}$ and $X$ in valuative criterion, without noticing that $Spec O_{X,P}$ is not of finite type over $k$, so in fact I couldn't replace $X$ with $Spec O_{X,P}$. Thanks! – Soarer Oct 5 '11 at 17:09

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