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(I suspect this question has nothing to do with the Valuation Criterion of Properness, but I don't know for sure - feel free to modify my title)

This question arises in section 2.4 of Fulton's book on Toric Varieties - in the proof that $\phi_* : X(\Delta') \to X(\Delta)$ is proper iff $\phi^{-1}(|\Delta|)=|\Delta'|$.

Let's say I give you a variety map $f: X \to Y$. To prove it's proper I must take any dvr R with fraction field $K$ and any commutative diagram where one path is $Spec(K) \to Spec(R) \to Y$ and another path is $Spec(K) \to X \to Y$ (the map $X \to Y$ is $f$), and tell you why there exists a unique map $Spec(R) \to X$ that makes both 'triangles' commute. (I don't know how to make pretty diagrams on math overflow, apologies.)

The claim that is made is: Let $U \subseteq X$ be your favorite open subset. If $X$ is irreducible, then to prove the above we may assume that $im(Spec(K) \to X) \subseteq U$.

Will someone please explain why this claim holds?


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I guess your two tags should be just one tag 'algebraic geometry'... – Matthieu Romagny Jun 23 '11 at 21:26

Assume that all diagrams like the one you describe, with $Spec(K)\to X$ mapping into $U$, may be completed with a map $Spec(R)\to X$. You want to prove that $f:X\to Y$ is proper. In order to prove this, you can replace $Y$ be the closed image of $f$ and hence assume that $Y$ is irreductible also.

By Chow's lemma, there exists a projective birational surjective morphism $g:X'\to X$ such that $f\circ g:X'\to Y$ is quasi-projective (see EGA2, thm. 5.6.1). It is easy to see that $X\to Y$ is proper if and only if $X'\to Y$ is proper, so you can replace $X$ by $X'$ and hence assume that $X$ is quasi-projective over $Y$.

Now let $j:X\to P$ be an open dense immersion into a projective $Y$-scheme. Let $x$ be a point of $P$. Since $U$ is dense in $X$ hence also in $P$, there exists a point $y\in U$, a discrete valuation ring $R$ with fraction field $K$, and a morphism $Spec(R)\to P$ mapping the generic point to $y$ and the closed point to $x$ (see EGA2, 7.1.9). By the original assumption, this map extends to a map $Spec(R)\to X$ making everything commutative. But since $X$ is separated, such an extension is unique : thus the image of the closed point, that is $x$, is in $X$. It follows that $j$ is surjective, hence $X=P$ is proper over $Y$.

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