Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Greetings,

(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?

Robert

share|improve this question
1  
I guess your two tags should be just one tag 'algebraic geometry'... –  Matthieu Romagny Jun 23 '11 at 21:26
    
See mathoverflow.net/questions/493/…. –  Matthieu Romagny Jun 24 '11 at 18:44

1 Answer 1

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 $Y$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.