Can the valuative criteria be checked "on a dense open"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:31:09Z http://mathoverflow.net/feeds/question/493 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/493/can-the-valuative-criteria-be-checked-on-a-dense-open Can the valuative criteria be checked "on a dense open"? Anton Geraschenko 2009-10-14T15:12:34Z 2009-10-15T10:06:21Z <p>The valuative criterion for separatedness (resp. properness) says that a noetherian scheme X is separated (resp. proper) if and only if</p> <blockquote> <p>for any DVR R, with fraction field K, any map Spec(K)&rarr;X extends in at most one way (resp. extends uniquely) to a map Spec(R)&rarr;X.</p> </blockquote> <p>If U&sube;X is a dense open subscheme, is it sufficient to check the valuative criteria only in the cases where the map Spec(K)&rarr;X lands in U?</p> <p>Intuitively, the valuative criteria are checking if it is possible to fill in a "missing point" on a curve on X (and if it is possible to fill it in in multiple ways). If there is a curve in Z=X\U with a missing point, it feels like it should be possible to find a curve in U that should have the same limit point. Also, I think it's common to verify that a compactified moduli space is separated by checking the valuative criterion using only families in the original moduli space (not "on the boundary").</p> http://mathoverflow.net/questions/493/can-the-valuative-criteria-be-checked-on-a-dense-open/526#526 Answer by Anton Geraschenko for Can the valuative criteria be checked "on a dense open"? Anton Geraschenko 2009-10-14T22:52:56Z 2009-10-14T22:52:56Z <p>Here's a proof by intimidation. On page 103 of Deligne and Mumford's <a href="http://archive.numdam.org/article/PMIHES%5F1969%5F%5F36%5F%5F75%5F0.pdf" rel="nofollow">The irreducibility of the space of curves of a given genus</a>, just after Theorem 4.19 (the valuative criterion for properness), they say say the answer is "yes". Their notation doesn't agree with mine, and they allow a field extension because they're dealing with algebraic stacks, but I'm pretty sure everything lines up:</p> <blockquote> <p>To prove a given f is proper, it suffices to verify the above criterion under the additional hypothesis that V is complete and has an algebraically closed residue field. Further, given a dense open subset U of T, it is enough to test only g's which factor through U.</p> </blockquote> <p>The bit about being about to being able to restrict to compete DVRs with algebraically closed residue field is EGA II, Remark 7.3.9 (i), as <a href="http://mathoverflow.net/questions/79/can-the-valuative-criteria-for-separatedness-properness-be-checked-formally/93#93" rel="nofollow">Jarod points out</a>. But I couldn't find any justification of the last part.</p> http://mathoverflow.net/questions/493/can-the-valuative-criteria-be-checked-on-a-dense-open/585#585 Answer by Jason Starr for Can the valuative criteria be checked "on a dense open"? Jason Starr 2009-10-15T10:06:21Z 2009-10-15T10:06:21Z <p>If you have a separated, finite type morphism of schemes which are finite type over a field (some more general bases are also valid), then you can get this reduction from Chow's lemma: every finite type, separated scheme over a field is the target of a projective, birational morphism whose domain is quasi-projective. </p> <p>First of all, properness is local on the target. Thus, you may replace your original target with a covering by open affines. So assume that the domain and target are both separated. </p> <p>Next apply Chow's lemma to the domain. This gives a new morphism with the same target as the original morphism, but the domain of the new morphism is quasi-projective. It is easy to see that the new morphism is proper if and only if the original morphism is proper. Thus you are reduced to the case when the domain is quasi-projective.</p> <p>Since the domain is quasi-projective it admits a (dense) open immersion into a projective scheme. So consider the "diagonal" map from the quasi-projective domain into the product of the target and this projective scheme. The diagonal map is certainly a locally closed immersion. In fact the original morphism is proper if and only if this diagonal map is a closed immersion. In other words, the original map is proper if and only if the image of the diagonal map equals its closure, i.e., if and only if the boundary is empty. </p> <p>But this can easily be checked with valuations / curves of the type you are discussing. If you prefer valuations, choose any irreducible component of the image of the diagonal map whose closure intersects the boundary. Next normalize the closure of that component, and form the inverse image of the boundary in that normalization. Next blow up this inverse image to produce an exceptional Cartier divisor. Finally, take the stalk of the blowing up at any generic point of the exceptional divisor to get a DVR of the type you want.</p> <p>If you prefer curves, do the same normalization and blowing up. You can arrange that everything is still quasi-projective. Now intersect with generic hyperplanes to produce an integral curve which intersects both the boundary and the strict transform of whatever dense open set you originally specified.</p>