What I had in mind was something like the following:

X is separated/proper iff for all curves C and all maps f : C \ c -> X, f extends to C in at most/exactly one way.

Is there a good reason why this cannot possibly be true?

Here X denotes a reduced scheme of finite type of a field k (I guess people usually call this prevariety). I am mostly interested in the case where k is algebraically closed.

  • $\begingroup$ I don't know any algebraic geometry, but I expect that the discussion at: mathoverflow.net/questions/493/… may have some insight. $\endgroup$ – Theo Johnson-Freyd Oct 17 '09 at 3:43
  • 1
    $\begingroup$ Doesn't the definition of a variety require it to be separated? $\endgroup$ – user332 Oct 17 '09 at 5:21
  • $\begingroup$ A variety is usually assumed to be reduced, irreducible (sometimes), separated (I assume you don't want this), and finite type over a field k. Are there any other conditions you're imposing? Supposing we know what a variety is, it's clear that if X is separated/proper, then it satisfies your condition (it's weaker than the usual valuative criteria). So it sounds like your question is, "is it enough to check DVRs whose residue field is k and whose field of fractions has transcendence degree 1 over k?" Do I have that right? $\endgroup$ – Anton Geraschenko Oct 17 '09 at 6:36
  • $\begingroup$ ... of course you're also allowing the residue field to be a finite extension of k. $\endgroup$ – Anton Geraschenko Oct 17 '09 at 6:48

If you make the statement

Fix an algebraically closed base field k and let X be a scheme of finite type over k. Then X/k is proper iff for all smooth quasi-projective curves C/k and all maps f: C\c -> X then f extends uniquely to f': C -> X.

it seems true to me. This should basically comes down to the fact that one can classify birational equivalence classes of curves over k in terms of the 'abstract curves' coming from all possible discrete valuations on dimension 1 function fields K/k. So using the fact that in this situation it is sufficient to check the valuative criterion on DVRs it seems like it should not be so hard to see the equivalence. The argument I had in mind is as follows:

The valuative criterion implies the statement above. For the converse it is sufficient to show that any f: Spec K -> X lifts to an open subset of the curve C_K determined by K/k by which I mean the unique nonsingular projective curve in the birational class corresponding to K/k (the distinguished point in the complement we think of as being removed is uniquely determined by the discrete valuation we pick so that is no problem). If f just hits a closed point we can just collapse C_K via the structure map to Spec k and this is fine since there is nothing to lift. If not, we hit a dimension 1 point whose closure with the reduced induced structure determines some curve C' birational to the corresponding C_K'. The map K' -> K induces a dominant morphism g: C_K -> C_K'. We thus get a map by taking a common (up to isomorphism) open in C_K' and C' , taking its preimage U in C_K and considering
U -> C' -> X which is the desired lift of f to the quasi-projective curve U.

I think there is also a slicker argument using the categorical definition of finite presentation.

For the same reasons this works for checking separatedness when one makes the obvious modifications to the statement.

Over other bases I am not sure at the moment... I can't remember if the birational classification is still that simple (although some people implicitly mean by variety that everything is over some fixed alg. closed base).

In the more general case (if your definition of variety doesn't include a finite type over a noetherian base hypothesis) where one needs non-noetherian valuation rings I think this interpretation is false - non-noetherian valuation rings can have arbitrary dimension.

| cite | improve this answer | |
  • $\begingroup$ This is the kind of answer I was looking for... do you know a reference? $\endgroup$ – Joel Kamnitzer Oct 17 '09 at 12:43
  • $\begingroup$ I don't know of a place where it is explicitly written like this. A reference for the birational models of curves corresponding to finitely generated transcendence degree 1 extensions of the base field is Hartshorne I.6. $\endgroup$ – Greg Stevenson Oct 17 '09 at 12:55
  • $\begingroup$ Added the argument I had in mind. $\endgroup$ – Greg Stevenson Oct 18 '09 at 2:25

I don't completely understand what kind of an answer you're looking for, but here's an observation.

It is not enough to consider smooth projective curves C over the ground field k, but for a dopey reason: you might need to remove more than one point of C to get any interesting maps to X.

For example, Let X be the complement of the coordinate axes in A2. We may regard X as an open subscheme of P2. Any rational map from C to X would extend to an actual map from C to P2, but any curve in P2 intersects the three coordinate lines in at least two points. So the only maps from C\{c} to X are constant. So this X satisfies the curvy version of the valuative criterion (were you remove a single point from a projective curve) for properness, but it isn't proper. You can double a point on X to get something non-separated which satisfies the curvy separatedness criterion, but isn't separated.

| cite | improve this answer | |
  • $\begingroup$ Can't you just allow incomplete curves to avoid this though? $\endgroup$ – Eric Wofsey Oct 17 '09 at 7:13
  • $\begingroup$ You are right, I fixed my answer to reflect this... I got a bit carried away translating from the generic point to the whole curve. $\endgroup$ – Greg Stevenson Oct 17 '09 at 7:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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