Is there a version of the valuative criteria for separateness/properness for varieties?

Question

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.

Answer 1

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.

Answer 2

I would like to answer my own question after nearly 12 years. I am teaching algebraic geometry this year and in the process I found the statement that I wanted as Corollary 15.10 in Gortz-Wedhorn (an excellent book).

Theorem Let $Y $ be a locally Noetherian scheme and let $ f : X \rightarrow Y $ be a morphism of finite type. Then $ f $ is proper if and only if for all normal absolute curves $C $ which are schemes over $Y $, for every closed point $ c \in C$, and for every $ Y $-morphism $ g : C \setminus \{c \} \rightarrow X $, there exists a morphism $ \bar g : C \rightarrow X $ such that $ \bar g|_{C \setminus \{c\}} = g $.

When I asked my question above, I was wondering if there was a more geometric version of the valuative criterion and that is provided by this theorem. In the special case where $ Y = Spec k $, then $ C $ will be a regular curve over $ k $.

Answer 3

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.