A weak version of Chow’s lemma

Suppose that X is a separated algebraic space of finite type over a ring A. Let W be an affine scheme and let f : W —> X be a surjective \’etale morphism. There exists an integer d such that all geometric fibres of f have ≤ d points. Picking d minimal we get a nonempty open U ⊂ X such that f^{-1}(U) —> U is finite etale of degree d. Let

V ⊂ W x_X W x_X … x_X W (d factors)

be the complement of all the diagonals. Choose an open immersion W ⊂ Y with Y projective over A (this is possible as W is of finite type over A). Let

Z ⊂ Y x_A Y x_A … x_A Y (d factors)

be the scheme theoretic closure of V. We obtain d morphisms g_i : Z —> Y. Then we consider

X’ = ⋃ g_i^{-1}(W) ⊂ Z

Claim: The morphism X’ —> X (coming from the g_i and W —> X) is projective.

The image of X’ —> X is closed and contains the open U. Replace X by X ∖ U and W by the complement of the inverse image of U; this decreases the integer d, so we can use induction. In this way we obtain the following weak version of Chow’s lemma: For X separated and of finite type over a ring A there exists a proper surjective morphism X’ —> X with X’ a quasi-projective scheme over A.

But this post is really about the proof of the claim. This claim comes up in the proof of Chow’s lemma in Mumford’s red book as well. I’ve never been able to see clearly why it holds, but now I think I have a good way to think about it.

It suffices to prove that X’ —> X is proper. To do this we may use the valuative criterion for properness. Since V is scheme theoretically dense in X’ it suffices to check liftability to X’ for diagrams

Spec(K) -------> V
  |              |
  v              v
Spec(R)------->  X

where R is a valuation ring with fraction field K. Note that the top horizontal map is given by d distinct K-valued points w_1, …, w_d of W and in fact this is a complete set of inverse images of the point x in X(K) coming from the diagram. OK, and now, since W —> X is surjective, we can, after possibly replacing R by an extension of valuation rings, lift the morphism Spec(R) —> X to a morphism w : Spec(R) —> W. Then since w_1, …, w_d is a complete collection of inverse images of x we see that w|_{Spec(K)} is equal to one of them, say w_i. Thus we see that we get a diagram

Spec(K) -------> Z
  |              | g_i
  v              v
Spec(R) --w--->  Y

and we can lift this to z : Spec(R) —> Z as g_i is projective. The image of z is in g_i^{-1}(W) ⊂ X’ and we win.

Update 10/23/12: This has now been added to the Stacks project. See Lemma Tag 089J.