# Chow's lemma

Not to be confused with Chow's theorem.

Chow's lemma, named after Wei-Liang Chow, is one of the foundational results in algebraic geometry. It roughly says that a proper morphism is fairly close to being a projective morphism. More precisely, a version of it states the following:[1]

If is a scheme that is proper over a noetherian base , then there exists a projective -scheme and a surjective -morphism that induces an isomorphism for some dense open .

## Proof

The proof here is a standard one (cf. EGA II, 5.6.1).

It is easy to reduce to the case when is irreducible, as follows. is noetherian since it is of finite type over a noetherian base. Then it's also topologically noetherian, and consists of a finite number of irreducible components , which are each proper over (because they're closed immersions in the scheme which is proper over ). If, within each of these irreducible components, there exists a dense open , then we can take . It is not hard to see that each of the disjoint pieces are dense in their respective , so the full set is dense in . In addition, it's clear that we can similarly find a morphism which satisfies the density condition.

Having reduced the problem, we now assume is irreducible. We recall that it must also be noetherian. Thus, we can find a finite open affine cover . are quasi-projective over ; there are open immersions over , into some projective -schemes . Put . is nonempty since is irreducible. Let

be given by 's restricted to over . Let

be given by and over . is then an immersion; thus, it factors as an open immersion followed by a closed immersion . Let be the immersion followed by the projection. We claim induces ; for that, it is enough to show . But this means that is closed in . factorizes as . is separated over and so the graph morphism is a closed immersion. This proves our contention.

It remains to show is projective over . Let be the closed immersion followed by the projection. Showing that is a closed immersion shows is projective over . This can be checked locally. Identifying with its image in we suppress from our notation.

Let where . We claim are an open cover of . This would follow from as sets. This in turn follows from on as functions on the underlying topological space. Thus it is enough to show that for each the map , denoted by , is a closed immersion (since the property of being a closed immersion is local on the base).

Fix . Let be the graph of . It is a closed subscheme of since is separated over . Let be the projections. We claim that factors through , which would imply is a closed immersion. But for we have:

The last equality holds and thus there is that satisfies the first equality. This proves our claim.