The proof here is a standard one. 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 . 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 where are quasi-projective over There are open immersions over into some projective -schemes Set Then is nonempty since is irreducible. Let be given by 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. Fix and 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.
Additional statements
In the statement of Chow's lemma, if is reduced, irreducible, or integral, we can assume that the same holds for. If both and are irreducible, then is a birational morphism..
