First off, I’d like to make a correction to the definition of an abstract variety: we’re going to need to assume both that is irreducible and that it is covered by *finitely many* affine varieties. I had both of these conditions in the back of my head when I wrote that, and fortunately the issues were brought up in the comments.

Anyway, today we’re going to take out abstract varieties and pull out the property of projective varieties that makes them so ridiculously useful: completeness.

We start with the definition: a variety is *complete* if the projection takes closed sets to closed sets for every variety

This property is, in the case of topological spaces, equivalent to being compact, so we want to think of complete varieties as compact topological spaces. So let’s take a moment and figure out which quasi-projective varieties are complete.

First, we check that projective varieties are complete. To see this, we first reduce to just checking , because closed subsets of a complete variety will be complete. As being closed is a local property, we can assume that is affine, so we have merely to check that for all affine , we have is a closed map. Even better, we can assume that . Now let be closed. Then it is defined by polynomials which are homogeneous in the variables on , . So the claim is that the set of with for some is closed. So we want to look at the locus where , the ideal generated by all monomials of degree . Each of these sets is closed, and so their intersection is. Thus, the map is closed.

So we can now use this fact to prove that the regular functions on a complete variety (and thus a projective variety) are all constant. Let be a regular function. We extend to a function . This defines a graph . The image of this under the projection to is closed, so the image of the regular function must be closed. It doesn’t contain the point at infinity, by construction, so it is a closed subset of with is still closed in . Thus, it is a finite set. And so for finitely many . This contradicts the irreducibility of , and so it must be a single point, so is constant on a complete variety.

This immediately shows us that affine and quasi-affine varieties cannot be complete. As for open subsets of projective varieties, we can see that they aren’t complete by taking a closed subset which is not closed in the closure of the quasi-projective variety, and doing a little bit of work that we won’t both with.

So of our old varieties, the projective ones are characterized by being complete. But the category of abstract varieties is bigger. There are now varieties which are complete but nonprojective, though examples are nontrivial.

To finish off, we’ll prove the following theorem, called Chow’s Lemma, which says the complete varieties aren’t all that different from projective varieties.

**Chow’s Lemma**: For any complete irreducible variety , there exists a projective variety and a surjective birational morphism .

Before proving it, we first say that that a morphism is *birational* if there exists a rational map such that, where defined, the composition of the two maps is the identity.

*Proof*: Take a finite affine cover . Define to be the closure of the ‘s in projective space. Then is projective.

Let and define to be the inclusion, and to be the composition of these inclusions. So we can now define by . Finally, we define to be the closure of in . Taking the projection , we get a map .

We will prove that this map is birational. To do this, we just need to check that , because on we have is the identity. In fact, for this we must merely show that , which is that the graph of . The morphism is surjective, then, because contains , which is dense in , and the map must have closed image.

So what’s left? We still need to show that is actually projective. We note that was projective, and look at the second projection , and restrict to . We will show that this is an embedding. Because this is a local property, we must only find open sets such that the union of the contains and such that the restrictions of to the are all closed embeddings.

To this end, we set to be the preimage of the th projection of . Then certainly cover .

All that is left is to show that is a closed embedding. Now, we note that , where . So then . Take to be the graph of the morphism which is the composite . Then is a closed subset of this and its projection onto is an isomorphism. Additionally, , and as is closed, we have closed in , so the map is a closed embedding. QED.

I would really appreciate it if you would set your RSS to “full-text” so that I can read your entire posts, instead of just the first few lines, in my reader.

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You argue that the cover . But couldn’t there be elements of that map to elements of ?

Thanks, and sorry for the late question.

Ok, so the proof I gave is following Shafarevich’s Basic Algebraic Geometry 2 (page 69-71) and now I’m feeling rather guilty for glossing over the fact that that’s a covering, because that’s one of the tricky technical points (though I was explicitly trying to not be too technical in this series…) and rather than try to fit the details into a comment, or do a large edit to the post, I’ll just link here: Chow’s Lemma

For the third time in the past year, I have found a fact discussed here that I was having trouble understanding…

Thanks again!