## Complete Varieties and Chow’s Lemma

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 $V$ 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 $X$ is complete if the projection $p:X\times Y\to Y$ takes closed sets to closed sets for every variety $Y.$

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 $\mathbb{P}^n$, because closed subsets of a complete variety will be complete. As being closed is a local property, we can assume that $Y$ is affine, so we have merely to check that for all affine $Y$, we have $\mathbb{P}^n\times Y\to Y$ is a closed map. Even better, we can assume that $Y=\mathbb{A}^m$. Now let $Z\subset \mathbb{P}^n\times\mathbb{A}^m$ be closed. Then it is defined by polynomials which are homogeneous in the variables on $\mathbb{P}^n$, $g_i(u,y)$. So the claim is that the set of $y_0$ with $g_i(u,y_0)=0$ for some $u$ is closed. So we want to look at the locus where $(g_1(u,y_0),\ldots,g_t(u,y_0))\not\supset I_s$, the ideal generated by all monomials of degree $s$. 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 $f:V\to \mathbb{A}^1$ be a regular function. We extend to a function $f:V\to \mathbb{P}^1$. This defines a graph $\Gamma_f=\{(v,a)|f(v)=a\}\subset V\times \mathbb{P}^1$. The image of this under the projection to $\mathbb{P}^1$ 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 $\mathbb{A}^1$ with is still closed in $\mathbb{P}^1$. Thus, it is a finite set. And so $X=\cup f^{-1}(x_i)$ for finitely many $x_i\in\mathbb{A}^1$. This contradicts the irreducibility of $X$, and so it must be a single point, so $f$ 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 $X$, there exists a projective variety $\bar{X}$ and a surjective birational morphism $f:\bar{X}\to X$.

Before proving it, we first say that that a morphism $f:X\to Y$ is birational if there exists a rational map $g:Y\dashrightarrow X$ such that, where defined, the composition of the two maps is the identity.

Proof: Take a finite affine cover $X=\cup U_i$. Define $Y_i$ to be the closure of the $U_i$‘s in projective space. Then $Y=\prod Y_i$ is projective.

Let $U=\cap U_i$ and define $\psi:U\to X$ to be the inclusion, and $\psi_i:U\to U_i\to Y_i$ to be the composition of these inclusions. So we can now define $\phi:U\to X\times Y$ by $\phi=\psi\times\prod\psi_i$. Finally, we define $\bar{X}$ to be the closure of $\phi(U)$ in $X\times Y$. Taking the projection $X\times Y\to X$, we get a map $f:\bar{X}\to X$.

We will prove that this map is birational. To do this, we just need to check that $f^{-1}(U)=\phi(U)$, because on $U$ we have $p_X\circ\phi$ is the identity. In fact, for this we must merely show that $(U\times Y)\cap \bar{X}=\phi(U)$, which is that the graph of $\prod \psi_i$. The morphism is surjective, then, because $f(\bar{X})$ contains $U$, which is dense in $X$, and the map must have closed image.

So what’s left? We still need to show that $\bar{X}$ is actually projective. We note that $Y$ was projective, and look at the second projection $g:X\times Y\to Y$, and restrict to $\bar{g}:\bar{X}\to Y$. We will show that this is an embedding. Because this is a local property, we must only find open sets $V_i\subset Y$ such that the union of the $g^{-1}(V_i)$ contains $\bar{X}$ and such that the restrictions of $\bar{g}$ to the $g^{-1}(V_i)$ are all closed embeddings.

To this end, we set $V_i$ to be the preimage of the $i$th projection $Y\to Y_i$ of $U_i$. Then $g^{-1}(V_i)$ certainly cover $\bar{X}$.

All that is left is to show that $\bar{g}:\bar{X}\cap g^{-1}(V_i)\to V_i$ is a closed embedding. Now, we note that $V_i=U_i\times \hat{Y}_i$, where $\hat{Y}_i=\prod_{i\neq j} Y_j$. So then $g^{-1}(V_i)=X\times U_i\times \hat{Y}_i$. Take $Z_i$ to be the graph of the morphism which is the composite $U_i\times\hat{Y}_i\to U_i\to X$. Then $Z_i$ is a closed subset of this and its projection onto $V_i=U_i\times \hat{Y}_i$ is an isomorphism. Additionally, $\phi(U)\subset Z_i$, and as $Z_i$ is closed, we have $\bar{X}\cap g^{-1}(V_i)$ closed in $Z_i$, so the map is a closed embedding. QED.

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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### 5 Responses to Complete Varieties and Chow’s Lemma

1. Vishal says:

2. Akhil Mathew says:

You argue that the $g^{-1}(V_i)$ cover $\bar{X}$. But couldn’t there be elements of $\bar{X}$ that map to elements of $\prod ( Y_i - U_i)$?

Thanks, and sorry for the late question.

3. 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

4. bebischof says:

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

Thanks again!