Chow's lemma

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 $X$ is a scheme that is proper over a noetherian base $S$, then there exists a projective $S$-scheme $X'$ and a surjective $S$-morphism $f\colon X'\to X$ that induces an isomorphism $f^{{-1}}(U)\simeq U$ for some dense open $U\subseteq X$.

Proof

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

It is easy to reduce to the case when $X$ is irreducible, as follows. $X$ 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 $X_{i}$, which are each proper over $S$ (because they're closed immersions in the scheme $X$ which is proper over $S$). If, within each of these irreducible components, there exists a dense open $U_{i}\subset X_{i}$, then we can take $U:=\bigsqcup (U_{i}\setminus \bigcup \limits _{{i\neq j}}X_{j})$. It is not hard to see that each of the disjoint pieces are dense in their respective $X_{i}$, so the full set $U$ is dense in $X$. In addition, it's clear that we can similarly find a morphism $g$ which satisfies the density condition.

Having reduced the problem, we now assume $X$ is irreducible. We recall that it must also be noetherian. Thus, we can find a finite open affine cover $X=\bigcup _{{i=1}}^{n}U_{i}$. $U_{i}$ are quasi-projective over $S$; there are open immersions over $S$, $\phi _{i}\colon U_{i}\to P_{i}$ into some projective $S$-schemes $P_{i}$. Put $U=\cap U_{i}$. $U$ is nonempty since $X$ is irreducible. Let

$\phi \colon U\to P=P_{1}\times _{S}\cdots \times _{S}P_{n}.$

be given by $\phi _{i}$'s restricted to $U$ over $S$. Let

$\psi \colon U\to X\times _{S}P.$

be given by $U\hookrightarrow X$ and $\phi$ over $S$. $\psi$ is then an immersion; thus, it factors as an open immersion followed by a closed immersion $X'\to X\times _{S}P$. Let $f\colon X'\to X$ be the immersion followed by the projection. We claim $f$ induces $f^{{-1}}(U)\simeq U$; for that, it is enough to show $f^{{-1}}(U)=\psi (U)$. But this means that $\psi (U)$ is closed in $U\times _{S}P$. $\psi$ factorizes as $U{\overset {\Gamma _{\phi }}\to }U\times _{S}P\to X\times _{S}P$. $P$ is separated over $S$ and so the graph morphism $\Gamma _{\phi }$ is a closed immersion. This proves our contention.

It remains to show $X'$ is projective over $S$. Let $g\colon X'\to P$ be the closed immersion followed by the projection. Showing that $g$ is a closed immersion shows $X'$ is projective over $S$. This can be checked locally. Identifying $U_{i}$ with its image in $P_{i}$ we suppress $\phi _{i}$ from our notation.

Let $V_{i}=p_{i}^{{-1}}(U_{i})$ where $p_{i}\colon P\to P_{i}$. We claim $g^{{-1}}(V_{i})$ are an open cover of $X'$. This would follow from $f^{{-1}}(U_{i})\subset g^{{-1}}(V_{i})$ as sets. This in turn follows from $f=p_{i}\circ g$ on $U_{i}$ as functions on the underlying topological space. Thus it is enough to show that for each $i$ the map $g\colon g^{{-1}}(V_{i})\to V_{i}$, denoted by $h$, is a closed immersion (since the property of being a closed immersion is local on the base).

Fix $i$. Let $Z$ be the graph of $u\colon V_{i}{\overset {p_{i}}\to }U_{i}\hookrightarrow X$. It is a closed subscheme of $X\times _{S}V_{i}$ since $X$ is separated over $S$. Let $q_{1}\colon X\times _{S}P\to X,\ q_{2}\colon X\times _{S}P\to P$ be the projections. We claim that $h$ factors through $Z$, which would imply $h$ is a closed immersion. But for $w\colon U'\to V_{i}$ we have:

$v=\Gamma _{u}\circ w\quad \Leftrightarrow \quad q_{1}\circ v=u\circ q_{2}\circ v\quad \Leftrightarrow \quad q_{1}\circ \psi =u\circ q_{2}\circ \psi \quad \Leftrightarrow \quad q_{1}\circ \psi =u\circ \phi .$

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

Additional statements

In the statement of Chow's lemma, if $X$ is reduced, irreducible, or integral, we can assume that the same holds for $X'$. If both $X$ and $X'$ are irreducible, then $f\colon X'\to X$ is a birational morphism. (cf. EGA II, 5.6).

References

1. Hartshorne, Ch II. Exercise 4.10

• Grothendieck, Alexandre; Dieudonné, Jean (1961). "Éléments de géométrie algébrique: II. Étude globale élémentaire de quelques classes de morphismes". Publications Mathématiques de l'IHÉS. 8. doi:10.1007/bf02699291. MR 0217084.
• Hartshorne, Robin (1977), Algebraic Geometry, Graduate Texts in Mathematics, 52, New York: Springer-Verlag, ISBN 978-0-387-90244-9, MR 0463157