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. Since $X$ is separated over $S$ and $U_i$ is dense, this is clear from looking at the relevant commutative diagram. Now, $X \times_S P \to P$ is closed since it is a base extension of the proper morphism $X \to S$. Thus, $g(X')$ is a closed subscheme covered by $V_i$, and so 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.

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" Check |url= value (help). 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