# Chow's lemma

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

## Proof

The proof here is a standard one (cf. Template:Harvnb).

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

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

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

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

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

Let ${\displaystyle V_{i}=p_{i}^{-1}(U_{i})}$ where ${\displaystyle p_{i}\colon P\to P_{i}}$. We claim ${\displaystyle g^{-1}(V_{i})}$ are an open cover of ${\displaystyle X'}$. This would follow from ${\displaystyle f^{-1}(U_{i})\subset g^{-1}(V_{i})}$ as sets. This in turn follows from ${\displaystyle f=p_{i}\circ g}$ on ${\displaystyle U_{i}}$ as functions on the underlying topological space. Since ${\displaystyle X}$ is separated over ${\displaystyle S}$ and ${\displaystyle U_{i}}$ is dense, this is clear from looking at the relevant commutative diagram. Now, ${\displaystyle X\times _{S}P\to P}$ is closed since it is a base extension of the proper morphism ${\displaystyle X\to S}$. Thus, ${\displaystyle g(X')}$ is a closed subscheme covered by ${\displaystyle V_{i}}$, and so it is enough to show that for each ${\displaystyle i}$ the map ${\displaystyle g\colon g^{-1}(V_{i})\to V_{i}}$, denoted by ${\displaystyle h}$, is a closed immersion.

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

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

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