# 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:
    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 $f:X'\to X$ that induces an isomorphism $f^{-1}(U)\simeq U$ for some dense open $U\subseteq X.$

## Proof

The proof here is a standard one.

### Reduction to the case of $X$ irreducible

We can first reduce to the case where $X$ is irreducible. To start, $X$ is noetherian since it is of finite type over a noetherian base. Therefore it has finitely many irreducible components $X_{i}$, and we claim that for each $X_{i}$ there is an irreducible proper $S$-scheme $Y_{i}$ so that $Y_{i}\to X$ has set-theoretic image $X_{i}$ and is an isomorphism on the open dense subset $X_{i}\setminus \cup _{j\neq i}X_{j}$ of $X_{i}$. To see this, define $Y_{i}$ to be the scheme-theoretic image of the open immersion
    $X\setminus \cup _{j\neq i}X_{j}\to X.$
Since $X\setminus \cup _{j\neq i}X_{j}$ is set-theoretically noetherian for each $i$, the map $X\setminus \cup _{j\neq i}X_{j}\to X$ is quasi-compact and we may compute this scheme-theoretic image affine-locally on $X$, immediately proving the two claims. If we can produce for each $Y_{i}$ a projective $S$-scheme $Y_{i}'$ as in the statement of the theorem, then we can take $X'$ to be the disjoint union $\coprod Y_{i}'$ and $f$ to be the composition $\coprod Y_{i}'\to \coprod Y_{i}\to X$: this map is projective, and an isomorphism over a dense open set of $X$, while $\coprod Y_{i}'$ is a projective $S$-scheme since it is a finite union of projective $S$-schemes. Since each $Y_{i}$ is proper over $S$, we've completed the reduction to the case $X$ irreducible.

### $X$ can be covered by finitely many quasi-projective $S$-schemes

Next, we will show that $X$ can be covered by a finite number of open subsets $U_{i}$ so that each $U_{i}$ is quasi-projective over $S$. To do this, we may by quasi-compactness first cover $S$ by finitely many affine opens $S_{j}$, and then cover the preimage of each $S_{j}$ in $X$ by finitely many affine opens $X_{jk}$ each with a closed immersion in to $\mathbb {A} _{S_{j}}^{n}$ since $X\to S$ is of finite type and therefore quasi-compact. Composing this map with the open immersions $\mathbb {A} _{S_{j}}^{n}\to \mathbb {P} _{S_{j}}^{n}$ and $\mathbb {P} _{S_{j}}^{n}\to \mathbb {P} _{S}^{n}$, we see that each $X_{ij}$ is a closed subscheme of an open subscheme of $\mathbb {P} _{S}^{n}$. As $\mathbb {P} _{S}^{n}$ is noetherian, every closed subscheme of an open subscheme is also an open subscheme of a closed subscheme, and therefore each $X_{ij}$ is quasi-projective over $S$.

### Construction of $X'$ and $f:X'\to X$

Now suppose $\{U_{i}\}$ is a finite open cover of $X$ by quasi-projective $S$-schemes, with $\phi _{i}:U_{i}\to P_{i}$ an open immersion in to a projective $S$-scheme. Set $U=\cap _{i}U_{i}$, which is nonempty as $X$ is irreducible. The restrictions of the $\phi _{i}$ to $U$ define a morphism
    $\phi :U\to P=P_{1}\times _{S}\cdots \times _{S}P_{n}$
so that $U\to U_{i}\to P_{i}=U{\stackrel {\phi }{\to }}P{\stackrel {p_{i}}{\to }}P_{i}$, where $U\to U_{i}$ is the canonical injection and $p_{i}:P\to P_{i}$ is the projection. Letting $j:U\to X$ denote the canonical open immersion, we define $\psi =(j,\phi )_{S}:U\to X\times _{S}P$, which we claim is an immersion. To see this, note that this morphism can be factored as the graph morphism $U\to U\times _{S}P$ (which is a closed immersion as $P\to S$ is separated) followed by the open immersion $U\times _{S}P\to X\times _{S}P$; as $X\times _{S}P$ is noetherian, we can apply the same logic as before to see that we can swap the order of the open and closed immersions.
Now let $X'$ be the scheme-theoretic image of $\psi$, and factor $\psi$ as
    $\psi :U{\stackrel {\psi '}{\to }}X'{\stackrel {h}{\to }}X\times _{S}P$
where $\psi '$ is an open immersion and $h$ is a closed immersion. Let $q_{1}:X\times _{S}P\to X$ and $q_{2}:X\times _{S}P\to P$ be the canonical projections. Set
    $f:X'{\stackrel {h}{\to }}X\times _{S}P{\stackrel {q_{1}}{\to }}X,$
    $g:X'{\stackrel {h}{\to }}X\times _{S}P{\stackrel {q_{2}}{\to }}P.$
We will show that $X'$ and $f$ satisfy the conclusion of the theorem.

### Verification of the claimed properties of $X'$ and $f$

To show $f$ is surjective, we first note that it is proper and therefore closed. As its image contains the dense open set $U\subset X$, we see that $f$ must be surjective. It is also straightforward to see that $f$ induces an isomorphism on $U$: we may just combine the facts that $f^{-1}(U)=h^{-1}(U\times _{S}P)$ and $\psi$ is an isomorphism on to its image, as $\psi$ factors as the composition of a closed immersion followed by an open immersion $U\to U\times _{S}P\to X\times _{S}P$. It remains to show that $X'$ is projective over $S$.
We will do this by showing that $g:X'\to P$ is an immersion. We define the following four families of open subschemes:
    $V_{i}=\phi _{i}(U_{i})\subset P_{i}$
    $W_{i}=p_{i}^{-1}(V_{i})\subset P$
    $U_{i}'=f^{-1}(U_{i})\subset X'$
    $U_{i}''=g^{-1}(W_{i})\subset X'.$
As the $U_{i}$ cover $X$, the $U_{i}'$ cover $X'$, and we wish to show that the $U_{i}''$ also cover $X'$. We will do this by showing that $U_{i}'\subset U_{i}''$ for all $i$. It suffices to show that $p_{i}\circ g|_{U_{i}'}:U_{i}'\to P_{i}$ is equal to $\phi _{i}\circ f|_{U_{i}'}:U_{i}'\to P_{i}$ as a map of topological spaces. Replacing $U_{i}'$ by its reduction, which has the same underlying topological space, we have that the two morphisms $(U_{i}')_{red}\to P_{i}$ are both extensions of the underlying map of topological space $U\to U_{i}\to P_{i}$, so by the reduced-to-separated lemma they must be equal as $U$ is topologically dense in $U_{i}$. Therefore $U_{i}'\subset U_{i}''$ for all $i$ and the claim is proven.
The upshot is that the $W_{i}$ cover $g(X')$, and we can check that $g$ is an immersion by checking that $g|_{U_{i}''}:U_{i}''\to W_{i}$ is an immersion for all $i$. For this, consider the morphism
    $u_{i}:W_{i}{\stackrel {p_{i}}{\to }}V_{i}{\stackrel {\phi _{i}^{-1}}{\to }}U_{i}\to X.$
Since $X\to S$ is separated, the graph morphism $\Gamma _{u_{i}}:W_{i}\to X\times _{S}W_{i}$ is a closed immersion and the graph $T_{i}=\Gamma _{u_{i}}(W_{i})$ is a closed subscheme of $X\times _{S}W_{i}$; if we show that $U\to X\times _{S}W_{i}$ factors through this graph (where we consider $U\subset X'$ via our observation that $f$ is an isomorphism over $f^{-1}(U)$ from earlier), then the map from $U_{i}''$ must also factor through this graph by construction of the scheme-theoretic image. Since the restriction of $q_{2}$ to $T_{i}$ is an isomorphism onto $W_{i}$, the restriction of $g$ to $U_{i}''$ will be an immersion into $W_{i}$, and our claim will be proven. Let $v_{i}$ be the canonical injection $U\subset X'\to X\times _{S}W_{i}$; we have to show that there is a morphism $w_{i}:U\subset X'\to W_{i}$ so that $v_{i}=\Gamma _{u_{i}}\circ w_{i}$. By the definition of the fiber product, it suffices to prove that $q_{1}\circ v_{i}=u_{i}\circ q_{2}\circ v_{i}$, or by identifying $U\subset X$ and $U\subset X'$, that $q_{1}\circ \psi =u_{i}\circ q_{2}\circ \psi$. But $q_{1}\circ \psi =j$ and $q_{2}\circ \psi =\phi$, so the desired conclusion follows from the definition of $\phi :U\to P$ and $g$ is an immersion. Since $X'\to S$ is proper, any $S$-morphism out of $X'$ is closed, and thus $g:X'\to P$ is a closed immersion, so $X'$ is projective. $\blacksquare$

## 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 $f:X'\to X$ is a birational morphism.