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 $X^{\prime }$ and a surjective $S$-morphism $f:X^{\prime }\to X$ 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 $X_i\setminus {\cup }_{j\ne i}{X}_j$ of $X_i$. To see this, define $Y_i$ to be the scheme-theoretic image of the open immersion

$X\setminus {\cup }_{j\ne i}{X}_j\to X.$

Since $X\setminus {\cup }_{j\ne i}{X}_j$ is set-theoretically noetherian for each $i$, the map $X\setminus {\cup }_{j\ne i}{X}_j\to X$ 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 ${Y_i}^{\prime }$ as in the statement of the theorem, then we can take $X^{\prime }$ to be the disjoint union $\coprod {Y_i}^{\prime }$ and $f$ to be the composition $\coprod {Y_i}^{\prime }\to \coprod Y_i\to X$: this map is projective, and an isomorphism over a dense open set of $X$, while $\coprod {Y_i}^{\prime }$ 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 ${\mathbb{𝔸}}_{S_j}^n\to {\mathbb{\mathbb{P}}}_{S_j}^n$ and ${\mathbb{\mathbb{P}}}_{S_j}^n\to {\mathbb{\mathbb{P}}}_S^n$, we see that each $X_{ij}$ is a closed subscheme of an open subscheme of ${\mathbb{\mathbb{P}}}_S^n$. As ${\mathbb{\mathbb{P}}}_S^n$ 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 $X^{\prime }$ and $f:X^{\prime }\to X$

Now suppose $\{U_i\}$ is a finite open cover of $X$ by quasi-projective $S$-schemes, with ${\varphi }_i:U_i\to P_i$ 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 ${\varphi }_i$ to $U$ define a morphism

$\varphi :U\to P=P_1{\times }_S\cdots {\times }_S{P}_n$

so that $U\to U_i\to P_i=U\stackrel{\varphi }{\to }P\stackrel{p_i}{\to }{P}_i$, 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 $\psi =(j,\varphi {)}_S:U\to X{\times }_{S}P$, 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 $X^{\prime }$ be the scheme-theoretic image of $\psi$, and factor $\psi$ as

$\psi :U\stackrel{{\psi }^{\prime }}{\to }{X}^{\prime }\stackrel{h}{\to }X{\times }_{S}P$

where ${\psi }^{\prime }$ 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

$f:X^{\prime }\stackrel{h}{\to }X{\times }_{S}P\stackrel{q_1}{\to }X,$

$g:X^{\prime }\stackrel{h}{\to }X{\times }_{S}P\stackrel{q_2}{\to }P.$

We will show that $X^{\prime }$ and $f$ satisfy the conclusion of the theorem.

Verification of the claimed properties of $X^{\prime }$ 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 $X^{\prime }$ is projective over $S$.

We will do this by showing that $g:X^{\prime }\to P$ is an immersion. We define the following four families of open subschemes:

$V_i={\varphi }_i(U_i)\subset P_i$

$W_i=p_i^{-1}(V_i)\subset P$

${U_i}^{\prime }=f^{-1}(U_i)\subset X^{\prime }$

${U_i}^{″}=g^{-1}(W_i)\subset X^{\prime }.$

As the $U_i$ cover $X$, the ${U_i}^{\prime }$ cover $X^{\prime }$, and we wish to show that the ${U_i}^{″}$ also cover $X^{\prime }$. We will do this by showing that ${U_i}^{\prime }\subset {U_i}^{″}$ for all $i$. It suffices to show that $p_i\circ g{|}_{{U_i}^{\prime }}:{U_i}^{\prime }\to P_i$ is equal to ${\varphi }_i\circ f{|}_{{U_i}^{\prime }}:{U_i}^{\prime }\to P_i$ as a map of topological spaces. Replacing ${U_i}^{\prime }$ by its reduction, which has the same underlying topological space, we have that the two morphisms $({U_i}^{\prime }{)}_{red}\to P_i$ 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 ${U_i}^{\prime }\subset {U_i}^{″}$ for all $i$ and the claim is proven.

The upshot is that the $W_i$ cover $g(X^{\prime })$, 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

$u_i:W_i\stackrel{p_i}{\to }{V}_i\stackrel{{\varphi }_i^{-1}}{\to }{U}_i\to X.$

Since $X\to S$ is separated, the graph morphism ${\mathrm{\Gamma }}_{u_i}:W_i\to X{\times }_S{W}_i$ is a closed immersion and the graph $T_i={\mathrm{\Gamma }}_{u_i}(W_i)$ 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 $U\subset X^{\prime }$ 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 $U\subset X^{\prime }\to X{\times }_S{W}_i$; we have to show that there is a morphism $w_i:U\subset X^{\prime }\to W_i$ so that $v_i={\mathrm{\Gamma }}_{u_i}\circ w_i$. 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 $U\subset X^{\prime }$, that $q_1\circ \psi =u_i\circ q_2\circ \psi$. But $q_1\circ \psi =j$ and $q_2\circ \psi =\varphi$, so the desired conclusion follows from the definition of $\varphi :U\to P$ and $g$ is an immersion. Since $X^{\prime }\to S$ is proper, any $S$-morphism out of $X^{\prime }$ is closed, and thus $g:X^{\prime }\to P$ is a closed immersion, so $X^{\prime }$ is projective. $◼$

Additional statements

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

Bibliography

• 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. http://www.numdam.org/item/PMIHES_1961__8__5_0. 
• Hartshorne, Robin (1977), Algebraic Geometry, Graduate Texts in Mathematics, 52, New York: Springer-Verlag, ISBN 978-0-387-90244-9