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 $$\mathbb{A}^n_{S_j}$$ since $$X\to S$$ is of finite type and therefore quasi-compact. Composing this map with the open immersions $$\mathbb{A}^n_{S_j}\to \mathbb{P}^n_{S_j}$$ and $$\mathbb{P}^n_{S_j} \to \mathbb{P}^n_S$$, we see that each $$X_{ij}$$ is a closed subscheme of an open subscheme of $$\mathbb{P}^n_S$$. As $$\mathbb{P}^n_S$$ 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.