Английская Википедия:Chow's lemma
Шаблон: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:Шаблон:Sfn
- If <math>X</math> is a scheme that is proper over a noetherian base <math>S</math>, then there exists a projective <math>S</math>-scheme <math>X'</math> and a surjective <math>S</math>-morphism <math>f: X' \to X</math> that induces an isomorphism <math>f^{-1}(U) \simeq U</math> for some dense open <math>U\subseteq X.</math>
Proof
The proof here is a standard one.Шаблон:Sfn
Reduction to the case of <math>X</math> irreducible
We can first reduce to the case where <math>X</math> is irreducible. To start, <math>X</math> is noetherian since it is of finite type over a noetherian base. Therefore it has finitely many irreducible components <math>X_i</math>, and we claim that for each <math>X_i</math> there is an irreducible proper <math>S</math>-scheme <math>Y_i</math> so that <math>Y_i\to X</math> has set-theoretic image <math>X_i</math> and is an isomorphism on the open dense subset <math>X_i\setminus \cup_{j\neq i} X_j</math> of <math>X_i</math>. To see this, define <math>Y_i</math> to be the scheme-theoretic image of the open immersion
- <math>X\setminus \cup_{j\neq i} X_j \to X.</math>
Since <math>X\setminus \cup_{j\neq i} X_j</math> is set-theoretically noetherian for each <math>i</math>, the map <math>X\setminus \cup_{j\neq i} X_j\to X</math> is quasi-compact and we may compute this scheme-theoretic image affine-locally on <math>X</math>, immediately proving the two claims. If we can produce for each <math>Y_i</math> a projective <math>S</math>-scheme <math>Y_i'</math> as in the statement of the theorem, then we can take <math>X'</math> to be the disjoint union <math>\coprod Y_i'</math> and <math>f</math> to be the composition <math>\coprod Y_i' \to \coprod Y_i\to X</math>: this map is projective, and an isomorphism over a dense open set of <math>X</math>, while <math>\coprod Y_i'</math> is a projective <math>S</math>-scheme since it is a finite union of projective <math>S</math>-schemes. Since each <math>Y_i</math> is proper over <math>S</math>, we've completed the reduction to the case <math>X</math> irreducible.
<math>X</math> can be covered by finitely many quasi-projective <math>S</math>-schemes
Next, we will show that <math>X</math> can be covered by a finite number of open subsets <math>U_i</math> so that each <math>U_i</math> is quasi-projective over <math>S</math>. To do this, we may by quasi-compactness first cover <math>S</math> by finitely many affine opens <math>S_j</math>, and then cover the preimage of each <math>S_j</math> in <math>X</math> by finitely many affine opens <math>X_{jk}</math> each with a closed immersion in to <math>\mathbb{A}^n_{S_j}</math> since <math>X\to S</math> is of finite type and therefore quasi-compact. Composing this map with the open immersions <math>\mathbb{A}^n_{S_j}\to \mathbb{P}^n_{S_j}</math> and <math>\mathbb{P}^n_{S_j} \to \mathbb{P}^n_S</math>, we see that each <math>X_{ij}</math> is a closed subscheme of an open subscheme of <math>\mathbb{P}^n_S</math>. As <math>\mathbb{P}^n_S</math> is noetherian, every closed subscheme of an open subscheme is also an open subscheme of a closed subscheme, and therefore each <math>X_{ij}</math> is quasi-projective over <math>S</math>.
Construction of <math>X'</math> and <math>f:X'\to X</math>
Now suppose <math>\{U_i\}</math> is a finite open cover of <math>X</math> by quasi-projective <math>S</math>-schemes, with <math>\phi_i:U_i\to P_i</math> an open immersion in to a projective <math>S</math>-scheme. Set <math>U=\cap_i U_i</math>, which is nonempty as <math>X</math> is irreducible. The restrictions of the <math>\phi_i</math> to <math>U</math> define a morphism
- <math>\phi: U \to P = P_1 \times_S \cdots \times_S P_n</math>
so that <math>U\to U_i\to P_i = U\stackrel{\phi}{\to} P \stackrel{p_i}{\to} P_i</math>, where <math>U\to U_i</math> is the canonical injection and <math>p_i:P\to P_i</math> is the projection. Letting <math>j:U\to X</math> denote the canonical open immersion, we define <math>\psi=(j,\phi)_S: U\to X\times_S P</math>, which we claim is an immersion. To see this, note that this morphism can be factored as the graph morphism <math>U\to U\times_S P</math> (which is a closed immersion as <math>P\to S</math> is separated) followed by the open immersion <math>U\times_S P\to X\times_S P</math>; as <math>X\times_S P</math> 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 <math>X'</math> be the scheme-theoretic image of <math>\psi</math>, and factor <math>\psi</math> as
- <math> \psi:U\stackrel{\psi'}{\to} X'\stackrel{h}{\to} X\times_S P</math>
where <math>\psi'</math> is an open immersion and <math>h</math> is a closed immersion. Let <math>q_1:X\times_S P\to X</math> and <math>q_2:X\times_S P\to P</math> be the canonical projections. Set
- <math>f:X'\stackrel{h}{\to} X\times_S P \stackrel{q_1}{\to} X,</math>
- <math>g:X'\stackrel{h}{\to} X\times_S P \stackrel{q_2}{\to} P.</math>
We will show that <math>X'</math> and <math>f</math> satisfy the conclusion of the theorem.
Verification of the claimed properties of <math>X'</math> and <math>f</math>
To show <math>f</math> is surjective, we first note that it is proper and therefore closed. As its image contains the dense open set <math>U\subset X</math>, we see that <math>f</math> must be surjective. It is also straightforward to see that <math>f</math> induces an isomorphism on <math>U</math>: we may just combine the facts that <math>f^{-1}(U)=h^{-1}(U\times_S P)</math> and <math>\psi</math> is an isomorphism on to its image, as <math>\psi</math> factors as the composition of a closed immersion followed by an open immersion <math>U\to U\times_S P \to X\times_S P</math>. It remains to show that <math>X'</math> is projective over <math>S</math>.
We will do this by showing that <math>g:X'\to P</math> is an immersion. We define the following four families of open subschemes:
- <math> V_i = \phi_i(U_i)\subset P_i </math>
- <math> W_i = p_i^{-1}(V_i)\subset P </math>
- <math> U_i' = f^{-1}(U_i)\subset X' </math>
- <math> U_i = g^{-1}(W_i)\subset X'. </math>
As the <math>U_i</math> cover <math>X</math>, the <math>U_i'</math> cover <math>X'</math>, and we wish to show that the <math>U_i</math> also cover <math>X'</math>. We will do this by showing that <math>U_i'\subset U_i</math> for all <math>i</math>. It suffices to show that <math>p_i\circ g|_{U_i'}:U_i'\to P_i</math> is equal to <math>\phi_i\circ f|_{U_i'}:U_i'\to P_i</math> as a map of topological spaces. Replacing <math>U_i'</math> by its reduction, which has the same underlying topological space, we have that the two morphisms <math>(U_i')_{red}\to P_i</math> are both extensions of the underlying map of topological space <math>U\to U_i\to P_i</math>, so by the reduced-to-separated lemma they must be equal as <math>U</math> is topologically dense in <math>U_i</math>. Therefore <math>U_i'\subset U_i</math> for all <math>i</math> and the claim is proven.
The upshot is that the <math>W_i</math> cover <math>g(X')</math>, and we can check that <math>g</math> is an immersion by checking that <math>g|_{U_i}:U_i\to W_i</math> is an immersion for all <math>i</math>. For this, consider the morphism
- <math> u_i:W_i\stackrel{p_i}{\to} V_i\stackrel{\phi_i^{-1}}{\to} U_i\to X.</math>
Since <math>X\to S</math> is separated, the graph morphism <math>\Gamma_{u_i}:W_i\to X\times_S W_i</math> is a closed immersion and the graph <math>T_i=\Gamma_{u_i}(W_i)</math> is a closed subscheme of <math>X\times_S W_i</math>; if we show that <math>U\to X\times_S W_i</math> factors through this graph (where we consider <math>U\subset X'</math> via our observation that <math>f</math> is an isomorphism over <math>f^{-1}(U)</math> from earlier), then the map from <math>U_i</math> must also factor through this graph by construction of the scheme-theoretic image. Since the restriction of <math>q_2</math> to <math>T_i</math> is an isomorphism onto <math>W_i</math>, the restriction of <math>g</math> to <math>U_i</math> will be an immersion into <math>W_i</math>, and our claim will be proven. Let <math>v_i</math> be the canonical injection <math>U\subset X' \to X\times_S W_i</math>; we have to show that there is a morphism <math>w_i:U\subset X'\to W_i</math> so that <math>v_i=\Gamma_{u_i}\circ w_i</math>. By the definition of the fiber product, it suffices to prove that <math>q_1\circ v_i= u_i\circ q_2\circ v_i</math>, or by identifying <math>U\subset X</math> and <math>U\subset X'</math>, that <math>q_1\circ\psi=u_i\circ q_2\circ \psi</math>. But <math>q_1\circ\psi = j</math> and <math>q_2\circ\psi=\phi</math>, so the desired conclusion follows from the definition of <math>\phi:U\to P</math> and <math>g</math> is an immersion. Since <math>X'\to S</math> is proper, any <math>S</math>-morphism out of <math>X'</math> is closed, and thus <math>g:X'\to P</math> is a closed immersion, so <math>X'</math> is projective. <math>\blacksquare</math>
Additional statements
In the statement of Chow's lemma, if <math>X</math> is reduced, irreducible, or integral, we can assume that the same holds for <math>X'</math>. If both <math>X</math> and <math>X'</math> are irreducible, then <math>f: X' \to X</math> is a birational morphism.Шаблон:Sfn
References
Bibliography
