Chow's lemma

Not to be confused with Chow's theorem.

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 ${\displaystyle f:X'\to X}$ 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. ^[2]

Reduction to the case of ${\displaystyle X}$ irreducible

We can first reduce to the case where ${\displaystyle X}$ is irreducible. To start, ${\displaystyle X}$ is noetherian since it is of finite type over a noetherian base. Therefore it has finitely many irreducible components ${\displaystyle X_{i}}$, and we claim that for each ${\displaystyle X_{i}}$ there is an irreducible proper ${\displaystyle S}$-scheme ${\displaystyle Y_{i}}$ so that ${\displaystyle Y_{i}\to X}$ has set-theoretic image ${\displaystyle X_{i}}$ and is an isomorphism on the open dense subset ${\displaystyle X_{i}\setminus \cup _{j\neq i}X_{j}}$ of ${\displaystyle X_{i}}$. To see this, define ${\displaystyle Y_{i}}$ to be the scheme-theoretic image of the open immersion

${\displaystyle X\setminus \cup _{j\neq i}X_{j}\to X.}$

Since ${\displaystyle X\setminus \cup _{j\neq i}X_{j}}$ is set-theoretically noetherian for each ${\displaystyle i}$, the map ${\displaystyle X\setminus \cup _{j\neq i}X_{j}\to X}$ is quasi-compact and we may compute this scheme-theoretic image affine-locally on ${\displaystyle X}$, immediately proving the two claims. If we can produce for each ${\displaystyle Y_{i}}$ a projective ${\displaystyle S}$-scheme ${\displaystyle Y_{i}'}$ as in the statement of the theorem, then we can take ${\displaystyle X'}$ to be the disjoint union ${\displaystyle \coprod Y_{i}'}$ and ${\displaystyle f}$ to be the composition ${\displaystyle \coprod Y_{i}'\to \coprod Y_{i}\to X}$: this map is projective, and an isomorphism over a dense open set of ${\displaystyle X}$, while ${\displaystyle \coprod Y_{i}'}$ is a projective ${\displaystyle S}$-scheme since it is a finite union of projective ${\displaystyle S}$-schemes. Since each ${\displaystyle Y_{i}}$ is proper over ${\displaystyle S}$, we've completed the reduction to the case ${\displaystyle X}$ irreducible.

${\displaystyle X}$ can be covered by finitely many quasi-projective ${\displaystyle S}$-schemes

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

Construction of ${\displaystyle X'}$ and ${\displaystyle f:X'\to X}$

Now suppose ${\displaystyle \{U_{i}\}}$ is a finite open cover of ${\displaystyle X}$ by quasi-projective ${\displaystyle S}$-schemes, with ${\displaystyle \phi _{i}:U_{i}\to P_{i}}$ an open immersion in to a projective ${\displaystyle S}$-scheme. Set ${\displaystyle U=\cap _{i}U_{i}}$, which is nonempty as ${\displaystyle X}$ is irreducible. The restrictions of the ${\displaystyle \phi _{i}}$ to ${\displaystyle U}$ define a morphism

${\displaystyle \phi :U\to P=P_{1}\times _{S}\cdots \times _{S}P_{n}}$

so that ${\displaystyle U\to U_{i}\to P_{i}=U{\stackrel {\phi }{\to }}P{\stackrel {p_{i}}{\to }}P_{i}}$, where ${\displaystyle U\to U_{i}}$ is the canonical injection and ${\displaystyle p_{i}:P\to P_{i}}$ is the projection. Letting ${\displaystyle j:U\to X}$ denote the canonical open immersion, we define ${\displaystyle \psi =(j,\phi )_{S}:U\to X\times _{S}P}$, which we claim is an immersion. To see this, note that this morphism can be factored as the graph morphism ${\displaystyle U\to U\times _{S}P}$ (which is a closed immersion as ${\displaystyle P\to S}$ is separated) followed by the open immersion ${\displaystyle U\times _{S}P\to X\times _{S}P}$; as ${\displaystyle X\times _{S}P}$ 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 ${\displaystyle X'}$ be the scheme-theoretic image of ${\displaystyle \psi }$, and factor ${\displaystyle \psi }$ as

${\displaystyle \psi :U{\stackrel {\psi '}{\to }}X'{\stackrel {h}{\to }}X\times _{S}P}$

where ${\displaystyle \psi '}$ is an open immersion and ${\displaystyle h}$ is a closed immersion. Let ${\displaystyle q_{1}:X\times _{S}P\to X}$ and ${\displaystyle q_{2}:X\times _{S}P\to P}$ be the canonical projections. Set

${\displaystyle f:X'{\stackrel {h}{\to }}X\times _{S}P{\stackrel {q_{1}}{\to }}X,}$

${\displaystyle g:X'{\stackrel {h}{\to }}X\times _{S}P{\stackrel {q_{2}}{\to }}P.}$

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

Verification of the claimed properties of ${\displaystyle X'}$ and ${\displaystyle f}$

To show ${\displaystyle f}$ is surjective, we first note that it is proper and therefore closed. As its image contains the dense open set ${\displaystyle U\subset X}$, we see that ${\displaystyle f}$ must be surjective. It is also straightforward to see that ${\displaystyle f}$ induces an isomorphism on ${\displaystyle U}$: we may just combine the facts that ${\displaystyle f^{-1}(U)=h^{-1}(U\times _{S}P)}$ and ${\displaystyle \psi }$ is an isomorphism on to its image, as ${\displaystyle \psi }$ factors as the composition of a closed immersion followed by an open immersion ${\displaystyle U\to U\times _{S}P\to X\times _{S}P}$. It remains to show that ${\displaystyle X'}$ is projective over ${\displaystyle S}$.

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

${\displaystyle V_{i}=\phi _{i}(U_{i})\subset P_{i}}$

${\displaystyle W_{i}=p_{i}^{-1}(V_{i})\subset P}$

${\displaystyle U_{i}'=f^{-1}(U_{i})\subset X'}$

${\displaystyle U_{i}''=g^{-1}(W_{i})\subset X'.}$

As the ${\displaystyle U_{i}}$ cover ${\displaystyle X}$, the ${\displaystyle U_{i}'}$ cover ${\displaystyle X'}$, and we wish to show that the ${\displaystyle U_{i}''}$ also cover ${\displaystyle X'}$. We will do this by showing that ${\displaystyle U_{i}'\subset U_{i}''}$ for all ${\displaystyle i}$. It suffices to show that ${\displaystyle p_{i}\circ g|_{U_{i}'}:U_{i}'\to P_{i}}$ is equal to ${\displaystyle \phi _{i}\circ f|_{U_{i}'}:U_{i}'\to P_{i}}$ as a map of topological spaces. Replacing ${\displaystyle U_{i}'}$ by its reduction, which has the same underlying topological space, we have that the two morphisms ${\displaystyle (U_{i}')_{red}\to P_{i}}$ are both extensions of the underlying map of topological space ${\displaystyle U\to U_{i}\to P_{i}}$, so by the reduced-to-separated lemma they must be equal as ${\displaystyle U}$ is topologically dense in ${\displaystyle U_{i}}$. Therefore ${\displaystyle U_{i}'\subset U_{i}''}$ for all ${\displaystyle i}$ and the claim is proven.

The upshot is that the ${\displaystyle W_{i}}$ cover ${\displaystyle g(X')}$, and we can check that ${\displaystyle g}$ is an immersion by checking that ${\displaystyle g|_{U_{i}''}:U_{i}''\to W_{i}}$ is an immersion for all ${\displaystyle i}$. For this, consider the morphism

${\displaystyle u_{i}:W_{i}{\stackrel {p_{i}}{\to }}V_{i}{\stackrel {\phi _{i}^{-1}}{\to }}U_{i}\to X.}$

Since ${\displaystyle X\to S}$ is separated, the graph morphism ${\displaystyle \Gamma _{u_{i}}:W_{i}\to X\times _{S}W_{i}}$ is a closed immersion and the graph ${\displaystyle T_{i}=\Gamma _{u_{i}}(W_{i})}$ is a closed subscheme of ${\displaystyle X\times _{S}W_{i}}$; if we show that ${\displaystyle U\to X\times _{S}W_{i}}$ factors through this graph (where we consider ${\displaystyle U\subset X'}$ via our observation that ${\displaystyle f}$ is an isomorphism over ${\displaystyle f^{-1}(U)}$ from earlier), then the map from ${\displaystyle U_{i}''}$ must also factor through this graph by construction of the scheme-theoretic image. Since the restriction of ${\displaystyle q_{2}}$ to ${\displaystyle T_{i}}$ is an isomorphism onto ${\displaystyle W_{i}}$, the restriction of ${\displaystyle g}$ to ${\displaystyle U_{i}''}$ will be an immersion into ${\displaystyle W_{i}}$, and our claim will be proven. Let ${\displaystyle v_{i}}$ be the canonical injection ${\displaystyle U\subset X'\to X\times _{S}W_{i}}$; we have to show that there is a morphism ${\displaystyle w_{i}:U\subset X'\to W_{i}}$ so that ${\displaystyle v_{i}=\Gamma _{u_{i}}\circ w_{i}}$. By the definition of the fiber product, it suffices to prove that ${\displaystyle q_{1}\circ v_{i}=u_{i}\circ q_{2}\circ v_{i}}$, or by identifying ${\displaystyle U\subset X}$ and ${\displaystyle U\subset X'}$, that ${\displaystyle q_{1}\circ \psi =u_{i}\circ q_{2}\circ \psi }$. But ${\displaystyle q_{1}\circ \psi =j}$ and ${\displaystyle q_{2}\circ \psi =\phi }$, so the desired conclusion follows from the definition of ${\displaystyle \phi :U\to P}$ and ${\displaystyle g}$ is an immersion. Since ${\displaystyle X'\to S}$ is proper, any ${\displaystyle S}$-morphism out of ${\displaystyle X'}$ is closed, and thus ${\displaystyle g:X'\to P}$ is a closed immersion, so ${\displaystyle X'}$ is projective. ${\displaystyle \blacksquare }$

Additional statements

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 ${\displaystyle f:X'\to X}$ is a birational morphism. ^[3]

References

1. Hartshorne 1977, Ch II. Exercise 4.10.
2. Grothendieck & Dieudonné 1961, 5.6.1.
3. Grothendieck & Dieudonné 1961, 5.6.

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. MR   0217084.
• Hartshorne, Robin (1977), Algebraic Geometry , Graduate Texts in Mathematics, vol. 52, New York: Springer-Verlag, ISBN   978-0-387-90244-9, MR   0463157