Chow's lemma

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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 is a scheme that is proper over a noetherian base , then there exists a projective -scheme and a surjective -morphism that induces an isomorphism for some dense open


The proof here is a standard one (cf. EGA II, 5.6.1).

Reduction to the case of irreducible[edit]

We can first reduce to the case where is irreducible. To start, is noetherian since it is of finite type over a noetherian base. Therefore it has finitely many irreducible components , and we claim that for each there is an irreducible proper -scheme so that has set-theoretic image and is an isomorphism on the open dense subset of . To see this, define to be the scheme-theoretic image of the open immersion

Since is set-theoretically noetherian for each , the map is quasi-compact and we may compute this scheme-theoretic image affine-locally on , immediately proving the two claims. If we can produce for each a projective -scheme as in the statement of the theorem, then we can take to be the disjoint union and to be the composition : this map is projective, and an isomorphism over a dense open set of , while is a projective -scheme since it is a finite union of projective -schemes. Since each is proper over , we've completed the reduction to the case irreducible.

can be covered by finitely many quasi-projective -schemes[edit]

Next, we will show that can be covered by a finite number of open subsets so that each is quasi-projective over . To do this, we may by quasi-compactness first cover by finitely many affine opens , and then cover the preimage of each in by finitely many affine opens each with a closed immersion in to since is of finite type and therefore quasi-compact. Composing this map with the open immersions and , we see that each is a closed subscheme of an open subscheme of . As is noetherian, every closed subscheme of an open subscheme is also an open subscheme of a closed subscheme, and therefore each is quasi-projective over .

Construction of and [edit]

Now suppose is a finite open cover of by quasi-projective -schemes, with an open immersion in to a projective -scheme. Set , which is nonempty as is irreducible. The restrictions of the to define a morphism

so that , where is the canonical injection and is the projection. Letting denote the canonical open immersion, we define , which we claim is an immersion. To see this, note that this morphism can be factored as the graph morphism (which is a closed immersion as is separated) followed by the open immersion ; as 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 be the scheme-theoretic image of , and factor as

where is an open immersion and is a closed immersion. Let and be the canonical projections. Set

We will show that and satsify the conclusion of the theorem.

Verification of the claimed properties of and [edit]

To show is surjective, we first note that it is proper and therefore closed. As its image contains the dense open set , we see that must be surjective. It is also straightforward to see that induces an isomorphism on : we may just combine the facts that and is an isomorphism on to its image, as factors as the composition of a closed immersion followed by an open immersion . It remains to show that is projective over .

We will do this by showing that is an immersion. We define the following four families of open subschemes:

As the cover , the cover , and we wish to show that the also cover . We will do this by showing that for all . It suffices to show that is equal to as a map of topological spaces. Replacing by its reduction, which has the same underlying topological space, we have that the two morphisms are both extensions of the underlying map of topological space , so by the reduced-to-separated lemma they must be equal as is topologically dense in . Therefore for all and the claim is proven.

The upshot is that the cover , and we can check that is an immersion by checking that is an immersion for all . For this, consider the morphism

Since is separated, the graph morphism is a closed immersion and the graph is a closed subscheme of ; if we show that factors through this graph (where we consider via our observation that is an isomorphism over from earlier), then the map from must also factor through this graph by construction of the scheme-theoretic image. Since the restriction of to is an isomorphism onto , the restriction of to will be an immersion into , and our claim will be proven. Let be the canonical injection ; we have to show that there is a morphism so that . By the definition of the fiber product, it suffices to prove that , or by identifying and , that . But and , so the desired conclusion follows from the definition of and is an immersion. Since is proper, any -morphism out of is closed, and thus is a closed immersion, so is projective.

Additional statements[edit]

In the statement of Chow's lemma, if is reduced, irreducible, or integral, we can assume that the same holds for . If both and are irreducible, then is a birational morphism. (cf. EGA II, 5.6).


  1. ^ Hartshorne, Ch II. Exercise 4.10
  • 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, 52, New York: Springer-Verlag, ISBN 978-0-387-90244-9, MR 0463157