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

Proof

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The proof here is a standard one.[2]

Reduction to the case of   irreducible

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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

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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  

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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   satisfy the conclusion of the theorem.

Verification of the claimed properties of   and  

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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

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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.[3]

References

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Bibliography

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  • 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