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In algebraic geometry, Chow's lemma, named after Wei-Liang Chow, 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 S-morphism ${\displaystyle f:X'\to X}$ such that ${\displaystyle f:f^{-1}(U)\to U}$ is an isomorphism for some open dense subset U.

Proof

The proof here is a standard one (cf. EGA II, 5.6.1 harvnb error: no target: CITEREFEGA_II (help)).

It is easy to reduce to the case when X is irreducible. X is noetherian since it is of finite type over a noetherian base. Thus, we can find an open affine cover ${\displaystyle X=\cup _{1}^{n}U_{i}}$. ${\displaystyle U_{i}}$ are quasi-projective over S; there are open immersions over S ${\displaystyle \phi _{i}:U_{i}\to P_{i}}$ into some projective S-schemes ${\displaystyle P_{i}}$. Put ${\displaystyle U=\cap U_{i}}$. U is nonempty since X is irreducible. By the universal property of the fiber product we get

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

Let

${\displaystyle \psi =(U\hookrightarrow X,\phi )_{S}:U\to X\times _{S}P.}$

${\displaystyle \psi }$ is then an immersion. Let ${\displaystyle X'}$ be the scheme-theoretic image of ${\displaystyle \psi }$ in ${\displaystyle X\times _{S}P}$. Let ${\displaystyle f:X'\to X}$ be the immersion followed by the projection. We claim ${\displaystyle f:f^{-1}(U)\to U}$ is an isomorphism; for this, it is enough to show ${\displaystyle f^{-1}(U)=\psi (U)}$. But this means that ${\displaystyle \psi (U)}$ is closed in ${\displaystyle U\times _{S}P}$. ${\displaystyle \psi }$ factorizes as ${\displaystyle U{\overset {\Gamma _{\phi }}{\to }}U\times _{S}P\to X\times P}$. ${\displaystyle P}$ is projective over S; in particular, separated and so the graph morphism ${\displaystyle \Gamma _{\phi }}$ is a closed immersion. This proves our contention.

It remains to show ${\displaystyle X'}$ is projective over S. Let ${\displaystyle g:X'\to P}$ be the immersion followed by the projection. We claim g is a closed immersion. This can be checked locally. Identifying ${\displaystyle U_{i}}$ with its image in ${\displaystyle P_{i}}$ we suppress ${\displaystyle \phi _{i}}$ from our notation. Let

${\displaystyle V_{i}=p_{i}^{-1}(U_{i})=P_{1}\times _{S}\cdots \times _{S}U_{i}\times _{S}\cdots \times _{S}P_{n}}$

where ${\displaystyle p_{i}:P\to P_{i}}$ is the canonical projection. Then ${\displaystyle g^{-1}(V_{i})}$ are an open cover of ${\displaystyle X'}$. Thus, it suffices to show for each i ${\displaystyle g:g^{-1}(V_{i})\to V_{i}}$ is a closed immersion. Let ${\displaystyle Z_{i}}$ be the graph of ${\displaystyle V_{i}{\overset {p_{i}}{\to }}U_{i}\hookrightarrow X}$. It is a closed subscheme of ${\displaystyle X\times _{S}V_{i}}$ since ${\displaystyle X}$ is separated over S. Let ${\displaystyle q_{2}:X\times _{S}P\to P}$ be the projection. Then ${\displaystyle q_{2}:Z_{i}{\overset {\simeq }{\to }}V_{i}}$. But ${\displaystyle g}$ is the restriction of ${\displaystyle q_{2}}$ to ${\displaystyle X'}$ and ${\displaystyle X'}$ is closed. Hence, this says that ${\displaystyle g}$ is a closed immersion.

References

1. Hartshorne, Ch II. Exercise 4.10 harvnb error: no target: CITEREFHartshorne (help)
2. Since f is quasi-compact, its set-theoretic image is dense in its scheme-theoretic image. But since X is irreducible, two images must coincide.

• 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

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