Parabolic induction

Template:Distinguish 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 $f:X'\to X$ that induces $f^{-1}(U)\simeq U$ for some open dense subset U.

Chow's lemma is one of the foundational results in algebraic geometry.

Proof

The proof here is a standard one (cf. Template:Harvnb).

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 a finite open affine cover $X=\bigcup _{i=1}^{n}U_{i}$. $U_{i}$ are quasi-projective over S; there are open immersions over S $\phi _{i}:U_{i}\to P_{i}$ into some projective S-schemes $P_{i}$. Put $U=\cap U_{i}$. U is nonempty since X is irreducible. Let

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

be given by $\phi _{i}$'s restricted to $U$ over S. Let

$\psi :U\to X\times _{S}P.$

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

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

Let $V_{i}=p_{i}^{-1}(U_{i})$ where $p_{i}:P\to P_{i}$. We claim $g^{-1}(V_{i})$ are an open cover of $X'$. This would follow from $f^{-1}(U_{i})\subset g^{-1}(V_{i})$ as sets. This in turn follows from $f=p_{i}\circ g$ on $U_{i}$ as functions on the underlying topological space. Since X is separated over S and $U_{i}$ is dense, this is clear from looking at the relevant commutative diagram. Now, $X\times _{S}P\to P$ is closed since it is a base extension of the proper morphism $X\to S$. Thus, $g(X')$ is a closed subscheme covered by $V_{i}$ and so it is enough to show for each i $g:g^{-1}(V_{i})\to V_{i}$, denoted by $h$, is a closed immersion.

Fix i. Let $Z$ be the graph of $u:V_{i}{\overset {p_{i}}{\to }}U_{i}\hookrightarrow X$. It is a closed subscheme of $X\times _{S}V_{i}$ since $X$ is separated over S. Let $q_{1}:X\times _{S}P\to X,q_{2}:X\times _{S}P\to P$ be the projections. We claim $h$ factors through $Z$, which would imply $h$ is a closed immersion. But for $w:U'\to V_{i}$ we have:

$v=\Gamma _{u}\circ w\Leftrightarrow q_{1}\circ v=u\circ q_{2}\circ v\Leftrightarrow q_{1}\circ \psi =u\circ q_{2}\circ \psi \Leftrightarrow q_{1}\circ \psi =u\circ \phi .$

The last equality holds and thus there is w that satisfies the first equality. This proves our claim. $\square$

References

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