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 $f:X^{\prime }\to X$ 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 ${\varphi }_i:U_i\to P_i$ into some projective S-schemes $P_i$. Put $U=\cap U_i$. U is nonempty since X is irreducible. Let

$\varphi :U\to P=P_1{\times }_S\cdots {\times }_S{P}_n.$

be given by ${\varphi }_i$'s restricted to $U$ over S. Let

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

be given by $U\hookrightarrow X$ and $\varphi$ over S. $\psi$ is then an immersion; thus, it factors as an open immersion followed by a closed immersion $X^{\prime }\to X{\times }_{S}P$. Let $f:X^{\prime }\to X$ 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 $U\stackrel{{\mathrm{\Gamma }}_{\varphi }}{\to }U{\times }_{S}P\to X{\times }_{S}P$. $P$ is separated over S and so the graph morphism ${\mathrm{\Gamma }}_{\varphi }$ is a closed immersion. This proves our contention.

It remains to show $X^{\prime }$ is projective over S. Let $g:X^{\prime }\to P$ be the closed immersion followed by the projection. Showing that g is a closed immersion shows $X^{\prime }$ is projective over S. This can be checked locally. Identifying $U_i$ with its image in $P_i$ we suppress ${\varphi }_i$ 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 $X^{\prime }$. 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, $g(X^{\prime })$ 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 $u:V_i\stackrel{p_i}{\to }{U}_i\hookrightarrow X$. 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 $w:U^{\prime }\to V_i$ we have:

$v={\mathrm{\Gamma }}_u\circ w\iff q_1\circ v=u\circ q_2\circ v\iff q_1\circ \psi =u\circ q_2\circ \psi \iff q_1\circ \psi =u\circ \varphi .$

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

References

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

• Template:EGA
• Template:Hartshorne AG