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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Chow's lemma ： ウィキペディア英語版 Chow's lemma 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: :If $X$ is a scheme that is proper over a noetherian base $S$, then there exists a projective $S$-scheme $X\text{'}$ and a surjective $S$-morphism $f\backslash colon\; X\text{'}\; \backslash to\; X$ that induces an isomorphism $f^(U)\; \backslash simeq\; U$ for some dense open $U\backslash subseteq\; X$. == Proof == The proof here is a standard one (cf. ). It is easy to reduce to the case when $X$ is irreducible, as follows. $X$ is noetherian since it is of finite type over a noetherian base. Then it's also topologically noetherian, and consists of a finite number of irreducible components $X\_i$, which are each proper over $S$ (because they're closed immersions in the scheme $X$ which is proper over $S$). If, within each of these irreducible components, there exists a dense open $U\_i\; \backslash subset\; X\_i$, then we can take $U\; :=\; \backslash bigsqcup\; (U\_i\; \backslash setminus\; \backslash bigcup\; \backslash limits\_\; X\_j)$. It is not hard to see that each of the disjoint pieces are dense in their respective $X\_i$, so the full set $U$ is dense in $X$. In addition, it's clear that we can similarly find a morphism $g$ which satisfies the density condition. Having reduced the problem, we now assume $X$ is irreducible. We recall that it must also be noetherian. Thus, we can find a finite open affine cover $X\; =\; \backslash bigcup\_^n\; U\_i$. $U\_i$ are quasi-projective over $S$; there are open immersions over $S$, $\backslash phi\_i\; \backslash colon\; U\_i\; \backslash to\; P\_i$ into some projective $S$-schemes $P\_i$. Put $U\; =\; \backslash cap\; U\_i$. $U$ is nonempty since $X$ is irreducible. Let :$\backslash phi\backslash colon\; U\; \backslash to\; P\; =\; P\_1\; \backslash times\_S\; \backslash cdots\; \backslash times\_S\; P\_n.$ be given by $\backslash phi\_i$'s restricted to $U$ over $S$. Let :$\backslash psi\backslash colon\; U\; \backslash to\; X\; \backslash times\_S\; P.$ be given by $U\; \backslash hookrightarrow\; X$ and $\backslash phi$ over $S$. $\backslash psi$ is then an immersion; thus, it factors as an open immersion followed by a closed immersion $X\text{'}\; \backslash to\; X\; \backslash times\_S\; P$. Let $f\backslash colon\; X\text{'}\; \backslash to\; X$ be the immersion followed by the projection. We claim $f$ induces $f^(U)\; \backslash simeq\; U$; for that, it is enough to show $f^(U)\; =\; \backslash psi(U)$. But this means that $\backslash psi(U)$ is closed in $U\; \backslash times\_S\; P$. $\backslash psi$ factorizes as $U\; \backslash overset\backslash to\; U\; \backslash times\_S\; P\; \backslash to\; X\; \backslash times\_S\; P$. $P$ is separated over $S$ and so the graph morphism $\backslash Gamma\_\backslash phi$ is a closed immersion. This proves our contention. It remains to show $X\text{'}$ is projective over $S$. Let $g\backslash colon\; X\text{'}\; \backslash to\; P$ be the closed immersion followed by the projection. Showing that $g$ is a closed immersion shows $X\text{'}$ is projective over $S$. This can be checked locally. Identifying $U\_i$ with its image in $P\_i$ we suppress $\backslash phi\_i$ from our notation. Let $V\_i\; =\; p\_i^(U\_i)$ where $p\_i\; \backslash colon\; P\; \backslash to\; P\_i$. We claim $g^(V\_i)$ are an open cover of $X\text{'}$. This would follow from $f^(U\_i)\; \backslash subset\; g^(V\_i)$ as sets. This in turn follows from $f\; =\; p\_i\; \backslash circ\; g$ 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, $X\; \backslash times\_S\; P\; \backslash to\; P$ is closed since it is a base extension of the proper morphism $X\; \backslash to\; S$. Thus, $g(X\text{'})$ is a closed subscheme covered by $V\_i$, and so it is enough to show that for each $i$ the map $g\backslash colon\; g^(V\_i)\; \backslash to\; V\_i$, denoted by $h$, is a closed immersion. Fix $i$. Let $Z$ be the graph of $u\backslash colon\; V\_i\; \backslash overset\backslash to\; U\_i\; \backslash hookrightarrow\; X$. It is a closed subscheme of $X\; \backslash times\_S\; V\_i$ since $X$ is separated over $S$. Let $q\_1\; \backslash colon\; X\; \backslash times\_S\; P\; \backslash to\; X,\backslash \; q\_2\; \backslash colon\; X\; \backslash times\_S\; P\; \backslash to\; P$ be the projections. We claim that $h$ factors through $Z$, which would imply $h$ is a closed immersion. But for $w\backslash colon\; U\text{'}\; \backslash to\; V\_i$ we have: :$v\; =\; \backslash Gamma\_u\; \backslash circ\; w\; \backslash quad\backslash Leftrightarrow\backslash quad\; q\_1\; \backslash circ\; v\; =\; u\; \backslash circ\; q\_2\; \backslash circ\; v\; \backslash quad\backslash Leftrightarrow\backslash quad\; q\_1\; \backslash circ\; \backslash psi\; =\; u\; \backslash circ\; q\_2\; \backslash circ\; \backslash psi\; \backslash quad\backslash Leftrightarrow\backslash quad\; q\_1\; \backslash circ\; \backslash psi\; =\; u\; \backslash circ\; \backslash phi.$ The last equality holds and thus there is $w$ that satisfies the first equality. This proves our claim. $\backslash square$ 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Chow's lemma」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.