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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 Prime avoidance lemma ： ウィキペディア英語版 Prime avoidance lemma In algebra, the prime avoidance lemma says that if an ideal ''I'' in a commutative ring ''R'' is contained in a union of finitely many prime ideals ''P''''i'''s, then it is contained in ''P''''i'' for some ''i''. There are many variations of the lemma (cf. Hochster); for example, if the ring ''R'' contains an infinite field or a finite field of sufficiently large cardinality, then the statement follows from a fact in linear algebra that a vector space over an infinite field or a finite field of large cardinality is not a finite union of its proper vector subspaces.〔Proof of the fact: suppose the vector space is a finite union of proper subspaces. Consider a finite product of linear functionals, each of which vanishes on a proper subspace that appears in the union; then it is a nonzero polynomial vanishing identically, a contradiction.〕 == Statement and proof == The following statement and argument are perhaps the most standard. Statement: Let ''E'' be a subset of ''R'' that is an additive subgroup of ''R'' and is multiplicatively closed. Let $I\_1,\; I\_2,\; \backslash dots,\; I\_n,\; n\; \backslash ge\; 1$ be ideals such that $I\_i$ are prime ideals for $i\; \backslash ge\; 3$. If ''E'' is not contained in any of $I\_i$'s, then ''E'' is not contained in the union $\backslash cup\; I\_i$. Proof by induction on ''n'': The idea is to find an element that is in ''E'' and not in any of $I\_i$'s. The basic case ''n'' = 1 is trivial. Next suppose ''n'' ≥ 2. For each ''i'' choose :$z\_i\; \backslash in\; E\; -\; \backslash cup\_\; I\_j$ where the set on the right is nonempty by inductive hypothesis. We can assume $z\_i\; \backslash in\; I\_i$ for all ''i''; otherwise, some $z\_i$ avoids all the $I\_i$'s and we are done. Put :$z\; =\; z\_1\; \backslash dots\; z\_\; +\; z\_n$. Then ''z'' is in ''E'' but not in any of $I\_i$'s. Indeed, if ''z'' is in $I\_i$ for some $i\; \backslash le\; n\; -\; 1$, then $z\_n$ is in $I\_i$, a contradiction. Suppose ''z'' is in $I\_n$. Then $z\_1\; \backslash dots\; z\_$ is in $I\_n$. If ''n'' is 2, we are done. If ''n'' > 2, then, since $I\_n$ is a prime ideal, some is in $I\_n$, a contradiction. 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「Prime avoidance 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.