翻訳と辞書 Words near each other ・ Krukówka, Podlaskie Voivodeship ・ Krukówka, Łódź Voivodeship ・ Krukówko ・ Krul ・ Krulak ・ Krulak Mendenhall mission ・ Krulj ・ Krull ・ Krull (film) ・ Krull (video game) ・ Krull dimension ・ Krull ring ・ Krull's principal ideal theorem ・ Krull's theorem ・ Krullsbach ・ Krull–Akizuki theorem ・ Krull–Schmidt category ・ Krull–Schmidt theorem ・ Krum ・ Krum (Bulgarian singer) ・ Krum (disambiguation) ・ Krum Bibishkov ・ Krum Georgiev ・ Krum High School ・ Krum Independent School District ・ Krum Lovkov ・ Krum Milev ・ Krum Rock ・ Krum Stoyanov ・ Krum Yanev Dictionary Lists mini英和辞書 mini和英辞書 Webster 1913 Latin-English FOLDOC Wikipedia English ウィキペディア

翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Krull–Akizuki theorem ： ウィキペディア英語版 Krull–Akizuki theorem In algebra, the Krull–Akizuki theorem states the following: let ''A'' be a one-dimensional reduced noetherian ring,〔In this article, a ring is commutative and has unity.〕 ''K'' its total ring of fractions. If ''B'' is a subring of a finite extension ''L'' of ''K'' containing ''A'' and is not a field, then ''B'' is a one-dimensional noetherian ring. Furthermore, for every nonzero ideal ''I'' of ''B'', $B/I$ is finite over ''A''. Note that the theorem does not say that ''B'' is finite over ''A''. The theorem does not extend to higher dimension. One important consequence of the theorem is that the integral closure of a Dedekind domain ''A'' in a finite extension of the field of fractions of ''A'' is again a Dedekind domain. This consequence does generalize to a higher dimension: the Mori–Nagata theorem states that the integral closure of a noetherian domain is a Krull domain. == Proof == Here, we give a proof when $L\; =\; K$. Let $\backslash mathfrak\_i$ be minimal prime ideals of ''A''; there are finitely many of them. Let $K\_i$ be the field of fractions of $A/$ and $I\_i$ the kernel of the natural map $B\; \backslash to\; K\; \backslash to\; K\_i$. Then we have: :$A/\; \backslash subset\; B/\; \backslash subset\; K\_i$. Now, if the theorem holds when ''A'' is a domain, then this implies that ''B'' is a one-dimensional noetherian domain since each $B/$ is and since $B\; =\; \backslash prod\; B/$. Hence, we reduced the proof to the case ''A'' is a domain. Let $0\; \backslash ne\; I\; \backslash subset\; B$ be an ideal and let ''a'' be a nonzero element in the nonzero ideal $I\; \backslash cap\; A$. Set $I\_n\; =\; a^nB\; \backslash cap\; A\; +\; aA$. Since $A/aA$ is a zero-dim noetherian ring; thus, artinian, there is an ''l'' such that $I\_n\; =\; I\_l$ for all $n\; \backslash ge\; l$. We claim :$a^l\; B\; \backslash subset\; a^B\; +\; A.$ Since it suffices to establish the inclusion locally, we may assume ''A'' is a local ring with the maximal ideal $\backslash mathfrak$. Let ''x'' be a nonzero element in ''B''. Then, since ''A'' is noetherian, there is an ''n'' such that $\backslash mathfrak^\; \backslash subset\; x^\; A$ and so $a^x\; \backslash in\; a^B\; \backslash cap\; A\; \backslash subset\; I\_$. Thus, :$a^n\; x\; \backslash in\; a^\; B\; \backslash cap\; A\; +\; A.$ Now, assume ''n'' is a minimum integer such that $n\; \backslash ge\; l$ and the last inclusion holds. If $n\; >\; l$, then we easily see that $a^n\; x\; \backslash in\; I\_$. But then the above inclusion holds for $n-1$, contradiction. Hence, we have $n\; =\; l$ and this establishes the claim. It now follows: :$B/\; \backslash simeq\; a^l\; B/a^\; B\; \backslash subset\; (a^B\; +\; A)/a^\; B\; \backslash simeq\; A/.$ Hence, $B/$ has finite length as ''A''-module. In particular, the image of ''I'' there is finitely generated and so ''I'' is finitely generated. Finally, the above shows that $B/$ has zero dimension and so ''B'' has dimension one. $\backslash square$ 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「Krull–Akizuki theorem」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.