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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 Fitting's lemma ： ウィキペディア英語版 Fitting lemma The Fitting lemma, named after the mathematician Hans Fitting, is a basic statement in abstract algebra. Suppose ''M'' is a module over some ring. If ''M'' is indecomposable and has finite length, then every endomorphism of ''M'' is either bijective or nilpotent. As an immediate consequence, we see that the endomorphism ring of every finite-length indecomposable module is local. A version of Fitting's lemma is often used in the representation theory of groups. This is in fact a special case of the version above, since every ''K''-linear representation of a group ''G'' can be viewed as a module over the group algebra ''KG''. To prove Fitting's lemma, we take an endomorphism ''f'' of ''M'' and consider the following two sequences of submodules. The first sequence is the descending sequence im(''f''), im(''f'' 2), im(''f'' 3),..., the second sequence is the ascending sequence ker(''f''), ker(''f'' 2), ker(''f'' 3),.... Because ''M'' has finite length, the first sequence cannot be ''strictly'' decreasing forever, so there exists some ''n'' with im(''f'' ''n'') = im(''f'' ''n''+1). Likewise (as ''M'' has finite length) the second sequence cannot be ''strictly'' increasing forever, so there exists some ''m'' with ker(''f'' ''m'') = ker(''f'' ''m''+1). It is easily seen that im(''f'' ''n'') = im(''f'' ''n''+1) yields im(''f'' ''n'') = im(''f'' ''n''+1) = im(''f'' ''n''+2) = ..., and that ker(''f'' ''m'') = ker(''f'' ''m''+1) yields ker(''f'' ''m'') = ker(''f'' ''m''+1) = ker(''f'' ''m''+2) = ... . Putting ''k'' = max(''m'',''n'' ), it now follows that im(''f'' ''k'') = im(''f'' 2''k'') and ker(''f'' ''k'') = ker(''f'' 2''k''). Hence, $\backslash mathrm\backslash left(f^k\backslash right)\; \backslash cap\; \backslash mathrm\backslash left(f^k\backslash right)\; =\; 0$ (because every $x\backslash in\; \backslash mathrm\backslash left(f^k\backslash right)\; \backslash cap\; \backslash mathrm\backslash left(f^k\backslash right)$ satisfies $x=f^k\backslash left(y\backslash right)$ for some $y\backslash in\; M$ but also $f^k\backslash left(x\backslash right)=0$, so that $0=f^k\backslash left(x\backslash right)=f^k\backslash left(f^k\backslash left(y\backslash right)\backslash right)=f^\backslash left(y\backslash right)$, therefore $y\backslash in\backslash mathrm\backslash left(f^\backslash right)=\backslash mathrm\backslash left(f^k\backslash right)$ and thus $0=f^k\backslash left(y\backslash right)=x$) and $\backslash mathrm\backslash left(f^k\backslash right)\; +\; \backslash mathrm\backslash left(f^k\backslash right)\; =\; M$ (since for every $x\backslash in\; M$, there exists some $y\backslash in\; M$ such that $f^k\backslash left(x\backslash right)=f^\backslash left(y\backslash right)$ (since $f^k\backslash left(x\backslash right)\backslash in\backslash mathrm\backslash left(f^k\backslash right)=\backslash mathrm\backslash left(f^\backslash right)$), and thus $f^k\backslash left(x-f^k\backslash left(y\backslash right)\backslash right)=f^k\backslash left(x\backslash right)-f^k\backslash left(f^k\backslash left(y\backslash right)\backslash right)=f^k\backslash left(x\backslash right)-f^\backslash left(y\backslash right)=0$, so that $x-f^k\backslash left(y\backslash right)\backslash in\backslash mathrm\backslash left(f^k\backslash right)$ and thus $x\backslash in\; \backslash mathrm\backslash left(f^k\backslash right)+f^k\backslash left(y\backslash right)\backslash subseteq\; \backslash mathrm\backslash left(f^k\backslash right)\; +\; \backslash mathrm\backslash left(f^k\backslash right)$). Consequently, ''M'' is the direct sum of im(''f'' ''k'') and ker(''f'' ''k''). Because ''M'' is indecomposable, one of those two summands must be equal to ''M'', and the other must be equal to . Depending on which of the two summands is zero, we find that ''f'' is bijective or nilpotent.〔Jacobson (2009), p. 113–114.〕 == Notes == 〔 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「Fitting 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.