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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 Algebraic independence ： ウィキペディア英語版 Algebraic independence In abstract algebra, a subset ''S'' of a field ''L'' is algebraically independent over a subfield ''K'' if the elements of ''S'' do not satisfy any non-trivial polynomial equation with coefficients in ''K''. In particular, a one element set is algebraically independent over ''K'' if and only if α is transcendental over ''K''. In general, all the elements of an algebraically independent set ''S'' over ''K'' are by necessity transcendental over ''K'', and over all of the field extensions over ''K'' generated by the remaining elements of ''S''. ==Example== The two real numbers $\backslash sqrt$ and $2\backslash pi+1$ are each transcendental numbers: they are not the roots of any nontrivial polynomial whose coefficients are rational numbers. Thus, each of the two singleton sets $\backslash$ and $\backslash$ are algebraically independent over the field $\backslash mathbb$ of rational numbers. However, the set $\backslash$ is ''not'' algebraically independent over the rational numbers, because the nontrivial polynomial :$P(x,y)=2x^2-y+1$ is zero when $x=\backslash sqrt$ and $y=2\backslash pi+1$. 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「Algebraic independence」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.