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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 linearly disjoint ： ウィキペディア英語版 linearly disjoint In mathematics, algebras ''A'', ''B'' over a field ''k'' inside some field extension $\backslash Omega$ of ''k'' (e.g., universal field) are said to be linearly disjoint over ''k'' if the following equivalent conditions are met: *(i) The map $A\; \backslash otimes\_k\; B\; \backslash to\; AB$ induced by $(x,\; y)\; \backslash mapsto\; xy$ is injective. *(ii) Any ''k''-basis of ''A'' remains linearly independent over ''B''. *(iii) If $u\_i,\; v\_j$ are ''k''-bases for ''A'', ''B'', then the products $u\_i\; v\_j$ are linearly independent over ''k''. Note that, since every subalgebra of $\backslash Omega$ is a domain, (i) implies $A\; \backslash otimes\_k\; B$ is a domain (in particular reduced). One also has: ''A'', ''B'' are linearly disjoint over ''k'' if and only if subfields of $\backslash Omega$ generated by $A,\; B$, resp. are linearly disjoint over ''k''. (cf. tensor product of fields) Suppose ''A'', ''B'' are linearly disjoint over ''k''. If $A\text{'}\; \backslash subset\; A$, $B\text{'}\; \backslash subset\; B$ are subalgebras, then $A\text{'}$ and $B\text{'}$ are linearly disjoint over ''k''. Conversely, if any finitely generated subalgebras of algebras ''A'', ''B'' are linearly disjoint, then ''A'', ''B'' are linearly disjoint (since the condition involves only finite sets of elements.) == See also == *Tensor product of fields 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「linearly disjoint」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.