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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 algebraically closed group ： ウィキペディア英語版 algebraically closed group In mathematics, in the realm of group theory, a group $A\backslash$ is algebraically closed if any finite set of equations and inequations that "make sense" in $A\backslash$ already have a solution in $A\backslash$. This idea will be made precise later in the article. ==Informal discussion== Suppose we wished to find an element $x\backslash$ of a group $G\backslash$ satisfying the conditions (equations and inequations): ::$x^2=1\backslash$ ::$x^3=1\backslash$ ::$x\backslash ne\; 1\backslash$ Then it is easy to see that this is impossible because the first two equations imply $x=1\backslash$. In this case we say the set of conditions are inconsistent with $G\backslash$. (In fact this set of conditions are inconsistent with any group whatsoever.) \ |- ! style="background: #ddffdd;"|$\backslash underline\; \backslash$ |$1\; \backslash$ |$a\; \backslash$ |- ! style="background: #ddffdd;"|$\backslash underline\; \backslash$ |$a\; \backslash$ |$1\; \backslash$ |} |} Now suppose $G\backslash$ is the group with the multiplication table: Then the conditions: ::$x^2=1\backslash$ ::$x\backslash ne\; 1\backslash$ have a solution in $G\backslash$, namely $x=a\backslash$. However the conditions: ::$x^4=1\backslash$ ::$x^2a^\; =\; 1\backslash$ Do not have a solution in $G\backslash$, as can easily be checked. \ ! style="background: #ddffdd;"|$\backslash underline\; \backslash$ ! style="background: #ddffdd;"|$\backslash underline\; \backslash$ |- ! style="background: #ddffdd;"|$\backslash underline\; \backslash$ |$1\; \backslash$ |$a\; \backslash$ |$b\; \backslash$ |$c\; \backslash$ |- ! style="background: #ddffdd;"|$\backslash underline\; \backslash$ |$a\; \backslash$ |$1\; \backslash$ |$c\; \backslash$ |$b\; \backslash$ |- ! style="background: #ddffdd;"|$\backslash underline\; \backslash$ |$b\; \backslash$ |$c\; \backslash$ |$a\; \backslash$ |$1\; \backslash$ |- ! style="background: #ddffdd;"|$\backslash underline\; \backslash$ |$c\; \backslash$ |$b\; \backslash$ |$1\; \backslash$ |$a\; \backslash$ |} |} However if we extend the group $G\; \backslash$ to the group $H\; \backslash$ with multiplication table: Then the conditions have two solutions, namely $x=b\; \backslash$ and $x=c\; \backslash$. Thus there are three possibilities regarding such conditions: * They may be inconsistent with $G\; \backslash$ and have no solution in any extension of $G\; \backslash$. * They may have a solution in $G\; \backslash$. * They may have no solution in $G\; \backslash$ but nevertheless have a solution in some extension $H\; \backslash$ of $G\; \backslash$. It is reasonable to ask whether there are any groups $A\; \backslash$ such that whenever a set of conditions like these have a solution at all, they have a solution in $A\; \backslash$ itself? The answer turns out to be "yes", and we call such groups algebraically closed groups. 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「algebraically closed group」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.