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In the subfield of algebra named field theory, a separable extension is an algebraic field extension such that for every , the minimal polynomial of over ''F'' is a separable polynomial (i.e., has distinct roots; see below for the definition in this context).〔Isaacs, p. 281〕 Otherwise, the extension is called inseparable. There are other equivalent definitions of the notion of a separable algebraic extension, and these are outlined later in the article. The importance of separable extensions lies in the fundamental role they play in Galois theory in finite characteristic. More specifically, a finite degree field extension is Galois if and only if it is both normal and separable.〔Isaacs, Theorem 18.13, p. 282〕 Since algebraic extensions of fields of characteristic zero, and of finite fields, are separable, separability is not an obstacle in most applications of Galois theory.〔Isaacs, Theorem 18.11, p. 281〕〔Isaacs, p. 293〕 For instance, every algebraic (in particular, finite degree) extension of the field of rational numbers is necessarily separable. Despite the ubiquity of the class of separable extensions in mathematics, its extreme opposite, namely the class of purely inseparable extensions, also occurs quite naturally. An algebraic extension is a purely inseparable extension if and only if for every , the minimal polynomial of over ''F'' is ''not'' a separable polynomial (i.e., ''does not'' have distinct roots).〔Isaacs, p. 298〕 For a field ''F'' to possess a non-trivial purely inseparable extension, it must necessarily be an infinite field of prime characteristic (i.e. specifically, imperfect), since any algebraic extension of a perfect field is necessarily separable.〔 ==Informal discussion== An arbitrary polynomial ''f'' with coefficients in some field ''F'' is said to have ''distinct roots'' if and only if it has deg(''f'') roots in some extension field . For instance, the polynomial ''g''(''X'')=''X''2+1 with real coefficients has precisely deg(''g'')=2 roots in the complex plane; namely the imaginary unit ''i'', and its additive inverse −''i'', and hence ''does have'' distinct roots. On the other hand, the polynomial ''h''(''X'')=(''X''−2)2 with real coefficients ''does not'' have distinct roots; only 2 can be a root of this polynomial in the complex plane and hence it has only one, and not deg(''h'')=2 roots. To test if a polynomial has distinct roots, it is not necessary to consider explicitly any field extension nor to compute the roots: a polynomial has distinct roots if and only if the greatest common divisor of the polynomial and its derivative is a constant. For instance, the polynomial ''g''(''X'')=''X''2+1 in the above paragraph, has 2''X'' as derivative, and, over a field of characteristic different of 2, we have ''g''(''X'') - (1/2 ''X'') 2''X'' = 1, which proves, by Bézout's identity, that the greatest common divisor is a constant. On the other hand, over a field where 2=0, the greatest common divisor is ''g'', and we have ''g''(''X'') = (''X''+1)2 has 1=-1 as double root. On the other hand, the polynomial ''h'' ''does not'' have distinct roots, whichever is the field of the coefficients, and indeed, ''h''(''X'')=(''X''−2)2, its derivative is 2 (''X''-2) and divides it, and hence ''does'' have a factor of the form for ). Although an arbitrary polynomial with rational or real coefficients may not have distinct roots, it is natural to ask at this stage whether or not there exists an ''irreducible polynomial'' with rational or real coefficients that does not have distinct roots. The polynomial ''h''(''X'')=(''X''−2)2 does not have distinct roots but it is not irreducible as it has a non-trivial factor (''X''−2). In fact, it is true that there is ''no irreducible polynomial with rational or real coefficients that does not have distinct roots''; in the language of field theory, every algebraic extension of or is separable and hence both of these fields are perfect. 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「Separable extension」の詳細全文を読む スポンサード リンク
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