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In the subfield of [[Abstract algebra|algebra]] named [[field theory (mathematics)|field theory]], a '''separable extension''' is an [[algebraic field extension]] <math>E\supset F</math> such that for every <math>\alpha\in E</math>, the [[minimal polynomial (field theory)|minimal polynomial]] of <math>\alpha</math> over ''F'' is a [[separable polynomial]] (i.e., has distinct [[Root of a polynomial|roots]]; see [[#Separable and inseparable polynomials|below]] for the definition in this context).<ref name="Isaacs281">Isaacs, p. 281</ref> 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.
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The importance of separable extensions lies in the fundamental role they play in [[Galois theory]] in [[characteristic (algebra)#Case of fields|finite characteristic]]. More specifically, a finite degree field extension is [[Galois extension|Galois]] if and only if it is both [[Normal extension|normal]] and separable.<ref>Isaacs, Theorem 18.13, p. 282</ref> 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]].<ref name="Isaacs18.11p281">Isaacs, Theorem 18.11, p. 281</ref><ref>Isaacs, p. 293</ref> 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 <math>E\supset F</math> is a purely inseparable extension if and only if for every <math>\alpha\in E\setminus F</math>, the minimal polynomial of <math>\alpha</math> over ''F'' is ''not'' a [[separable polynomial]] (i.e., ''does not'' have distinct roots).<ref name="Isaacs298">Isaacs, p. 298</ref> 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 field|imperfect]]), since any algebraic extension of a [[perfect field]] is necessarily separable.<ref name="Isaacs18.11p281"/>
 
==Informal discussion==
''The reader may wish to assume that, in what follows, F is the field of rational, real or complex numbers, unless otherwise stated.''
 
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]] <math>E\supseteq F</math>. For instance, the polynomial ''g''(''X'')=''X''<sup>2</sup>+1 with real coefficients has precisely deg(''g'')=2 roots in the complex plane; namely the [[imaginary unit]] ''i'', and its additive inverse &minus;''i'', and hence ''does have'' distinct roots. On the other hand, the polynomial ''h''(''X'')=(''X''&minus;2)<sup>2</sup> 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 [[polynomial greatest common divisor|greatest common divisor]] of the polynomial and its [[derivative]] is a constant.
For instance, the polynomial ''g''(''X'')=''X''<sup>2</sup>+1 in the above paragraph, has 2''X'' as derivative, and, over a field of [[characteristic of a field|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)<sup>2</sup> 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''&minus;2)<sup>2</sup>, its derivative is 2 (''X''-2) and divides it, and hence ''does'' have a factor of the form <math>(X-\alpha)^2</math> for <math>\alpha=2</math>).
 
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''&minus;2)<sup>2</sup> does not have distinct roots but it is not irreducible as it has a non-trivial factor (''X''&minus;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 <math>\mathbb{Q}</math> or <math>\mathbb{R}</math> is separable and hence both of these fields are [[Perfect field|perfect]].
 
==Separable and inseparable polynomials==
A polynomial ''f'' in ''F''[''X''] is a ''separable polynomial'' if and only if every irreducible factor of ''f'' in ''F''[''X''] has distinct roots.<ref>Isaacs, p. 280</ref> The separability of a polynomial depends on the field in which its coefficients are considered to lie; for instance, if ''g'' is an inseparable polynomial in ''F''[''X''], and one considers a [[splitting field]], ''E'', for ''g'' over ''F'', ''g'' is necessarily separable in ''E''[''X''] since an arbitrary irreducible factor of ''g'' in ''E''[''X''] is linear and hence has distinct roots.<ref name="Isaacs281"/> Despite this, a separable polynomial ''h'' in ''F''[''X''] must necessarily be separable over ''every'' extension field of ''F''.<ref>Isaacs, Lemma 18.10, p. 281</ref>
 
Let ''f'' in ''F''[''X''] be an irreducible polynomial and ''f''<nowiki>'</nowiki> its [[formal derivative]]. Then the following are equivalent conditions for ''f'' to be separable; that is, to have distinct roots:
* If <math>E\supseteq F</math> and <math>\alpha\in E</math>, then <math>(X-\alpha)^2</math> does not divide ''f'' in ''E''[''X''].<ref name=IsaacsLem18.7>Isaacs, Lemma 18.7, p. 280</ref>
* There exists <math>K \supseteq F</math> such that ''f'' has deg(''f'') roots in ''K''.<ref name=IsaacsLem18.7/>
* ''f'' and ''f''<nowiki>'</nowiki> do not have a common root in any extension field of ''F''.<ref>Isaacs, Theorem 19.4, p. 295</ref>
* ''f''<nowiki>'</nowiki> is not the zero polynomial.<ref>Isaacs, Corollary 19.5, p. 296</ref>
 
By the last condition above, if an irreducible polynomial does not have distinct roots, its derivative must be zero. Since the formal derivative of a positive degree polynomial can be zero only if the field has prime characteristic, for an irreducible polynomial to not have distinct roots its coefficients must lie in a field of prime characteristic. More generally, if an irreducible (non-zero) polynomial ''f'' in ''F''[''X''] does not have distinct roots, not only must the characteristic of ''F'' be a (non-zero) prime number ''p'', but also ''f''(''X'')=''g''(''X''<sup>''p''</sup>) for some ''irreducible'' polynomial ''g'' in ''F''[''X''].<ref>Isaacs, Corollary 19.6, p. 296</ref> By repeated application of this property, it follows that in fact, <math>f(X)=g(X^{p^n})</math> for a non-negative integer ''n'' and some ''separable irreducible'' polynomial ''g'' in ''F''[''X''] (where ''F'' is assumed to have prime characteristic ''p'').<ref>Isaacs, Corollary 19.9, p. 298</ref>
 
By the property noted in the above paragraph, if ''f'' is an irreducible (non-zero) polynomial with coefficients in the field ''F'' of prime characteristic ''p'', and does not have distinct roots, it is possible to write ''f''(''X'')=''g''(''X''<sup>''p''</sup>). Furthermore, if <math>g(X)=\sum a_iX^i</math>, and if the [[Frobenius endomorphism]] of ''F'' is an [[automorphism]], ''g'' may be written as <math>g(X)=\sum b_i^{p}X^i</math>, and in particular, <math>f(X)=g(X^p)=\sum b_i^{p}X^{pi}=(\sum b_iX^i)^p</math>; a contradiction of the irreducibility of ''f''. Therefore, if ''F''[''X''] possesses an inseparable irreducible (non-zero) polynomial, then the Frobenius endomorphism of ''F'' cannot be an automorphism (where ''F'' is assumed to have prime characteristic ''p'').<ref>Isaacs, Theorem 19.7, p. 297</ref>
 
If ''K'' is a finite field of prime characteristic ''p'', and if ''X'' is an indeterminant, then the field of rational functions over ''K'', ''K''(''X''), is necessarily [[Imperfect field|imperfect]]. Furthermore, the polynomial ''f''(''Y'')=''Y''<sup>''p''</sup>&minus;''X'' is inseparable.<ref name="Isaacs281"/> (To see this, note that there is some extension field <math>E\supseteq K(X)</math> in which ''f'' has a root <math>\alpha</math>; necessarily, <math>\alpha^{p}=X</math> in ''E''. Therefore, working over ''E'', <math>f(Y)=Y^p-X=Y^p-\alpha^{p}=(Y-\alpha)^p</math> (the final equality in the sequence follows from [[freshman's dream]]), and ''f'' does not have distinct roots.) More generally, if ''F'' is any field of (non-zero) prime characteristic for which the [[Frobenius endomorphism]] is not an automorphism, ''F'' possesses an inseparable algebraic extension.<ref name="Isaacs299">Isaacs, p. 299</ref>
 
A field ''F'' is [[Perfect field|perfect]] if and only if all of its algebraic extensions are separable (in fact, all algebraic extensions of ''F'' are separable if and only if all finite degree extensions of ''F'' are separable). By the argument outlined in the above paragraphs, it follows that ''F'' is perfect if and only if ''F'' has characteristic zero, or ''F'' has (non-zero) prime characteristic ''p'' and the [[Frobenius endomorphism]] of ''F'' is an automorphism.
 
==Properties==
*If <math>E\supseteq F</math> is an algebraic field extension, and if <math>\alpha,\beta\in E</math> are separable over ''F'', then <math>\alpha+\beta</math> and <math>\alpha\beta</math> are separable over ''F''. In particular, the set of all elements in ''E'' separable over ''F'' forms a field.<ref>Isaacs, Lemma 19.15, p. 300</ref>
*If <math>E\supseteq L\supseteq F</math> is such that <math>E\supseteq L</math> and <math>L\supseteq F</math> are separable extensions, then <math>E\supseteq F</math> is separable.<ref>Isaacs, Corollary 19.17, p. 301</ref> Conversely, if <math>E\supseteq F</math> is a separable algebraic extension, and if ''L'' is any intermediate field, then <math>E\supseteq L</math> and <math>L\supseteq F</math> are separable extensions.<ref>Isaacs, Corollary 18.12, p. 281</ref>
*If <math>E\supseteq F</math> is a finite degree separable extension, then it has a primitive element; i.e., there exists <math>\alpha\in E</math> with <math>E=F[\alpha]</math>. This fact is also known as the ''[[primitive element theorem]]'' or ''Artin's theorem on primitive elements''.
 
==Separable extensions within algebraic extensions==
Separable extensions occur quite naturally within arbitrary algebraic field extensions. More specifically, if <math>E\supseteq F</math> is an algebraic extension and if <math>S=\{\alpha\in E|\alpha \mbox{ is separable over } F\}</math>, then ''S'' is the unique intermediate field that is ''separable'' over ''F'' and over which ''E'' is ''purely inseparable''.<ref>Isaacs, Theorem 19.14, p. 300</ref> If <math>E\supseteq F</math> is a finite degree extension, the degree [''S'' : ''F''] is referred to as the '''separable part''' of the degree of the extension <math>E\supseteq F</math> (or the '''separable degree''' of ''E''/''F''), and is often denoted by [''E'' : ''F'']<sub>sep</sub> or [''E'' : ''F'']<sub>s</sub>.<ref name="Isaacs302">Isaacs, p. 302</ref> The '''inseparable degree''' of ''E''/''F'' is the quotient of the degree by the separable degree. When the characteristic of ''F'' is ''p''&nbsp;>&nbsp;0, it is a power of ''p''.<ref>{{harvnb|Lang|2002|loc=Corollary V.6.2}}</ref> Since the extension <math>E\supseteq F</math> is separable if and only if <math>S=E</math>, it follows that for separable extensions, [''E'' : ''F'']=[''E'' : ''F'']<sub>sep</sub>, and conversely. If <math>E\supseteq F</math> is not separable (i.e., inseparable), then [''E'' : ''F'']<sub>sep</sub> is necessarily a non-trivial divisor of [''E'' : ''F''], and the quotient is necessarily a power of the characteristic of ''F''.<ref name="Isaacs302"/>
 
On the other hand, an arbitrary algebraic extension <math>E\supseteq F</math> may not possess an intermediate extension ''K'' that is ''purely inseparable'' over ''F'' and over which ''E'' is ''separable'' (however, such an intermediate extension does exist when <math>E\supseteq F</math> is a finite degree normal extension (in this case, ''K'' can be the fixed field of the Galois group of ''E'' over ''F'')). If such an intermediate extension does exist, and if [''E'' : ''F''] is finite, then if ''S'' is defined as in the previous paragraph, [''E'' : ''F'']<sub>sep</sub>=[''S'' : ''F'']=[''E'' : ''K''].<ref>Isaacs, Theorem 19.19, p. 302</ref> One known proof of this result depends on the [[primitive element theorem]], but there does exist a proof of this result independent of the primitive element theorem (both proofs use the fact that if <math>K\supseteq F</math> is a purely inseparable extension, and if ''f'' in ''F''[''X''] is a separable irreducible polynomial, then ''f'' remains irreducible in ''K''[''X'']<ref>Isaacs, Lemma 19.20, p. 302</ref>). The equality above ([''E'' : ''F'']<sub>sep</sub>=[''S'' : ''F'']=[''E'' : ''K'']) may be used to prove that if <math>E\supseteq U\supseteq F</math> is such that [''E'' : ''F''] is finite, then [''E'' : ''F'']<sub>sep</sub>=[''E'' : ''U'']<sub>sep</sub>[''U'' : ''F'']<sub>sep</sub>.<ref>Isaacs, Corollary 19.21, p. 303</ref>
 
If ''F'' is any field, the '''separable closure''' ''F''<sup>sep</sup> of ''F'' is the field of all elements in an [[algebraic closure]] of ''F'' that are separable over ''F''. This is the maximal [[Galois extension]] of ''F''.  By definition, ''F'' is perfect if and only if its separable and algebraic closures coincide (in particular, the notion of a separable closure is only interesting for imperfect fields).
 
== The definition of separable non-algebraic extension fields ==
Although many important applications of the theory of separable extensions stem from the context of algebraic field extensions, there are important instances in mathematics where it is profitable to study (not necessarily algebraic) separable field extensions.
 
Let <math>F/k</math> be a field extension and let ''p'' be the [[characteristic exponent of a field|characteristic exponent]] of <math>k</math>.<ref>The characteristic exponent of ''k'' is ''1'' if ''k'' has characteristic zero; otherwise, it is the characteristic of ''k''.</ref> For any field extension ''L'' of ''k'', we write <math>F_L = L \otimes_k F</math> (cf. [[Tensor product of fields]].) Then ''F'' is said to be ''separable over <math>k</math>'' if the following equivalent conditions are met:
*<math>F^p</math> and <math>k</math> are [[linearly disjoint]] over <math>k^p</math>
*<math>F_{k^{1/p}}</math> is reduced.
*<math>F_L</math> is reduced for all field extensions ''L'' of ''k''.
(In other words, ''F'' is separable over ''k'' if ''F'' is a [[separable algebra|separable ''k''-algebra]].)
 
Suppose there is some field extension ''L'' of ''k'' such that <math>F_L</math> is a domain. Then <math>F</math> is separable over ''k'' if and only if the field of fractions of <math>F_L</math> is separable over ''L''.
 
An algebraic element of ''F'' is said to be ''separable over <math>k</math>'' if its minimal polynomial is separable. If <math>F/k</math> is an algebraic extension, then the following are equivalent.
*''F'' is separable over ''k''.
*''F'' consists of elements that are separable over ''k''.
*Every subextension of ''F/k'' is separable.
*Every finite subextension of ''F/k'' is separable.
 
If <math>F/k</math> is finite extension, then the following are equivalent.
*(i) ''F'' is separable over ''k''.
*(ii) <math>F = k(a_1, ..., a_r)</math> where <math>a_1, ..., a_r</math> are separable over ''k''.
*(iii) In (ii), one can take <math>r = 1.</math>
*(iv) For some very large field <math>\Omega</math>, there are precisely <math>[F : k]</math> ''k''-isomorphisms from <math>F</math> to <math>\Omega</math>.
In the above, (iii) is known as the ''[[primitive element theorem]]''.
 
Fix the algebraic closure <math>\overline{k}</math>, and denote by <math>k_s</math> the set of all elements of <math>\overline{k}</math> that are separable over ''k''. <math>k_s</math> is then separable algebraic over ''k'' and any separable algebraic subextension of <math>\overline{k}</math> is contained in <math>k_s</math>; it is called the '''separable closure''' of ''k'' (inside <math>\overline{k}</math>). <math>\overline{k}</math> is then purely inseparable over <math>k_s</math>. Put in another way, ''k'' is perfect if and only if <math>\overline{k} = k_s</math>.
 
== Differential criteria ==
The separability can be studied with the aid of derivations and [[Kähler differential]]s. Let <math>F</math> be a [[finitely generated field extension]] of a field <math>k</math>. Then
:<math>\dim_F \operatorname{Der}_k(F, F) \ge \operatorname{tr.deg}_k F</math>
where the equality holds if and only if ''F'' is separable over ''k''.
 
In particular, if <math>F/k</math> is an algebraic extension, then <math>\operatorname{Der}_k(F, F) = 0</math> if and only if <math>F/k</math> is separable.
 
Let <math>D_1, ..., D_m</math> be a basis of <math>\operatorname{Der}_k(F, F)</math> and <math>a_1, ..., a_m \in F</math>. Then <math>F</math> is separable algebraic over <math>k(a_1, ..., a_m)</math> if and only if the matrix <math>D_i(a_j)</math> is invertible. In particular, when <math>m = \operatorname{tr.deg}_k F</math>, <math>\{ a_1, ..., a_m \}</math> above is called the ''separating transcendence basis.''
 
==See also==
*[[Purely inseparable extension]]
*[[Separable polynomial]]
*[[Perfect field]]
*[[Primitive element theorem]]
*[[Normal extension]]
*[[Galois extension]]
*[[Algebraic closure]]
*[[Derivation (algebra)|Derivation]]
 
==Notes==
{{reflist|30em}}
 
==References==
* Borel, A. ''Linear algebraic groups'', 2nd ed.
* P.M. Cohn (2003). Basic algebra
* {{cite book
| author = I. Martin Isaacs
| year = 1993
| title = Algebra, a graduate course
| edition = 1st
| publisher = Brooks/Cole Publishing Company
| isbn = 0-534-19002-2
}}
* M. Nagata (1985). Commutative field theory: new edition, Shokado. (Japanese) [http://www.shokabo.co.jp/mybooks/ISBN978-4-7853-1309-8.htm]
*{{cite book |last=Silverman |first=Joseph |title=The Arithmetic of Elliptic Curves |year=1993 |publisher=Springer |isbn=0-387-96203-4}}
 
==External links==
*{{Springer|id=s/s084470|title=separable extension of a field k}}
 
[[Category:Field extensions]]
 
[[de:Körpererweiterung#Separable Erweiterungen]]

Revision as of 03:37, 24 February 2014

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