Editing Kummer theory (section)

== Kummer extensions ==
A '''Kummer extension''' is a field extension ''L''/''K'', where for some given integer ''n'' > 1 we have

*''K'' contains ''n'' distinct ''n''th [[root of unity|roots of unity]] (i.e., roots of ''X''<sup>''n''</sup> − 1)
*''L''/''K'' has [[abelian group|abelian]] [[Galois group]] of [[exponent (group theory)|exponent]] ''n''.

For example, when ''n'' = 2, the first condition is always true if ''K'' has [[characteristic (algebra)|characteristic]] &ne; 2. The Kummer extensions in this case include '''quadratic extensions''' <math>L= K(\sqrt{a})</math> where ''a'' in ''K'' is a non-square element. By the usual solution of [[quadratic equation]]s, any extension of degree 2 of ''K'' has this form. The Kummer extensions in this case also include '''biquadratic extensions''' and more general '''multiquadratic extensions'''. When ''K'' has characteristic 2, there are no such Kummer extensions.

Taking ''n'' = 3, there are no degree 3 Kummer extensions of the [[rational number]] field '''Q''', since for three cube roots of 1 [[complex number]]s are required. If one takes ''L'' to be the splitting field of ''X''<sup>3</sup> − ''a'' over '''Q''', where ''a'' is not a cube in the rational numbers, then ''L'' contains a subfield ''K'' with three cube roots of 1; that is because if &alpha; and &beta; are roots of the cubic polynomial, we shall have (&alpha;/&beta;)<sup>3</sup> =1 and the cubic is a [[separable polynomial]]. Then ''L''/''K'' is a Kummer extension.

More generally, it is true that when ''K'' contains ''n'' distinct ''n''th roots of unity, which implies that the characteristic of ''K'' doesn't divide ''n'', then adjoining to ''K'' the ''n''th root of any element ''a'' of ''K'' creates a Kummer extension (of degree ''m'', for some ''m'' dividing ''n''). As the [[splitting field]] of the polynomial ''X<sup>n</sup>'' − ''a'', the Kummer extension is necessarily [[Galois extension|Galois]], with Galois group that is [[cyclic group|cyclic]] of order ''m''. It is easy to track the Galois action via the root of unity in front of <math>\sqrt[n]{a}.</math>

'''Kummer theory''' provides converse statements. When ''K'' contains ''n'' distinct ''n''th roots of unity, it states that any [[abelian extension]] of ''K'' of exponent dividing ''n'' is formed by extraction of roots of elements of ''K''. Further, if ''K''<sup>×</sup> denotes the multiplicative group of non-zero elements of ''K'', abelian extensions of ''K'' of exponent ''n'' correspond bijectively with subgroups of

:<math>K^{\times}/(K^{\times})^n,</math>

that is, elements of ''K''<sup>×</sup> [[Modular arithmetic|modulo]] ''n''th powers. The correspondence can be described explicitly as follows. Given a subgroup

:<math>\Delta \subseteq K^{\times}/(K^{\times})^n,</math>

the corresponding extension is given by

:<math>K \left (\Delta^{\frac{1}{n}} \right),</math>

where 

:<math>\Delta^{\frac{1}{n}} = \left \{ \sqrt[n]{a}:a\in K^{\times}, a \cdot \left (K^{\times} \right )^n \in \Delta \right \}.</math> 

In fact it suffices to adjoin ''n''th root of one representative of each element of any set of generators of the group &Delta;. Conversely, if ''L'' is a Kummer extension of ''K'', then &Delta; is recovered by the rule

:<math>\Delta = \left (K^\times \cap (L^\times)^n \right )/(K^{\times})^n.</math>

In this case there is an isomorphism

:<math>\Delta \cong \operatorname{Hom}_{\text{c}}(\operatorname{Gal}(L/K), \mu_n)</math>

given by

:<math>a \mapsto \left(\sigma \mapsto \frac{\sigma(\alpha)}{\alpha}\right),</math>

where &alpha; is any ''n''th root of ''a'' in ''L''. Here <math>\mu_n</math> denotes the multiplicative group of ''n''th roots of unity (which belong to ''K'') and <math>\operatorname{Hom}_{\text{c}}(\operatorname{Gal}(L/K), \mu_n)</math> is the group of continuous homomorphisms from <math>\operatorname{Gal}(L/K)</math> equipped with [[Krull topology]] to <math>\mu_n</math> with discrete topology (with group operation given by pointwise multiplication). This group (with discrete topology) can also be viewed as [[Pontryagin dual]] of <math>\operatorname{Gal}(L/K)</math>, assuming we regard <math>\mu_n</math> as a subgroup of [[circle group]]. If the extension ''L''/''K'' is finite, then <math>\operatorname{Gal}(L/K)</math> is a finite discrete group and we have

:<math>\Delta \cong \operatorname{Hom}(\operatorname{Gal}(L/K), \mu_n) \cong \operatorname{Gal}(L/K),</math>

however the last isomorphism isn't [[Natural homomorphism|natural]].

=== Recovering {{math|1= ''a''<sup>1/''n''</sup> }} from a primitive element ===
For <math>p</math> prime, let <math>K</math> be a field containing <math>\zeta_p</math> and <math>K(\beta)/K</math> a degree <math>p</math> Galois extension. Note the Galois group is cyclic, generated by <math>\sigma</math>. Let 

:<math>\alpha= \sum_{l=0}^{p-1} \zeta_p^{l} \sigma^l(\beta) \in K(\beta)</math>

Then 

:<math>\zeta_p \sigma(\alpha) = \sum_{l=0}^{p-1} \zeta_p^{l+1} \sigma^{l+1}(\beta) = \alpha.</math> 

Since <math>\alpha\ne \sigma(\alpha), K(\alpha) = K(\beta)</math> and 

:<math>\alpha^p = \pm \prod_{l=0}^{p-1} \zeta_p^{-l} \alpha = \pm \prod_{l=0}^{p-1} \sigma^l(\alpha) = \pm N_{K(\beta)/K}(\alpha) \in K</math>,
where the <math>\pm</math> sign is <math>+</math> if <math>p</math> is odd and <math>-</math> if <math>p=2</math>.

When <math>L/K</math> is an abelian extension of degree <math>n= \prod_{j=1}^m p_j</math> square-free such that <math>\zeta_n \in K</math>, apply the same argument to the subfields <math>K(\beta_j)/K</math> Galois of degree <math>p_j</math> to obtain 

:<math>L = K \left (a_1^{1/p_1},\ldots,a_m^{1/p_m} \right ) = K \left (A^{1/p_1},\ldots,A^{1/p_m} \right )= K \left (A^{1/n} \right )</math> 

where 

:<math>A = \prod_{j=1}^m a_j^{n/p_j} \in K</math>.