Kummer theory

Kummer theory is basic, for example, in class field theory and in general in understanding abelian extensions; it says that in the presence of enough roots of unity, cyclic extensions can be understood in terms of extracting roots. The main burden in class field theory is to dispense with extra roots of unity ('descending' back to smaller fields); which is something much more serious.

Kummer extensions

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

• K contains n distinct nth roots of unity (i.e., roots of Xⁿ − 1)
• L/K has abelian Galois group of exponent n.

For example, when n = 2, the first condition is always true if K has characteristic ≠ 2. The Kummer extensions in this case include quadratic extensions ${\displaystyle L=K(\sqrt{a})}$ where a in K is a non-square element. By the usual solution of quadratic equations, 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 numbers are required. If one takes L to be the splitting field of X³ − 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 α and β are roots of the cubic polynomial, we shall have (α/β)³ =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 nth roots of unity, which implies that the characteristic of K doesn't divide n, then adjoining to K the nth 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 Xⁿ − a, the Kummer extension is necessarily Galois, with Galois group that is cyclic of order m. It is easy to track the Galois action via the root of unity in front of ${\displaystyle \sqrt[n]{a}.}$

Kummer theory provides converse statements. When K contains n distinct nth 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^× denotes the multiplicative group of non-zero elements of K, abelian extensions of K of exponent n correspond bijectively with subgroups of

${\displaystyle K^{\times }/(K^{\times }{)}^n,}$

that is, elements of K^× modulo nth powers. The correspondence can be described explicitly as follows. Given a subgroup

${\displaystyle \mathrm{\Delta }\subseteq K^{\times }/(K^{\times }{)}^n,}$

the corresponding extension is given by

${\displaystyle K\left({\mathrm{\Delta }}^{\frac{1}{n}}\right),}$

where

${\displaystyle {\mathrm{\Delta }}^{\frac{1}{n}}=\left\{\sqrt[n]{a}:a\in K^{\times },a\cdot {\left(K^{\times }\right)}^n\in \mathrm{\Delta }\right\}.}$

In fact it suffices to adjoin nth root of one representative of each element of any set of generators of the group Δ. Conversely, if L is a Kummer extension of K, then Δ is recovered by the rule

${\displaystyle \mathrm{\Delta }=\left(K^{\times }\cap (L^{\times }{)}^n\right)/(K^{\times }{)}^n.}$

In this case there is an isomorphism

${\displaystyle \mathrm{\Delta }\cong {\mathrm{Hom}}_{\text{c}}(\mathrm{Gal}(L/K),{\mu }_n)}$

given by

${\displaystyle a\mapsto \left(\sigma \mapsto \frac{\sigma (\alpha )}{\alpha }\right),}$

where α is any nth root of a in L. Here ${\displaystyle {\mu }_n}$ denotes the multiplicative group of nth roots of unity (which belong to K) and ${\displaystyle {\mathrm{Hom}}_{\text{c}}(\mathrm{Gal}(L/K),{\mu }_n)}$ is the group of continuous homomorphisms from ${\displaystyle \mathrm{Gal}(L/K)}$ equipped with Krull topology to ${\displaystyle {\mu }_n}$ with discrete topology (with group operation given by pointwise multiplication). This group (with discrete topology) can also be viewed as Pontryagin dual of ${\displaystyle \mathrm{Gal}(L/K)}$, assuming we regard ${\displaystyle {\mu }_n}$ as a subgroup of circle group. If the extension L/K is finite, then ${\displaystyle \mathrm{Gal}(L/K)}$ is a finite discrete group and we have

${\displaystyle \mathrm{\Delta }\cong \mathrm{Hom}(\mathrm{Gal}(L/K),{\mu }_n)\cong \mathrm{Gal}(L/K),}$

however the last isomorphism isn't natural.

Recovering a1/n from a primitive element

For ${\displaystyle p}$ prime, let ${\displaystyle K}$ be a field containing ${\displaystyle {\zeta }_p}$ and ${\displaystyle K(\beta )/K}$ a degree ${\displaystyle p}$ Galois extension. Note the Galois group is cyclic, generated by ${\displaystyle \sigma }$. Let

${\displaystyle \alpha =\sum _{l=0}^{p-1}{\zeta }_p^l{\sigma }^l(\beta )\in K(\beta )}$

Then

${\displaystyle {\zeta }_p\sigma (\alpha )=\sum _{l=0}^{p-1}{\zeta }_p^{l+1}{\sigma }^{l+1}(\beta )=\alpha .}$

Since ${\displaystyle \alpha \ne \sigma (\alpha ),K(\alpha )=K(\beta )}$ and

${\displaystyle {\alpha }^p=\pm \prod _{l=0}^{p-1}{\zeta }_p^{-l}\alpha =\pm \prod _{l=0}^{p-1}{\sigma }^l(\alpha )=\pm N_{K(\beta )/K}(\alpha )\in K}$,

where the ${\displaystyle \pm }$ sign is ${\displaystyle +}$ if ${\displaystyle p}$ is odd and ${\displaystyle -}$ if ${\displaystyle p=2}$.

When ${\displaystyle L/K}$ is an abelian extension of degree ${\displaystyle n=\prod _{j=1}^m{p}_j}$ square-free such that ${\displaystyle {\zeta }_n\in K}$, apply the same argument to the subfields ${\displaystyle K({\beta }_j)/K}$ Galois of degree ${\displaystyle p_j}$ to obtain

${\displaystyle L=K\left(a_1^{1/p_1},\dots ,a_m^{1/p_m}\right)=K\left(A^{1/p_1},\dots ,A^{1/p_m}\right)=K\left(A^{1/n}\right)}$

where

${\displaystyle A=\prod _{j=1}^m{a}_j^{n/p_j}\in K}$.