---
tag: 6RXEC9CT6M
kind: definition
name: Kummer theory
msc: [11R20, 12F10]
defines: XZQENYV15Y
uses: [04C5ZBG1WT, 0YV2CKP2AV, 0QMZAQ6QQ4, 082GHNPJ53, 0XTY86BZKY, 0KVGH0J5B3]
sources:
  - url: https://en.wikipedia.org/wiki/Kummer_theory
---

## Definition

Let $n \ge 1$ be an integer, and let $K$ be a [[field|04C5ZBG1WT]] whose characteristic does not divide $n$.
Assume that $K$ contains the group $\mu_n$ of $n$ distinct $n$-th [[roots of unity|0YV2CKP2AV]].
Fix an algebraic closure $\overline{K}$ of $K$.
For a subgroup $B$ of $K^\times$ that contains $K^{\times n} = \set{x^n : x \in K^\times}$, let $K(\sqrt[n]{B})$ be the subfield of $\overline{K}$ generated over $K$ by all $\beta \in \overline{K}$ with $\beta^n \in B$.
A *Kummer extension* of exponent $n$ of $K$ is an extension of the form $K(\sqrt[n]{B})$.
For such an extension $L = K(\sqrt[n]{B})$, the *Kummer pairing* is the map
$$\Gal(L/K) \times B/K^{\times n} \to \mu_n, \qquad \langle \sigma, b \rangle = \frac{\sigma(\beta)}{\beta},$$
where $\beta \in L$ is any element with $\beta^n = b$.
Kummer theory is the study of these extensions through this pairing.

## Remarks

The pairing is well defined: two choices of $\beta$ differ by a factor in $\mu_n \subseteq K$, which $\sigma$ fixes.
The main theorem of Kummer theory: $L/K$ is a [[Galois extension|0QMZAQ6QQ4]] with [[abelian|082GHNPJ53]] Galois group of exponent dividing $n$, and if $B/K^{\times n}$ is finite the pairing is perfect, so $\Gal(L/K) \cong \Hom(B/K^{\times n}, \mu_n)$.
The map $B \mapsto K(\sqrt[n]{B})$ is a bijection from the subgroups $B$ with $K^{\times n} \subseteq B \subseteq K^\times$ and $B/K^{\times n}$ finite onto the finite abelian extensions of $K$ in $\overline{K}$ of exponent dividing $n$.
In particular, for $a \in K^\times$, the extension $K(\sqrt[n]{a})/K$ is [[cyclic|0KVGH0J5B3]] of degree equal to the order of $a$ in $K^\times / K^{\times n}$.

Example: $K = \Q$, $n = 2$, $\mu_2 = \set{\pm 1}$, and $B$ generated by $\Q^{\times 2}$, $2$ and $3$.
Then $L = \Q(\sqrt{2}, \sqrt{3})$, and $B/\Q^{\times 2}$ has order $4$, so $\Gal(L/\Q) \cong (\Z/2\Z)^2$, as expected.

Non-example: $K = \Q$ and $n = 3$.
The field $\Q$ does not contain the three cube roots of unity, so the definition does not apply, and indeed $\Q(\sqrt[3]{2})/\Q$ is not even Galois.