Kummer theory

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 with abelian 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 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.