 Solution

The hypotheses for Kummer theory are that $n\ge 1$ is prime to the characteristic of $K$ and that all $n$th roots of unity lie in $K$. Under these hypotheses, adjoining radicals produces abelian extensions of exponent dividing $n$, and conversely every such finite extension is obtained this way. The roots-of-unity hypothesis matters: without it, adjoining one radical need not give a Galois extension.

We first prove Hilbert's theorem 90, including its cocycle form. Distinct field automorphisms are linearly independent as functions into the field. Indeed, in a shortest nontrivial relation ${\sum }_i{c}_i{\tau }_i(y)=0$, choose $x$ with ${\tau }_1(x)\ne {\tau }_r(x)$. Subtract ${\tau }_r(x)$ times the relation at $y$ from the relation at $xy$. This eliminates the last term while leaving the first nonzero, contradicting minimality. This is linear independence of distinct field embeddings.

Let $L/K$ be finite Galois with group $G$, and suppose $a_{\sigma }\in L^{\ast }$ satisfy $a_{\sigma \tau }=a_{\sigma }\sigma (a_{\tau })$. By independence of the automorphisms, choose $c\in L$ for whichThenThus every multiplicative cocycle is a coboundary:For a cyclic group generated by $\sigma$ of order $m$, an element $a$ of norm one defines the cocycle $a_{{\sigma }^i}={\prod }_{j=0}^{i-1}{\sigma }^j(a)$; the condition at $i=m$ is exactly $N_{L/K}(a)=1$. HenceThe reverse implication follows by telescoping the product of conjugates.

Now let $\Delta$ be a finite subgroup of $K^{\ast }/K^{\ast n}$ and put $L=K(\sqrt[n]{\Delta })$. Finitely many representatives generate it; since ${\mu }_n\subset K$, the polynomials $X^n-a$ split in $L$ and are separable. Thus $L/K$ is finite Galois. For $\sigma \in G$ and $[a]\in \Delta$, choose ${\alpha }^n=a$ and define the Kummer pairingChanging the root or multiplying $a$ by an $n$th power does not change this value. It is multiplicative in both arguments. Since the radicals generate $L$, it embeds $G$ in $\mathrm{Hom}(\Delta ,{\mu }_n)$. Consequently $G$ is abelian of exponent dividing $n$.

For completeness, establish perfection rather than just this injection. Let ${\Delta }_L$ consist of all classes in $K^{\ast }/K^{\ast n}$ having an $n$th root in $L$. The same construction gives an injective map ${\Delta }_L\to \mathrm{Hom}(G,{\mu }_n)$: the character is trivial precisely when the root lies in $K$. It is surjective by Hilbert 90. Given $\chi :G\to {\mu }_n$, apply the cocycle theorem to $a_{\sigma }=\chi (\sigma {)}^{-1}$; it gives $\sigma (b)/b=\chi (\sigma )$, so $b^n\in K$ and its class has the required character. A finite abelian group of exponent dividing $n$ has as many ${\mu }_n$-characters as elements, by decomposition into cyclic factors. ThereforeIt follows that $\Delta ={\Delta }_L$ andConversely, for any finite abelian $L/K$ of exponent dividing $n$, Hilbert 90 supplies eigenvectors for every character. Their $n$th powers lie in $K$, and the intersection of their character kernels is trivial, so they generate $L$. This proves the Kummer classification and the inclusion-reversing duality between subgroups and intermediate extensions.

The simplest case is a cyclic extension of degree $m\mid n$: it has a generator $\alpha$ with ${\alpha }^m\in K$, and its automorphisms multiply $\alpha$ by $m$th roots of unity. Quadratic extensions in characteristic different from two are the case $n=2$. The cohomological summary is the exact sequence $1\to {\mu }_n\to K_s^{\ast }\stackrel{z\mapsto z^n}{\to }{K}_s^{\ast }\to 1$ together with Hilbert 90, giving $H^1(G_K,{\mu }_n)\cong K^{\ast }/K^{\ast n}$.