Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 25 / 1

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 $H^1(G_K,{\mu }_n)\cong K^{\ast }/K^{\ast n}$.

Normalize each finite-place valuation to have image $\mathbb{Z}$. For a radical class $[a]$, the residue of $v(a)$ modulo $n$ is well-defined, because replacing $a$ by $a{b}^n$ adds $nv(b)$. A suitable finite set isOne can compute the last union from any finite generating set of $\Delta$. Each representative has only finitely many nonzero valuations, so this indeed defines a finite set depending only on $n$ and the classes, not on a choice of representatives. Including all infinite places safely allows possible real-to-complex ramification.

Fix $v\notin S$ and work over the completion $K_v$. A representative has $v(a)=nk$, so $a={\pi }^{nk}u$ with $u$ a unit. Adjoining an $n$th root of $a$ is the same as adjoining an $n$th root of $u$. Since $n$ is invertible in the residue field, $X^n-\bar{u}$ is separable: its derivative $n{X}^{n-1}$ has no common zero with it.

Take a finite extension of the finite residue field containing all roots of all these finitely many reduced polynomials. The corresponding finite unramified extension of $K_v$ exists by the classification proved in Question 2. Hensel's lemma lifts each simple residue root to a root of the original polynomial in that unramified extension. Thus all the required radicals, and every conjugate root, lie in an unramified local extension. Every completion of the radical field at $v$ is therefore unramified. This proves finite ramification support for radical extensions.

The local argument does not assume ${\mu }_n\subset K$: adjoining all roots proves the stronger unramified assertion for the splitting field, and hence for any field obtained by choosing the radicals. For $n=1$ the radical extension is trivial; the displayed set is still valid.