Kummer Theory of Number Fields

Suppose $ K$ is a number field and fix a positive integer $ n$. Consider the exact sequence

$\displaystyle 1 \to \mu_n \to \overline{K}^* \xrightarrow{n} \overline{K}^* \to 1.

The long exact sequence is

$\displaystyle 1 \to \mu_n(K) \to K^* \xrightarrow{n} K^* \to \H^1(K,\mu_n) \to \H^1(K,\overline{K}^*) =0,

where $ \H^1(K,\overline{K}^*) = 0$ by Theorem 11.4.2.

Assume now that the group $ \mu_n$ of $ n$th roots of unity is contained in $ K$. Using Galois cohomology we obtain a relatively simple classification of all abelian extensions of $ K$ with Galois group cyclic of order dividing $ n$. Moreover, since the action of $ \Gal (\overline{K}/K)$ on $ \mu_n$ is trivial, by our hypothesis that $ \mu_n\subset K$, we see that

$\displaystyle \H^1(K,\mu_n) = \Hom (\Gal (\overline{K}/K),\mu_n).

Thus we obtain an exact sequence

$\displaystyle 1 \to \mu_n \to K^* \xrightarrow{n} K^* \to \Hom (\Gal (\overline{K}/K),\mu_n) \to 1,

or equivalently, an isomorphism

$\displaystyle K^*/(K^*)^n \cong \Hom (\Gal (\overline{K}/K),\mu_n),

By Galois theory, homomorphisms $ \Gal (\overline{K}/K)\to \mu_n$ (up to automorphisms of $ \mu_n$) correspond to cyclic abelian extensions of $ K$ with Galois group a subgroup of the cyclic group $ \mu_n$ of order $ n$. Unwinding the definitions, what this says is that every cyclic abelian extension of $ K$ of degree dividing $ n$ is of the form $ K(a^{1/n})$ for some element $ a\in K$.

One can prove via calculations with discriminants, etc. that $ K(a^{1/n})$ is unramified outside $ n$ and and the primes that divide $ \Norm (a)$. Moreover, and this is a much bigger result, one can combine this with facts about class groups and unit groups to prove the following theorem:

Theorem 12.1.1   Suppose $ K$ is a number field with $ \mu_n\subset K$, where $ n$ is a positive integer. Then the maximal abelian exponent $ n$ extension $ L$ of $ K$ unramified outside a finite set $ S$ of primes is of finite degree.

Proof. [Sketch of Proof] We may enlarge $ S$, because if an extension is unramified outside a set larger than $ S$, then it is unramified outside $ S$.

We first argue that we can enlarge $ S$ so that the ring

$\displaystyle \O_{K,S} = \{a \in K^* : \ord _\mathfrak{p}(a\O_K) \geq 0$    all $\displaystyle \mathfrak{p}\not\in S\}
\cup \{0\}

is a principal ideal domain. Note that for any $ S$, the ring $ \O_{K,S}$ is a Dedekind domain. Also, the condition $ \ord _\mathfrak{p}(a\O_K) \geq 0$ means that in the prime ideal factorization of the fractional ideal $ a\O_K$, we have that  $ \mathfrak{p}$ occurs to a nonnegative power. Thus we are allowing denominators at the primes in $ S$. Since the class group of $ \O_K$ is finite, there are primes $ \mathfrak{p}_1,\ldots, \mathfrak{p}_r$ that generate the class group as a group (for example, take all primes with norm up to the Minkowski bound). Enlarge $ S$ to contain the primes $ \mathfrak{p}_i$. Note that the ideal $ \mathfrak{p}_i\O_{K,S}$ is the unit ideal (we have $ \mathfrak{p}_i^m=(\alpha)$ for some $ m\geq 1$; then $ 1/\alpha \in \O_{K,S}$, so $ (\mathfrak{p}_i\O_{K,S})^m$ is the unit ideal, hence $ \mathfrak{p}_i\O_{K,S}$ is the unit ideal by unique factorization in the Dedekind domain $ \O_{K,S}$.) Then $ \O_{K,S}$ is a principal ideal domain, since every ideal of $ \O_{K,S}$ is equivalent modulo a principal ideal to a product of ideals $ \mathfrak{p}_i\O_{K,S}$. Note that we have used that the class group of $ \O_K$ is finite.

Next enlarge $ S$ so that all primes over $ n\O_K$ are in $ S$. Note that $ \O_{K,S}$ is still a PID. Let

$\displaystyle K(S,n) = \{a \in K^*/(K^*)^n : n \mid \ord _\mathfrak{p}(a)$    all $\displaystyle \mathfrak{p}\not\in S\}.

Then a refinement of the arguments at the beginning of this section show that $ L$ is generated by all $ n$th roots of the elements of $ K(S,n)$. It thus sufficies to prove that $ K(S,n)$ is finite.

There is a natural map

$\displaystyle \phi: \O_{K,S}^* \to K(S,n).

Suppose $ a\in K^*$ is a representative of an element in $ K(S,n)$. The ideal $ a \O_{K,S}$ has factorization which is a product of $ n$th powers, so it is an $ n$th power of an ideal. Since $ \O_{K,S}$ is a PID, there is $ b\in \O_{K,S}$ and $ u\in \O_{K,S}^*$ such that

$\displaystyle a = b^n \cdot u.

Thus $ u\in \O_{K,S}^*$ maps to $ [a] \in K(S,n)$. Thus $ \phi$ is surjective.

Recall that we proved Dirichlet's unit theorem (see Theorem 8.1.2), which asserts that the group $ \O_K^*$ is a finitely generated abelian group of rank $ r+s-1$. More generally, we now show that $ \O_{K,S}^*$ is a finitely generated abelian group of rank $ r+s+\char93 S -1$. Once we have shown this, then since $ K(S,n)$ is torsion group that is a quotient of a finitely generated group, we will conclude that $ K(S,n)$ is finite, which will prove the theorem.

Thus it remains to prove that $ \O_{K,S}^*$ has rank $ r+s-1 + \char93 S$. Let $ \mathfrak{p}_1,\ldots, \mathfrak{p}_n$ be the primes in $ S$. Define a map $ \phi: \O_{K,S}^* \to \mathbf{Z}^n$ by

$\displaystyle \phi(u) = (\ord _{\mathfrak{p}_1}(u), \ldots, \ord _{\mathfrak{p}_n}(u)).

First we show that $ \Ker (\phi) = \O_K^*$. We have that $ u\in \Ker (\phi)$ if and only if $ u\in \O_{K,S}^*$ and $ \ord _{\mathfrak{p}_i}(u) = 0$ for all $ i$; but the latter condition implies that $ u$ is a unit at each prime in $ S$, so $ u\in \O_K^*$. Thus we have an exact sequence

$\displaystyle 1 \to \O_K^* \to \O_{K,S}^* \xrightarrow{\phi} \mathbf{Z}^n.

Next we show that the image of $ \phi$ has finite index in $ \mathbf{Z}^n$. Let $ h$ be the class number of $ \O_K$. For each $ i$ there exists $ \alpha_i \in \O_K$ such that $ \mathfrak{p}_i^h = (\alpha_i)$. But $ \alpha_i \in \O_{K,S}^*$ since $ \ord _{\mathfrak{p}}(\alpha_i) = 0$ for all $ \mathfrak{p}\not\in S$ (by unique factorization). Then

$\displaystyle \phi(\alpha_i) = (0,\ldots, 0, h, 0,\ldots, 0).

It follows that $ (h\mathbf{Z})^n \subset$   Im$ (\phi)$, so the image of $ \phi$ has finite index in $ \mathbf{Z}^n$. It follows that $ \O_{K,S}^*$ has rank equal to $ r+s-1 + \char93 S$. $ \qedsymbol$

William Stein 2012-09-24