Kummer Theory of Number Fields

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

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

The long exact sequence is

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

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

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

Thus we obtain an exact sequence

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

or equivalently, an isomorphism

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $r+s-1 + \char93 S$. $\qedsymbol$