# Kummer Theory of Number Fields

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

The long exact sequence is

where by Theorem 11.4.2.

Assume now that the group of th roots of unity is contained in . Using Galois cohomology we obtain a relatively simple classification of all abelian extensions of  with Galois group cyclic of order dividing . Moreover, since the action of on is trivial, by our hypothesis that , we see that

Thus we obtain an exact sequence

or equivalently, an isomorphism

By Galois theory, homomorphisms (up to automorphisms of ) correspond to cyclic abelian extensions of  with Galois group a subgroup of the cyclic group of order . Unwinding the definitions, what this says is that every cyclic abelian extension of of degree dividing  is of the form for some element .

One can prove via calculations with discriminants, etc. that is unramified outside and and the primes that divide . 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 is a number field with , where is a positive integer. Then the maximal abelian exponent  extension of  unramified outside a finite set  of primes is of finite degree.

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

We first argue that we can enlarge  so that the ring

all

is a principal ideal domain. Note that for any , the ring is a Dedekind domain. Also, the condition means that in the prime ideal factorization of the fractional ideal , we have that  occurs to a nonnegative power. Thus we are allowing denominators at the primes in . Since the class group of is finite, there are primes that generate the class group as a group (for example, take all primes with norm up to the Minkowski bound). Enlarge  to contain the primes . Note that the ideal is the unit ideal (we have for some ; then , so is the unit ideal, hence is the unit ideal by unique factorization in the Dedekind domain .) Then is a principal ideal domain, since every ideal of is equivalent modulo a principal ideal to a product of ideals . Note that we have used that the class group of is finite.

Next enlarge so that all primes over are in . Note that is still a PID. Let

all

Then a refinement of the arguments at the beginning of this section show that is generated by all th roots of the elements of . It thus sufficies to prove that is finite.

There is a natural map

Suppose is a representative of an element in . The ideal has factorization which is a product of th powers, so it is an th power of an ideal. Since is a PID, there is and such that

Thus maps to . Thus is surjective.

Recall that we proved Dirichlet's unit theorem (see Theorem 8.1.2), which asserts that the group is a finitely generated abelian group of rank . More generally, we now show that is a finitely generated abelian group of rank . Once we have shown this, then since is torsion group that is a quotient of a finitely generated group, we will conclude that is finite, which will prove the theorem.

Thus it remains to prove that has rank . Let be the primes in . Define a map by

First we show that . We have that if and only if and for all ; but the latter condition implies that is a unit at each prime in , so . Thus we have an exact sequence

Next we show that the image of has finite index in . Let be the class number of . For each there exists such that . But since for all (by unique factorization). Then

It follows that    Im, so the image of has finite index in . It follows that has rank equal to .

William Stein 2012-09-24