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:
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
.