To complete the proof we show that is not divisible by any , or equivalently, that each exactly divides . First we show that divides . Because , there exists such that . Since and , it follows that , so divides . Now assume for the sake of contradiction that divides ); then , which contradicts that we chose . Thus does not divide , as claimed.

Suppose is a nonzero ideal of . As an abelian group
is free of rank equal to the degree
of , and is of
finite index in , so can be generated as an abelian group,
hence as an ideal, by
generators. The following proposition
asserts something much better, namely that can be generated *as an ideal* in by at most two elements.

Let be *any* nonzero element of the integral ideal . We
will show that there is some such that . Let
. By Lemma 5.2.1, there exists such that
is coprime to . Since , we have and , so
. Suppose
with
prime and . Then
and
, so
, since is coprime to . We have
and
, so
. Thus by unique factorization of ideals in we
have that
. Sine
we conclude
that , as claimed.

We can also use Theorem 5.1.4 to determine the -module structure of .

It remains to show that is surjective, and this is where we will use Theorem 5.1.4. Suppose . By Theorem 5.1.4 there exists such that

and

We have
since
and
by the second displayed condition, so
since
, we have
, hence
. Finally

William Stein 2012-09-24