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
William Stein 2012-09-24