# Structural Applications of the CRT

The next lemma is an application of the Chinese Remainder Theorem. We will use it to prove that every ideal of can be generated by two elements. Suppose that  is a nonzero integral ideals of . If , then , so  divides  and the quotient is an integral ideal. The following lemma asserts that  can be chosen so the quotient is coprime to any given ideal.

Lemma 5.2.1   If and are nonzero integral ideals in , then there exists an such that the integral ideal is coprime to .

Before we give the proof in general, note that the lemma is trivial when is principal, since if , just take , and then is coprime to every ideal.

Proof. Let be the prime divisors of . For each , let be the largest power of that divides . Since , we can choose an element that is not in . By Theorem 5.1.4 applied to the coprime integral ideals

there exists such that

for all and also

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.

Proposition 5.2.2   Suppose is a fractional ideal in the ring of integers of a number field. Then there exist such that .

Proof. If , then is generated by element and we are done. If is not an integral ideal, then there is such that is an integral ideal, and the number of generators of is the same as the number of generators of , so we may assume that is an integral ideal.

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 .

Proposition 5.2.3   Let be a nonzero prime ideal of , and let be an integer. Then as -modules.

Proof. 5.1Since , by unique factorization, there is an element such that . Let be the -module morphism defined by . The kernel of is since clearly and if then , so , so , since does not divide . Thus  induces an injective -module homomorphism .

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

so is surjective.

William Stein 2012-09-24