Norms of Ideals

In this section we extend the notion of norm to ideals. This will be helpful in the next chapter, where we will prove that the group of fractional ideals modulo principal fractional ideals of a number field is finite by showing that every ideal is equivalent to an ideal with norm at most some bound. This is enough, because as we will see below there are only finitely many ideals of bounded norm.

Definition 6.3.1 (Lattice Index)   If and are two lattices in a vector space , then the lattice index is by definition the absolute value of the determinant of any linear automorphism of such that .

For example, if and , then

since multiplies onto .

The lattice index has the following properties:

• If , then .
• If are any lattices in , then

Definition 6.3.2 (Norm of Fractional Ideal)   Suppose is a fractional ideal of . The norm of  is the lattice index

or 0 if .

Note that if is an integral ideal, then .

Lemma 6.3.3   Suppose and is an integral ideal. Then

Proof. By properties of the lattice index mentioned above we have

Here we have used that , which is because left multiplication by is an automorphism of that sends onto , so

Proposition 6.3.4   If and are fractional ideals, then

Proof. By Lemma 6.3.3, it suffices to prove this when and are integral ideals. If and are coprime, then Theorem 5.1.4 (the Chinese Remainder Theorem) implies that . Thus we reduce to the case when and for some prime ideal and integers . By Proposition 5.2.3, which is a consequence of CRT, the filtration of given by powers of  has successive quotients isomorphic to . Thus we see that , which proves that .

Example 6.3.5   We compute some ideal norms using SAGE.
sage: K.<a> = NumberField(x^2 - 5)
sage: I = K.fractional_ideal(a)
sage: I.norm()
5
sage: J = K.fractional_ideal(17)
sage: J.norm()
289


We can also use functional notation:

sage: norm(I*J)
1445


We will use the following proposition in the next chapter when we prove finiteness of class groups.

Proposition 6.3.6   Fix a number field . Let be a positive integer. There are only finitely many integral ideals of with norm at most .

Proof. An integral ideal is a subgroup of of index equal to the norm of . If is any finitely generated abelian group, then there are only finitely many subgroups of of index at most , since the subgroups of index dividing an integer are all subgroups of that contain , and the group is finite.

William Stein 2012-09-24