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 $L$ and $M$ are two lattices in a vector space $V$, then the lattice index $[L:M]$ is by definition the absolute value of the determinant of any linear automorphism $A$ of $V$ such that $A(L)=M$.

For example, if $L=2\mathbf{Z}$ and $M=10\mathbf{Z}$, then

$$[L:M] = [2\mathbf{Z}: 10\mathbf{Z}] = \det([5]) = 5,$$

since $5$ multiplies $2\mathbf{Z}$ onto $10\mathbf{Z}$.

The lattice index has the following properties:

• If $M\subset L$, then $[L:M]=\char93 (L/M)$.
• If $M, L, N$ are any lattices in $V$, then
  $$[L:N] = [L:M]\cdot [M:N].$$

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

$$\Norm (I) = [\O_K : I] \in \mathbf{Q}_{\geq 0},$$

or 0 if $I=0$.

Note that if $I$ is an integral ideal, then $\Norm (I)=\char93 (\O_K/I)$.

Lemma 6.3.3   Suppose $a\in K$ and $I$ is an integral ideal. Then

$$\Norm (a I) = \vert\Norm _{K/\mathbf{Q}}(a)\vert \Norm (I).$$

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

$$[\O_K : aI] = [\O_K : I] \cdot [I:aI] = \Norm (I) \cdot \vert\Norm _{K/\mathbf{Q}}(a)\vert.$$

Here we have used that $[I:aI]=\vert\Norm _{K/\mathbf{Q}}(a)\vert$, which is because left multiplication $\ell_a$ by $a$ is an automorphism of $K$ that sends $I$ onto $aI$, so

$$[I:aI]=\vert\det(\ell_a)\vert=\vert\Norm _{K/\mathbf{Q}}(a)\vert.$$

$\qedsymbol$

Proposition 6.3.4   If $I$ and $J$ are fractional ideals, then

$$\Norm (IJ) = \Norm (I)\cdot \Norm (J).$$

Proof. By Lemma 6.3.3, it suffices to prove this when $I$ and $J$ are integral ideals. If $I$ and $J$ are coprime, then Theorem 5.1.4 (the Chinese Remainder Theorem) implies that $\Norm (IJ) = \Norm (I)\cdot \Norm (J)$. Thus we reduce to the case when $I=\mathfrak{p}^m$ and $J=\mathfrak{p}^k$ for some prime ideal $\mathfrak{p}$ and integers $m,k$. By Proposition 5.2.3, which is a consequence of CRT, the filtration of $\O_K/\mathfrak{p}^{n}$ given by powers of  $\mathfrak{p}$ has successive quotients isomorphic to $\O_K/\mathfrak{p}$. Thus we see that $\char93 (\O_K/\mathfrak{p}^{n}) = \char93 (\O_K/\mathfrak{p})^{n}$, which proves that $\Norm (\mathfrak{p}^n)=\Norm (\mathfrak{p})^n$. $\qedsymbol$

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 $K$. Let $B$ be a positive integer. There are only finitely many integral ideals $I$ of $\O_K$ with norm at most $B$.

Proof. An integral ideal $I$ is a subgroup of $\O_K$ of index equal to the norm of $I$. If $G$ is any finitely generated abelian group, then there are only finitely many subgroups of $G$ of index at most $B$, since the subgroups of index dividing an integer $n$ are all subgroups of $G$ that contain $nG$, and the group $G/nG$ is finite. $\qedsymbol$