Norms of Ideals

In this section we extend the notion of norm to ideals. This will be helpful in proving of class groups in the next section. For example, 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 a priori bound.

Definition 10.3.1 (Lattice Index)   If $ L$ and $ M$ are two lattices in vector space $ V$, then the $ [L:M]$ is by definition the absolute value of the determinant of any linear automorphism $ A$ of $ V$ such that $ A(L)=M$.

The lattice index has the following properties:

Definition 10.3.2 (Norm of Fractional Ideal)   Suppose $ I$ is a fractional ideal of $ \O _K$. The of $ I$ is the lattice index

$\displaystyle \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 10.3.3   Suppose $ a\in K$ and $ I$ is an integral ideal. Then

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

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

$\displaystyle [\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$ is an automorphism of $ K$ that sends $ I$ onto $ aI$, so $ [I:aI]=\vert{\mathrm{Det}}(\ell_a)\vert=\vert\Norm _{K/\mathbf{Q}}(a)\vert$. $ \qedsymbol$

Proposition 10.3.4   If $ I$ and $ J$ are fractional ideals, then

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

Proof. By Lemma 10.3.3, it suffices to prove this when $ I$ and $ J$ are integral ideals. If $ I$ and $ J$ are coprime, then Theorem 9.1.3 (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 9.1.8 (consequence of CRT that $ \O _K/\mathfrak{p}\cong \mathfrak{p}^n/\mathfrak{p}^{n+1}$), the filtration of $ \O _K/\mathfrak{p}^{n}$ given by powers of  $ \mathfrak{p}$ has successive quotients isomorphic to $ \O _K/\mathfrak{p}$, so 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$

Lemma 10.3.5   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. This proves the lemma. $ \qedsymbol$

William Stein 2004-05-06