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:
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
.
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