Before discussing norms and traces we introduce some notation for
field extensions. If
are number fields, we let
denote the dimension of viewed as a -vector space. If is a
number field and
, let be the number field
generated by , which is the smallest number field that
contains . If
then has a minimal polynomial
, and the of are the roots
of . For example the element has minimal polynomial
and the Galois conjugates of are
.
Suppose
is an inclusion of number fields and let . Then left multiplication by defines a -linear
transformation
. (The transformation is
-linear because is commutative.)

**Definition 5.2.1** (Norm and Trace)
The

and

of

from

to

are

and

It is standard from linear algebra that
determinants are multiplicative
and traces are additive, so for we have
and
Note that if
is the characteristic polynomial of ,
then the constant term of is
, and the
coefficient of
is
.

*Proof*.
We prove the proposition by computing the characteristic
polynomial

of

. Let

be the minimal polynomial
of

over

, and note that

has distinct roots (since it is the
polynomial in

of least degree that is satisfied by

).
Since

is irreducible,

, and

satisfies a
polynomial if and only if

does, the characteristic polynomial
of

acting on

is

. Let

be a basis
for

over

and note that

is a basis for

, where

. Then

is a basis for

over

, and left multiplication by

acts the same way on the span of

as on the span of

, for any pair

. Thus the matrix of

on

is a block direct sum of copies of the matrix of

acting
on

, so the characteristic polynomial of

on

is

. The proposition follows because the roots of

are exactly the images

, with multiplicity

(since each embedding of

into

extends in
exactly

ways to

by Exercise

9).

The following corollary asserts that the norm and trace behave well in
towers.

**Corollary 5.2.3**
*
Suppose
is a tower of number fields, and
let . Then
*
and

*Proof*.
For the first equation, both sides are the product of

,
where

runs through the embeddings of

into

. To see
this, suppose

fixes

. If

is an
extension of

to

, and

are the
embeddings of

into

that fix

, then

are exactly the extensions of

to

. For the second statement, both sides are the sum of
the

.

The norm and trace down to
of an algebraic integer is an
element of
, because the minimal polynomial of has integer
coefficients, and the characteristic polynomial of is a power of the
minimal polynomial, as we saw in the proof of
Proposition 5.2.2.

**Proposition 5.2.4**
*
Let be a number field. The ring of integers is a lattice
in , i.e.,
and is an abelian group of rank
.*
*Proof*.
We saw in Lemma

5.1.8 that

. Thus there exists a
basis

for

, where each

is in

.
Suppose that as

varies over all elements of

the denominators of the coefficients

are arbitrarily
large. Then subtracting off integer multiples of the

, we see
that as

varies over elements of

with

between

0 and

, the denominators of the

are also
arbitrarily large. This implies that there are infinitely many elements
of

in the bounded subset

Thus for any

, there are elements

such that the
coefficients of

are all less than

(otherwise the elements
of

would all be a ``distance'' of least

from each other, so only finitely
many of them would fit in

).

As mentioned above, the norms of elements of are integers.
Since the norm of an element is the determinant of left multiplication
by that element, the norm is a homogenous polynomial of degree in
the indeterminate coefficients . If the get arbitrarily
small for elements of , then the values of the norm polynomial
get arbitrarily small, which would imply that there are elements of
with positive norm too small to be in
, a contradiction.
So the set contains only finitely many elements of . Thus
the denominators of the are bounded, so for some , we have
that has finite index in
. Since is isomorphic to
, it follows
from the structure theorem for finitely generated abelian groups that
is isomorphic as a
-module to
, as claimed.

*Proof*.
By Proposition

5.2.4, the ring

is
finitely generated as a module over

, so it is certainly
finitely generated as a ring over

. By the Hilbert
Basis Theorem,

is Noetherian.

William Stein
2004-05-06