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
![$ a$](img163.png)
from
![$ L$](img13.png)
to
![$ K$](img9.png)
are
![$\displaystyle \Norm _{L/K}(a)={\mathrm{Det}}(\ell_a)$](img290.png)
and
It is standard from linear algebra that
determinants are multiplicative
and traces are additive, so for
we have
and
Note that if
is the characteristic polynomial of
,
then the constant term of
is
, and the
coefficient of
is
.
Proof.
We prove the proposition by computing the characteristic
polynomial
![$ F$](img302.png)
of
![$ a$](img163.png)
. Let
![$ f\in K[x]$](img303.png)
be the minimal polynomial
of
![$ a$](img163.png)
over
![$ K$](img9.png)
, and note that
![$ f$](img162.png)
has distinct roots (since it is the
polynomial in
![$ K[x]$](img304.png)
of least degree that is satisfied by
![$ a$](img163.png)
).
Since
![$ f$](img162.png)
is irreducible,
![$ [K(a):K]=\deg(f)$](img305.png)
, and
![$ a$](img163.png)
satisfies a
polynomial if and only if
![$ \ell_a$](img289.png)
does, the characteristic polynomial
of
![$ \ell_a$](img289.png)
acting on
![$ K(a)$](img283.png)
is
![$ f$](img162.png)
. Let
![$ b_1,\ldots,b_n$](img306.png)
be a basis
for
![$ L$](img13.png)
over
![$ K(a)$](img283.png)
and note that
![$ 1,\ldots, a^m$](img307.png)
is a basis for
![$ K(a)/K$](img308.png)
, where
![$ m=\deg(f)-1$](img309.png)
. Then
![$ a^i b_j$](img310.png)
is a basis for
![$ L$](img13.png)
over
![$ K$](img9.png)
, and left multiplication by
![$ a$](img163.png)
acts the same way on the span of
![$ b_j, a b_j, \ldots, a^m b_j$](img311.png)
as on the span of
![$ b_k, a b_k, \ldots,
a^m b_k$](img312.png)
, for any pair
![$ j, k\leq n$](img313.png)
. Thus the matrix of
![$ \ell_a$](img289.png)
on
![$ L$](img13.png)
is a block direct sum of copies of the matrix of
![$ \ell_a$](img289.png)
acting
on
![$ K(a)$](img283.png)
, so the characteristic polynomial of
![$ \ell_a$](img289.png)
on
![$ L$](img13.png)
is
![$ f^{[L:K(a)]}$](img314.png)
. The proposition follows because the roots of
![$ f^{[L:K(a)]}$](img314.png)
are exactly the images
![$ \sigma_i(a)$](img315.png)
, with multiplicity
![$ [L:K(a)]$](img316.png)
(since each embedding of
![$ K(a)$](img283.png)
into
![$ \overline{\mathbf{Q}}$](img201.png)
extends in
exactly
![$ [L:K(a)]$](img316.png)
ways to
![$ L$](img13.png)
by Exercise
9).
The following corollary asserts that the norm and trace behave well in
towers.
Corollary 5.2.3
Suppose
is a tower of number fields, and
let
. Then
![$\displaystyle \Norm _{M/K}(a) = \Norm _{L/K}(\Norm _{M/L}(a))$](img319.png)
and
Proof.
For the first equation, both sides are the product of
![$ \sigma_i(a)$](img315.png)
,
where
![$ \sigma_i$](img321.png)
runs through the embeddings of
![$ M$](img92.png)
into
![$ K$](img9.png)
. To see
this, suppose
![$ \sigma:L\to \overline{\mathbf{Q}}$](img322.png)
fixes
![$ K$](img9.png)
. If
![$ \sigma'$](img323.png)
is an
extension of
![$ \sigma$](img324.png)
to
![$ M$](img92.png)
, and
![$ \tau_1,\ldots, \tau_d$](img325.png)
are the
embeddings of
![$ M$](img92.png)
into
![$ \overline{\mathbf{Q}}$](img201.png)
that fix
![$ L$](img13.png)
, then
![$ \tau_1\sigma',\ldots,\tau_d\sigma'$](img326.png)
are exactly the extensions of
![$ \sigma$](img324.png)
to
![$ M$](img92.png)
. For the second statement, both sides are the sum of
the
![$ \sigma_i(a)$](img315.png)
.
The norm and trace down to
of an algebraic integer
is an
element of
, because the minimal polynomial of
has integer
coefficients, and the characteristic polynomial of
is a power of the
minimal polynomial, as we saw in the proof of
Proposition 5.2.2.
Proposition 5.2.4
Let
be a number field. The ring of integers
is a lattice
in
, i.e.,
and
is an abelian group of rank
.
Proof.
We saw in Lemma
5.1.8 that
![$ \mathbf{Q}\O _K = K$](img264.png)
. Thus there exists a
basis
![$ a_1,\ldots, a_n$](img133.png)
for
![$ K$](img9.png)
, where each
![$ a_i$](img115.png)
is in
![$ \O _K$](img200.png)
.
Suppose that as
![$ x=\sum c_i a_i\in \O _K$](img328.png)
varies over all elements of
![$ \O _K$](img200.png)
the denominators of the coefficients
![$ c_i$](img138.png)
are arbitrarily
large. Then subtracting off integer multiples of the
![$ a_i$](img115.png)
, we see
that as
![$ x=\sum c_i a_i\in \O _K$](img328.png)
varies over elements of
![$ \O _K$](img200.png)
with
![$ c_i$](img138.png)
between
0 and
![$ 1$](img147.png)
, the denominators of the
![$ c_i$](img138.png)
are also
arbitrarily large. This implies that there are infinitely many elements
of
![$ \O _K$](img200.png)
in the bounded subset
Thus for any
![$ \varepsilon >0$](img330.png)
, there are elements
![$ a,b\in \O _K$](img331.png)
such that the
coefficients of
![$ a-b$](img332.png)
are all less than
![$ \varepsilon $](img333.png)
(otherwise the elements
of
![$ \O _K$](img200.png)
would all be a ``distance'' of least
![$ \varepsilon $](img333.png)
from each other, so only finitely
many of them would fit in
![$ S$](img121.png)
).
As mentioned above, the norms of elements of
are integers.
Since the norm of an element is the determinant of left multiplication
by that element, the norm is a homogenous polynomial of degree
in
the indeterminate coefficients
. If the
get arbitrarily
small for elements of
, then the values of the norm polynomial
get arbitrarily small, which would imply that there are elements of
with positive norm too small to be in
, a contradiction.
So the set
contains only finitely many elements of
. Thus
the denominators of the
are bounded, so for some
, we have
that
has finite index in
. Since
is isomorphic to
, it follows
from the structure theorem for finitely generated abelian groups that
is isomorphic as a
-module to
, as claimed.
Proof.
By Proposition
5.2.4, the ring
![$ \O _K$](img200.png)
is
finitely generated as a module over
![$ \mathbf{Z}$](img40.png)
, so it is certainly
finitely generated as a ring over
![$ \mathbf{Z}$](img40.png)
. By the Hilbert
Basis Theorem,
![$ \O _K$](img200.png)
is Noetherian.
William Stein
2004-05-06