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 extension of
generated by
, which is the smallest number field that contains
both
and
. If
then
has a minimal polynomial
, and the Galois conjugates of
are the roots
of
. The are called the Galois conjugates because the are the orbit
of
under the action of
.
We create the extension
in SAGE.
sage: L.<cuberoot2> = CyclotomicField(3).extension(x^3 - 2) sage: cuberoot2^3 2
Then we list the Galois conjugates of
.
sage: cuberoot2.galois_conjugates() [cuberoot2, (-zeta3 - 1)*cuberoot2, zeta3*cuberoot2]
Note that
:
sage: zeta3 = L.base_field().0 sage: zeta3^2 -zeta3 - 1
Suppose
is an inclusion of number fields and let
. Then left multiplication by
defines a
-linear
transformation
. (The transformation
is
-linear because
is commutative.)
Note that if
is the characteristic polynomial of
,
then the constant term of
is
, and the
coefficient of
is
.
It is important in Proposition 2.4.3 that
the product and sum be over all the images
,
not over just the distinct images. For example, if
, then
, whereas the sum of the distinct conjugates
of
is
.
The following corollary asserts that the norm and trace behave well in towers.
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 2.4.3.
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
, which is 0 only on the
element 0. 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.
William Stein 2012-09-24