Discriminants
Suppose
are a basis for as a
-module,
which we view as a
-vector space. Let
be
the embedding
, where
are the distinct embeddings of
into
. Let be the matrix whose rows are
. The quantity depends on the ordering of the
, and need not be an integer.
If we consider instead, we obtain a number that does
not depend on ordering; moreover, as we will see, it is an integer.
Note that
so can be defined purely in terms of the trace without
mentioning the embeddings . Also, changing our choice of
basis for is the same as left multiplying by an integer
matrix of determinant ; this does not change the squared
determinant, since
.
Thus
is well defined as a quantity associated to .
If we view as a
-vector space, then
defines a bilinear pairing
on , which we call
the trace pairing. The following lemma asserts that this
pairing is nondegenerate, so
hence
.
Proof.
If the trace pairing is degenerate, then there exists
such
that for every
we have
. In particularly, taking
we see that
, which is
absurd.
Definition 6.2.2 (Discriminant)
Suppose
is any
-basis of
. The
discriminant
of
is
The
discriminant
of an order
in
is
the discriminant of any basis for
.
The
discriminant
of the number field
is the discriminant of
.
Note that these discriminants are all nonzero
by Lemma
6.2.1.
Remark 6.2.3
It is also standard to define the discriminant of a monic polynomial
to be the product of the differences of the roots. If
with
of finite index in
, and
is the
minimal polynomial of
, then
.
To see this, note that if we choose the basis
for
, then both
discriminants are the square of the same Vandermonde determinant.
Example 6.2.4
In
SAGE, we compute the discriminant of a number field or order
using the discriminant command:
sage: K.<a> = NumberField(x^2 - 5)
sage: K.discriminant()
5
This also works for orders (notice the square factor
below, which will be explained
by Proposition 6.2.5):
sage: R = K.order([7*a]); R
Order in Number Field in a with defining polynomial x^2 - 5
sage: factor(R.discriminant())
2^2 * 5 * 7^2
Warning: In is defined to be the
discriminant of the polynomial you happened to use to define .
> K := NumberField(x^2-5);
> Discriminant(K);
20
This is an intentional choice done for efficiency reasons, since
computing the maximal order can take a long time. Nonetheless, it
conflicts with standard mathematical usage, so beware.
The following proposition asserts that the discriminant of an order
in is bigger than
by a factor of the square
of the index.
Proposition 6.2.5
Suppose is an order in . Then
Proof.
Let
be a matrix whose rows are the images via
of a basis
for
, and let
be a matrix whose rows are the images via
of a basis for
. Since
has finite
index, there is an integer matrix
such that
,
and
. Then
Example 6.2.6
Let
be a number field and consider the quantity
and
One might hope that
is equal to the discriminant
of
, but this is not the case in general. Recall
Example
4.3.2, in which we considered the field
generated
by a root of
. In that example, the
discriminant of
is
with
prime:
sage: K.<a> = NumberField(x^3 + x^2 - 2*x + 8)
sage: factor(K.discriminant())
-1 * 503
For every
, we have
, since
fails to be monogenic at
. By
Proposition
6.2.5, the discriminant of
is
divisible by
for all
, so
is also
divisible by
. This is why
is called an ``inessential
discriminant divisor''.
Proposition 6.2.5 gives an algorithm for computing ,
albeit a slow one. Given , find some order
, and compute
. Factor , and use the factorization
to write
, where is the largest square that
divides . Then the index of in is a divisor of ,
and we (tediously) can enumerate all rings with
and
, until we find the largest one all of
whose elements are integral. A much better algorithm is to proceed
exactly as just described, except use the ideas
of Section 4.3.3 to find a -maximal order for each prime
divisor of , then add these -maximal orders together.
Example 6.2.7
Consider the ring
of integers of
. The discriminant of the basis
is
Let
be the order generated by
.
Then
has basis
, so
hence
.
Example 6.2.8
Consider the cubic field
, and
let
be the order
.
Relative to the base
for
,
the matrix of the trace pairing is
Thus
Suppose we do not know that the ring of integers
is equal to
. By Proposition
6.2.5,
we have
so
, and
.
Thus to prove
it suffices to prove
that
is
-maximal and
-maximal,
which could be accomplished as described in
Section
4.3.3.
William Stein
2012-09-24