Discriminants
Suppose
are a basis for a number field
, 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 is a well-defined integer which can
be either positive or negative. Note that
so
can be defined purely in terms of the trace without
mentioning the embeddings
. Also, changing the basis for
is the same as left multiplying
by an integer matrix
of
determinant
, which does not change the squared determinant,
since
. Thus
is well defined, and does not depend on the choice of basis.
If we view
as a
-vector space, then
defines a bilinear pairing
on
, which we call
the . The following lemma asserts that this
pairing is nondegenerate, so
hence
.
Proof.
If the trace pairing is degenerate, then there exists
![$ a\in K$](img371.png)
such
that for every
![$ b\in K$](img738.png)
we have
![$ \Tr (ab)=0$](img739.png)
. In particularly, taking
![$ b=a^{-1}$](img740.png)
we see that
![$ 0=\Tr (a a^{-1})=\Tr (1)=[K:\mathbf{Q}]>0$](img741.png)
, which is
absurd.
Definition 10.2.2 (Discriminant)
Suppose
![$ a_1,\ldots, a_n$](img133.png)
is any
![$ \mathbf{Q}$](img10.png)
-basis of
![$ K$](img9.png)
. The
of
![$ a_1,\ldots, a_n$](img133.png)
is
The
![$ \Disc (\O )$](img743.png)
of an order
![$ \O$](img543.png)
in
![$ \O _K$](img200.png)
is
the discriminant of any basis for
![$ \O$](img543.png)
.
The
![$ d_K=\Disc (K)$](img744.png)
of the number field
![$ K$](img9.png)
is the discrimimant of
![$ \O _K$](img200.png)
.
Note that the discriminants defined above are all nonzero
by Lemma 10.2.1.
Warning: In
is defined to be the discriminant of the
polynomial you happened to use to define
, which is (in my opinion)
a poor choice and goes against most of the literature.
The following proposition asserts that the discriminant of an order
in
is bigger than
by a factor of the square
of the index.
Proposition 10.2.3
Suppose
is an order in
. Then
Proof.
Let
![$ A$](img60.png)
be a matrix whose rows are the images via
![$ \sigma$](img324.png)
of a basis
for
![$ \O _K$](img200.png)
, and let
![$ B$](img82.png)
be a matrix whose rows are the images via
![$ \sigma$](img324.png)
of a basis for
![$ \O$](img543.png)
. Since
![$ \O\subset \O _K$](img748.png)
has finite
index, there is an integer matrix
![$ C$](img749.png)
such that
![$ CA=B$](img750.png)
,
and
![$ \vert{\mathrm{Det}}(C)\vert= [\O _K:\O ]$](img751.png)
. Then
This result is enough to give an algorithm for computing
,
albeit a potentially slow one. Given
, find some order
, and compute
. Factor
, and use the factorization
to write
, where
is the largest square that
divides
. Then the index of
in
is a divisor of
,
and we (tediously) can enumerate all rings
with
and
, until we find the largest one all of
whose elements are integral.
Example 10.2.4
Consider the ring
![$ \O _K = \mathbf{Z}[(1+\sqrt{5})/2]$](img250.png)
of integers of
![$ K=\mathbf{Q}(\sqrt{5})$](img249.png)
. The discriminant of the basis
![$ 1,a=(1+\sqrt{5})/2$](img759.png)
is
Let
![$ \O =\mathbf{Z}[\sqrt{5}]$](img761.png)
be the order generated by
![$ \sqrt{5}$](img762.png)
.
Then
![$ \O$](img543.png)
has basis
![$ 1,\sqrt{5}$](img763.png)
, so
William Stein
2004-05-06