Definition 7.1.1 (Class Group)
Let
be the ring of integers of a number field
. The
class group of
is the group of fractional ideals
modulo the sugroup of principal fractional ideals
, for
.
Note that if we let
denote the group of fractional
ideals, then we have an exact sequence
That the class group is finite follows from the first part of
the following theorem and the fact that there are only finitely many
ideals of norm less than a given integer (Proposition 6.3.6).
The explicit bound in the theorem is called the Minkowski
bound. There are other better bounds, but they depend on unproven
conjectures.
The following two examples illustrate how to apply
Theorem 7.1.2 to compute in
simple cases.
Example 7.1.3
Let
. Then
,
, and
, so the Minkowski bound is
Thus every fractional ideal is equivalent to an ideal of norm
.
Since
is the only ideal of norm
, every ideal is principal,
so
is trivial.
Example 7.1.4
Let
.
We have
,
so
,
,
, and the
Minkowski bound is
We compute the Minkowski bound in
SAGE as follows:
sage: K = QQ[sqrt(10)]; K
Number Field in sqrt10 with defining polynomial x^2 - 10
sage: B = K.minkowski_bound(); B
sqrt(10)
sage: B.n()
3.16227766016838
Theorem
7.1.2 implies that every ideal class has a
representative that is an integral ideal of norm
,
, or
.
The ideal
is ramified in
, so
If
were principal, say
, then
would have norm
.
Then the equation
|
(7.1) |
would have an integer solution. But the squares mod
are
, so (
7.1.1) has no solutions.
Thus
defines a nontrivial element of the class group,
and it has order
since its square is the principal ideal
.
Thus
.
To find the integral ideals of norm , we
factor modulo , and see that
If either of the prime divisors of
were principal,
then the equation
would have an integer
solution. Since it does not have one mod
, the prime divisors
of
are both nontrivial elements of the class
group.
Let
Then
so the classes over
are equal.
In summary, we now know that every element of is equivalent to one of
or
Thus the class group is a group of order at most
that contains an
element of order
. Thus it must have order
. We verify this in
SAGE below, where we also check that
generates the
class group.
sage: K.<sqrt10> = QQ[sqrt(10)]; K
Number Field in sqrt10 with defining polynomial x^2 - 10
sage: G = K.class_group(); G
Class group of order 2 with structure C2 of Number Field ...
sage: G.0
Fractional ideal class (3, sqrt10 + 1)
sage: G.0^2
Trivial principal fractional ideal class
sage: G.0 == G( (3, 2 + sqrt10) )
True
Before proving Theorem 7.1.2, we prove a few
lemmas. The strategy of the proof is to start with any nonzero
ideal , and prove that there is some nonzero , with very
small norm, such that is an integral ideal. Then
will be small, since
is small. The trick is to determine precisely
how small an we can choose subject to the condition that
is an integral ideal, i.e., that
.
Let be a subset of
. Then is convex if
whenever then the line connecting and lies entirely
in . We say that is symmetric about the origin if
whenever then also. If is a lattice in the
real vector space
, then the volume of is the
volume of the compact real manifold , which is the same thing as
the absolute value of the determinant of any matrix whose rows form a
basis for .
Proof.
Let
be an automorphism of
such that
. Then
defines an isomorphism of real manifolds
that changes
volume by a factor of
. The claimed
formula then follows, since
, by definition.
Fix a number field with ring of integers .
Let
be the real embeddings
of and
be half
the complex embeddings of , with one representative of
each pair of complex conjugate embeddings.
Let
be the embedding
Note that this is not exactly the same as the one
at the beginning of Section 6.2 if .
Proof.
Let
.
From a basis
for
we obtain a matrix
whose
th row is
and whose determinant has absolute value equal to the volume
of
. By doing the following three column operations,
we obtain a matrix whose rows are exactly the images of
the
under
all embeddings of
into
, which
is the matrix that came up when we defined
in Section
6.2.
- Add
times each column with entries
Im
to the column with entries
Re.
- Multiply all columns with entries
Im
by , thus changing the determinant by .
- Add each columns that now has entries
ReIm
to the the column with entries
Im
to obtain columns
ReIm.
Recalling the definition of discriminant, we see that if
is the matrix constructed by doing the above three
operations to
, then
.
Thus
Proof.
Since
has rank
as an abelian group, and
Lemma
7.1.7 implies that
also spans
,
it follows that
is a lattice in
.
For some nonzero integer
we have
,
so
is also a lattice in
.
To prove the displayed volume
formula, combine Lemmas
7.1.6-
7.1.7 to get
Proof.
[Proof of Theorem
7.1.2]
Let
be a number field with ring of integers
,
let
be as above,
and let
be the function defined by
Notice that if
then
,
and for any
,
Let
be a fixed choice of closed, bounded, convex, subset with
positive volume that is
symmetric with respect to the origin and has positive volume. Since is closed
and bounded,
exists.
Suppose is any fractional ideal of . Our goal
is to prove that there is an integral ideal with small norm. We
will do this by finding an appropriate
.
By Lemma 7.1.8,
Let
, where
.
Then
so by Lemma
7.1.5 there exists
.
Let
be such that
.
Since
is the largest norm of an element of
, the largest norm
of an element of
is at most
,
so
Since
, we have
, so
is an integral ideal of
that is equivalent to
, and
Notice that the right hand side is independent of
. It
depends only on
,
,
, and our choice of
.
This completes the proof of the theorem, except for
the assertion that
can be chosen to give the claim
at the end of the theorem, which we leave as an exercise.
Corollary 7.1.9
Suppose that
is a number field. Then .
Proof.
Applying Theorem
7.1.2 to the unit ideal,
we get the bound
Thus
and the right hand quantity is strictly bigger than
for
any
and any
(exercise).
A prime ramifies in if and only if ,
so the corollary implies that every nontrivial extension of
is ramified at some prime.
William Stein
2012-09-24