We have seen examples in which is not a unique factorization
domain. If is a principal ideal domain, then it is a unique
factorization domain, so it is of interest to understand how badly
fails to be a principal ideal domain. The class group of
measures this failure. As one sees in a course on Class Field
Theory, the class group and its generalizations also yield deep
insight into the possible abelian Galois extensions of .
be the ring of integers of a number field
is the group of nonzero fractional ideals
modulo the sugroup of principal fractional ideals
Note that if we let denote the group of nonzero fractional
ideals, then there is an exact sequence
A basic theorem in algebraic number theory is that the class group
is finite, which 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.
The explicit bound in the theorem is called the Minkowski bound, and I think it
is the best known unconditional general bound (though there are better
bounds in certain special cases).
Before proving Theorem 10.4.2, we prove a few
lemmas. The strategy of the proof will be 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
be an integral ideal, i.e., that
Let be a subset of
. Then is
if whenever then the line connecting and
lies entirely in . We say that is if whenever then also.
If is a lattice in , then the 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 .
be an automorphism of
defines an isomorphism of real manifolds
volume by a factor of
. The claimed
formula then follows.
Fix a number field with ring of integers .
be the embedding
are the real embeddings
the complex embeddings of , with one representative of
each pair of complex conjugate embeddings.
Note that this is not exactly the same as the one
at the beginning of Section 10.2.
From a basis
we obtain a matrix
th row is
and whose determinant has absolute value equal to the volume
. By doing the following three column operations,
we obtain a matrix whose rows are exactly the images of
is the matrix that came up when we defined
times each column with entries
to the column with entries
- Multiply all columns
by , thus changing the determinant by .
- Add each columns with entries
to the the column with entries
Recalling the definition of discriminant, we see that if
is the matrix constructed by the above three
We know that
is a nonzero rational number.
is a lattice in
as abelian group and spans
is also a lattice in
. For the volume
formula, combine Lemmas 10.4.4
[Proof of Theorem 10.4.2
be a number field with ring of integers
be as above,
be the function defined by
Notice that if
be any closed, bounded, convex, subset that is
symmetric with respect to the origin and has positive volume. Since is closed
Suppose is any nonzero fractional ideal of . Our goal
is to prove there is an integral ideal with small norm. We
will do this by finding an appropriate
By Lemma 10.4.6,
so by Lemma 10.4.3
is the largest norm of an element of
, the largest norm
of an element of
is at most
, we have
is an integral ideal of
that is equivalent to
Notice that the right hand side is independent of
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.
is a number field. Then .
Applying Theorem 10.4.2
to the unit ideal,
we get the bound
and the right hand quantity is strictly bigger than