associated to a number field
group of elements of
that have an inverse in
We prove the theorem by defining a map
showing that the kernel of is finite and the image of
is a lattice in a hyperplane in
. The trickiest part of the
proof is showing that the image of spans a hyperplane, and we
do this by a clever application of Blichfeldt's lemma (that if is
closed, bounded, symmetric, etc., and has volume at least
, then contains a nonzero element).
is due to Dirichlet who lived 1805-1859.
Thomas Hirst described Dirichlet as follows:
He is a rather tall, lanky-looking man, with moustache and beard about
to turn grey with a somewhat harsh voice and rather deaf. He was
unwashed, with his cup of coffee and cigar. One of his failings is
forgetting time, he pulls his watch out, finds it past three, and runs
out without even finishing the sentence.
Koch wrote that:
... important parts of mathematics were influenced by Dirichlet. His
proofs characteristically started with surprisingly simple
observations, followed by extremely sharp analysis of the remaining
I think Koch's observation nicely describes the proof we will give of
The following proposition explains how to think about units in terms
of the norm.
is a unit, then
is also a
. Since both
are integers, it follows that
, then the equation
is the product of the images of
is also a product of
, hence a product of algebraic integers,
hence an algebraic integer. Thus
, which proves
is a unit.
Let be the number of real and the number of complex conjugate
embeddings of into
Define a map
The image of lies in the hyperplane
by Proposition 12.1.4
Taking logs of both sides proves the lemma.
is the bounded subset of
of elements all of whose coordinates
have absolute value at most
is a lattice
(see Proposition 5.2.4
is finite, so
The kernel of is a finite cyclic group.
It is a general fact that any finite subgroup of the multiplicative
group of a field is cyclic. [Homework.]
To prove Theorem 12.1.2, it suffices to proove that
Im is a lattice in the hyperplane from
(12.1.1), which we view as a vector space of dimension
Define an embedding
where we view
Note that this is exactly the same as the embedding
from before, except that we have re-ordered the last imaginary
components to be next to their corresponding real parts.
The image of is discrete in
is any bounded subset of
. Then for any
the coordinates of
are bounded in terms
is an increasing function). Thus
a bounded subset of
is a lattice in
, it follows that
is finite. Since
is finite, and
has finite kernel, so
is finite, which implies that
To finish the proof of Theorem 12.1.2, we will show that the
image of spans . Let be the
-span of the image
, and note that is a subspace of . We will show
that indirectly by showing that if
where is with respect to the dot product on
. This will show that
, hence that
, as required.
Define a function
To show that
we show that there exists some with
Choose any positive real numbers
Then is closed, bounded, convex, symmetric with respect to the
origin, and of dimension , since is a product of intervals
and discs, each of which has these properties.
Viewing as a product of intervals and discs, we see that the volume of is
Recall that if is a lattice and is closed,
bounded, etc., and has volume at least
contains a nonzero element. To apply this lemma, we
, where is as in (12.1.2).
We showed, when proving finiteness of the class group, that
. To check the hypothesis
to Blichfeld's lemma, note that
Thus there exists a nonzero element
, i.e., a nonzero
We then have
is nonzero, we also have
Moreover, if for any , we have
a contradiction, so
Our strategy is to use an appropriately chosen to construct a unit
. First, let
representative generators for the finitely many nonzero principal
ideals of of norm at most . Since
we have , for some , so there is a unit
(12.1.3) above. We first show that
The amazing thing about (12.1.4) is that the bound on the right
hand side does not depend on the .
Suppose we can choose positive real numbers such that
is such that . Then
would imply that , which is exactly what we aimed to
prove. It is possible to choose such , by proceeding as follows.
If , then we are trying to prove that
is a lattice
, which is automatically true, so assume .
Then there are at least two distinct . Let be such that
(which exists since ). Then
, so we choose very large and the other
, for , in any way we want subject to the condition
Since it is possible to choose the as needed, it is possible
to find a unit such that . We conclude that
which finishes the proof Theorem 12.1.2.