We prove the theorem by defining a map , and 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).

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 problem.I think Koch's observation nicely describes the proof we will give of Theorem 12.1.2.

The following proposition explains how to think about units in terms of the norm.

Let be the number of real and the number of complex conjugate embeddings of into , so . Define a map

for all | ||

where is the bounded subset of of elements all of whose coordinates have absolute value at most . Since is a lattice (see Proposition 5.2.4), the intersection is finite, so is finite.

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

given by , where we view via . Note that this is exactly the same as the embedding

Re Re Im Im |

from before, except that we have re-ordered the last imaginary components to be next to their corresponding real parts.

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 , then . This will show that , hence that , as required.

Thus suppose . Define a function by

To show that we show that there exists some with .

Let

for | ||

for |

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
, then
contains a nonzero element. To apply this lemma, we
take
, 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
*

Since is nonzero, we also have

for and for

Our strategy is to use an appropriately chosen to construct a unit such . First, let be representative generators for the finitely many nonzero principal ideals of of norm at most . Since , we have , for some , so there is a unit such that .

Let

We have

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

William Stein 2004-05-06