# Finishing the proof of Dirichlet's Unit Theorem

We begin by finishing Dirichlet's proof that the group of units of is isomorphic to , where  is the number of real embeddings,  is half the number of complex embeddings, and  is the number of roots of unity in . Recall that we defined a map by

Without much trouble, we proved that the kernel of if finite and the image is discrete, and in the last section we were finishing the proof that the image of spans the subspace of elements of that are orthogonal to , where of the entries are 's and of them are 's. The somewhat indirect route we followed was to suppose

i.e., that is not a multiple of , and prove that is not orthogonal to some element of . Writing , this would show that , so . We ran into two problems: (1) we ran out of time, and (2) the notes contained an incomplete argument that a quantity can be chosen to be arbitrarily large. We will finish going through a complete proof, then compute many examples of unit groups using .

Recall that was defined by

(dot product)

and our goal is to show that there is a such that .

Our strategy is to use an appropriately chosen  to construct a unit such . Recall that we used Blichfeld's lemma to find an such that , and

 for     and     for (12.5)

Let be representative generators for the finitely many nonzero principal ideals of of norm at most . Modify the to have the property that is minimal among generators of (this is possible because ideals are discrete). Note that the set depends only on . Since , we have , for some , so there is a unit  such that .

Let

Lemma 12.2.1   We have

and depends only on and our fixed choice of .

Proof. By properties of logarithms, . We next use the triangle inequality in various ways, properties of logarithms, and the bounds (12.2.1) in the following computation:

The inequality of the lemma now follows. That only depends on and our choice of follows from the formula for and how we chose the .

The amazing thing about Lemma 12.2.1 is that the bound on the right hand side does not depend on the . Suppose we could somehow cleverly choose the positive real numbers in such a way that

and

Then the facts that and would together imply that (since is closer to than is to 0), which is exactly what we aimed to prove. We finish the proof by showing that it is possible to choose such . Note that if we change the , then  could change, hence the such that is a unit could change, but the don't change, just the subscript . Also note that if , then we are trying to prove that is a lattice in , which is automatically true, so we may assume that .

Lemma 12.2.2   Assume . Then there is a choice of such that

Proof. It is easier if we write

where and for , and and for ,

The condition that is that the are not all the same, and in our new coordinates the lemma is equivalent to showing that , subject to the condition that . Order the so that . By hypothesis there exists a such that , and again re-ordering we may assume that . Set . Then and , so

Since , we have as .

William Stein 2004-05-06