# The Group of Units

Definition 12.1.1 (Unit Group)   The associated to a number field  is the group of elements of that have an inverse in .

Theorem 12.1.2 (Dirichlet)   The group is the product of a finite cyclic group of roots of unity with a free abelian group of rank , where  is the number of real embeddings of  and  is the number of complex conjugate pairs of embeddings.

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).

Remark 12.1.3   Theorem 12.1.2 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 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.

Proposition 12.1.4   An element is a unit if and only if .

Proof. Write . If is a unit, then is also a unit, and . Since both and are integers, it follows that . Conversely, if and , then the equation implies that . But is the product of the images of in by all embeddings of into , so is also a product of images of  in , hence a product of algebraic integers, hence an algebraic integer. Thus , which proves that  is a unit.

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

by

Lemma 12.1.5   The image of lies in the hyperplane

 (12.1)

Proof. If , then by Proposition 12.1.4,

Taking logs of both sides proves the lemma.

Lemma 12.1.6   The kernel of is finite.

Proof. We have

 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.

Lemma 12.1.7   The kernel of is a finite cyclic group.

Proof. 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

 (12.2)

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.

Lemma 12.1.8   The image of is discrete in .

Proof. Suppose is any bounded subset of . Then for any the coordinates of are bounded in terms of (since is an increasing function). Thus is a bounded subset of . Since , and is a lattice in , it follows that is finite. Since is injective, is finite, and has finite kernel, so is finite, which implies that is discrete.

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

 (12.3)

To show that we show that there exists some with .

Let

Choose any positive real numbers such that

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

Thus there exists a nonzero element , i.e., a nonzero such that for . We then have

Since is nonzero, we also have

Moreover, if for any , we have , then

a contradiction, so for . Likewise, , for . Rewriting this we 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

and recall defined in (12.1.3) above. We first show that

 (12.4)

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

and 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 in , which is automatically true, so assume . Then there are at least two distinct . Let be such that (which exists since ). Then as , 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 , so , whence , which finishes the proof Theorem 12.1.2.

William Stein 2004-05-06