# The Group of Units

Definition 8.1.1 (Unit Group)   The group of units associated to a number field  is the group of elements of that have an inverse in .

Theorem 8.1.2 (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.

(Note that we will prove a generalization of Theorem 8.1.2 in Section 12.1 below.)

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 Blichfeld's Lemma 7.1.5.

Remark 8.1.3   Theorem 8.1.2 is due to Dirichlet who lived 1805-1859. Thomas Hirst described Dirichlet thus:
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 8.1.2.

Units have a simple characterization in terms of their norm.

Proposition 8.1.4   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 the log embedding

by

(Here is the usual absolute value of , so .)

Lemma 8.1.5   The image of lies in the hyperplane

 (8.1)

Proof. If , then by Proposition 8.1.4,

Taking logs of both sides proves the lemma.

Lemma 8.1.6   The kernel of is finite.

Proof. We have

 for

where is the bounded subset of of elements all of whose coordinates have absolute value at most . Since is a lattice (see Proposition 2.4.5), the intersection is finite, so is finite.

Lemma 8.1.7   The kernel of  is a finite cyclic group.

Proof. Lemma 8.1.6 implies that is a finite group. It is a general fact that any finite subgroup of the multiplicative group of a field is cyclic. (Proof: If  is the exponent of , then every element of  is a root of the polynomial . A polynomial of degree  over a field has at most  roots, so has order at most , hence  is cyclic of order .)

To prove Theorem 8.1.2, it suffices to prove that Im is a lattice in the hyperplane  of (8.1.1), which we view as a vector space of dimension .

Define an embedding

 (8.2)

given by , where we view via . Thus this is the embedding

 Re   Im   Re   Im

Lemma 8.1.8   The image is discrete.

Proof. We will show that for any bounded subset  of , the intersection is finite. If is bounded, then for any the coordinates of are bounded, since is bounded on bounded subsets of . Thus is a bounded subset of . Since , and is a lattice in , it follows that is finite; moreover,  is injective, so is finite. Thus is finite.

We will use the following lemma in our proof of Theorem 8.1.2.

Lemma 8.1.9   Let be an integer, suppose are not all equal, and suppose are positive. Then there exist such that

and .

Proof. Order the so that . By hypothesis there exists a such that , and again re-ordering we may assume that . Set . Suppose are any positive real numbers with . Since ,

Since , we have as . It is thus possible to choose the as in the lemma.

Proof. [Proof of Theorem 8.1.2] By Lemma 8.1.8, the image is discrete, so it remains to show that spans . Let  be the -span of the image , and note that  is a subspace of , by Lemma 8.1.5. We will show that indirectly by showing that if , where  is the orthogonal complement with respect to the dot product on , then . This will show that , hence that , as required.

Thus suppose . Define a function by

 (8.3)

Note that if and only if , so 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 Blichfeldt's Lemma 7.1.5, which asserts 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 (8.1.2). By Lemma 7.1.7, we have . To check the hypothesis of Blichfeld's lemma, note that

Thus there exists a nonzero element in . Let with , then , so 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 (8.4)

Recall that our overall 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 (8.1.3) above. We first show that

We have

In the last step we use (8.1.4).

Let , and note that does not depend on the choice of the ; in fact, it only depends on the field . Moreover, for any choice of the as above, we have

If we can choose positive real numbers  such that

then the fact that would then imply that , which is exactly what we aimed to prove.

If , then we are trying to prove that is a lattice in , which is automatically true, so assume . To finish the proof, we explain how to use Lemma 8.1.9 to choose such that . We have

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 . But this is exactly what Lemma 8.1.9 shows. It is thus possible to find a unit  such that . Thus , so , whence , which finishes the proof Theorem 8.1.2.

William Stein 2012-09-24