As usual, let be a global field. Now we do the same thing that we did last time, but for the ideles.

**1. Ideles **

First of all, we have to define the ideles. These are only a group, and are defined as the restricted direct product

relative to the unit subgroups of -units (which are defined to be if is archimedean). In other words, an idele is required to satisfy for almost all .

If is a finite set of places containing the archimedean ones, we can define the subset ; this has the product topology and is an open subgroup of . These are called the **-ideles**. As we will see, they form an extremely useful filtration on the whole idele group.

Dangerous bend: Note incidentally that while the ideles are a subset of the adeles, the induced topology on is **not** the -topology. For instance, take . Consider the sequence of ideles where is at (where is the -th prime) and 1 everywhere else. Then but not in .

However, we still do have a canonical “diagonal” embedding , since any nonzero element of is a unit almost everywhere. This is analogous to the embedding .

**2. Idele classes **

In analogy to the whole business of adele classes modulo the ground field, we will consider the **idele class group** . This will be super-important in class field theory. In fact, class field theory exhibits a bijection between the open finite-index subgroups of and the abelian extensions of .

First, though, let’s prove that is discrete as a subset of . Cassels-Frohlich suggest that, rather than an actual proof, a metamathematical argument is more convincing: it is impossible to imagine any natural topology on the global field other than the discrete one!

But, let’s give a legitimate proof. Consider the unit idele . We can find a small neighborhood of it of the form

and if belonged to , then would not satisfy the product formula, contradiction. This completes the proof of the claim.

**3. Compactness **

In the adele case, we had compact. This doesn’t work in the idele case. Indeed, first define the **idele norm**

so that by the product formula, for . This is evidently continuous since it is continuous on each , and it factors to become a continuous map whose image is easily seen to be unbounded.

However, we do have:

Theorem 1The group of idele classes of norm 1 is compact in the induced topology. More generally, of -ideles of norm 1 modulo -units is compact whenever contains the archimedean places.

We have an embedding in an obvious manner, so it suffices to prove the first claim. However, it even suffices to prove that the topological group of ideles of norm (where ) is compact for some . We can take very large if we want. This is because the groups and are topologically isomorphic, the isomorphism being given by multiplication by an idele which is at one archimedean place and 1 everywhere else.

Lemma 2There is so large such that if has norm , then there is with for all .

This will be an application of the business of counting lattice points in parallelotopes, i.e. the “Riemann-Roch problem” in number fields.

Note that the family for any idele is a **divisor** in the sense of that post, and is just the norm of . So if is large enough (depending only on ), say at least , then there exists with

which means that for each . In addition, since each ,

So if , we get the lemma with .

We can finish the proof of the compactness theorem. Now consider the group

with the product topology. It is evidently compact, by Tychonoff’s theorem. (Incidentally, does this mean algebraic number theory depends on the axiom of choice?) I claim that it is actually a topological subgroup of the idele group . But this is because, for almost all , the inequality implies by the way these absolute values are normalized. Then surjects on , so is compact, proving the theorem.

**4. Application I: The unit theorem **

We can now finish the proof of the unit theorem (started here). Consider the continuous map sending

The image is contained in the subspace of vectors such that the sum of the coordinates is zero. Also, it is easy to see that contains a lattice of maximal rank in (and much more, actually). The image of the -units is a lattice in this subspace, as we discussed already. Now is compact (as the continuous image of the compact set . This means that is compact, but it is also discrete ( being a lattice), which means that must be of the same rank as . In particular, has maximal rank in . Since all that was left in the prooof of the unit theorem was determining the rank of , we have completed the proof.

**5. Application II: Finiteness of the class number **

This is a big theorem:

Theorem 3If is a number field, then the ideal class group of is finite.

Indeed, we have a map from (subgroup of ideles of norm 1) to the group of fractional ideals of . This sends to If we endow with the discrete topology, then is easily seen to be continuous. Moreover, becomes a continuous map , where is the subgroup of principal ideals.

But is surjective. In fact, since the values at archimedean places don’t affect , it is even surjective as a map . Since is compact, its image (i.e. the ideal class group) is both compact and discrete, hence finite.

## Leave a Reply