Adele – c’est un nom si belle. (Oops, that’s bad French, isn’t it?)

I actually will not be able to finish the proof of the unit theorem here, because I don’t get to the *ideles* in this post. That will come next time (there are some of the same themes as here).

Let be a global field, i.e. a finite extension of either or . Then we can consider the set absolute values on . In the number field case, these are extensions (up to a power) of the archimedean absolute value on or the -adic absolute values by a theorem of Ostrowski classifying absolute values on . In the function field case, we need another result.

Here’s how we define the adele ring. It is the restricted direct product

where restricted means that any vector is required to satisfy for almost all . This becomes a topological ring if we take a basis of the form

where are open and is the ring of integers, and is a finite set containing the archimedean places. It is clear that addition and multiplication are continuous, and that is locally compact. For finite and containing the archimedean absolute values , there is a subring , and is the union of these subrings.

Since any is contained in for almost all (this is analogous to a rational function on a curve having only finitely many poles), there is an injective homomorphism .

Next, we may define a Haar measure on by taking the product of the Haar measures on , normalized such that for . Thus one gets a (i.e., *the*) Haar measure on itself.

**1. Investigation of **

We want to prove compactness results about the adele ring, and it is easiest to start with the case .

Theorem 1is a discrete subset of , and , with the quotient topology, is compact.

First, recall that is a restricted direct product

We will find a closed neighborhood of zero containing no other element of . For this, consider

Since any nonzero satisfies the **product formula** (where ranges over all normalized absolute values of ), we cannot have or else . This proves discreteness. The proof generalizes, since the product formula, and shows that is discrete in for any number field .

Now, we show that any can be represented as where and where is the compact set as above. Indeed, let be the finite set of nonarchimedean absolute values where . Then set . Then clearly

since for . Also, by subtracting an integer , we can arrange it so that (since ). Then take . Since is compact (by Tychonoff), it follows that is compact as well.

**2. Changing fields **

We now want to prove an extension of the previous result, namely:

Theorem 2If is a number field, then is discrete and is compact.

The fact about discreteness follows from the product formula in the same way. Compactness is tricker. We will need a discussion about tensor products and completions first with which we can reduce to the case .

Let be a field with a valuation (possibly archimedean) and a finite separable extension.

Proposition 3We have an isomorphism of -algebras and topological rings,where denote completions and ranges over all extensions of to .

First, is a direct product of fields. Indeed, we can write for suitable by the primitive element theorem. Let split in as ; there are no repeated factors by separability. Then

So is isomorphic to a direct product (i.e. direct sum) of fields .

I claim now that the fields are precisely the for prolonging . Indeed, first note that , as a finite extension of the complete field , has a unique extension valuation with respect to which it is complete. The composite map induces a map of and a valuation on extending . is dense in , since gets mapped to the image of in . So is the completion of with respect to the valuation of induced on it.

What we now need to show is that the valuations induced on are distinct and that every -valuation prolonging can be realized in this way. Start with the first task.

is a finite-dimensional space over , so it has a canonical topology induced by a (non-unique) norm, and the topologies on the are the subspace topologies. Also, is dense in it. If the morphisms induced the same valuation on for different , then we would not be able to approximate a vector in whose -th coordinate was 1 and whose -th coordinate was zero arbitrarily closely by elements of .

Suppose we have a valuation prolonging on . Then extends to a multiplicative function on by continuity. This must be the norm on one of the , since it is not identically zero. So we can get any by pulling back from some .

Corollary 4For a finite separable extension ,

In the case of a discrete valuation, this is just the equality.

Corollary 5Let be an extension of number fields. As topological rings and -algebras,

Here is as a topological group . This is the restricted direct product

Now, we shall prove that is compact for any number field . As a group, this is topologically isomorphic to under the identification though, and this last group is compact.

## Leave a Reply