Let be a global field. For each normalized valuation of , let denote the completion of . If is non-archimedean, let denote the ring of integers of .

Definition 18.3.1 (Adele Ring)   The adele ring of is the topological ring whose underlying topological space is the restricted topological product of the with respect to the , and where addition and multiplication are defined componentwise:

 for (18.4)

It is readily verified that (i) this definition makes sense, i.e., if , then and , whose components are given by (18.3.1), are also in , and (ii) that addition and multiplication are continuous in the -topology, so is a topological ring, as asserted. Also, Lemma 18.2.4 implies that is locally compact because the are locally compact (Corollary 15.1.6), and the are compact (Theorem 15.1.4).

There is a natural continuous ring inclusion

 (18.5)

that sends to the adele every one of whose components is . This is an adele because for almost all , by Lemma 18.1.2. The map is injective because each map is an inclusion.

Definition 18.3.2 (Principal Adeles)   The image of (18.3.2) is the ring of principal adeles.

It will cause no trouble to identify with the principal adeles, so we shall speak of  as a subring of .

Formation of the adeles is compatibility with base change, in the following sense.

Lemma 18.3.3   Suppose is a finite (separable) extension of the global field . Then

 (18.6)

both algebraically and topologically. Under this isomorphism,

maps isomorphically onto .

Proof. Let be a basis for and let run through the normalized valuations on . The left hand side of (18.3.3), with the tensor product topology, is the restricted product of the tensor products

with respect to the integers

 (18.7)

(An element of the left hand side is a finite linear combination of adeles and coefficients , and there is a natural isomorphism from the ring of such formal sums to the restricted product of the .)

We proved before (Theorem 17.1.8) that

where are the normalizations of the extensions of to . Furthermore, as we proved using discriminants (see Lemma 18.1.6), the above identification identifies (18.3.4) with

for almost all . Thus the left hand side of (18.3.3) is the restricted product of the with respect to the . But this is canonically isomorphic to the restricted product of all completions with respect to , which is the right hand side of (18.3.3). This establishes an isomorphism between the two sides of (18.3.3) as topological spaces. The map is also a ring homomorphism, so the two sides are algebraically isomorphic, as claimed.

Corollary 18.3.4   Let denote the topological group obtained from the additive structure on . Suppose is a finite seperable extension of . Then

summands

In this isomorphism the additive group of the principal adeles is mapped isomorphically onto .

Proof. For any nonzero , the subgroup of is isomorphic as a topological group to (the isomorphism is multiplication by ). By Lemma 18.3.3, we have isomorphisms

If , write , with . Then maps via the above map to

where denotes the principal adele defined by . Under the final map, maps to the tuple

The dimensions of and of over are the same, so this proves the final claim of the corollary.

Theorem 18.3.5   The global field is discrete in and the quotient of additive groups is compact in the quotient topology.

At this point Cassels remarks
It is impossible to conceive of any other uniquely defined topology on . This metamathematical reason is more persuasive than the argument that follows!''

Proof. Corollary 18.3.4, with for and or for , shows that it is enough to verify the theorem for or , and we shall do it here for .

To show that is discrete in it is enough, because of the group structure, to find an open set that contains , but which contains no other elements of . (If , then is an open subset of whose intersection with is .) We take for the set of with

and   (all $p$)

where and are respectively the -adic and the usual archimedean absolute values on  . If , then in the first place because for all , and then because . This proves that is discrete in . (If we leave out one valuation, as we will see later (Theorem 18.4.4), this theorem is false--what goes wrong with the proof just given?)

Next we prove that the quotient is compact. Let consist of the with

and   for all primes $p$

We show that every adele is of the form

which will imply that the compact set maps surjectively onto . Fix an adele . Since is an adele, for each prime we can find a rational number

with   and

such that

and

almost all $p$

More precisely, for the finitely many such that

choose to be a rational number that is the value of an appropriate truncation of the -adic expansion of , and when just choose . Hence is well defined. The for do not mess up the inequality since the valuation is non-archimedean and the do not have any in their denominator:

Now choose such that

Then and do what is required, since has the desired property (since and the -adic valuations are non-archimedean).

Hence the continuous map induced by the quotient map is surjective. But is compact (being the topological product of the compact spaces and the for all ), hence is also compact.

Corollary 18.3.6   There is a subset of defined by inequalities of the type , where for almost all , such that every can be put in the form

i.e., .

Proof. We constructed such a set for when proving Theorem 18.3.5. For general  the coming from the proof determines compenent-wise a subset of that is a subset of a set with the properties claimed by the corollary.

As already remarked, is a locally compact group, so it has an invariant Haar measure. In fact one choice of this Haar measure is the product of the Haar measures on the , in the sense of Definition 18.2.5.

Corollary 18.3.7   The quotient has finite measure in the quotient measure induced by the Haar measure on .

Remark 18.3.8   This statement is independent of the particular choice of the multiplicative constant in the Haar measure on . We do not here go into the question of finding the measure in terms of our explicitly given Haar measure. (See Tate's thesis, [Cp86, Chapter XV].)

Proof. This can be reduced similarly to the case of or which is immediate, e.g., the defined above has measure for our Haar measure.

Alternatively, finite measure follows from compactness. To see this, cover with the translates of , where is a nonempty open set with finite measure. The existence of a finite subcover implies finite measure.

Remark 18.3.9   We give an alternative proof of the product formula for nonzero . We have seen that if , then multiplication by magnifies the Haar measure in by a factor of . Hence if , then multiplication by magnifies the Haar measure in by . But now multiplication by takes into , so gives a well-defined bijection of onto which magnifies the measure by the factor . Hence Corollary 18.3.7. (The point is that if is the measure of , then , so because is finite we must have .)

William Stein 2012-09-24