Let be a global field. For each normalized valuation
of ,
let denote the completion of . If
is
nonarchimedean, 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.
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
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
falsewhat goes wrong with the proof just given?)
Next we prove that the quotient
is compact.
Let
consist of the
with
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
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 nonarchimedean 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
nonarchimedean).
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.
Proof.
We constructed such a set for
when proving
Theorem
18.3.5. For general
the
coming from the
proof determines compenentwise 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 welldefined 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
20120924