Adele ring explained

In mathematics, the adele ring of a global field (also adelic ring, ring of adeles or ring of adèles[1]) is a central object of class field theory, a branch of algebraic number theory. It is the restricted product of all the completions of the global field and is an example of a self-dual topological ring.

An adele derives from a particular kind of idele. "Idele" derives from the French "idèle" and was coined by the French mathematician Claude Chevalley. The word stands for 'ideal element' (abbreviated: id.el.). Adele (French: "adèle") stands for 'additive idele' (that is, additive ideal element).

The ring of adeles allows one to describe the Artin reciprocity law, which is a generalisation of quadratic reciprocity, and other reciprocity laws over finite fields. In addition, it is a classical theorem from Weil that

G

-bundles on an algebraic curve over a finite field can be described in terms of adeles for a reductive group

G

. Adeles are also connected with the adelic algebraic groups and adelic curves.

The study of geometry of numbers over the ring of adeles of a number field is called adelic geometry.

Definition

Let

K

be a global field (a finite extension of

Q

or the function field of a curve
X/Fq
over a finite field). The adele ring of

K

is the subring

AK =\prod(K\nu,l{O}\nu)\subseteq\prodK\nu

consisting of the tuples

(a\nu)

where

a\nu

lies in the subring

l{O}\nu\subsetK\nu

for all but finitely many places

\nu

. Here the index

\nu

ranges over all valuations of the global field

K

,

K\nu

is the completion at that valuation and

l{O}\nu

the corresponding valuation ring.[2]

Motivation

The ring of adeles solves the technical problem of "doing analysis on the rational numbers

Q

." The classical solution was to pass to the standard metric completion

R

and use analytic techniques there. But, as was learned later on, there are many more absolute values other than the Euclidean distance, one for each prime number

p\inZ

, as was classified by Ostrowski. The Euclidean absolute value, denoted

||infty

, is only one among many others,

||p

, but the ring of adeles makes it possible to comprehend and . This has the advantage of enabling analytic techniques while also retaining information about the primes, since their structure is embedded by the restricted infinite product.

The purpose of the adele ring is to look at all completions of

K

at once. The adele ring is defined with the restricted product, rather than the Cartesian product. There are two reasons for this:

K

the valuations are zero for almost all places, i.e., for all places except a finite number. So, the global field can be embedded in the restricted product.

Why the restricted product?

The restricted infinite product is a required technical condition for giving the number field

Q

a lattice structure inside of

AQ

, making it possible to build a theory of Fourier analysis (cf. Harmonic analysis) in the adelic setting. This is analogous to the situation in algebraic number theory where the ring of integers of an algebraic number field embeds

l{O}K\hookrightarrowK

as a lattice. With the power of a new theory of Fourier analysis, Tate was able to prove a special class of L-functions and the Dedekind zeta functions were meromorphic on the complex plane. Another natural reason for why this technical condition holds can be seen by constructing the ring of adeles as a tensor product of rings. If defining the ring of integral adeles

AZ

as the ring

AZ=R x \hat{Z

} = \mathbf\times \prod_p \mathbf_p,
then the ring of adeles can be equivalently defined as

\begin{align} AQ&=QZAZ\\ &=QZ\left(R x \prodpZp\right). \end{align}

The restricted product structure becomes transparent after looking at explicit elements in this ring. The image of an element

b/c(r,(ap))\inAQ

inside of the unrestricted product \mathbf\times \prod_p \mathbf_p is the element
\left(br
c

,\left(

bap
c

\right)\right).

The factor

bap/c

lies in

Zp

whenever

p

is not a prime factor of

c

, which is the case for all but finitely many primes

p

.[3]

Origin of the name

The term "idele" (French: idèle) is an invention of the French mathematician Claude Chevalley (1909–1984) and stands for "ideal element" (abbreviated: id.el.). The term "adele" (French: French: adèle) stands for additive idele. Thus, an adele is an additive ideal element.

Examples

Ring of adeles for the rational numbers

The rationals

K=\bold{Q}

have a valuation for every prime number

p

, with

(K\nu,l{O}\nu)=(Qp,Zp)

, and one infinite valuation with

Qinfty=R

. Thus an element of

AQ =R x \prodp(Qp,Zp)

is a real number along with a p-adic rational for each

p

of which all but finitely many are p-adic integers.

Ring of adeles for the function field of the projective line

Secondly, take the function field

1)=F
K=F
q(t)
of the projective line over a finite field. Its valuations correspond to points

x

of

X=P1

, i.e. maps over

SpecFq

x:SpecF
qn

\longrightarrowP1.

For instance, there are

q+1

points of the form

SpecFq\longrightarrowP1

. In this case

l{O}\nu=\widehat{l{O}}X,x

is the completed stalk of the structure sheaf at

x

(i.e. functions on a formal neighbourhood of

x

) and

K\nu=KX,x

is its fraction field. Thus
A
1)
F
q(P

 = \prodx\in(l{K}X,x,\widehat{l{O}}X,x).

The same holds for any smooth proper curve
X/Fq
over a finite field, the restricted product being over all points of

x\inX

.

Related notions

The group of units in the adele ring is called the idele group

IK=

x
A
K
.The quotient of the ideles by the subgroup

K x \subseteqIK

is called the idele class group

CK = I

x .
K/K
The integral adeles are the subring

OK = \prodO\nu\subseteqAK.

Applications

Stating Artin reciprocity

The Artin reciprocity law says that for a global field

K

,

\widehat{CK}=

x /K
\widehat{A
K

x }\simeqGal(Kab/K)

where

Kab

is the maximal abelian algebraic extension of

K

and

\widehat{(...)}

means the profinite completion of the group.

Giving adelic formulation of Picard group of a curve

If

X/Fq
is a smooth proper curve then its Picard group is[4]

Pic(X) =K x \backslash

x
A
X/O
x
X
and its divisor group is
x
Div(X)=A
X/O
x
X
. Similarly, if

G

is a semisimple algebraic group (e.g. SL_n, it also holds for

GLn

) then Weil uniformisation says that[5]

BunG(X) =G(K)\backslashG(AX)/G(OX).

Applying this to

G=Gm

gives the result on the Picard group.

Tate's thesis

There is a topology on

AK

for which the quotient

AK/K

is compact, allowing one to do harmonic analysis on it. John Tate in his thesis "Fourier analysis in number fields and Hecke Zeta functions" proved results about Dirichlet L-functions using Fourier analysis on the adele ring and the idele group. Therefore, the adele ring and the idele group have been applied to study the Riemann zeta function and more general zeta functions and the L-functions.

Proving Serre duality on a smooth curve

If

X

is a smooth proper curve over the complex numbers, one can define the adeles of its function field

C(X)

exactly as the finite fields case. John Tate proved[6] that Serre duality on

X

H1(X,l{L}) \simeqH

-1
Xl{L}

)*

can be deduced by working with this adele ring

AC(X)

. Here L is a line bundle on

X

.

Notation and basic definitions

Global fields

Throughout this article,

K

is a global field, meaning it is either a number field (a finite extension of

\Q

) or a global function field (a finite extension of
F
pr

(t)

for

p

prime and

r\in\N

). By definition a finite extension of a global field is itself a global field.

Valuations

v

of

K

it can be written

Kv

for the completion of

K

with respect to

v.

If

v

is discrete it can be written

Ov

for the valuation ring of

Kv

and

ak{m}v

for the maximal ideal of

Ov.

If this is a principal ideal denoting the uniformising element by

\piv.

A non-Archimedean valuation is written as

v<infty

or

v\nmidinfty

and an Archimedean valuation as

v|infty.

Then assume all valuations to be non-trivial.

There is a one-to-one identification of valuations and absolute values. Fix a constant

C>1,

the valuation

v

is assigned the absolute value

||v,

defined as:

\forallx\inK:|x|v:= \begin{cases} C-v(x)&x0\\ 0&x=0 \end{cases}

Conversely, the absolute value

||

is assigned the valuation

v||,

defined as:

\forallx\inK x :v||(x):=-logC(|x|).

A place of

K

is a representative of an equivalence class of valuations (or absolute values) of

K.

Places corresponding to non-Archimedean valuations are called finite, whereas places corresponding to Archimedean valuations are called infinite. Infinite places of a global field form a finite set, which is denoted by

Pinfty.

Define

style\widehat{O}:=\prodvOv

and let

\widehat{O} x

be its group of units. Then

style\widehat{O} x =\prodv

x
O
v

.

Finite extensions

Let

L/K

be a finite extension of the global field

K.

Let

w

be a place of

L

and

v

a place of

K.

If the absolute value

||w

restricted to

K

is in the equivalence class of

v

, then

w

lies above

v,

which is denoted by

w|v,

and defined as:

\begin{align} Lv&:=\prodwLw,\\ \widetilde{Ov}&:=\prodwOw. \end{align}

(Note that both products are finite.)

If

w|v

,

Kv

can be embedded in

Lw.

Therefore,

Kv

is embedded diagonally in

Lv.

With this embedding

Lv

is a commutative algebra over

Kv

with degree

\sumw|v[Lw:Kv]=[L:K].

The adele ring

The set of finite adeles of a global field

K,

denoted

AK,fin,

is defined as the restricted product of

Kv

with respect to the

Ov:

AK,fin:={\prodv<infty

}^' K_v = \left \.

It is equipped with the restricted product topology, the topology generated by restricted open rectangles, which have the following form:

U=\prodvUv x \prodvOv\subset{\prodv<infty

}^' K_v,

where

E

is a finite set of (finite) places and

Uv\subsetKv

are open. With component-wise addition and multiplication

AK,fin

is also a ring.

The adele ring of a global field

K

is defined as the product of

AK,fin

with the product of the completions of

K

at its infinite places. The number of infinite places is finite and the completions are either

\R

or

\C.

In short:

AK:=AK,fin x \prodvKv={\prodv

}^' K_v \times \prod_K_v.

With addition and multiplication defined as component-wise the adele ring is a ring. The elements of the adele ring are called adeles of

K.

In the following, it is written as

AK=

'
{\prod
v}

Kv,

although this is generally not a restricted product.

Remark. Global function fields do not have any infinite places and therefore the finite adele ring equals the adele ring.

Lemma. There is a natural embedding of

K

into

AK

given by the diagonal map:

a\mapsto(a,a,\ldots).

Proof. If

a\inK,

then

a\in

x
O
v
for almost all

v.

This shows the map is well-defined. It is also injective because the embedding of

K

in

Kv

is injective for all

v.

Remark. By identifying

K

with its image under the diagonal map it is regarded as a subring of

AK.

The elements of

K

are called the principal adeles of

AK.

Definition. Let

S

be a set of places of

K.

Define the set of the

S

-adeles of

K

as

AK,S:={\prodv

}^' K_v.

Furthermore, if

S
A
K

:={\prodv

}^' K_v

the result is:

AK=AK,S x

S.
A
K

The adele ring of rationals

By Ostrowski's theorem the places of

\Q

are

\{p\in\N:pprime\}\cup\{infty\},

it is possible to identify a prime

p

with the equivalence class of the

p

-adic absolute value and

infty

with the equivalence class of the absolute value

||infty

defined as:

\forallx\in\Q:|x|infty:=\begin{cases} x&x\geq0\\ -x&x<0\end{cases}

The completion of

\Q

with respect to the place

p

is

\Qp

with valuation ring

\Zp.

For the place

infty

the completion is

\R.

Thus:

\begin{align} A\Q,fin&={\prodp

}^' \Q_p \\\mathbb_ &= \left(^' \Q_p \right) \times \R \end

Or for short

A\Q={\prodp

}^' \Q_p,\qquad \Q_\infty:=\R.

the difference between restricted and unrestricted product topology can be illustrated using a sequence in

A\Q

:

Lemma. Consider the following sequence in

A\Q

:
\begin{align} x
1&=\left(1
2
,1,1,\ldots\right)\\ x
2&=\left(1,1
3
,1,\ldots\right)\\ x
3&=\left(1,1,1
5
,1,\ldots\right)\\ x
4&=\left(1,1,1,1
7

,1,\ldots\right)\\ &\vdots \end{align}

In the product topology this converges to

(1,1,\ldots)

, but it does not converge at all in the restricted product topology.Proof. In product topology convergence corresponds to the convergence in each coordinate, which is trivial because the sequences become stationary. The sequence doesn't converge in restricted product topology. For each adele

a=(ap)p\inA\Q

and for each restricted open rectangle

styleU=\prodpUp x \prodp\Zp,

it has:

\tfrac{1}{p}-ap\notin\Zp

for

ap\in\Zp

and therefore

\tfrac{1}{p}-ap\notin\Zp

for all

p\notinF.

As a result

xn-a\notinU

for almost all

n\in\N.

In this consideration,

E

and

F

are finite subsets of the set of all places.

Alternative definition for number fields

Definition (profinite integers). The profinite integers are defined as the profinite completion of the rings

\Z/n\Z

with the partial order

n\geqm\Leftrightarrowm|n,

i.e.,

\widehat{\Z}:=\varprojlimn\Z/n\Z,

Lemma.

style\widehat{\Z}\cong\prodp\Zp.

Proof. This follows from the Chinese Remainder Theorem.

Lemma.

A\Q,=\widehat{\Z}\Z\Q.

Proof. Use the universal property of the tensor product. Define a

\Z

-bilinear function

\begin{cases}\Psi:\widehat{\Z} x \Q\toA\Q,fin\\left((ap)p,q\right)\mapsto(apq)p\end{cases}

This is well-defined because for a given

q=\tfrac{m}{n}\in\Q

with

m,n

co-prime there are only finitely many primes dividing

n.

Let

M

be another

\Z

-module with a

\Z

-bilinear map

\Phi:\widehat{\Z} x \Q\toM.

It must be the case that

\Phi

factors through

\Psi

uniquely, i.e., there exists a unique

\Z

-linear map

\tilde{\Phi}:A\Q,fin\toM

such that

\Phi=\tilde{\Phi}\circ\Psi.

\tilde{\Phi}

can be defined as follows: for a given

(up)p

there exist

u\in\N

and

(vp)p\in\widehat{\Z}

such that

up=\tfrac{1}{u}vp

for all

p.

Define

\tilde{\Phi}((up)p):=\Phi((vp)p,\tfrac{1}{u}).

One can show

\tilde{\Phi}

is well-defined,

\Z

-linear, satisfies

\Phi=\tilde{\Phi}\circ\Psi

and is unique with these properties.

Corollary. Define

A\Z:=\widehat{\Z} x \R.

This results in an algebraic isomorphism

A\Q\congA\Z\Z\Q.

Proof.

A\Z\Z\Q=\left(\widehat{\Z} x \R\right)\Z\Q\cong\left(\widehat{\Z}\Z\Q\right) x (\R\Z\Q)\cong\left(\widehat{\Z}\Z\Q\right) x \R=A\Q,fin x \R=A\Q.

Lemma. For a number field

K,AK=A\Q\QK.

Remark. Using

A\Q\QK\congA\Q...A\Q,

where there are

[K:\Q]

summands, give the right side receives the product topology and transport this topology via the isomorphism onto

A\Q\QK.

The adele ring of a finite extension

If

L/K

be a finite extension, then

L

is a global field. Thus

AL

is defined, and

styleAL=

'
{\prod
v}

Lv.

AK

can be identified with a subgroup of

AL.

Map

a=(av)v\inAK

to

a'=(a'w)w\inAL

where

a'w=av\inKv\subsetLw

for

w|v.

Then

a=(aw)w\inAL

is in the subgroup

AK,

if

aw\inKv

for

w|v

and

aw=aw'

for all

w,w'

lying above the same place

v

of

K.

Lemma. If

L/K

is a finite extension, then

AL\congAKKL

both algebraically and topologically.

With the help of this isomorphism, the inclusion

AK\subsetAL

is given by

\begin{cases} AK\toAL\\ \alpha\mapsto\alphaK1 \end{cases}

Furthermore, the principal adeles in

AK

can be identified with a subgroup of principal adeles in

AL

via the map

\begin{cases} K\to(KKL)\congL\\ \alpha\mapsto1K\alpha \end{cases}

Proof.[7] Let

\omega1,\ldots,\omegan

be a basis of

L

over

K.

Then for almost all

v,

\widetilde{Ov}\congOv\omega1Ov\omegan.

Furthermore, there are the following isomorphisms:

Kv\omega1Kv\omegan\congKvKL\congLv=\prod\nolimitswLw

For the second use the map:

\begin{cases}KvKL\toLv\\\alphava\mapsto(\alphav(\tauw(a)))w\end{cases}

in which

\tauw:L\toLw

is the canonical embedding and

w|v.

The restricted product is taken on both sides with respect to

\widetilde{Ov}:

\begin{align} AKKL&=\left(

'
{\prod
v}

Kv\right)KL\\ &\cong

'
{\prod
v}

(Kv\omega1Kv\omegan)\\ &\cong

'
{\prod
v}

(KvKL)\\ &\cong

'
{\prod
v}

Lv\\ &=AL \end{align}

Corollary. As additive groups

AL\congAKAK,

where the right side has

[L:K]

summands.

The set of principal adeles in

AL

is identified with the set

KK,

where the left side has

[L:K]

summands and

K

is considered as a subset of

AK.

The adele ring of vector-spaces and algebras

Lemma. Suppose

P\supsetPinfty

is a finite set of places of

K

and define

AK(P):=\prodvKv x \prodvOv.

Equip

AK(P)

with the product topology and define addition and multiplication component-wise. Then

AK(P)

is a locally compact topological ring.

Remark. If

P'

is another finite set of places of

K

containing

P

then

AK(P)

is an open subring of

AK(P').

Now, an alternative characterisation of the adele ring can be presented. The adele ring is the union of all sets

AK(P)

:

AK=

cup
P\supsetPinfty,|P|<infty

AK(P).

Equivalently

AK

is the set of all

x=(xv)v

so that

|xv|v\leq1

for almost all

v<infty.

The topology of

AK

is induced by the requirement that all

AK(P)

be open subrings of

AK.

Thus,

AK

is a locally compact topological ring.

Fix a place

v

of

K.

Let

P

be a finite set of places of

K,

containing

v

and

Pinfty.

Define

AK'(P,v):=\prodw

} K_w \times \prod_ O_w.

Then:

AK(P)\congKv x AK'(P,v).

Furthermore, define

AK'(v):=cup

P\supsetPinfty\cup\{v\
} \mathbb_K'(P,v),

where

P

runs through all finite sets containing

Pinfty\cup\{v\}.

Then:

AK\congKv x AK'(v),

via the map

(aw)w\mapsto(av,(aw)w).

The entire procedure above holds with a finite subset

\widetilde{P}

instead of

\{v\}.

By construction of

AK'(v),

there is a natural embedding:

Kv\hookrightarrowAK.

Furthermore, there exists a natural projection

AK\twoheadrightarrowKv.

The adele ring of a vector-space

Let

E

be a finite dimensional vector-space over

K

and

\{\omega1,\ldots,\omegan\}

a basis for

E

over

K.

For each place

v

of

K

:

\begin{align} Ev&:=EKKv\congKv\omega1Kv\omegan\\ \widetilde{Ov}&:=Ov\omega1Ov\omegan \end{align}

The adele ring of

E

is defined as

AE:=

'
{\prod
v}

Ev.

This definition is based on the alternative description of the adele ring as a tensor product equipped with the same topology that was defined when giving an alternate definition of adele ring for number fields. Next,

AE

is equipped with the restricted product topology. Then

AE=EKAK

and

E

is embedded in

AE

naturally via the map

e\mapstoe1.

An alternative definition of the topology on

AE

can be provided. Consider all linear maps:

E\toK.

Using the natural embeddings

E\toAE

and

K\toAK,

extend these linear maps to:

AE\toAK.

The topology on

AE

is the coarsest topology for which all these extensions are continuous.

The topology can be defined in a different way. Fixing a basis for

E

over

K

results in an isomorphism

E\congKn.

Therefore fixing a basis induces an isomorphism
n
(A
K)

\congAE.

The left-hand side is supplied with the product topology and transport this topology with the isomorphism onto the right-hand side. The topology doesn't depend on the choice of the basis, because another basis defines a second isomorphism. By composing both isomorphisms, a linear homeomorphism which transfers the two topologies into each other is obtained. More formally

\begin{align} AE&=EKAK\\ &\cong(KKAK)(KKAK)\\ &\congAKAK \end{align}

where the sums have

n

summands. In case of

E=L,

the definition above is consistent with the results about the adele ring of a finite extension

L/K.

[8]

The adele ring of an algebra

Let

A

be a finite-dimensional algebra over

K.

In particular,

A

is a finite-dimensional vector-space over

K.

As a consequence,

AA

is defined and

AA\congAKKA.

Since there is multiplication on

AK

and

A,

a multiplication on

AA

can be defined via:

\forall\alpha,\beta\inAKand\foralla,b\inA:    (\alphaKa)(\betaKb):=(\alpha\beta)K(ab).

As a consequence,

AA

is an algebra with a unit over

AK.

Let

l{B}

be a finite subset of

A,

containing a basis for

A

over

K.

For any finite place

v

,

Mv

is defined as the

Ov

-module generated by

l{B}

in

Av.

For each finite set of places,

P\supsetPinfty,

define

AA(P,\alpha)=\prodvAv x \prodvMv.

One can show there is a finite set

P0,

so that

AA(P,\alpha)

is an open subring of

AA,

if

P\supsetP0.

Furthermore

AA

is the union of all these subrings and for

A=K,

the definition above is consistent with the definition of the adele ring.

Trace and norm on the adele ring

Let

L/K

be a finite extension. Since

AK=AKKK

and

AL=AKKL

from the Lemma above,

AK

can be interpreted as a closed subring of

AL.

For this embedding, write

\operatorname{con}L/K

. Explicitly for all places

w

of

L

above

v

and for any

\alpha\inAK,(\operatorname{con}L/K(\alpha))w=\alphav\inKv.

Let

M/L/K

be a tower of global fields. Then:

\operatorname{con}M/K(\alpha)=\operatorname{con}M/L(\operatorname{con}L/K(\alpha))    \forall\alpha\inAK.

Furthermore, restricted to the principal adeles

\operatorname{con}

is the natural injection

K\toL.

Let

\{\omega1,\ldots,\omegan\}

be a basis of the field extension

L/K.

Then each

\alpha\inAL

can be written as
n
style\sum
j=1

\alphaj\omegaj,

where

\alphaj\inAK

are unique. The map

\alpha\mapsto\alphaj

is continuous. Define

\alphaij

depending on

\alpha

via the equations:

\begin{align} \alpha\omega1

n
&=\sum
j=1

\alpha1j\omegaj\\ &\vdots\\ \alpha\omegan

n
&=\sum
j=1

\alphanj\omegaj \end{align}

Now, define the trace and norm of

\alpha

as:

\begin{align} \operatorname{Tr}L/K(\alpha)&:=\operatorname{Tr}((\alphaij)i,j

n
)=\sum
i=1

\alphaii\\ NL/K(\alpha)&:=N((\alphaij)i,j)=\det((\alphaij)i,j) \end{align}

These are the trace and the determinant of the linear map

\begin{cases}AL\toAL\x\mapsto\alphax\end{cases}

They are continuous maps on the adele ring, and they fulfil the usual equations:

\begin{align} \operatorname{Tr}L/K(\alpha+\beta)&=\operatorname{Tr}L/K(\alpha)+\operatorname{Tr}L/K(\beta)&&\forall\alpha,\beta\inAL\\ \operatorname{Tr}L/K(\operatorname{con}(\alpha))&=n\alpha&&\forall\alpha\inAK\\ NL/K(\alpha\beta)&=NL/K(\alpha)NL/K(\beta)&&\forall\alpha,\beta\inAL\\ NL/K(\operatorname{con}(\alpha))&=\alphan&&\forall\alpha\inAK \end{align}

Furthermore, for

\alpha\inL,

\operatorname{Tr}L/K(\alpha)

and

NL/K(\alpha)

are identical to the trace and norm of the field extension

L/K.

For a tower of fields

M/L/K,

the result is:

\begin{align} \operatorname{Tr}L/K(\operatorname{Tr}M/L(\alpha))&=\operatorname{Tr}M/K(\alpha)&&\forall\alpha\inAM\\ NL/K(NM/L(\alpha))&=NM/K(\alpha)&&\forall\alpha\inAM \end{align}

Moreover, it can be proven that:[9]

\begin{align} \operatorname{Tr}L/K(\alpha)&=\left(\sumw

\operatorname{Tr}
Lw/Kv

(\alphaw)\right)v&&\forall\alpha\inAL\\ NL/K(\alpha)&=\left(\prodw

N
Lw/Kv

(\alphaw)\right)v&&\forall\alpha\inAL \end{align}

Properties of the adele ring

Theorem.[10] For every set of places

S,AK,S

is a locally compact topological ring.

Remark. The result above also holds for the adele ring of vector-spaces and algebras over

K.

Theorem.[11]

K

is discrete and cocompact in

AK.

In particular,

K

is closed in

AK.

Proof. Prove the case

K=\Q.

To show

\Q\subsetA\Q

is discrete it is sufficient to show the existence of a neighbourhood of

0

which contains no other rational number. The general case follows via translation. Define

U:=\left\{(\alphap)p\left|\forallp<infty:|\alphap|p\leq1and|\alphainfty|infty<1\right.\right\}=\widehat{\Z} x (-1,1).

U

is an open neighbourhood of

0\inA\Q.

It is claimed that

U\cap\Q=\{0\}.

Let

\beta\inU\cap\Q,

then

\beta\in\Q

and

|\beta|p\leq1

for all

p

and therefore

\beta\in\Z.

Additionally,

\beta\in(-1,1)

and therefore

\beta=0.

Next, to show compactness, define:

W:=\left\{(\alphap)p\left|\forallp<infty:|\alphap|p\leq1and|\alphainfty|infty\leq

1
2

\right.\right\}=\widehat{\Z} x \left[-

1
2,1
2

\right].

Each element in

A\Q/\Q

has a representative in

W,

that is for each

\alpha\inA\Q,

there exists

\beta\in\Q

such that

\alpha-\beta\inW.

Let

\alpha=(\alphap)p\inA\Q,

be arbitrary and

p

be a prime for which

|\alphap|>1.

Then there exists

rp=z

xp
p/p
with

zp\in\Z,xp\in\N

and

|\alphap-rp|\leq1.

Replace

\alpha

with

\alpha-rp

and let

qp

be another prime. Then:

\left|\alphaq-rp\right|q\leqmax\left\{|aq|q,|rp|q\right\}\leqmax\left\{|aq|q,1\right\}\leq1.

Next, it can be claimed that:

|\alphaq-rp|q\leq1\Longleftrightarrow|\alphaq|q\leq1.

The reverse implication is trivially true. The implication is true, because the two terms of the strong triangle inequality are equal if the absolute values of both integers are different. As a consequence, the (finite) set of primes for which the components of

\alpha

are not in

\Zp

is reduced by 1. With iteration, it can be deduced that there exists

r\in\Q

such that

\alpha-r\in\widehat{\Z} x \R.

Now select

s\in\Z

such that

\alphainfty-r-s\in\left[-\tfrac{1}{2},\tfrac{1}{2}\right].

Then

\alpha-(r+s)\inW.

The continuous projection

\pi:W\toA\Q/\Q

is surjective, therefore

A\Q/\Q,

as the continuous image of a compact set, is compact.

Corollary. Let

E

be a finite-dimensional vector-space over

K.

Then

E

is discrete and cocompact in

AE.

Theorem. The following are assumed:

A\Q=\Q+A\Z.

\Z=\Q\capA\Z.

A\Q/\Z

is a divisible group.[12]

\Q\subsetA\Q,fin

is dense.

Proof. The first two equations can be proved in an elementary way.

By definition

A\Q/\Z

is divisible if for any

n\in\N

and

y\inA\Q/\Z

the equation

nx=y

has a solution

x\inA\Q/\Z.

It is sufficient to show

A\Q

is divisible but this is true since

A\Q

is a field with positive characteristic in each coordinate.

For the last statement note that

A\Q,fin=\Q\widehat{\Z},

because the finite number of denominators in the coordinates of the elements of

A\Q,fin

can be reached through an element

q\in\Q.

As a consequence, it is sufficient to show

\Z\subset\widehat{\Z}

is dense, that is each open subset

V\subset\widehat{\Z}

contains an element of

\Z.

Without loss of generality, it can be assumed that

V=\prodp

lp
\left(a
p+p

\Zp\right) x \prodp\Zp,

because

m\Z
(p
p)

m

is a neighbourhood system of

0

in

\Zp.

By Chinese Remainder Theorem there exists

l\in\Z

such that

l\equivap\bmod

lp
p

.

Since powers of distinct primes are coprime,

l\inV

follows.

Remark.

A\Q/\Z

is not uniquely divisible. Let

y=(0,0,\ldots)+\Z\inA\Q/\Z

and

n\geq2

be given. Then

\begin{align} x1&=(0,0,\ldots)+\Z\\ x2&=\left(\tfrac{1}{n},\tfrac{1}{n},\ldots\right)+\Z \end{align}

both satisfy the equation

nx=y

and clearly

x1x2

(

x2

is well-defined, because only finitely many primes divide

n

). In this case, being uniquely divisible is equivalent to being torsion-free, which is not true for

A\Q/\Z

since

nx2=0,

but

x20

and

n0.

Remark. The fourth statement is a special case of the strong approximation theorem.

Haar measure on the adele ring

Definition. A function

f:AK\to\C

is called simple if

stylef=\prodvfv,

where

fv:Kv\to\C

are measurable and

fv=

1
Ov
for almost all

v.

Theorem.[13] Since

AK

is a locally compact group with addition, there is an additive Haar measure

dx

on

AK.

This measure can be normalised such that every integrable simple function

stylef=\prodvfv

satisfies:
\int
AK

fdx=\prodv

\int
Kv

fvdxv,

where for

v<infty,dxv

is the measure on

Kv

such that

Ov

has unit measure and

dxinfty

is the Lebesgue measure. The product is finite, i.e., almost all factors are equal to one.

The idele group

Definition. Define the idele group of

K

as the group of units of the adele ring of

K,

that is

IK:=

x
A
K

.

The elements of the idele group are called the
ideles of

K.

Remark.

IK

is equipped with a topology so that it becomes a topological group. The subset topology inherited from

AK

is not a suitable candidate since the group of units of a topological ring equipped with subset topology may not be a topological group. For example, the inverse map in

A\Q

is not continuous. The sequence

\begin{align} x1&=(2,1,\ldots)\\ x2&=(1,3,1,\ldots)\\ x3&=(1,1,5,1,\ldots)\\ &\vdots \end{align}

converges to

1\inA\Q.

To see this let

U

be neighbourhood of

0,

without loss of generality it can be assumed:

U=\prodpUp x \prodp\Zp

Since

(xn)p-1\in\Zp

for all

p,

xn-1\inU

for

n

large enough. However, as was seen above the inverse of this sequence does not converge in

A\Q.

Lemma. Let

R

be a topological ring. Define:

\begin{cases} \iota:R x \toR x R\\ x\mapsto(x,x-1) \end{cases}

Equipped with the topology induced from the product on topology on

R x R

and

\iota,R x

is a topological group and the inclusion map

R x \subsetR

is continuous. It is the coarsest topology, emerging from the topology on

R,

that makes

R x

a topological group.

Proof. Since

R

is a topological ring, it is sufficient to show that the inverse map is continuous. Let

U\subsetR x

be open, then

U x U-1\subsetR x R

is open. It is necessary to show

U-1\subsetR x

is open or equivalently, that

U-1 x (U-1)-1=U-1 x U\subsetR x R

is open. But this is the condition above.

The idele group is equipped with the topology defined in the Lemma making it a topological group.

Definition. For

S

a subset of places of

K

set:

IK,S

x ,
:=A
K,S
S)
I
K

x .

Lemma. The following identities of topological groups hold:

\begin{align} IK,S&={\prodv

}^' K_v^\\I_K^S&= ^' K_v^\\I_K&= ^' K_v^\end

where the restricted product has the restricted product topology, which is generated by restricted open rectangles of the form

\prodvUv x \prodv

x
O
v

,

where

E

is a finite subset of the set of all places and

Uv\subset

x
K
v
are open sets.

Proof. Prove the identity for

IK

; the other two follow similarly. First show the two sets are equal:

\begin{align} IK&=\{x=(xv)v\inAK:\existsy=(yv)v\inAK:xy=1\}\\ &=\{x=(xv)v\inAK:\existsy=(yv)v\inAK:xvyv=1\forallv\}\\ &=\{x=(xv)v:xv\in

x
K
v

\forallvandxv\in

x
O
v

foralmostallv\}\\ &=

'
{\prod
v}
x \end{align}
K
v

In going from line 2 to 3,

x

as well as

x-1=y

have to be in

AK,

meaning

xv\inOv

for almost all

v

and
-1
x
v

\inOv

for almost all

v.

Therefore,

xv\in

x
O
v
for almost all

v.

Now, it is possible to show the topology on the left-hand side equals the topology on the right-hand side. Obviously, every open restricted rectangle is open in the topology of the idele group. On the other hand, for a given

U\subsetIK,

which is open in the topology of the idele group, meaning

U x U-1\subsetAK x AK

is open, so for each

u\inU

there exists an open restricted rectangle, which is a subset of

U

and contains

u.

Therefore,

U

is the union of all these restricted open rectangles and therefore is open in the restricted product topology.

Lemma. For each set of places,

S,IK,S

is a locally compact topological group.

Proof. The local compactness follows from the description of

IK,S

as a restricted product. It being a topological group follows from the above discussion on the group of units of a topological ring.

A neighbourhood system of

1\inAK(P

x
infty)
is a neighbourhood system of

1\inIK.

Alternatively, take all sets of the form:

\prodvUv,

where

Uv

is a neighbourhood of

1\in

x
K
v
and

Uv=O

x
v
for almost all

v.

Since the idele group is a locally compact, there exists a Haar measure

d x x

on it. This can be normalised, so that
\int
IK,fin

1\widehat{O

}\, d^\times x =1.

This is the normalisation used for the finite places. In this equation,

IK,fin

is the finite idele group, meaning the group of units of the finite adele ring. For the infinite places, use the multiplicative lebesgue measure

\tfrac{dx}{|x|}.

The idele group of a finite extension

Lemma. Let

L/K

be a finite extension. Then:

IL=

'
{\prod
v}
x .
L
v

where the restricted product is with respect to

x
\widetilde{O
v}

.

Lemma. There is a canonical embedding of

IK

in

IL.

Proof. Map

a=(av)v\inIK

to

a'=(a'w)w\inIL

with the property

a'w=av\in

x
K
v

\subset

x
L
w
for

w|v.

Therefore,

IK

can be seen as a subgroup of

IL.

An element

a=(aw)w\inIL

is in this subgroup if and only if his components satisfy the following properties:

aw\in

x
K
v
for

w|v

and

aw=aw'

for

w|v

and

w'|v

for the same place

v

of

K.

The case of vector spaces and algebras

[14]

The idele group of an algebra

Let

A

be a finite-dimensional algebra over

K.

Since
x
A
A
is not a topological group with the subset-topology in general, equip
x
A
A
with the topology similar to

IK

above and call
x
A
A
the idele group. The elements of the idele group are called idele of

A.

Proposition. Let

\alpha

be a finite subset of

A,

containing a basis of

A

over

K.

For each finite place

v

of

K,

let

\alphav

be the

Ov

-module generated by

\alpha

in

Av.

There exists a finite set of places

P0

containing

Pinfty

such that for all

v\notinP0,

\alphav

is a compact subring of

Av.

Furthermore,

\alphav

contains
x .
A
v
For each

v,

x
A
v
is an open subset of

Av

and the map

x\mapstox-1

is continuous on
x
A
v

.

As a consequence

x\mapsto(x,x-1)

maps
x
A
v
homeomorphically on its image in

Av x Av.

For each

v\notinP0,

the
x
\alpha
v
are the elements of
x ,
A
v
mapping in

\alphav x \alphav

with the function above. Therefore,
x
\alpha
v
is an open and compact subgroup of
x .
A
v
[15]

Alternative characterisation of the idele group

Proposition. Let

P\supsetPinfty

be a finite set of places. Then

AA(P,\alpha) x :=\prodv

x
A
v

x \prodv

x
\alpha
v

is an open subgroup of

x
A
A

,

where
x
A
A
is the union of all

AA(P,\alpha) x .

[16]

Corollary. In the special case of

A=K

for each finite set of places

P\supsetPinfty,

x
A
K(P)

=\prodv

x
K
v

x \prodv

x
O
v

is an open subgroup of

x
A
K

=IK.

Furthermore,

IK

is the union of all
x
A
K(P)

.

Norm on the idele group

The trace and the norm should be transfer from the adele ring to the idele group. It turns out the trace can't be transferred so easily. However, it is possible to transfer the norm from the adele ring to the idele group. Let

\alpha\inIK.

Then

\operatorname{con}L/K(\alpha)\inIL

and therefore, it can be said that in injective group homomorphism

\operatorname{con}L/K:IK\hookrightarrowIL.

Since

\alpha\inIL,

it is invertible,

NL/K(\alpha)

is invertible too, because

(NL/K(\alpha))-1=NL/K(\alpha-1).

Therefore

NL/K(\alpha)\inIK.

As a consequence, the restriction of the norm-function introduces a continuous function:

NL/K:IL\toIK.

The Idele class group

Lemma. There is natural embedding of

K x

into

IK,S

given by the diagonal map:

a\mapsto(a,a,a,\ldots).

Proof. Since

K x

is a subset of
x
K
v
for all

v,

the embedding is well-defined and injective.

Corollary.

A x

is a discrete subgroup of
x
A
A

.

Defenition. In analogy to the ideal class group, the elements of

K x

in

IK

are called principal ideles of

IK.

The quotient group

CK:=

x
I
K/K
is called idele class group of

K.

This group is related to the ideal class group and is a central object in class field theory.

Remark.

K x

is closed in

IK,

therefore

CK

is a locally compact topological group and a Hausdorff space.

Lemma.[17] Let

L/K

be a finite extension. The embedding

IK\toIL

induces an injective map:

\begin{cases} CK\toCL\\ \alphaK x \mapsto\alphaL x \end{cases}

Properties of the idele group

Absolute value on the idele group of K and 1-idele

Definition. For

\alpha=(\alphav)v\inIK

define:

style|\alpha|:=\prodv|\alphav|v.

Since

\alpha

is an idele this product is finite and therefore well-defined.

Remark. The definition can be extended to

AK

by allowing infinite products. However, these infinite products vanish and so

||

vanishes on

AK\setminusIK.

||

will be used to denote both the function on

IK

and

AK.

Theorem.

||:IK\to\R+

is a continuous group homomorphism.

Proof. Let

\alpha,\beta\inIK.

\begin{align} |\alpha\beta|&=\prodv|(\alpha\beta)v|v\\ &=\prodv|\alphav\betav|v\\ &=\prodv(|\alphav|v|\betav|v)\\ &=\left(\prodv|\alphav|v\right)\left(\prodv|\betav|v\right)\\ &=|\alpha||\beta| \end{align}

where it is used that all products are finite. The map is continuous which can be seen using an argument dealing with sequences. This reduces the problem to whether

||

is continuous on

Kv.

However, this is clear, because of the reverse triangle inequality.

Definition. The set of

1

-idele can be defined as:
1:=\{x
I
K

\inIK:|x|=1\}=\ker(||).

1
I
K
is a subgroup of

IK.

Since
1=||
I
K

-1(\{1\}),

it is a closed subset of

AK.

Finally the

AK

-topology on
1
I
K
equals the subset-topology of

IK

on
1.
I
K
[18] [19]

Artin's Product Formula.

|k|=1

for all

k\inK x .

Proof.[20] Proof of the formula for number fields, the case of global function fields can be proved similarly. Let

K

be a number field and

a\inK x .

It has to be shown that:

\prodv|a|v=1.

For finite place

v

for which the corresponding prime ideal

ak{p}v

does not divide

(a)

,

v(a)=0

and therefore

|a|v=1.

This is valid for almost all

ak{p}v.

There is:

\begin{align} \prodv|a|v&=\prodp\prodv||a|v\\ &=\prodp\prodv|

|N
Kv/\Qp

(a)|p\\ &=\prodp|NK(a)|p \end{align}

In going from line 1 to line 2, the identity

|a|w=|N

Lw/Kv

(a)|v,

was used where

v

is a place of

K

and

w

is a place of

L,

lying above

v.

Going from line 2 to line 3, a property of the norm is used. The norm is in

\Q

so without loss of generality it can be assumed that

a\in\Q.

Then

a

possesses a unique integer factorisation:

a=\pm\prodp

vp
p

,

where

vp\in\Z

is

0

for almost all

p.

By Ostrowski's theorem all absolute values on

\Q

are equivalent to the real absolute value

||infty

or a

p

-adic absolute value. Therefore:

\begin{align} |a|&=\left(\prodp|a|p\right)|a|infty\\ &=\left(\prodp

-vp
p

\right)\left(\prodp

vp
p

\right)\\ &=1 \end{align}

Lemma.[21] There exists a constant

C,

depending only on

K,

such that for every

\alpha=(\alphav)v\inAK

satisfying

style\prodv|\alphav|v>C,

there exists

\beta\inK x

such that

|\betav|v\leq|\alphav|v

for all

v.

Corollary. Let

v0

be a place of

K

and let

\deltav>0

be given for all

vv0

with the property

\deltav=1

for almost all

v.

Then there exists

\beta\inK x ,

so that

|\beta|\leq\deltav

for all

vv0.

Proof. Let

C

be the constant from the lemma. Let

\piv

be a uniformising element of

Ov.

Define the adele

\alpha=(\alphav)v

via

\alphav:=\pi

kv
v
with

kv\in\Z

minimal, so that

|\alphav|v\leq\deltav

for all

vv0.

Then

kv=0

for almost all

v.

Define
\alpha
v0
k
v0
:=\pi
v0
with
k
v0

\in\Z,

so that

style\prodv|\alphav|v>C.

This works, because

kv=0

for almost all

v.

By the Lemma there exists

\beta\inK x ,

so that

|\beta|v\leq|\alphav|v\leq\deltav

for all

vv0.

Theorem.

K x

is discrete and cocompact in
1.
I
K

Proof.[22] Since

K x

is discrete in

IK

it is also discrete in
1.
I
K
To prove the compactness of
1/K
I
K

x

let

C

is the constant of the Lemma and suppose

\alpha\inAK

satisfying

style\prodv|\alphav|v>C

is given. Define:

W\alpha:=\left\{\xi=(\xiv)v\inAK||\xiv|v\leq|\alphav|vforallv\right\}.

Clearly

W\alpha

is compact. It can be claimed that the natural projection

W\alpha\cap

1
I
K

\to

1/K
I
K

x

is surjective. Let

\beta=(\betav)v\in

1
I
K
be arbitrary, then:

|\beta|=\prodv|\betav|v=1,

and therefore

\prodv

-1
|\beta
v

|v=1.

It follows that

\prodv

-1
|\beta
v

\alphav|v=\prodv|\alphav|v>C.

By the Lemma there exists

η\inK x

such that

|η|v\leq

-1
|\beta
v

\alphav|v

for all

v,

and therefore

η\beta\inW\alpha

proving the surjectivity of the natural projection. Since it is also continuous the compactness follows.

Theorem.[23] There is a canonical isomorphism

1/\Q
I
\Q

x \cong\widehat{\Z} x .

Furthermore,

\widehat{\Z} x x \{1\}\subset

1
I
\Q
is a set of representatives for
1/\Q
I
\Q

x

and

\widehat{\Z} x x (0,infty)\subsetI\Q

is a set of representatives for

I\Q/\Q x .

Proof. Consider the map

\begin{cases}\phi:\widehat{\Z} x \to

1/\Q
I
\Q

x \(ap)p\mapsto((ap)

x
p,1)\Q

\end{cases}

This map is well-defined, since

|ap|p=1

for all

p

and therefore

style\left(\prodp<infty|ap|p\right)1=1.

Obviously

\phi

is a continuous group homomorphism. Now suppose

((ap)

x =((b
p)
x .
p,1)\Q
Then there exists

q\in\Q x

such that

((ap)p,1)q=((bp)p,1).

By considering the infinite place it can be seen that

q=1

proves injectivity. To show surjectivity let

((\betap)p,\betainfty)\Q x \in

1/\Q
I
\Q

x .

The absolute value of this element is

1

and therefore

|\betainfty|

infty=1
\prodp|\betap|p

\in\Q.

Hence

\betainfty\in\Q

and there is:

((\betap)p,\betainfty)\Q x =\left(\left(

\betap
\betainfty

\right)p,1\right)\Q x .

Since

\forallp:    \left|

\betap
\betainfty

\right|p=1,

It has been concluded that

\phi

is surjective.

Theorem.[23] The absolute value function induces the following isomorphisms of topological groups:

\begin{align} I\Q&\cong

1
I
\Q

x (0,infty)

1
\\ I
\Q

&\congI\Q, x \{\pm1\}. \end{align}

Proof. The isomorphisms are given by:

\begin{cases}\psi:I\Q\to

1
I
\Q

x (0,infty)\a=(afin,ainfty)\mapsto\left

(a
fin,ainfty
|a|

,|a|\right)\end{cases}    and    \begin{cases}\widetilde{\psi}:I\Q, x \{\pm1\}\to

1
I
\Q

\\(afin,\varepsilon)\mapsto\left(afin,

\varepsilon
|afin|

\right)\end{cases}

Relation between ideal class group and idele class group

Theorem. Let

K

be a number field with ring of integers

O,

group of fractional ideals

JK,

and ideal class group

\operatorname{Cl}K

x .
=J
K/K
Here's the following isomorphisms

\begin{align} JK&\congIK,fin/\widehat{O} x \\ \operatorname{Cl}K&\congCK,fin/\widehat{O} x K x \\ \operatorname{Cl}K&\congCK/\left(\widehat{O} x x \prodv

x
K
v

\right)K x \end{align}

where

CK,fin:=IK,fin/K x

has been defined.

Proof. Let

v

be a finite place of

K

and let

||v

be a representative of the equivalence class

v.

Define

ak{p}v:=\{x\inO:|x|v<1\}.

Then

ak{p}v

is a prime ideal in

O.

The map

v\mapstoak{p}v

is a bijection between finite places of

K

and non-zero prime ideals of

O.

The inverse is given as follows: a prime ideal

ak{p}

is mapped to the valuation

vak{p},

given by

\begin{align} vak{p}(x)&:=max\{k\in\N0:x\inak{p}k\}\forallx\inO x

\\ v
ak{p}\left(x
y

\right)&:=vak{p}(x)-vak{p}(y)\forallx,y\inO x \end{align}

The following map is well-defined:

\begin{cases} ():IK,fin\toJK\\ \alpha=(\alphav)v\mapsto\prodv

v(\alphav)
ak{p}
v

, \end{cases}

The map

()

is obviously a surjective homomorphism and

\ker(())=\widehat{O} x .

The first isomorphism follows from fundamental theorem on homomorphism. Now, both sides are divided by

K x .

This is possible, because

\begin{align} (\alpha)&=((\alpha,\alpha,...c))\\ &=\prodv

v(\alpha)
ak{p}
v

\\ &=(\alpha)&&forall\alpha\inK x . \end{align}

Please, note the abuse of notation: On the left hand side in line 1 of this chain of equations,

()

stands for the map defined above. Later, the embedding of

K x

into

IK,fin

is used. In line 2, the definition of the map is used. Finally, use that

O

is a Dedekind domain and therefore each ideal can be written as a product of prime ideals. In other words, the map

()

is a

K x

-equivariant group homomorphism. As a consequence, the map above induces a surjective homomorphism

\begin{cases} \phi:CK,fin\to\operatorname{Cl}K\\ \alphaK x \mapsto(\alpha)K x \end{cases}

To prove the second isomorphism, it has to be shown that

\ker(\phi)=\widehat{O} x K x .

Consider

\xi=(\xiv)v\in\widehat{O} x .

Then

style\phi(\xiK x )=\prodv

v(\xiv)
ak{p}
v

K x =K x ,

because

v(\xiv)=0

for all

v.

On the other hand, consider

\xiK x \inCK,fin

with

\phi(\xiK x )=OK x ,

which allows to write

style\prodvak{p}

v(\xiv)
v

K x =OK x .

As a consequence, there exists a representative, such that:

style\prodv

v(\xi'v)
ak{p}
v

=O.

Consequently,

\xi'\in\widehat{O} x

and therefore

\xiK x =\xi'K x \in\widehat{O} x K x .

The second isomorphism of the theorem has been proven.

For the last isomorphism note that

\phi

induces a surjective group homomorphism

\widetilde{\phi}:CK\to\operatorname{Cl}K

with

\ker(\widetilde{\phi})=\left(\widehat{O} x x \prodv

x
K
v

\right)K x .

Remark. Consider

IK,fin

with the idele topology and equip

JK,

with the discrete topology. Since

(\{ak{a}\})-1

is open for each

ak{a}\inJK,()

is continuous. It stands, that

(\{ak{a}\})-1=\alpha\widehat{O} x

is open, where

\alpha=(\alphav)v\inAK,fin,

so that

styleak{a}=\prodv

v(\alphav)
ak{p}
v

.

Decomposition of the idele group and idele class group of K

Theorem.

\begin{align} IK&\cong

1
I
K

x M:\begin{cases}M\subsetIKdiscreteandM\cong\Z&\operatorname{char}(K)>0\M\subsetIKclosedandM\cong\R+&\operatorname{char}(K)=0\end{cases}\\ CK&\cong

1/K
I
K

x x N:\begin{cases}N=\Z&\operatorname{char}(K)>0\N=\R+&\operatorname{char}(K)=0\end{cases} \end{align}

Proof.

\operatorname{char}(K)=p>0.

For each place

v

of

K,\operatorname{char}(Kv)=p,

so that for all

x\in

x
K
v

,

|x|v

belongs to the subgroup of

\R+,

generated by

p.

Therefore for each

z\inIK,

|z|

is in the subgroup of

\R+,

generated by

p.

Therefore the image of the homomorphism

z\mapsto|z|

is a discrete subgroup of

\R+,

generated by

p.

Since this group is non-trivial, it is generated by

Q=pm

for some

m\in\N.

Choose

z1\inIK,

so that

|z1|=Q,

then

IK

is the direct product of
1
I
K
and the subgroup generated by

z1.

This subgroup is discrete and isomorphic to

\Z.

\operatorname{char}(K)=0.

For

λ\in\R+

define:

z(λ)=(zv)v,zv=\begin{cases}1&v\notinPinfty\ λ&v\inPinfty\end{cases}

The map

λ\mapstoz(λ)

is an isomorphism of

\R+

in a closed subgroup

M

of

IK

and

IK\congM x

1.
I
K
The isomorphism is given by multiplication:

\begin{cases} \phi:M x

1
I
K

\toIK,\\ ((\alphav)v,(\betav)v)\mapsto(\alphav\betav)v \end{cases}

Obviously,

\phi

is a homomorphism. To show it is injective, let

(\alphav\betav)v=1.

Since

\alphav=1

for

v<infty,

it stands that

\betav=1

for

v<infty.

Moreover, it exists a

λ\in\R+,

so that

\alphav

for

v|infty.

Therefore,
-1
\beta
v
for

v|infty.

Moreover

style\prodv|\betav|v=1,

implies

λn=1,

where

n

is the number of infinite places of

K.

As a consequence

λ=1

and therefore

\phi

is injective. To show surjectivity, let

\gamma=(\gammav)v\inIK.

It is defined that
1
n
λ:=|\gamma|
and furthermore,

\alphav=1

for

v<infty

and

\alphav

for

v|infty.

Define
style\beta=\gamma
\alpha

.

It stands, that
style|\beta|=|\gamma|=
|\alpha|
λn
λn

=1.

Therefore,

\phi

is surjective.

The other equations follow similarly.

Characterisation of the idele group

Theorem.[24] Let

K

be a number field. There exists a finite set of places

S,

such that:

IK=\left(IK,S x \prodv

x
O
v

\right)K x =\left(\prodv

x
K
v

x \prodv

x \right)
O
v

K x .

Proof. The class number of a number field is finite so let

ak{a}1,\ldots,ak{a}h

be the ideals, representing the classes in

\operatorname{Cl}K.

These ideals are generated by a finite number of prime ideals

ak{p}1,\ldots,ak{p}n.

Let

S

be a finite set of places containing

Pinfty

and the finite places corresponding to

ak{p}1,\ldots,ak{p}n.

Consider the isomorphism:

IK/\left(\prodv<

x
O
v

x \prodv

x \right)
K
v

\congJK,

induced by

(\alphav)v\mapsto\prodv

v(\alphav)
ak{p}
v

.

At infinite places the statement is immediate, so the statement has been proved for finite places. The inclusion ″

\supset

″ is obvious. Let

\alpha\inIK,fin.

The corresponding ideal

style(\alpha)=\prodv<

v(\alphav)
ak{p}
v
belongs to a class
x
ak{a}
iK

,

meaning

(\alpha)=ak{a}i(a)

for a principal ideal

(a).

The idele

\alpha'=\alphaa-1

maps to the ideal

ak{a}i

under the map

IK,fin\toJK.

That means

styleak{a}i=\prodv<

v(\alpha'v)
ak{p}
v

.

Since the prime ideals in

ak{a}i

are in

S,

it follows

v(\alpha'v)=0

for all

v\notinS,

that means

\alpha'v\in

x
O
v
for all

v\notinS.

It follows, that

\alpha'=\alphaa-1\inIK,S,

therefore

\alpha\inIK,SK x .

Applications

Finiteness of the class number of a number field

In the previous section the fact that the class number of a number field is finite had been used. Here this statement can be proved:

Theorem (finiteness of the class number of a number field). Let

K

be a number field. Then

|\operatorname{Cl}K|<infty.

Proof. The map

\begin{cases}

1
I
K

\toJK\\left((\alphav)v,(\alphav)v\right)\mapsto\prodv<infty

v(\alphav)
ak{p}
v

\end{cases}

is surjective and therefore

\operatorname{Cl}K

is the continuous image of the compact set
1/K
I
K

x .

Thus,

\operatorname{Cl}K

is compact. In addition, it is discrete and so finite.

Remark. There is a similar result for the case of a global function field. In this case, the so-called divisor group is defined. It can be shown that the quotient of the set of all divisors of degree

0

by the set of the principal divisors is a finite group.[25]

Group of units and Dirichlet's unit theorem

Let

P\supsetPinfty

be a finite set of places. Define

\begin{align} \Omega(P)&:=\prodv\in

x
K
v

x \prodv

x
O
v
x \\ E(P)&:=K
=(A
K(P))

x \cap\Omega(P) \end{align}

Then

E(P)

is a subgroup of

K x ,

containing all elements

\xi\inK x

satisfying

v(\xi)=0

for all

v\notinP.

Since

K x

is discrete in

IK,

E(P)

is a discrete subgroup of

\Omega(P)

and with the same argument,

E(P)

is discrete in

\Omega1(P):=\Omega(P)\cap

1.
I
K

An alternative definition is:

E(P)=K(P) x ,

where

K(P)

is a subring of

K

defined by

K(P):=K\cap\left(\prodv\inKv x \prodvOv\right).

As a consequence,

K(P)

contains all elements

\xi\inK,

which fulfil

v(\xi)\geq0

for all

v\notinP.

Lemma 1. Let

0<c\leqC<infty.

The following set is finite:

\left\{η\inE(P):\left.\begin{cases}|ηv|v=1&\forallv\notinP\c\leq|ηv|v\leqC&\forallv\inP\end{cases}\right\}\right\}.

Proof. Define

W:=\left\{(\alphav)v:\left.\begin{cases}|\alphav|v=1&\forallv\notinP\c\leq|\alphav|v\leqC&\forallv\inP\end{cases}\right\}\right\}.

W

is compact and the set described above is the intersection of

W

with the discrete subgroup

K x

in

IK

and therefore finite.

Lemma 2. Let

E

be set of all

\xi\inK

such that

|\xi|v=1

for all

v.

Then

E=\mu(K),

the group of all roots of unity of

K.

In particular it is finite and cyclic.

Proof. All roots of unity of

K

have absolute value

1

so

\mu(K)\subsetE.

For converse note that Lemma 1 with

c=C=1

and any

P

implies

E

is finite. Moreover

E\subsetE(P)

for each finite set of places

P\supsetPinfty.

Finally suppose there exists

\xi\inE,

which is not a root of unity of

K.

Then

\xin1

for all

n\in\N

contradicting the finiteness of

E.

Unit Theorem.

E(P)

is the direct product of

E

and a group isomorphic to

\Zs,

where

s=0,

if

P=\emptyset

and

s=|P|-1,

if

P\emptyset.

[26]

Dirichlet's Unit Theorem. Let

K

be a number field. Then

O x \cong\mu(K) x \Zr+s-1,

where

\mu(K)

is the finite cyclic group of all roots of unity of

K,r

is the number of real embeddings of

K

and

s

is the number of conjugate pairs of complex embeddings of

K.

It stands, that

[K:\Q]=r+2s.

Remark. The Unit Theorem generalises Dirichlet's Unit Theorem. To see this, let

K

be a number field. It is already known that

E=\mu(K),

set

P=Pinfty

and note

|Pinfty|=r+s.

Then there is:

\begin{align} E x \Zr+s-1=

x
E(P
infty)&=K

\cap\left(\prodv

x
K
v

x \prodv

x \right)
O
v

\\ &\congK x \cap\left(\prodv

x
O
v

\right)\\ &\congO x \end{align}

Approximation theorems

Weak Approximation Theorem.[27] Let

||1,\ldots,||N,

be inequivalent valuations of

K.

Let

Kn

be the completion of

K

with respect to

||n.

Embed

K

diagonally in

K1 x x KN.

Then

K

is everywhere dense in

K1 x x KN.

In other words, for each

\varepsilon>0

and for each

(\alpha1,\ldots,\alphaN)\inK1 x x KN,

there exists

\xi\inK,

such that:

\foralln\in\{1,\ldots,N\}:|\alphan-\xi|n<\varepsilon.

Strong Approximation Theorem.[28] Let

v0

be a place of

K.

Define

V:=

{\prod
vv0
}^' K_v.

Then

K

is dense in

V.

Remark. The global field is discrete in its adele ring. The strong approximation theorem tells us that, if one place (or more) is omitted, the property of discreteness of

K

is turned into a denseness of

K.

Hasse principle

Hasse-Minkowski Theorem. A quadratic form on

K

is zero, if and only if, the quadratic form is zero in each completion

Kv.

Remark. This is the Hasse principle for quadratic forms. For polynomials of degree larger than 2 the Hasse principle isn't valid in general. The idea of the Hasse principle (also known as local–global principle) is to solve a given problem of a number field

K

by doing so in its completions

Kv

and then concluding on a solution in

K.

Characters on the adele ring

Definition. Let

G

be a locally compact abelian group. The character group of

G

is the set of all characters of

G

and is denoted by

\widehat{G}.

Equivalently

\widehat{G}

is the set of all continuous group homomorphisms from

G

to

T:=\{z\in\C:|z|=1\}.

Equip

\widehat{G}

with the topology of uniform convergence on compact subsets of

G.

One can show that

\widehat{G}

is also a locally compact abelian group.

Theorem. The adele ring is self-dual:

AK\cong\widehat{AK}.

Proof. By reduction to local coordinates, it is sufficient to show each

Kv

is self-dual. This can be done by using a fixed character of

Kv.

The idea has been illustrated by showing

\R

is self-dual. Define:

\begin{cases}einfty:\R\toT\einfty(t):=\exp(2\piit)\end{cases}

Then the following map is an isomorphism which respects topologies:

\begin{cases}\phi:\R\to\widehat{\R}\s\mapsto\begin{cases}\phis:\R\toT\\phis(t):=einfty(ts)\end{cases}\end{cases}

Theorem (algebraic and continuous duals of the adele ring).[29] Let

\chi

be a non-trivial character of

AK,

which is trivial on

K.

Let

E

be a finite-dimensional vector-space over

K.

Let

E\star

and
\star
A
E
be the algebraic duals of

E

and

AE.

Denote the topological dual of

AE

by

AE'

and use

\langle,\rangle

and

[{},{}]

to indicate the natural bilinear pairings on

AE x AE'

and

AE x

\star
A
E

.

Then the formula

\langlee,e'\rangle=\chi([e,e\star])

for all

e\inAE

determines an isomorphism

e\star\mapstoe'

of
\star
A
E
onto

AE',

where

e'\inAE'

and

e\star\in

\star.
A
E
Moreover, if

e\star\in

\star
A
E
fulfils

\chi([e,e\star])=1

for all

e\inE,

then

e\star\inE\star.

Tate's thesis

With the help of the characters of

AK,

Fourier analysis can be done on the adele ring.[30] John Tate in his thesis "Fourier analysis in Number Fields and Hecke Zeta Functions" proved results about Dirichlet L-functions using Fourier analysis on the adele ring and the idele group. Therefore, the adele ring and the idele group have been applied to study the Riemann zeta function and more general zeta functions and the L-functions. Adelic forms of these functions can be defined and represented as integrals over the adele ring or the idele group, with respect to corresponding Haar measures. Functional equations and meromorphic continuations of these functions can be shown. For example, for all

s\in\C

with

\Re(s)>1,

\int\widehat{\Z

} |x|^s d^\times x = \zeta(s),

where

d x x

is the unique Haar measure on

I\Q,fin

normalised such that

\widehat{\Z} x

has volume one and is extended by zero to the finite adele ring. As a result, the Riemann zeta function can be written as an integral over (a subset of) the adele ring.[31]

Automorphic forms

The theory of automorphic forms is a generalisation of Tate's thesis by replacing the idele group with analogous higher dimensional groups. To see this note:

\begin{align} I\Q&=\operatorname{GL}(1,A\Q)

1
\\ I
\Q

&=(\operatorname{GL}(1,

1:=\{x
A
\Q))

\in\operatorname{GL}(1,A\Q):|x|=1\}\\ \Q x &=\operatorname{GL}(1,\Q)\end{align}

Based on these identification a natural generalisation would be to replace the idele group and the 1-idele with:

\begin{align} I\Q&\leftrightsquigarrow\operatorname{GL}(2,A\Q)

1
\\ I
\Q

&\leftrightsquigarrow(\operatorname{GL}(2,

1:=\{x
A
\Q))

\in\operatorname{GL}(2,A\Q):|\det(x)|=1\}\\ \Q&\leftrightsquigarrow\operatorname{GL}(2,\Q)\end{align}

And finally

\Q x \backslash

1
I
\Q

\cong\Q x \backslashI\Q\leftrightsquigarrow(\operatorname{GL}(2,\Q)\backslash(\operatorname{GL}(2,

1
A
\Q))

\cong(\operatorname{GL}(2,\Q)Z\R)\backslash\operatorname{GL}(2,A\Q),

where

Z\R

is the centre of

\operatorname{GL}(2,\R).

Then it define an automorphic form as an element of

L2((\operatorname{GL}(2,\Q)Z\R)\backslash\operatorname{GL}(2,A\Q)).

In other words an automorphic form is a function on

\operatorname{GL}(2,A\Q)

satisfying certain algebraic and analytic conditions. For studying automorphic forms, it is important to know the representations of the group

\operatorname{GL}(2,A\Q).

It is also possible to study automorphic L-functions, which can be described as integrals over

\operatorname{GL}(2,A\Q).

[32]

Generalise even further is possible by replacing

\Q

with a number field and

\operatorname{GL}(2)

with an arbitrary reductive algebraic group.

Further applications

A generalisation of Artin reciprocity law leads to the connection of representations of

\operatorname{GL}(2,AK)

and of Galois representations of

K

(Langlands program).

The idele class group is a key object of class field theory, which describes abelian extensions of the field. The product of the local reciprocity maps in local class field theory gives a homomorphism of the idele group to the Galois group of the maximal abelian extension of the global field. The Artin reciprocity law, which is a sweeping generalisation of the Gauss quadratic reciprocity law, states that the product vanishes on the multiplicative group of the number field. Thus, the global reciprocity map of the idele class group to the abelian part of the absolute Galois group of the field will be obtained.

The self-duality of the adele ring of the function field of a curve over a finite field easily implies the Riemann–Roch theorem and the duality theory for the curve.

Sources

External links

Notes and References

  1. Groechenig. Michael. August 2017. Adelic Descent Theory. Compositio Mathematica. 153. 8. 1706–1746. 10.1112/S0010437X17007217. 0010-437X. 1511.06271. 54016389.
  2. Book: Sutherland, Andrew . 18.785 Number theory I Lecture #22 . 1 December 2015 . . 4.
  3. Web site: ring of adeles in nLab. ncatlab.org.
  4. .
  5. .
  6. .
  7. This proof can be found in
  8. The definitions are based on
  9. See or .
  10. For proof see, theorem 5.2.1.
  11. See, Theorem, or, Theorem 2.
  12. The next statement can be found in .
  13. See, Theorem 5.2.2 for the rational case.
  14. This section is based on
  15. A proof of this statement can be found in
  16. A proof of this statement can be found in
  17. For a proof see .
  18. This statement can be found in .
  19. 1
    A
    K
    is also used for the set of the

    1

    -idele but
    1
    I
    K
    is used in this example.
  20. There are many proofs for this result. The one shown below is based on
  21. For a proof see
  22. This proof can be found in or in .
  23. Part of Theorem 5.3.3 in .
  24. The general proof of this theorem for any global field is given in
  25. For more information, see .
  26. A proof can be found in or in .
  27. A proof can be found in .
  28. A proof can be found in
  29. A proof can be found in .
  30. For more see .
  31. A proof can be found, Theorem 5.3.4. See also p. 139 for more information on Tate's thesis.
  32. For further information see Chapters 7 and 8 in .