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 G
The study of geometry of numbers over the ring of adeles of a number field is called adelic geometry.
Let
K
Q
X/Fq |
K
AK = \prod(K\nu,l{O}\nu) \subseteq \prodK\nu
(a\nu)
a\nu
l{O}\nu\subsetK\nu
\nu
\nu
K
K\nu
l{O}\nu
The ring of adeles solves the technical problem of "doing analysis on the rational numbers
Q
R
p\inZ
| ⋅ |infty
| ⋅ |p
The purpose of the adele ring is to look at all completions of
K
K
The restricted infinite product is a required technical condition for giving the number field
Q
AQ
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 adelesl{O}K\hookrightarrowK
AZ
then the ring of adeles can be equivalently defined as} = \mathbf\times \prod_p \mathbf_p,AZ=R x \hat{Z
The restricted product structure becomes transparent after looking at explicit elements in this ring. The image of an element\begin{align} AQ&=Q ⊗ ZAZ\\ &=Q ⊗ Z\left(R x \prodpZp\right). \end{align}
b/c ⊗ (r,(ap))\inAQ
The factor
\left( br c ,\left(
bap c \right)\right).
bap/c
Zp
p
c
p
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.
The rationals
K=\bold{Q}
p
(K\nu,l{O}\nu)=(Qp,Zp)
Qinfty=R
AQ = R x \prodp(Qp,Zp)
p
Secondly, take the function field
1)=F | |
K=F | |
q(t) |
x
X=P1
SpecFq
x : SpecF | |
qn |
\longrightarrow P1.
q+1
SpecFq \longrightarrow P1
l{O}\nu=\widehat{l{O}}X,x
x
x
K\nu=KX,x
A | |||||||
|
= \prodx\in(l{K}X,x,\widehat{l{O}}X,x).
X/Fq |
x\inX
The group of units in the adele ring is called the idele group
IK=
x | |
A | |
K |
K x \subseteqIK
CK = I
x . | |
K/K |
OK = \prodO\nu \subseteq AK.
The Artin reciprocity law says that for a global field
K
\widehat{CK}=
x /K | |
\widehat{A | |
K |
x } \simeq Gal(Kab/K)
Kab
K
\widehat{(...)}
If
X/Fq |
Pic(X) = K x \backslash
x | |
A | |
X/O |
x | |
X |
x | |
Div(X)=A | |
X/O |
x | |
X |
G
GLn
BunG(X) = G(K)\backslashG(AX)/G(OX).
G=Gm
There is a topology on
AK
AK/K
If
X
C(X)
X
H1(X,l{L}) \simeq H
-1 | |
X ⊗ l{L} |
)*
AC(X)
X
Throughout this article,
K
\Q
F | |
pr |
(t)
p
r\in\N
v
K
Kv
K
v.
v
Ov
Kv
ak{m}v
Ov.
\piv.
v<infty
v\nmidinfty
v|infty.
There is a one-to-one identification of valuations and absolute values. Fix a constant
C>1,
v
| ⋅ |v,
\forallx\inK: |x|v:= \begin{cases} C-v(x)&x ≠ 0\\ 0&x=0 \end{cases}
Conversely, the absolute value
| ⋅ |
v| ⋅ |,
\forallx\inK x : v| ⋅ |(x):=-logC(|x|).
A place of
K
K.
Pinfty.
Define
style\widehat{O}:=\prodvOv
\widehat{O} x
style\widehat{O} x =\prodv
x | |
O | |
v |
.
Let
L/K
K.
w
L
v
K.
| ⋅ |w
K
v
w
v,
w|v,
\begin{align} Lv&:=\prodwLw,\\ \widetilde{Ov}&:=\prodwOw. \end{align}
(Note that both products are finite.)
If
w|v
Kv
Lw.
Kv
Lv.
Lv
Kv
\sumw|v[Lw:Kv]=[L:K].
The set of finite adeles of a global field
K,
AK,fin,
Kv
Ov:
AK,fin:={\prodv<infty
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
where
E
Uv\subsetKv
AK,fin
The adele ring of a global field
K
AK,fin
K
\R
\C.
AK:=AK,fin x \prodvKv={\prodv
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.
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
AK
a\mapsto(a,a,\ldots).
Proof. If
a\inK,
a\in
x | |
O | |
v |
v.
K
Kv
v.
Remark. By identifying
K
AK.
K
AK.
Definition. Let
S
K.
S
K
AK,S:={\prodv
Furthermore, if
S | |
A | |
K |
:={\prodv
the result is:
AK=AK,S x
S. | |
A | |
K |
By Ostrowski's theorem the places of
\Q
\{p\in\N:pprime\}\cup\{infty\},
p
p
infty
| ⋅ |infty
\forallx\in\Q: |x|infty:=\begin{cases} x&x\geq0\\ -x&x<0\end{cases}
The completion of
\Q
p
\Qp
\Zp.
infty
\R.
\begin{align} A\Q,fin&={\prodp
Or for short
A\Q={\prodp
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 | ||||
|
,1,1,\ldots\right)\\ x | ||||
|
,1,\ldots\right)\\ x | ||||
|
,1,\ldots\right)\\ x | ||||
|
,1,\ldots\right)\\ &\vdots \end{align}
In the product topology this converges to
(1,1,\ldots)
a=(ap)p\inA\Q
styleU=\prodpUp x \prodp\Zp,
\tfrac{1}{p}-ap\notin\Zp
ap\in\Zp
\tfrac{1}{p}-ap\notin\Zp
p\notinF.
xn-a\notinU
n\in\N.
E
F
Definition (profinite integers). The profinite integers are defined as the profinite completion of the rings
\Z/n\Z
n\geqm\Leftrightarrowm|n,
\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
\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
m,n
n.
M
\Z
\Z
\Phi:\widehat{\Z} x \Q\toM.
\Phi
\Psi
\Z
\tilde{\Phi}:A\Q,fin\toM
\Phi=\tilde{\Phi}\circ\Psi.
\tilde{\Phi}
(up)p
u\in\N
(vp)p\in\widehat{\Z}
up=\tfrac{1}{u} ⋅ vp
p.
\tilde{\Phi}((up)p):=\Phi((vp)p,\tfrac{1}{u}).
\tilde{\Phi}
\Z
\Phi=\tilde{\Phi}\circ\Psi
Corollary. Define
A\Z:=\widehat{\Z} x \R.
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,
[K:\Q]
A\Q ⊗ \QK.
If
L/K
L
AL
styleAL=
' | |
{\prod | |
v} |
Lv.
AK
AL.
a=(av)v\inAK
a'=(a'w)w\inAL
a'w=av\inKv\subsetLw
w|v.
a=(aw)w\inAL
AK,
aw\inKv
w|v
aw=aw'
w,w'
v
K.
Lemma. If
L/K
AL\congAK ⊗ KL
With the help of this isomorphism, the inclusion
AK\subsetAL
\begin{cases} AK\toAL\\ \alpha\mapsto\alpha ⊗ K1 \end{cases}
Furthermore, the principal adeles in
AK
AL
\begin{cases} K\to(K ⊗ KL)\congL\\ \alpha\mapsto1 ⊗ K\alpha \end{cases}
Proof.[7] Let
\omega1,\ldots,\omegan
L
K.
v,
\widetilde{Ov}\congOv\omega1 ⊕ … ⊕ Ov\omegan.
Furthermore, there are the following isomorphisms:
Kv\omega1 ⊕ … ⊕ Kv\omegan\congKv ⊗ KL\congLv=\prod\nolimitswLw
For the second use the map:
\begin{cases}Kv ⊗ KL\toLv\\\alphav ⊗ a\mapsto(\alphav ⋅ (\tauw(a)))w\end{cases}
in which
\tauw:L\toLw
w|v.
\widetilde{Ov}:
\begin{align} AK ⊗ KL&=\left(
' | |
{\prod | |
v} |
Kv\right) ⊗ KL\\ &\cong
' | |
{\prod | |
v} |
(Kv\omega1 ⊕ … ⊕ Kv\omegan)\\ &\cong
' | |
{\prod | |
v} |
(Kv ⊗ KL)\\ &\cong
' | |
{\prod | |
v} |
Lv\\ &=AL \end{align}
Corollary. As additive groups
AL\congAK ⊕ … ⊕ AK,
[L:K]
The set of principal adeles in
AL
K ⊕ … ⊕ K,
[L:K]
K
AK.
Lemma. Suppose
P\supsetPinfty
K
AK(P):=\prodvKv x \prodvOv.
Equip
AK(P)
AK(P)
Remark. If
P'
K
P
AK(P)
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
x=(xv)v
|xv|v\leq1
v<infty.
AK
AK(P)
AK.
AK
Fix a place
v
K.
P
K,
v
Pinfty.
AK'(P,v):=\prodw
Then:
AK(P)\congKv x AK'(P,v).
Furthermore, define
AK'(v):=cup
P\supsetPinfty\cup\{v\ |
where
P
Pinfty\cup\{v\}.
AK\congKv x AK'(v),
via the map
(aw)w\mapsto(av,(aw)w).
\widetilde{P}
\{v\}.
By construction of
AK'(v),
Kv\hookrightarrowAK.
AK\twoheadrightarrowKv.
Let
E
K
\{\omega1,\ldots,\omegan\}
E
K.
v
K
\begin{align} Ev&:=E ⊗ KKv\congKv\omega1 ⊕ … ⊕ Kv\omegan\\ \widetilde{Ov}&:=Ov\omega1 ⊕ … ⊕ Ov\omegan \end{align}
The adele ring of
E
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
AE=E ⊗ KAK
E
AE
e\mapstoe ⊗ 1.
An alternative definition of the topology on
AE
E\toK.
E\toAE
K\toAK,
AE\toAK.
AE
The topology can be defined in a different way. Fixing a basis for
E
K
E\congKn.
n | |
(A | |
K) |
\congAE.
\begin{align} AE&=E ⊗ KAK\\ &\cong(K ⊗ KAK) ⊕ … ⊕ (K ⊗ KAK)\\ &\congAK ⊕ … ⊕ AK \end{align}
where the sums have
n
E=L,
L/K.
Let
A
K.
A
K.
AA
AA\congAK ⊗ KA.
AK
A,
AA
\forall\alpha,\beta\inAKand\foralla,b\inA: (\alpha ⊗ Ka) ⋅ (\beta ⊗ Kb):=(\alpha\beta) ⊗ K(ab).
As a consequence,
AA
AK.
l{B}
A,
A
K.
v
Mv
Ov
l{B}
Av.
P\supsetPinfty,
AA(P,\alpha)=\prodvAv x \prodvMv.
One can show there is a finite set
P0,
AA(P,\alpha)
AA,
P\supsetP0.
AA
A=K,
Let
L/K
AK=AK ⊗ KK
AL=AK ⊗ KL
AK
AL.
\operatorname{con}L/K
w
L
v
\alpha\inAK,(\operatorname{con}L/K(\alpha))w=\alphav\inKv.
Let
M/L/K
\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}
K\toL.
Let
\{\omega1,\ldots,\omegan\}
L/K.
\alpha\inAL
n | |
style\sum | |
j=1 |
\alphaj\omegaj,
\alphaj\inAK
\alpha\mapsto\alphaj
\alphaij
\alpha
\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
\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)
NL/K(\alpha)
L/K.
M/L/K,
\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}
Theorem.[10] For every set of places
S,AK,S
Remark. The result above also holds for the adele ring of vector-spaces and algebras over
K.
Theorem.[11]
K
AK.
K
AK.
Proof. Prove the case
K=\Q.
\Q\subsetA\Q
0
U:=\left\{(\alphap)p\left|\forallp<infty:|\alphap|p\leq1 and |\alphainfty|infty<1\right.\right\}=\widehat{\Z} x (-1,1).
U
0\inA\Q.
U\cap\Q=\{0\}.
\beta\inU\cap\Q,
\beta\in\Q
|\beta|p\leq1
p
\beta\in\Z.
\beta\in(-1,1)
\beta=0.
W:=\left\{(\alphap)p\left|\forallp<infty:|\alphap|p\leq1 and |\alphainfty|infty\leq
1 | |
2 |
\right.\right\}=\widehat{\Z} x \left[-
1 | |||
|
\right].
Each element in
A\Q/\Q
W,
\alpha\inA\Q,
\beta\in\Q
\alpha-\beta\inW.
\alpha=(\alphap)p\inA\Q,
p
|\alphap|>1.
rp=z
xp | |
p/p |
zp\in\Z,xp\in\N
|\alphap-rp|\leq1.
\alpha
\alpha-rp
q ≠ p
\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
\Zp
r\in\Q
\alpha-r\in\widehat{\Z} x \R.
s\in\Z
\alphainfty-r-s\in\left[-\tfrac{1}{2},\tfrac{1}{2}\right].
\alpha-(r+s)\inW.
\pi:W\toA\Q/\Q
A\Q/\Q,
Corollary. Let
E
K.
E
AE.
Theorem. The following are assumed:
A\Q=\Q+A\Z.
\Z=\Q\capA\Z.
A\Q/\Z
\Q\subsetA\Q,fin
Proof. The first two equations can be proved in an elementary way.
By definition
A\Q/\Z
n\in\N
y\inA\Q/\Z
nx=y
x\inA\Q/\Z.
A\Q
A\Q
For the last statement note that
A\Q,fin=\Q\widehat{\Z},
A\Q,fin
q\in\Q.
\Z\subset\widehat{\Z}
V\subset\widehat{\Z}
\Z.
V=\prodp
lp | |
\left(a | |
p+p |
\Zp\right) x \prodp\Zp,
because
m\Z | |
(p | |
p) |
m
0
\Zp.
l\in\Z
l\equivap\bmod
lp | |
p |
.
l\inV
Remark.
A\Q/\Z
y=(0,0,\ldots)+\Z\inA\Q/\Z
n\geq2
\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
x1 ≠ x2
x2
n
A\Q/\Z
nx2=0,
x2 ≠ 0
n ≠ 0.
Remark. The fourth statement is a special case of the strong approximation theorem.
Definition. A function
f:AK\to\C
stylef=\prodvfv,
fv:Kv\to\C
fv=
1 | |
Ov |
v.
Theorem.[13] Since
AK
dx
AK.
stylef=\prodvfv
\int | |
AK |
fdx=\prodv
\int | |
Kv |
fvdxv,
where for
v<infty,dxv
Kv
Ov
dxinfty
Definition. Define the idele group of
K
K,
IK:=
x | |
A | |
K |
.
K.
Remark.
IK
AK
A\Q
\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.
U
0,
U=\prodpUp x \prodp\Zp
Since
(xn)p-1\in\Zp
p,
xn-1\inU
n
A\Q.
Lemma. Let
R
\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
\iota,R x
R x \subsetR
R,
R x
Proof. Since
R
U\subsetR x
U x U-1\subsetR x R
U-1\subsetR x
U-1 x (U-1)-1=U-1 x U\subsetR x R
The idele group is equipped with the topology defined in the Lemma making it a topological group.
Definition. For
S
K
IK,S
x , | |
:=A | |
K,S |
S) | |
I | |
K |
x .
Lemma. The following identities of topological groups hold:
\begin{align} IK,S&={\prodv
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
Uv\subset
x | |
K | |
v |
Proof. Prove the identity for
IK
\begin{align} IK&=\{x=(xv)v\inAK:\existsy=(yv)v\inAK:xy=1\}\\ &=\{x=(xv)v\inAK:\existsy=(yv)v\inAK:xv ⋅ yv=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
x-1=y
AK,
xv\inOv
v
-1 | |
x | |
v |
\inOv
v.
xv\in
x | |
O | |
v |
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,
U x U-1\subsetAK x AK
u\inU
U
u.
U
Lemma. For each set of places,
S,IK,S
Proof. The local compactness follows from the description of
IK,S
A neighbourhood system of
1\inAK(P
x | |
infty) |
1\inIK.
\prodvUv,
where
Uv
1\in
x | |
K | |
v |
Uv=O
x | |
v |
v.
Since the idele group is a locally compact, there exists a Haar measure
d x x
\int | |
IK,fin |
1\widehat{O
This is the normalisation used for the finite places. In this equation,
IK,fin
\tfrac{dx}{|x|}.
Lemma. Let
L/K
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
IL.
Proof. Map
a=(av)v\inIK
a'=(a'w)w\inIL
a'w=av\in
x | |
K | |
v |
\subset
x | |
L | |
w |
w|v.
IK
IL.
a=(aw)w\inIL
aw\in
x | |
K | |
v |
w|v
aw=aw'
w|v
w'|v
v
K.
Let
A
K.
x | |
A | |
A |
x | |
A | |
A |
IK
x | |
A | |
A |
A.
Proposition. Let
\alpha
A,
A
K.
v
K,
\alphav
Ov
\alpha
Av.
P0
Pinfty
v\notinP0,
\alphav
Av.
\alphav
x . | |
A | |
v |
v,
x | |
A | |
v |
Av
x\mapstox-1
x | |
A | |
v |
.
x\mapsto(x,x-1)
x | |
A | |
v |
Av x Av.
v\notinP0,
x | |
\alpha | |
v |
x , | |
A | |
v |
\alphav x \alphav
x | |
\alpha | |
v |
x . | |
A | |
v |
Proposition. Let
P\supsetPinfty
AA(P,\alpha) x :=\prodv
x | |
A | |
v |
x \prodv
x | |
\alpha | |
v |
is an open subgroup of
x | |
A | |
A |
,
x | |
A | |
A |
AA(P,\alpha) x .
Corollary. In the special case of
A=K
P\supsetPinfty,
x | |
A | |
K(P) |
=\prodv
x | |
K | |
v |
x \prodv
x | |
O | |
v |
is an open subgroup of
x | |
A | |
K |
=IK.
IK
x | |
A | |
K(P) |
.
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.
\operatorname{con}L/K(\alpha)\inIL
\operatorname{con}L/K:IK\hookrightarrowIL.
Since
\alpha\inIL,
NL/K(\alpha)
(NL/K(\alpha))-1=NL/K(\alpha-1).
NL/K(\alpha)\inIK.
NL/K:IL\toIK.
Lemma. There is natural embedding of
K x
IK,S
a\mapsto(a,a,a,\ldots).
Proof. Since
K x
x | |
K | |
v |
v,
Corollary.
A x
x | |
A | |
A |
.
Defenition. In analogy to the ideal class group, the elements of
K x
IK
IK.
CK:=
x | |
I | |
K/K |
K.
Remark.
K x
IK,
CK
Lemma.[17] Let
L/K
IK\toIL
\begin{cases} CK\toCL\\ \alphaK x \mapsto\alphaL x \end{cases}
Definition. For
\alpha=(\alphav)v\inIK
style|\alpha|:=\prodv|\alphav|v.
\alpha
Remark. The definition can be extended to
AK
| ⋅ |
AK\setminusIK.
| ⋅ |
IK
AK.
Theorem.
| ⋅ |:IK\to\R+
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
| ⋅ |
Kv.
Definition. The set of
1
1:=\{x | |
I | |
K |
\inIK:|x|=1\}=\ker(| ⋅ |).
1 | |
I | |
K |
IK.
1=| ⋅ | | |
I | |
K |
-1(\{1\}),
AK.
AK
1 | |
I | |
K |
IK
1. | |
I | |
K |
Artin's Product Formula.
|k|=1
k\inK x .
Proof.[20] Proof of the formula for number fields, the case of global function fields can be proved similarly. Let
K
a\inK x .
\prodv|a|v=1.
For finite place
v
ak{p}v
(a)
v(a)=0
|a|v=1.
ak{p}v.
\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,
v
K
w
L,
v.
\Q
a\in\Q.
a
a=\pm\prodp
vp | |
p |
,
where
vp\in\Z
0
p.
\Q
| ⋅ |infty
p
\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,
K,
\alpha=(\alphav)v\inAK
style\prodv|\alphav|v>C,
\beta\inK x
|\betav|v\leq|\alphav|v
v.
Corollary. Let
v0
K
\deltav>0
v ≠ v0
\deltav=1
v.
\beta\inK x ,
|\beta|\leq\deltav
v ≠ v0.
Proof. Let
C
\piv
Ov.
\alpha=(\alphav)v
\alphav:=\pi
kv | |
v |
kv\in\Z
|\alphav|v\leq\deltav
v ≠ v0.
kv=0
v.
\alpha | |
v0 |
| |||||
:=\pi | |||||
v0 |
k | |
v0 |
\in\Z,
style\prodv|\alphav|v>C.
kv=0
v.
\beta\inK x ,
|\beta|v\leq|\alphav|v\leq\deltav
v ≠ v0.
Theorem.
K x
1. | |
I | |
K |
Proof.[22] Since
K x
IK
1. | |
I | |
K |
1/K | |
I | |
K |
x
C
\alpha\inAK
style\prodv|\alphav|v>C
W\alpha:=\left\{\xi=(\xiv)v\inAK||\xiv|v\leq|\alphav|vforallv\right\}.
Clearly
W\alpha
W\alpha\cap
1 | |
I | |
K |
\to
1/K | |
I | |
K |
x
\beta=(\betav)v\in
1 | |
I | |
K |
|\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
|η|v\leq
-1 | |
|\beta | |
v |
\alphav|v
v,
η\beta\inW\alpha
Theorem.[23] There is a canonical isomorphism
1/\Q | |
I | |
\Q |
x \cong\widehat{\Z} x .
\widehat{\Z} x x \{1\}\subset
1 | |
I | |
\Q |
1/\Q | |
I | |
\Q |
x
\widehat{\Z} x x (0,infty)\subsetI\Q
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
p
style\left(\prodp<infty|ap|p\right) ⋅ 1=1.
\phi
((ap)
x =((b | |
p) |
x . | |
p,1)\Q |
q\in\Q x
((ap)p,1)q=((bp)p,1).
q=1
((\betap)p,\betainfty)\Q x \in
1/\Q | |
I | |
\Q |
x .
1
|\betainfty|
|
\in\Q.
Hence
\betainfty\in\Q
((\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
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 | ||||
|
,|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}
Theorem. Let
K
O,
JK,
\operatorname{Cl}K
x . | |
=J | |
K/K |
\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
Proof. Let
v
K
| ⋅ |v
v.
ak{p}v:=\{x\inO:|x|v<1\}.
Then
ak{p}v
O.
v\mapstoak{p}v
K
O.
ak{p}
vak{p},
\begin{align} vak{p}(x)&:=max\{k\in\N0:x\inak{p}k\} \forallx\inO x
\\ v | ||||
|
\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
( ⋅ )
\ker(( ⋅ ))=\widehat{O} x .
K x .
\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,
( ⋅ )
K x
IK,fin
O
( ⋅ )
K x
\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 .
\xi=(\xiv)v\in\widehat{O} x .
style\phi(\xiK x )=\prodv
v(\xiv) | |
ak{p} | |
v |
K x =K x ,
v(\xiv)=0
v.
\xiK x \inCK,fin
\phi(\xiK x )=OK x ,
style\prodvak{p}
v(\xiv) | |
v |
K x =OK x .
style\prodv
v(\xi'v) | |
ak{p} | |
v |
=O.
\xi'\in\widehat{O} x
\xiK x =\xi'K x \in\widehat{O} x K x .
For the last isomorphism note that
\phi
\widetilde{\phi}:CK\to\operatorname{Cl}K
\ker(\widetilde{\phi})=\left(\widehat{O} x x \prodv
x | |
K | |
v |
\right)K x .
Remark. Consider
IK,fin
JK,
(\{ak{a}\})-1
ak{a}\inJK,( ⋅ )
(\{ak{a}\})-1=\alpha\widehat{O} x
\alpha=(\alphav)v\inAK,fin,
styleak{a}=\prodv
v(\alphav) | |
ak{p} | |
v |
.
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.
v
K,\operatorname{char}(Kv)=p,
x\in
x | |
K | |
v |
,
|x|v
\R+,
p.
z\inIK,
|z|
\R+,
p.
z\mapsto|z|
\R+,
p.
Q=pm
m\in\N.
z1\inIK,
|z1|=Q,
IK
1 | |
I | |
K |
z1.
\Z.
\operatorname{char}(K)=0.
λ\in\R+
z(λ)=(zv)v, zv=\begin{cases}1&v\notinPinfty\ λ&v\inPinfty\end{cases}
The map
λ\mapstoz(λ)
\R+
M
IK
IK\congM x
1. | |
I | |
K |
\begin{cases} \phi:M x
1 | |
I | |
K |
\toIK,\\ ((\alphav)v,(\betav)v)\mapsto(\alphav\betav)v \end{cases}
Obviously,
\phi
(\alphav\betav)v=1.
\alphav=1
v<infty,
\betav=1
v<infty.
λ\in\R+,
\alphav=λ
v|infty.
-1 | |
\beta | |
v=λ |
v|infty.
style\prodv|\betav|v=1,
λn=1,
n
K.
λ=1
\phi
\gamma=(\gammav)v\inIK.
| ||||
λ:=|\gamma| |
\alphav=1
v<infty
\alphav=λ
v|infty.
style\beta= | \gamma |
\alpha |
.
style|\beta|= | |\gamma| | = |
|\alpha| |
λn | |
λn |
=1.
\phi
The other equations follow similarly.
Theorem.[24] Let
K
S,
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
\operatorname{Cl}K.
ak{p}1,\ldots,ak{p}n.
S
Pinfty
ak{p}1,\ldots,ak{p}n.
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
\alpha\inIK,fin.
style(\alpha)=\prodv<
v(\alphav) | |
ak{p} | |
v |
x | |
ak{a} | |
iK |
,
(\alpha)=ak{a}i(a)
(a).
\alpha'=\alphaa-1
ak{a}i
IK,fin\toJK.
styleak{a}i=\prodv<
v(\alpha'v) | |
ak{p} | |
v |
.
ak{a}i
S,
v(\alpha'v)=0
v\notinS,
\alpha'v\in
x | |
O | |
v |
v\notinS.
\alpha'=\alphaa-1\inIK,S,
\alpha\inIK,SK x .
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
|\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
1/K | |
I | |
K |
x .
\operatorname{Cl}K
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
Let
P\supsetPinfty
\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)
K x ,
\xi\inK x
v(\xi)=0
v\notinP.
K x
IK,
E(P)
\Omega(P)
E(P)
\Omega1(P):=\Omega(P)\cap
1. | |
I | |
K |
An alternative definition is:
E(P)=K(P) x ,
K(P)
K
K(P):=K\cap\left(\prodv\inKv x \prodvOv\right).
As a consequence,
K(P)
\xi\inK,
v(\xi)\geq0
v\notinP.
Lemma 1. Let
0<c\leqC<infty.
\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
W
K x
IK
Lemma 2. Let
E
\xi\inK
|\xi|v=1
v.
E=\mu(K),
K.
Proof. All roots of unity of
K
1
\mu(K)\subsetE.
c=C=1
P
E
E\subsetE(P)
P\supsetPinfty.
\xi\inE,
K.
\xin ≠ 1
n\in\N
E.
Unit Theorem.
E(P)
E
\Zs,
s=0,
P=\emptyset
s=|P|-1,
P ≠ \emptyset.
Dirichlet's Unit Theorem. Let
K
O x \cong\mu(K) x \Zr+s-1,
\mu(K)
K,r
K
s
K.
[K:\Q]=r+2s.
Remark. The Unit Theorem generalises Dirichlet's Unit Theorem. To see this, let
K
E=\mu(K),
P=Pinfty
|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}
Weak Approximation Theorem.[27] Let
| ⋅ |1,\ldots,| ⋅ |N,
K.
Kn
K
| ⋅ |n.
K
K1 x … x KN.
K
K1 x … x KN.
\varepsilon>0
(\alpha1,\ldots,\alphaN)\inK1 x … x KN,
\xi\inK,
\foralln\in\{1,\ldots,N\}: |\alphan-\xi|n<\varepsilon.
Strong Approximation Theorem.[28] Let
v0
K.
V:=
{\prod | |
v ≠ v0 |
Then
K
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
K.
Hasse-Minkowski Theorem. A quadratic form on
K
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
Kv
K.
Definition. Let
G
G
G
\widehat{G}.
\widehat{G}
G
T:=\{z\in\C:|z|=1\}.
\widehat{G}
G.
\widehat{G}
Theorem. The adele ring is self-dual:
AK\cong\widehat{AK}.
Proof. By reduction to local coordinates, it is sufficient to show each
Kv
Kv.
\R
\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
AK,
K.
E
K.
E\star
\star | |
A | |
E |
E
AE.
AE
AE'
\langle ⋅ , ⋅ \rangle
[{ ⋅ },{ ⋅ }]
AE x AE'
AE x
\star | |
A | |
E |
.
\langlee,e'\rangle=\chi([e,e\star])
e\inAE
e\star\mapstoe'
\star | |
A | |
E |
AE',
e'\inAE'
e\star\in
\star. | |
A | |
E |
e\star\in
\star | |
A | |
E |
\chi([e,e\star])=1
e\inE,
e\star\inE\star.
With the help of the characters of
AK,
s\in\C
\Re(s)>1,
\int\widehat{\Z
where
d x x
I\Q,fin
\widehat{\Z} x
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
\operatorname{GL}(2,\R).
L2((\operatorname{GL}(2,\Q)Z\R)\backslash\operatorname{GL}(2,A\Q)).
\operatorname{GL}(2,A\Q)
\operatorname{GL}(2,A\Q).
\operatorname{GL}(2,A\Q).
Generalise even further is possible by replacing
\Q
\operatorname{GL}(2)
A generalisation of Artin reciprocity law leads to the connection of representations of
\operatorname{GL}(2,AK)
K
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.
1 | |
A | |
K |
1
1 | |
I | |
K |