Adele ring

In mathematics, the adele ring (also adelic ring or ring of adeles) is defined in class field theory, a branch of algebraic number theory. It allows one to elegantly describe the Artin reciprocity law. The adele ring is a self-dual topological ring, which is built on a global field. It is the restricted product of all the completions of the global field and therefore contains all the completions of the global field.

The idele class group, which is the quotient group of the group of units of the adele ring by the group of units of the global field, is a central object in class field theory.

Contents

Notation

Throughout this article, $K$ is a global field. That is, $K$ is an algebraic number field or a global function field. In the first case, $K/\mathbb {Q}$ is a finite field extension; in the second case $K/\mathbb {F} _{p^{r}}(t)$ is a finite field extension. We denote by $v$ a place (a representative of an equivalence class of valuations) of $K$. The trivial valuation and the corresponding trivial value aren't allowed in the whole article. A finite/non-Archimedean valuation is written as $v<\infty$ or $v\nmid \infty$ and an infinite/Archimedean valuation as $v\mid \infty .$ We write $P_{\infty }$ for the finite set of all infinite places of $K$ and $P$ for a finite subset of all places of $K,$ which contains $P_{\infty }.$ In addition, we write $K_{v}$ for the completion of $K$ with respect to the valuation $v.$ If the valuation $v$ is discrete, then we write ${\mathcal {O}}_{v}$ for the valuation ring of $K_{v}.$ We write ${\mathfrak {m}}_{v}$ for the maximal ideal of ${\mathcal {O}}_{v}.$ If this is a principal ideal, then we write $\pi _{v}$ for a uniformizing element. By fixing a suitable constant $C>1,$ there is a one-to-one identification of valuations and absolute values: The valuation $v$ is assigned the absolute value $|\cdot |_{v},$ which is defined as:

$|x|_{v}:={\begin{cases}C^{-v(x)}&,{\text{ if }}x\neq 0\\0&,{\text{ if }}x=0\end{cases}}\quad \forall x\in K.$

Conversely, the absolute value $|\cdot |$ is assigned the valuation $v_{|\cdot |},$ which is defined as: $v_{|\cdot |}(x):=-\log _{C}(|x|)\quad \forall x\in K^{\times }.$ This will be used throughout the article.

Origin of the name

In local class field theory, the group of units of the local field plays a central role. In global class field theory, the idele class group takes this role (see also the definition of the idele class group). The term "idele" is a variation of the term ideal. Both terms have a relation, see the theorem about the relation between the ideal class group and the idele class group. 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" (adèle) stands for additive idele.

The idea of the adele ring is that we want to have a look on all completions of $K$ at once. A first glance, the Cartesian product could be a good candidate. However, the adele ring is defined with the restricted product (see next section). There are two reasons for this:

• For each element of the global field ${K},$ the valuations are zero for almost all places, which means for all places except a finite number. So, the global field can be embedded in the restricted product.
• The restricted product is a locally compact space, while the Cartesian product is not. Therefore, we can't apply harmonic analysis on the Cartesian product.

Definition of the adele ring of a global field ${K}$

Definition: the set of the finite adeles of a global field $K$

The set of the finite adeles of a global field $K,$ named $\mathbb {A} _{K,{\text{fin}}},$ is defined as the restricted product of $K_{v}$ with respect to the ${\mathcal {O}}_{v},$ which means

$\mathbb {A} _{K,{\text{fin}}}:={\widehat {\prod \limits _{v\nmid \infty }}}^{{\mathcal {O}}_{v}}K_{v}.$

This means, that the set of the finite adeles contains all $\textstyle (x_{v})_{v}\in \prod _{v\nmid \infty }K_{v},$ so that $x_{v}\in {\mathcal {O}}_{v}$ for almost all $v.$ Addition and multiplication are defined component-wise. In this way $\mathbb {A} _{K,{\text{fin}}}$ is a ring. The topology is the restricted product topology. That means that the topology is generated by the so-called restricted open rectangles, which have the following form:

$U=\prod _{v\in E}U_{v}\times \prod _{v\notin E}{\mathcal {O}}_{v},$

where $E$ is a finite subset of the set of all places of $K,$ containing $P_{\infty }$ and $U_{v}\subset K_{v}$ is open. In the following, we will use the term finite adele ring of $K$ as a synonym for $\mathbb {A} _{K,{\text{fin}}}.$

Definition: the adele ring of a global field ${K}$

The adele ring of a global field ${K},$ named $\mathbb {A} _{K},$ is defined as the product of the set of the finite adeles with the product of the completions at the infinite valuations. These are $\mathbb {R}$ or $\mathbb {C} ,$ their number is finite and they appear only in case, when $K$ is an algebraic number field. That means

$\mathbb {A} _{K}:=\mathbb {A} _{K,{\text{fin}}}\times \prod \limits _{v\mid \infty }K_{v}={\widehat {\prod \limits _{v\nmid \infty }}}^{{\mathcal {O}}_{v}}K_{v}\times \prod \limits _{v\mid \infty }K_{v}.$

In case of a global function field, the finite adele ring equals the adele ring. We define addition and multiplication component-wise. As a result, the adele ring is a ring. The elements of the adele ring are called adeles of $K.$ In the following, we write

$\mathbb {A} _{K}={\widehat {\prod _{v}}}K_{v},$

although this is generally not a restricted product.

Definition: the set of the $S$-adeles of a global field $K$

Let $K$ be a global field and $S$ a subset of the set of places of $K.$ Define the set of the $S$-adeles of $K$ as

${\begin{aligned}\mathbb {A} _{K,S}:={\widehat {\prod \limits _{v\in S}}}^{{\mathcal {O}}_{v}}K_{v}.\end{aligned}}$

If there are infinite valuations in ${S},$ they are added as usual without any restricting conditions.

Furthermore, define

$\mathbb {A} _{K}^{S}:={\widehat {\prod \limits _{v\notin S}}}^{{\mathcal {O}}_{v}}K_{v}.$

Thus, $\mathbb {A} _{K}=\mathbb {A} _{K,S}\times \mathbb {A} _{K}^{S}.$

Example: the rational adele ring $\mathbb {A} _{\mathbb {Q} }$

We consider the case $K=\mathbb {Q} .$ Due to Ostrowski's theorem, we can identify the set of all places of $\mathbb {Q}$ with $\{p\in \mathbb {N} :p{\text{ prime}}\}\cup \{\infty \},$ where we identify the prime number $p$ with the equivalence class of the $p$-adic absolute value and we identify $\infty$ with the equivalence class of the absolute value $|\cdot |_{\infty }$ on $\mathbb {Q} ,$ defined as follows:

$|x|_{\infty }:={\begin{cases}x&,{\text{ if }}x\geq 0\\-x&,{\text{ if }}x<0\end{cases}}\quad \forall x\in \mathbb {Q} .$

Next, we note, that the completion of $\mathbb {Q}$ with respect to the places $p$ is the field of the p-adic numbers $\mathbb {Q} _{p}$ to which the valuation ring $\mathbb {Z} _{p}$ belongs. For the place $\infty$ the completion is $\mathbb {R} .$ Thus, the finite adele ring of the rational numbers is

$\mathbb {A} _{\mathbb {Q} ,{\text{fin}}}={\widehat {\prod \limits _{p<\infty }}}^{\mathbb {Z} _{p}}\mathbb {Q} _{p}.$

As a consequence, the rational adele ring is

$\mathbb {A} _{\mathbb {Q} }=\left({\widehat {\prod \limits _{p<\infty }}}^{\mathbb {Z} _{p}}\mathbb {Q} _{p}\right)\times \mathbb {R} .$

We denote in short

$\mathbb {A} _{\mathbb {Q} }={\widehat {\prod \limits _{p\leq \infty }}}\mathbb {Q} _{p},$

for the adele ring of $\mathbb {Q}$ with the convention $\mathbb {Q} _{\infty }:=\mathbb {R} .$

Lemma: the difference between restricted and unrestricted product topology

The sequence in $\mathbb {A} _{\mathbb {Q} }$

${\begin{aligned}x_{1}&=\left({\frac {1}{2}},1,1,\dotsc \right)\\x_{2}&=\left(1,{\frac {1}{3}},1,\dotsc \right)\\x_{3}&=\left(1,1,{\frac {1}{5}},1,\dotsc \right)\\x_{4}&=\left(1,1,1,{\frac {1}{7}},1,\dotsc \right)\\&\,\,\,\vdots \end{aligned}}$

converges in the product topology with limit $x=(1,1,\dotsc ),$ however, it doesn't converge in the restricted product topology.

Proof: The convergence in the product topology corresponds to the convergence in each coordinate. The convergence in each coordinate is trivial, because the sequences become stationary. The sequence doesn't converge in the restricted product topology because for each adele $a=(a_{p})_{p}\in \mathbb {A} _{\mathbb {Q} }$ and for each restricted open rectangle $\textstyle U=\prod _{p\in E}U_{p}\times \prod _{p\notin E}\mathbb {Z} _{p},$ we have the result: $\textstyle {\frac {1}{p}}-a_{p}\notin \mathbb {Z} _{p}$ for $a_{p}\in \mathbb {Z} _{p}$ and therefore $\textstyle {\frac {1}{p}}-a_{p}\notin \mathbb {Z} _{p}$ for all $p\notin F.$ As a result, it stands, that $(x_{n}-a)\notin U$ for almost all $n\in \mathbb {N} .$ In this consideration, $E$ and $F$ are finite subsets of the set of all places.

The adele ring does not have the subspace topology, because otherwise the adele ring would not be a locally compact group (see the theorem below).

Lemma: diagonal embedding of $K$ in $\mathbb {A} _{K}$

Let $K$ be a global field. There is a natural diagonal embedding of $K$ into its adele ring $\mathbb {A} _{K}:$

${\begin{aligned}&K\hookrightarrow \mathbb {A} _{K},\\&a\mapsto (a,a,a,\dotsc ).\end{aligned}}$

This embedding is well-defined, because for each $\alpha \in K,$ it stands, that $\alpha \in {\mathcal {O}}_{v}^{\times }$ for almost all $v.$ The embedding is injective, because the embedding of ${K}$ in $K_{v}$ is injective for each $v.$ As a consequence, we can view $K$ as a subgroup of $\mathbb {A} _{K}.$ In the following, $K$ is a subring of its adele ring. The elements of $K\subset \mathbb {A} _{K}$ are the so-called principal adeles of $\mathbb {A} _{K}.$

Alternative definition of the adele ring of an algebraic number field

Definition: profinite integers

Define

${\widehat {\mathbb {Z} }}:=\lim \limits _{\overleftarrow {n}}\mathbb {Z} /n\mathbb {Z} ,$

that means ${\widehat {\mathbb {Z} }}$ is the profinite completion of the rings $\mathbb {Z} /n\mathbb {Z}$ with the partial order $n\geq m:\Leftrightarrow m\mid n.$

With the Chinese Remainder Theorem, it can be shown that the profinite integers are isomorphic to the product of the integer p-adic numbers. It stands:

${\widehat {\mathbb {Z} }}\cong \prod \limits _{p{\text{ prime }}}\mathbb {Z} _{p}.$

Lemma: alternative definition of the adele ring of an algebraic number field

Define the ring

$\mathbb {A} _{\mathbb {Z} }:={\widehat {\mathbb {Z} }}\times \mathbb {R} .$

With the help of this ring the adele ring of the rational numbers can be written as:

$\mathbb {A} _{\mathbb {Q} }\cong \mathbb {A} _{\mathbb {Z} }\otimes _{\mathbb {Z} }\mathbb {Q} .$

This is an algebraic isomorphism. For an algebraic number field $K$ it stands:

$\mathbb {A} _{K}=\mathbb {A} _{\mathbb {Q} }\otimes _{\mathbb {Q} }K,$

where we install on the right hand side the following topology: It stands, that $\mathbb {A} _{\mathbb {Q} }\otimes _{\mathbb {Q} }K\cong \mathbb {A} _{\mathbb {Q} }\oplus \dots \oplus \mathbb {A} _{\mathbb {Q} },$ where the right hand side has $n:=[K:\mathbb {Q} ]$ summands. We give the right hand side the product topology of $(\mathbb {A} _{\mathbb {Q} })^{n}$ and transport this topology via the isomorphism onto $\mathbb {A} _{\mathbb {Q} }\otimes _{\mathbb {Q} }K.$

Proof: We will first prove the equation about the rational adele ring. Thus, we have to show that $\mathbb {A} _{\mathbb {Q} }\cong \mathbb {A} _{\mathbb {Z} }\otimes _{\mathbb {Z} }\mathbb {Q} .$ It stands $\mathbb {A} _{\mathbb {Z} }\otimes _{\mathbb {Z} }\mathbb {Q} =({\widehat {\mathbb {Z} }}\times \mathbb {R} )\otimes _{\mathbb {Z} }\mathbb {Q} \cong ({\widehat {\mathbb {Z} }}\otimes _{\mathbb {Z} }\mathbb {Q} )\times (\mathbb {R} \otimes _{\mathbb {Z} }\mathbb {Q} )\cong ({\widehat {\mathbb {Z} }}\otimes _{\mathbb {Z} }\mathbb {Q} )\times \mathbb {R} .$ As a result, it is sufficient to show that $\mathbb {A} _{\mathbb {Q} ,{\text{fin}}}={\widehat {\mathbb {Z} }}\otimes _{\mathbb {Z} }\mathbb {Q} .$ We will prove the universal property of the tensor product: Define a $\mathbb {Z}$-bilinear function $\Psi :{\widehat {\mathbb {Z} }}\times \mathbb {Q} \rightarrow \mathbb {A} _{\mathbb {Q} ,{\text{fin}}}$ via $((a_{p})_{p},q)\mapsto (a_{p}\cdot q)_{p}.$ This function is obviously well-defined, because only a finite number of prime numbers divide the denominator of $q\in \mathbb {Q} .$ It stands that $\Psi$ is $\mathbb {Z}$-bilinear.

Let $Z$ be another $\mathbb {Z}$-module together with a $\mathbb {Z}$-bilinear function $\varphi :{\widehat {\mathbb {Z} }}\times \mathbb {Q} \rightarrow Z.$ We have to show that there exists one and only one $\mathbb {Z}$-linear function $\theta :\mathbb {A} _{\mathbb {Q} ,{\text{fin}}}\rightarrow Z$ with the property: $\theta \circ \Psi =\varphi .$ We define the function $\theta$ in the following way: For a given $(u_{p})_{p}$ there exists a $u\in \mathbb {N}$ and a $(v_{p})_{p}\in {\widehat {\mathbb {Z} }},$ such that $\textstyle u_{p}={\frac {1}{u}}\cdot v_{p}$ for all $p.$ Define $\textstyle \theta ((u_{p})_{p}):=\varphi ((v_{p})_{p},{\frac {1}{u}}).$ It can be shown that $\theta$ is well-defined, $\mathbb {Z}$-linear and satisfies $\theta \circ \Psi =\varphi .$ Furthermore, $\theta$ is unique with these properties. The general statement can be shown similarly and will be proved in the following section in general formulation.

The adele ring $\mathbb {A} _{L}$ in case of a field extension $L/K$

Lemma: alternative description of the adele ring in case of $L/K$

Let $K$ be a global field. Let $L/K$ be a finite field extension. In case K is an algebraic number field the extension is separable. If K is a global function field, it can be assumed as separable as well. In any case, $L$ is a global field and thus $\mathbb {A} _{L}$ is defined. For a place $w$ of $L$ and a place $v$ of $K,$ we define

$w\mid v$

if the absolute value $|\cdot |_{w},$ restricted to $K,$ is in the equivalence class of $v.$ We say the place $w$ lies above the place $v.$ Define

${\begin{aligned}L_{v}&:=\prod _{w\mid v}L_{w},\\{\widetilde {{\mathcal {O}}_{v}}}&:=\prod _{w\mid v}{\mathcal {O}}_{w}.\end{aligned}}$

Here $v$ denotes a place of $K$ and $w$ denotes a place of $L.$ Furthermore, both products are finite.

Remark: We can embed $K_{v}$ in $L_{w},$ if $w\mid v.$ Therefore, we can embed $K_{v}$ diagonal in $L_{v}.$ With this embedding the set $L_{v}$ is a commutative algebra over $K_{v}$ with degree $\textstyle \sum _{w\mid v}[L_{w}:K_{v}]=[L:K].$

It is valid, that

$\mathbb {A} _{L}={\widehat {\prod \limits _{v}}}^{{\mathcal {O}}_{v}}L_{v}.$

This can be shown with elementary properties of the restricted product.

The adeles of $K$ can be canonically embedded in the adeles of $L:$ The adele $a=(a_{v})_{v}\in \mathbb {A} _{K}$ is assigned to the adele $a'=(a'_{w})_{w}\in \mathbb {A} _{L}$ with $a'_{w}=a_{v}\in K_{v}\subset L_{w}$ for $w\mid v.$ Therefore, $\mathbb {A} _{K}$ can be seen as a subgroup of $\mathbb {A} _{L}.$ An element $a=(a_{w})_{w}\in \mathbb {A} _{L}$ is in the subgroup $\mathbb {A} _{K},$ if $a_{w}\in K_{v}$ for $w\mid v$ and if $a_{w}=a_{w'}$ for all $w\mid v$ and $w'\mid v$ for the same place $v$ of $K.$

Lemma: the adele ring as a tensor product

Let $K$ be a global field and let $L/K$ be a finite field extension. It stands:

$\mathbb {A} _{L}\cong \mathbb {A} _{K}\otimes _{K}L.$

This is an algebraic and topological isomorphism and we install the same topology on the tensor product as we defined it in the lemma about the alternative definition of the adele ring. In order to do this, we need the condition $[L:K]<\infty .$ With the help of this isomorphism, the inclusion $\mathbb {A} _{K}\subset \mathbb {A} _{L}$ is given via the function

${\begin{aligned}\mathbb {A} _{K}&\hookrightarrow \mathbb {A} _{L},\\\alpha &\mapsto \alpha \otimes _{K}1.\end{aligned}}$

Furthermore, the principal adeles of $K$ can be identified with a subgroup of the principal adeles of $L$ via the map

${\begin{aligned}K&\hookrightarrow (K\otimes _{K}L)\cong L,\\\alpha &\mapsto 1\otimes _{K}\alpha .\end{aligned}}$

Proof: Let $\omega _{1},\dotsc ,\omega _{n}$ be a basis of $L$ over $K.$ It stands, that

${\widetilde {{\mathcal {O}}_{v}}}\cong {\mathcal {O}}_{v}\omega _{1}\oplus \dotsc \oplus {\mathcal {O}}_{v}\omega _{n}$

Furthermore, there are the following isomorphisms:

${\begin{aligned}K_{v}\omega _{1}\oplus \dotsc \oplus K_{v}\omega _{n}\cong K_{v}\otimes _{K}L&{\xrightarrow {\cong }}L_{v}=\prod _{w\mid v}L_{w},\\\alpha _{v}\otimes a&\mapsto (\alpha _{v}\cdot (\tau _{w}(a)))_{w}\end{aligned}}$

where $\tau _{w}$ is the canonical embedding $\tau _{w}:L\rightarrow L_{w}$ and as usual $w\mid v.$ We take on both sides the restricted product with restriction condition ${\widetilde {{\mathcal {O}}_{v}}}:$

${\begin{aligned}\mathbb {A} _{K}\otimes _{K}L&=({\widehat {\prod \limits _{v}}}^{{\mathcal {O}}_{v}}K_{v})\otimes _{K}L\\&\cong {\widehat {\prod \limits _{v}}}^{({\mathcal {O}}_{v}\omega _{1}\oplus \cdot \oplus {\mathcal {O}}_{v}\omega _{n})}(K_{v}\omega _{1}\oplus \dotsc \oplus K_{v}\omega _{n})\\&\cong {\widehat {\prod \limits _{v}}}^{\widetilde {{\mathcal {O}}_{v}}}(K_{v}\otimes _{K}L)\\&\cong {\widehat {\prod \limits _{v}}}^{\widetilde {{\mathcal {O}}_{v}}}L_{v}\\&=\mathbb {A} _{L}.\end{aligned}}$

Thus we arrive at the desired result.

Corollary: the adele ring of $L$ as an additive group

Viewed as additive groups, the following is true:

$\mathbb {A} _{L}\cong \mathbb {A} _{K}\oplus \dotsc \oplus \mathbb {A} _{K},$

where the left side has $n:=[L:K]$ summands. The set of principal adeles in $\mathbb {A} _{L}$ are identified with the set $K\oplus \dotsc \oplus K,$ where the left side has $n$ summands and we consider $K$ as a subset of $\mathbb {A} _{K}.$

Definition of the adele ring of a vector-space $E$ over $K$ and an algebra $A$ over $K$

Lemma: alternative description of the adele ring

Let $K$ be a global field. Let $P$ be a finite subset of the set of all places of $K,$ which contains $P_{\infty }.$ As usual, we write $P_{\infty }$ for the set of all infinite places of $K.$ Define

$\mathbb {A} _{K}(P):=\prod _{v\in P}K_{v}\times \prod _{v\notin P}{\mathcal {O}}_{v}.$

We define addition and multiplication component-wise and we install the product topology on this ring. Then $\mathbb {A} _{K}(P)$ is a locally compact, topological ring. In other words, we can describe $\mathbb {A} _{K}(P)$ as the set of all $\textstyle x=(x_{v})_{v}\in \prod _{v}K_{v},$ where $x_{v}\in {\mathcal {O}}_{v}$ for all $v\notin P.$ That means $|x_{v}|_{v}\leq 1$ for all $v\notin P.$

Remark: Is $P'$ another subset of the set of places of $K$ with the property $P\subset P',$ we note, that $\mathbb {A} _{K}(P)$ is an open subring of $\mathbb {A} _{K}(P').$

Now, we are able to give an alternative characterisation of the adele ring. The adele ring is the union of all the sets $\mathbb {A} _{K}(P),$ where $P$ passes all the finite subsets of the whole set of places of $K,$ which contains $P_{\infty }.$ In other words:

$\mathbb {A} _{K}=\bigcup _{P\supset P_{\infty }, \atop P{\text{ finite }}}\mathbb {A} _{K}(P).$

That means, that $\mathbb {A} _{K}$ is the set of all $x=(x_{v})_{v}$ so that $|x_{v}|_{v}\leq 1$ for almost all $v<\infty .$ The topology of $\mathbb {A} _{K}$ is induced by the requirement, that all $\mathbb {A} _{K}(P)$ become open subrings of $\mathbb {A} _{K}.$ Thus, $\mathbb {A} _{K}$ is a locally compact, topological ring.

Let's fix a place $v$ of $K.$ Let $P$ be a finite subset of the set of all places of $K,$ containing $v$ and $P_{\infty }.$ It stands:

$\mathbb {A} _{K}(P)=\prod _{w\in P}K_{w}\times \prod _{w\notin P}{\mathcal {O}}_{w}.$

Define

$\mathbb {A} _{K}'(P,v):=\prod _{w\in P\setminus \{v\}}K_{w}\times \prod _{w\notin P}{\mathcal {O}}_{w}.$

It stands:

$\mathbb {A} _{K}(P)\cong K_{v}\times \mathbb {A} _{K}'(P,v).$

Furthermore, define

$\mathbb {A} _{K}'(v):=\bigcup _{P\supset (P_{\infty }\cup \{v\})}\mathbb {A} _{K}'(P,v),$

where $P$ runs through all finite sets fulfilling $P\supset (P_{\infty }\cup \{v\}).$ Obviously it stands:

$\mathbb {A} _{K}\cong K_{v}\times \mathbb {A} _{K}'(v),$

via the map $(a_{w})_{w}\mapsto (a_{v},(a_{w})_{w\neq v}).$ The entire procedure above can be performed also with a finite subset ${\widetilde {P}}$ instead of $\{v\}.$

By construction of $\mathbb {A} _{K}'(v),$ there is a natural embedding of $K_{v}$ in $\mathbb {A} _{K}:$ $K_{v}\hookrightarrow \mathbb {A} _{K}.$ Furthermore, there exists a natural projection $\mathbb {A} _{K}\twoheadrightarrow K_{v}.$

Definition: the adele ring of a vector-space $E$ over $K$

The two following definitions are based on Weil 1967, p. 60. Let $K$ be a global field. Let $E$ be a $n$-dimensional vector-space over $K,$ where $n<\infty .$ We fix a basis $\omega _{1},\dotsc ,\omega _{n}$ of $E$ over $K.$ For each place $v$ of $K,$ we write $E_{v}:=E\otimes _{K}K_{v}\cong K_{v}\omega _{1}\oplus \dotsc \oplus K_{v}\omega _{n}$ and ${\widetilde {{\mathcal {O}}_{v}}}:={\mathcal {O}}_{v}\omega _{1}\oplus \dotsc \oplus {\mathcal {O}}_{v}\omega _{n}.$ We define the adele ring of $E$ as

$\mathbb {A} _{E}:={\widehat {\prod \limits _{v}}}^{\widetilde {{\mathcal {O}}_{v}}}E_{v}.$

This definition is based on the alternative description of the adele ring as a tensor product. On the tensor product we install the same topology we defined in the lemma about the alternative definition of the adele ring. In order to do this, we need the condition $\operatorname {dim} _{K}(E)=n<\infty .$ We install the restricted product topology on the adele ring $\mathbb {A} _{E}.$

We receive the result, that $\mathbb {A} _{E}=E\otimes _{K}\mathbb {A} _{K}.$ We can embed $E$ naturally in $\mathbb {A} _{E}$ via the function $e\mapsto e\otimes 1.$

In the following, we give an alternative definition of the topology on the adele ring $\mathbb {A} _{E}.$ The topology on $\mathbb {A} _{E}$ is given as the coarsest topology, for which all linear forms (linear functionals) on $E,$ that means linear maps $\lambda :E\rightarrow K,$ extending to linear functionals of $\mathbb {A} _{E}$ to $\mathbb {A} _{K},$ are continuous. We use the natural embedding of $E$ into $\mathbb {A} _{E}$ respectively of $K$ into $\mathbb {A} _{K},$ to extend the linear forms.

We can define the topology in a different way: Take a basis $\varepsilon$ of $E$ over $K.$ This results in an isomorphism of $K^{n}$ to $E.$ As a consequence the basis $\varepsilon$ induces an isomorphism of $(\mathbb {A} _{K})^{n}$ to $\mathbb {A} _{E}.$ We supply the left hand side 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, we obtain a linear homeomorphism. This homeomorphism transfers the two topologies into each other.

In a formal way, it stands:

${\begin{aligned}\mathbb {A} _{E}&=E\otimes _{K}\mathbb {A} _{K}\\&\cong (K\otimes _{K}\mathbb {A} _{K})\oplus \dotsc \oplus (K\otimes _{K}\mathbb {A} _{K})\\&\cong \mathbb {A} _{K}\oplus \dotsc \oplus \mathbb {A} _{K}\\&=(\mathbb {A} _{K})^{n},\end{aligned}}$

where the sums have $n$ summands. In case of $E=L,$ the definition above is consistent with the results about the adele ring in case of a field extension $L/K.$

Definition: the adele ring of an algebra $A$ over $K$

Let $K$ be a global field and let $A$ be a finite-dimensional algebra over $K.$ In particular, $A$ is a finite-dimensional vector-space over $K.$ As a consequence, $\mathbb {A} _{A}$ is defined. We establish a multiplication on $\mathbb {A} _{A},$ based on the multiplication of $A:$

It stands, that $\mathbb {A} _{A}\cong \mathbb {A} _{K}\otimes _{K}A.$ Since, we have a multiplication on $\mathbb {A} _{K}$ and on $A,$ we can define a multiplication on $\mathbb {A} _{A}$ via

$(a\otimes _{K}b)\cdot (c\otimes _{K}d):=(ac)\otimes _{K}(bd)\quad \forall a,c\in \mathbb {A} _{K}{\text{ and }}\forall b,d\in A.$

Alternatively, we fix a basis $\alpha _{1},\dotsc ,\alpha _{n}$ of $A$ over $K.$ To describe the multiplication of $A,$ it is sufficient to know, how we multiply two elements of the basis. There are $\beta _{i,j,k}\in K,$ so that

$\alpha _{i}\alpha _{j}=\sum _{k=1}^{n}\beta _{i,j,k}\alpha _{k}\quad \forall i,j.$

With the help of the $\beta _{i,j,k},$ we can define a multiplication on $K^{n}\cong A:$

$e_{i}e_{j}:=\sum _{k=1}^{n}\beta _{i,j,k}e_{k}\quad \forall i,j.$

In addition to that, we can define a multiplication on $(K_{v})^{n}\cong A_{v}$ and therefore on $(\mathbb {A} _{K})^{n}\cong \mathbb {A} _{A}.$

As a consequence, $\mathbb {A} _{A}$ is an algebra with 1 over $\mathbb {A} _{K}.$ Let $\alpha$ be a finite subset of $A,$ containing a basis of $A$ over $K.$ We define $\alpha _{v}$ as the ${\mathcal {O}}_{v}$-modul generated by $\alpha$ in $A_{v},$ where $v$ is a finite place of $K.$ For each finite subset $P$ of the set of all places, containing $P_{\infty },$ we define

$\mathbb {A} _{A}(P,\alpha )=\prod \limits _{v\in P}A_{v}\times \prod \limits _{v\notin P}\alpha _{v}.$

It can be shown, that there is a finite set $P_{0},$ so that $\mathbb {A} _{A}(P,\alpha )$ is an open subring of $\mathbb {A} _{A},$ if $P$ contains $P_{0}.$ Furthermore, it stands, that $\mathbb {A} _{A}$ is the union of all these subrings. It can be shown, that in case of $A=K,$ the definition above is consistent with the definition of the adele ring.

Trace and norm on the adele ring

Definition: trace and norm on the adele ring

Let $L$ be a finite extension of the global field $K.$ It stands $\mathbb {A} _{L}=\mathbb {A} _{K}\otimes _{K}L.$ Furthermore, it stands $\mathbb {A} _{K}=\mathbb {A} _{K}\otimes _{K}K.$ As a consequence, we can interpret $\mathbb {A} _{K}$ as a closed subring of $\mathbb {A} _{L}.$ We write $\operatorname {con} _{L/K}$ for this embedding. Explicitly, it stands: $(\operatorname {con} _{L/K}(\alpha ))_{w}=\alpha _{v}\in K_{v}$ and this is true for all places $w$ of $L$ above $v$ and for any $\alpha \in \mathbb {A} _{K}.$

Now, let $M/L/K$ be a tower of global fields. It stands:

$\operatorname {con} _{M/K}(\alpha )=\operatorname {con} _{M/L}(\operatorname {con} _{L/K}(\alpha ))\qquad \forall \alpha \in \mathbb {A} _{K}.$

Furthermore, if we restrict the map $\operatorname {con}$ to the principal adeles, $\operatorname {con}$ becomes the natural injection $K\hookrightarrow L.$

Let $\omega _{1},\dotsc ,\omega _{n}$ be a basis of the field extension $L/K.$ That means, that each $\alpha \in \mathbb {A} _{L}$ can be written as $\textstyle \sum _{j=1}^{n}\alpha _{j}\omega _{j},$ where the $\alpha _{j}\in \mathbb {A} _{K}$ are unique. The map $\alpha \mapsto \alpha _{j}$ is continuous. We define $\alpha _{i,j},$ depending on $\alpha ,$ via the equations

$\alpha \omega _{i}=\sum _{j=1}^{n}\alpha _{i,j}\omega _{j}\qquad \forall i.$

Now, we define the trace and norm of $\alpha$ as:

${\begin{aligned}\operatorname {Tr} _{L/K}(\alpha )&:=\operatorname {Tr} ((\alpha _{i,j})_{i,j})=\sum _{i=1}^{n}\alpha _{i,i}\qquad {\text{and}}\\N_{L/K}(\alpha )&:=N((\alpha _{i,j})_{i,j})=\det((\alpha _{i,j})_{i,j}).\end{aligned}}$

These are the trace and the determinant of the linear map $\mathbb {A} _{L}\rightarrow \mathbb {A} _{L},$ $x\mapsto \alpha x.$ They are continuous maps on the adele ring.

Lemma: properties of trace and norm

Trace and norm fulfil the usual equations:

${\begin{aligned}\operatorname {Tr} _{L/K}(\alpha +\beta )&=\operatorname {Tr} _{L/K}(\alpha )+\operatorname {Tr} _{L/K}(\beta )\qquad \forall \alpha ,\beta \in \mathbb {A} _{L},\\\operatorname {Tr} _{L/K}(\operatorname {con} (\alpha ))&=n\alpha \qquad \qquad \qquad \qquad \qquad \forall \alpha \in \mathbb {A} _{K},\\N_{L/K}(\alpha \cdot \beta )&=N_{L/K}(\alpha )\cdot N_{L/K}(\beta )\qquad \forall \alpha ,\beta \in \mathbb {A} _{L},\\N_{L/K}(\operatorname {con} (\alpha ))&=\alpha ^{n}\qquad \qquad \qquad \qquad \qquad \forall \alpha \in \mathbb {A} _{K}.\end{aligned}}$

Furthermore, we note that for an $\alpha \in L$ the trace $\operatorname {Tr} _{L/K}(\alpha )$ and the norm $N_{L/K}(\alpha )$ are identical to the trace and norm of the field extension $L/K.$ For a tower of fields $M/L/K,$ it stands:

${\begin{aligned}\operatorname {Tr} _{L/K}(\operatorname {Tr} _{M/L}(\alpha ))&=\operatorname {Tr} _{M/K}(\alpha )\qquad \forall \alpha \in \mathbb {A} _{M},\\N_{L/K}(N_{M/L}(\alpha ))&=N_{M/K}(\alpha )\qquad \forall \alpha \in \mathbb {A} _{M}.\end{aligned}}$

Moreover, it can be shown, that

${\begin{aligned}\operatorname {Tr} _{L/K}(\alpha )&=(\sum _{w\mid v}\operatorname {Tr} _{L_{w}/K_{v}}(\alpha _{w}))_{v}\qquad \forall \alpha \in \mathbb {A} _{L},\\N_{L/K}(\alpha )&=(\prod _{w\mid v}N_{L_{w}/K_{v}}(\alpha _{w}))_{v}\qquad \forall \alpha \in \mathbb {A} _{L}.\end{aligned}}$

Remark: The last two equations are not obvious.

Properties of the adele ring

In principle, to prove the following statements, we can reduce the situation to the case $K=\mathbb {Q}$ or $K=\mathbb {F} _{p}(t).$ The generalisation for global fields is often trivial.

Theorem: the adele ring is a locally compact, topological ring

Let $K$ be a global field. It stands, that $\mathbb {A} _{A,S}$ is a topological ring for every subset $S$ of the set of all places. Furthermore, $(\mathbb {A} _{A,S},+)$ is a locally compact group, that means, that the set $\mathbb {A} _{K,S}$ is locally compact and the group operation is continuous, that means that the map

${\begin{aligned}+:\mathbb {A} _{K}\times \mathbb {A} _{K}&\rightarrow \mathbb {A} _{K},\\(a,b)&\mapsto a+b\end{aligned}}$

is continuous and the map of the inverse is continuous, too, resulting in the continuous map

${\begin{aligned}i:\mathbb {A} _{K}&\rightarrow \mathbb {A} _{K},\\a&\mapsto -a.\end{aligned}}$

A neighbourhood system of $0$ in $\mathbb {A} _{K}(P_{\infty })$ is a neighbourhood system of $0$ in the adele ring. Alternatively, we can take all sets of the form $\textstyle \prod _{v}U_{v},$ where $U_{v}$ is a neighbourhood of $0$ in $K_{v}$ and $U_{v}={\mathcal {O}}_{v}$ for almost all $v.$

Idea of proof: The set $\mathbb {A} _{A,S}$ is locally compact, because all the ${\mathcal {O}}_{v}$ are compact and the adele ring is a restricted product. The continuity of the group operations can be shown with the continuity of the group operations in each component of the restricted product.

Remark: The result above can be shown similarly for the adele ring of a vector-space $E$ over $K$ and an algebra $A$ over $K.$

Theorem: the global field is a discrete, cocompact subgroup in its adele ring

The adele ring contains the global field as a discrete, cocompact subgroup. This means that $K\subset \mathbb {A} _{K}$ is discrete and $\mathbb {A} _{K}/K$ is compact in the topology of the quotient. In particular, $K$ is closed in $\mathbb {A} _{K}.$

Proof: in the following, we reflect the proof for the case $K=\mathbb {Q} :$

We have to show that $\mathbb {Q}$ is discrete in $\mathbb {A} _{\mathbb {Q} }.$ It is sufficient to show that there exists a neighbourhood of $0$ which contains no more rational numbers. The general case follows via translation. Define

$U:=\{(\alpha _{p})_{p}:|\alpha _{p}|_{p}\leq 1\quad \forall p{\text{ and }}|\alpha _{\infty }|_{\infty }<1\}=\prod _{p<\infty }\mathbb {Z} _{p}\times (-1,1).$

Then $U$ is an open neighbourhood of $0$ in $\mathbb {A} _{\mathbb {Q} }.$ We have to show that $U\cap \mathbb {Q} =\{0\}.$ Let $\beta$ be in $U\cap \mathbb {Q} .$ It follows that $\beta \in \mathbb {Q}$ and $|\beta |_{p}\leq 1$ for all $p$ and therefore $\beta \in \mathbb {Z} .$ Additionally, we have that $\beta \in (-1,1)$ and therefore $\beta =0.$

Next, we show that $\mathbb {A} _{K}/K$ is compact. Define the set

$W:=\{(\alpha _{p})_{p}:|\alpha _{p}|_{p}\leq 1\quad \forall p{\text{ and }}|\alpha _{\infty }|_{\infty }\leq 1/2\}=\prod _{p<\infty }\mathbb {Z} _{p}\times \left[-{\frac {1}{2}},{\frac {1}{2}}\right].$

We show that each element in $\mathbb {A} _{\mathbb {Q} }/\mathbb {Q}$ has a representative in $W.$ In other words, we need to show that for each adele $\alpha \in \mathbb {A} _{\mathbb {Q} }$, there exists $\beta \in \mathbb {Q}$ such that $\alpha -\beta \in W.$ Take an arbitrary $\alpha =(\alpha _{p})_{p}\in \mathbb {A} _{\mathbb {Q} },$ and let $p$ be a prime number for which $|\alpha _{p}|>1.$ Then there exists $r_{p}=z_{p}/p^{x_{p}}$ with $z_{p}\in \mathbb {Z} ,$ $x_{p}\in \mathbb {N}$ and $|\alpha _{p}-r_{p}|\leq 1.$ We replace $\alpha$ by $\alpha -r_{p}.$ This replacement changes the other places as follows:

Let $q\neq p$ be another prime number. One has the following: $|\alpha _{q}-r_{p}|_{q}\leq \max\{|a_{q}|_{q},|r_{p}|_{q}\}\leq \max\{|a_{q}|_{q},1\}\leq 1.$ It follows that $|\alpha _{q}-r_{p}|_{q}\leq 1\Leftrightarrow |\alpha _{q}|_{q}\leq 1$ (″$\Rightarrow$″ 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 prime numbers for which the components of $\alpha$ aren't in $\mathbb {Z} _{p}$ is reduced by 1. With an iteration, we arrive at the result that $r\in \mathbb {Q}$ exists with the property that $\alpha -r\in {\widehat {\mathbb {Z} }}\times \mathbb {R} .$ Now we select $s\in \mathbb {Z}$ such that $\alpha _{\infty }-r-s$ is in $[-1/2,1/2].$ Since $s$ is in $\mathbb {Z} ,$ it follows that $\alpha -\beta \in W$ for $\beta :=r+s\in \mathbb {Q} .$ We consider the continuous projection $\pi :W\twoheadrightarrow \mathbb {A} _{\mathbb {Q} }/\mathbb {Q} .$ The projection is surjective. Therefore, $\mathbb {A} _{\mathbb {Q} }/\mathbb {Q}$ is the continuous image of a compact set, and thus compact by itself.

The last statement is a lemma about topological groups.

Corollary: Let $K$ be a global field and let $E$ be a finite-dimensional vector-space over $K.$ Then $E$ is discrete and cocompact in $\mathbb {A} _{E}.$

Lemma: properties of the rational adele ring

In a previous section, we defined $\mathbb {A} _{\mathbb {Z} }={\widehat {\mathbb {Z} }}\times \mathbb {R} =\prod _{p}\mathbb {Z} _{p}\times \mathbb {R} .$ It stands

${\begin{aligned}\mathbb {A} _{\mathbb {Q} }&=\mathbb {Q} +\mathbb {A} _{\mathbb {Z} }{\text{ and }}\\\mathbb {Z} &=\mathbb {Q} \cap \mathbb {A} _{\mathbb {Z} }.\end{aligned}}$

Furthermore, it stands, that $\mathbb {A} _{\mathbb {Q} }/\mathbb {Z}$ is unlimited divisible, which is equivalent to the statement, that the equation $nx=y$ has a solution $x\in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z}$ for each $n\in \mathbb {N}$ and for each $y\in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} .$ This solution is generally not unique.

Furthermore, it stands, that $\mathbb {Q} \subset \mathbb {A} _{\mathbb {Q} ,{\text{fin}}}$ is dense in $\mathbb {A} _{\mathbb {Q} ,{\text{fin}}}.$ This statement is a special case of the strong approximation theorem.

Proof: The first two equations can be proved in an elementary way. The next statement can be found in Neukirch 2007, p. 383. We will prove it. Let $n\in \mathbb {N}$ and $y\in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z}$ be given. We need to show the existence of a $x\in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z}$ with the property: $nx=y.$ It is sufficient to show this statement for $\mathbb {A} _{\mathbb {Q} }.$ This is easily seen, because $\mathbb {A} _{\mathbb {Q} }$ is a field with characteristic unequal zero in each coordinate. In the following, we give a counter example, showing, that $\mathbb {A} _{\mathbb {Q} }/\mathbb {Z}$ isn't uniquely reversible. Let $y=(0,0,\dotsc ,0)+\mathbb {Z} \in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z}$ and $n\geq 2$ be given. Then $x_{1}=(0,0,\dotsc ,0)+\mathbb {Z} \in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z}$ fulfils the equation $nx=y.$ In addition, $x_{2}=(1/n,1/n,\dotsc ,1/n)+\mathbb {Z} \in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z}$ fulfils this equations as well, because $nx=(1,1,\dotsc ,1)+\mathbb {Z} =y.$ It stands, that $x_{2}$ is well-defined, because there exists only a finite number of prime numbers, dividing $n.$ However, it stands, that $x_{1}\neq x_{2},$ because by considering the last coordinate, we obtain $1/n\neq 0+z\,\,\forall z\in \mathbb {Z} .$

Remark: In this case, the unique reversibility is equivalent to the torsion freedom, which is not provided here: $n\cdot x_{2}=0,$ but $x_{2}\neq 0$ and $n\neq 0.$

We now prove the last statement. It stands: $\mathbb {A} _{\mathbb {Q} ,{\text{fin}}}=\mathbb {Q} {\widehat {\mathbb {Z} }},$ as we can reach the finite number of denominators in the coordinates of the elements of $\mathbb {A} _{\mathbb {Q} ,{\text{fin}}}$ through an element $q\in \mathbb {Q} .$ As a consequence, it is sufficient to show, that $\mathbb {Z}$ is dense in ${\widehat {\mathbb {Z} }}.$ For this purpose, we have to show, that each open subset $V$ of ${\widehat {\mathbb {Z} }}$ contains an element of $\mathbb {Z} .$ Without loss of generality, we can assume

$V=\prod _{p\in E}(a_{p}+p^{l_{p}}\mathbb {Z} _{p})\times \prod _{p\notin E}\mathbb {Z} _{p},$

because $(p^{m}\mathbb {Z} _{p})_{m\in \mathbb {N} }$ is a neighbourhood system of $0$ in $\mathbb {Z} _{p}.$

With the help of the Chinese Remainder Theorem, we can prove the existence of a $l\in \mathbb {Z}$ with the property: $l\equiv a_{p}\;{\bmod {\;}}p^{l_{p}},$ because the powers of different prime numbers are coprime integers. Thus, $l\in V$ follows.

Definition: Haar measure on the adele ring

Let $K$ be a global field. We have seen that $(\mathbb {A} _{K},+)$ is a locally compact group. Therefore, there exists a Haar measure $dx$ on $(\mathbb {A} _{K},+).$ We can normalise $dx$ as follows: Let $f$ be a simple function on $\mathbb {A} _{K},$ that means $\textstyle f=\prod _{v}f_{v},$ where $f_{v}:K_{v}\rightarrow \mathbb {C} ,$ measurable and $f_{v}=\mathbf {1} _{{\mathcal {O}}_{v}}$ for almost all $v.$ The Haar measure $dx$ on $\mathbb {A} _{K}$ can be normalised so that for each simple, integrable function $\textstyle f=\prod _{v}f_{v},$ the following product formula is satisfied:

$\int _{\mathbb {A} }f(x)\,dx=\prod _{v}\int _{K_{v}}f_{v}\,dx_{v},$

where for each finite place, one has that $\textstyle \int _{{\mathcal {O}}_{v}}1\,dx_{v}=1.$ At the infinite places we choose Lebesgue measure.

We construct this measure by defining it on simple sets $\textstyle \prod _{v}A_{v},$ where $A_{v}\subset K_{v}$ is open and $A_{v}={\mathcal {O}}_{v}$ for almost all $v.$ Since the simple sets generate the entire Borel $\sigma$-algebra, the measure can be defined on the entire $\sigma$-algebra.

Finitely, it can be shown that $\mathbb {A} _{K}/K$ has finite total measure under the quotient measure induced by the Haar measure on $\mathbb {A} _{K}.$ The finiteness of this measure is a corollary of the theorem above, since compactness implies finite total measure.

Idele group

Definition of the idele group of a global field $K$

Definition and lemma: topology on the group of units of a topological ring

Let $R$ be a topological ring. The group of units $R^{\times },$ together with the subspace topology, aren't a topological group in general. Therefore, we define a coarser topology on $R^{\times },$ which means that less sets are open. This is done in the following way: Let $\iota$ be the inclusion map:

${\begin{aligned}\iota :R^{\times }&\rightarrow R\times R,\\x&\mapsto (x,x^{-1}).\end{aligned}}$

We define the topology on $R^{\times }$ as the topology induced by the subset topology on $R\times R.$ That means, on $\iota (R^{\times })$ we consider the subset topology of the product topology. A set $U\subset R^{\times }$ is open in the new topology if and only if $\iota (U)$ is open in the subset topology. With this new topology $(R^{\times },\cdot )$ is a topological group and the inclusion map $R^{\times }\hookrightarrow R$ is continuous. It is the coarsest topology, emerging from the topology on $R,$ that makes $R^{\times }$ a topological group.

Proof: We consider the topological ring $\mathbb {A} _{\mathbb {Q} }.$ The inversion map isn't continuous. To demonstrate this, we consider the sequence

${\begin{aligned}x_{1}&=(2,1,\dotsc )\\x_{2}&=(1,3,1,\dotsc )\\x_{3}&=(1,1,5,1,\dotsc )\\&\,\,\,\vdots \end{aligned}}$

This sequence converges in the topology of $\mathbb {A} _{\mathbb {Q} }$ with the limit $1\in \mathbb {A} _{\mathbb {Q} }.$ The reason for this is, that for a given neighbourhood $U$ of $0,$ it stands, that without loss of generality we can assume, that $U$ is of form:

$U=\prod _{p{\text{ prime }} \atop p\leq N}U_{p}\times \prod _{p{\text{ prime }} \atop p>N}\mathbb {Z} _{p}$

Furthermore, it stands, that $(x_{n})_{p}-1\in \mathbb {Z} _{p}$ for all $p.$ Therefore, it stands, that $x_{n}-1\in U$ for all $n$ big enough. The inversion of this sequence does not converge in the subset-topology of $\mathbb {A} _{\mathbb {Q} }.$ We have shown this in the lemma about the difference between the restricted and the unrestricted product topology. In our new topology, the sequence and its inverse do not converge. This example shows that the two topologies are different in general. Now we show, that $R^{\times }$ is a topological group with the topology defined above. Since $R$ is a topological ring, it is sufficient to show, that the function $x\mapsto x^{-1}$ is continuous. Let $U$ be an open subset of $R^{\times }$ in our new topology, i.e. $U\times U^{-1}\subset R\times R$ is open. We have to show, that $U^{-1}\subset R^{\times }$ is open or equivalently, that $U^{-1}\times (U^{-1})^{-1}=U^{-1}\times U\subset R\times R$ is open. But this is the condition above.

Definition: the idele group of a global field $K$

Let $K$ be a global field. We define the idele group of $K$ as the group of units of the adele ring of $K,$ which we write in the following as:

$I_{K}:=\mathbb {A} _{K}^{\times }$

Furthermore, we define

${\begin{aligned}I_{K,S}&:=\mathbb {A} _{K,S}^{\times }\\{\text{and}}\qquad I_{K}^{S}&:=(\mathbb {A} _{K}^{S})^{\times }.\end{aligned}}$

We provide the idele group with the topology defined above. Thus, the idele group is a topological group. The elements of the idele group are called the ideles of $K.$

Lemma: the idele group as a restricted product

Let $K$ be a global field. It stands

${\begin{aligned}I_{K,S}&={\widehat {\prod \limits _{v\in S}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times },\\I_{K}^{S}&={\widehat {\prod \limits _{v\notin S}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times },\\{\text{and}}\qquad I_{K}&={\widehat {\prod \limits _{v}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times },\end{aligned}}$

where these are identities of topological groups. The restricted product has the restricted product topology, which is generated by restricted open rectangles having the form

${\begin{aligned}\prod _{v\in E}U_{v}\times \prod _{v\notin E}{\mathcal {O}}_{v}^{\times },\end{aligned}}$

where $E$ is a finite subset of the set of all places and $U_{v}\subset K_{v}^{\times }$ are open sets.

Proof: We will give a proof for the equation with $I_{K}$. The other two equations follow similarly. First we show that the two sets are equal:

${\begin{aligned}I_{K}:=\mathbb {A} _{K}^{\times }:&=\{x=(x_{v})_{v}\in \mathbb {A} _{K}:\exists y=(y_{v})_{v}\in \mathbb {A} _{K}:xy=1\}\\&=\{x=(x_{v})_{v}\in \mathbb {A} _{K}:\exists y=(y_{v})_{v}\in \mathbb {A} _{K}:x_{v}\cdot y_{v}=1\quad \forall v\}\\&=\{x=(x_{v})_{v}:x_{v}\in K_{v}^{\times }\,\,\forall v{\text{ and }}x_{v}\in {\mathcal {O}}_{v}^{\times }\,\,{\text{ for almost all }}v\}\\&={\widehat {\prod \limits _{v}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times }.\end{aligned}}$

Note, that in going from line 2 to 3, $x$ as well as $x^{-1}=y$ have to be in $\mathbb {A} _{K},$ meaning $x_{v}\in {\mathcal {O}}_{v}$ for almost all $v$ and $x_{v}^{-1}\in {\mathcal {O}}_{v}$ for almost all $v.$ Therefore, $x_{v}\in {\mathcal {O}}_{v}^{\times }$ for almost all $v.$

Now, we can show that 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 $U\subset I_{K},$ which is open in the topology of the idele group, meaning $U\times U^{-1}\subset \mathbb {A} _{K}\times \mathbb {A} _{K}$ is open, it stands that for each $u\in U$ 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 rectangle and is therefore open in the restricted product topology.

Further definitions:

Define

${\widehat {\mathcal {O}}}:=\prod _{v\nmid \infty }{\mathcal {O}}_{v}=\prod _{v<\infty }{\mathcal {O}}_{v}$

and ${\widehat {\mathcal {O}}}^{\times }$ as the group of units of ${\widehat {\mathcal {O}}}.$ It stands

${\widehat {\mathcal {O}}}^{\times }=\prod _{v<\infty }{\mathcal {O}}_{v}^{\times }.$

The idele group $I_{L}$ in case $L/K$

This section is based on the corresponding section about the adele ring.

Lemma: alternative description of the idele group in case $L/K$

Let $K$ be a global field and let $L/K$ be a finite field extension. It stands, that $L$ is a global field and therefore the idele group $I_{L}$ is defined. Define

${\begin{aligned}L_{v}^{\times }&:=\prod _{w\mid v}L_{w}^{\times },\\{\widetilde {{\mathcal {O}}_{v}}}^{\times }&:=\prod _{w\mid v}{\mathcal {O}}_{w}^{\times }.\end{aligned}}$

Note, that both products are finite. It stands:

${\begin{aligned}I_{L}={\widehat {\prod \limits _{v}}}^{{\widetilde {{\mathcal {O}}_{v}}}^{\times }}L_{v}^{\times }.\end{aligned}}$

Lemma: embedding of $I_{K}$ in $I_{L}$

There is a canonical embedding of the idele group of $K$ in the idele group of $L:$ We assign an idele $a=(a_{v})_{v}\in I_{K}$ the idele $a'=(a'_{w})_{w}\in I_{L}$ with the property $a'_{w}=a_{v}\in K_{v}^{\times }\subset L_{w}^{\times }$ for $w\mid v.$ Therefore, $I_{K}$ can be seen as a subgroup of $I_{L}.$ An element $a=(a_{w})_{w}\in I_{L}$ is in this subgroup if and only if his components satisfy the following properties: $a_{w}\in K_{v}^{\times }$ for $w\mid v$ and $a_{w}=a_{w'}$ for $w\mid v$ and $w'\mid v$ for the same place $v$ of $K.$

The case of a vector-space $E$ over $K$ and an algebra $A$ over $K$

The following section is based on Weil 1967, p. 71.

Definition: $\operatorname {End} (E)$

Let $E$ be a finite-dimensional vector-space over $K,$ where $K$ is a global field. Define:

$\operatorname {End} (E):=\{\varphi :E\rightarrow E,\varphi \,\,{\text{ is a }}\,\,K{\text{-linear map }}\}.$

This is an algebra over $K.$ It stands, that $\operatorname {End} (E)^{\times }=\operatorname {Aut} (E),$ where for a linear map the inverse function exists if and only if the determinant is not equal to $0.$ Since $K$ is a global field, which in particular means that $K$ is a topological field, $\operatorname {Aut} (E)$ is an open subset of $\operatorname {End} (E).$ The reason for this is, that $\operatorname {End} (E)\setminus \operatorname {Aut} (E)=\det ^{-1}(\{0\}).$ Since $\{0\}$ is closed and the determinant $\det$ is continuous, $\operatorname {Aut} (E)$ is open.

Definition and proposition: the idele group of an algebra $A$ over $K$

Let $A$ be a finite-dimensional algebra over $K,$ where $K$ is global field. We consider the group of units of $\mathbb {A} _{A}.$ The map $x\mapsto x^{-1}$ is generally not continuous with the subset-topology. Therefore, the group of units is not a topological group in general. On $\mathbb {A} _{A}^{\times }$ we install the topology we defined in the section about the group of units of a topological ring. With this topology, we call the group of units of $\mathbb {A} _{A}$ the idele group $\mathbb {A} _{A}^{\times }.$ The elements of the idele group are called idele of $A.$

Let $\alpha$ be a finite subset of $A,$ containing a basis of $A$ over $K.$ For each finite place $v$ of $K,$ we call $\alpha _{v}$ the ${\mathcal {O}}_{v}$-modul generated by $\alpha$ in $A_{v}.$ As before, there exists a finite subset $P_{0}$ of the set of all places, containing $P_{\infty },$ so that it stands for all $v\notin P_{0},$ that $\alpha _{v}$ is a compact subring of $A_{v}.$ Furthermore, $\alpha _{v}$ contains the group of units of $A_{v}.$ In addition to that, it stands, that $A_{v}^{\times }$ is an open subset of $A_{v}$ for each $v$ and that the map $x\mapsto x^{-1}$ is continuous on $A_{v}^{\times }.$ As a consequence, the function $x\mapsto (x,x^{-1})$ maps $A_{v}^{\times }$ homeomorphic on the image of this map in $A_{v}\times A_{v}.$ For each $v\notin P_{0},$ it stands, that the $\alpha _{v}^{\times }$ are the elements of $A_{v}^{\times },$ mapping in $\alpha _{v}\times \alpha _{v}$ with the function above. Therefore, $\alpha _{v}^{\times }$ is an open and compact subgroup of $A_{v}^{\times }.$

Proposition: alternative characterisation of the idele group

We consider the same situation as before. Let $P$ be a finite subset of the set of all places containing $P_{0}.$ It stands, that

$\mathbb {A} _{A}(P,\alpha )^{\times }:=\prod \limits _{v\in P}A_{v}^{\times }\times \prod \limits _{v\notin P}\alpha _{v}^{\times }$

is an open subgroup of $\mathbb {A} _{A}^{\times },$ where $\mathbb {A} _{A}^{\times }$ is the union of all the $\mathbb {A} _{A}(P,\alpha )^{\times }.$

Corollary: the case $A=K$

We consider the case $A=K.$ For each finite subset of the set of all places of $K,$ containing $P_{\infty },$ it stands, that the group

$\mathbb {A} _{K}(P)^{\times }=\prod \limits _{v\in P}K_{v}^{\times }\times \prod \limits _{v\notin P}{\mathcal {O}}_{v}^{\times }$

is an open subgroup of $\mathbb {A} _{K}^{\times }=I_{K}.$ Furthermore, it stands, that $I_{K}$ is the union of all these subgroups $\mathbb {A} _{K}(P)^{\times }.$

Norm on the idele group

We want to transfer the trace and the norm from the adele ring to the idele group. It turns out, that 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 $\alpha$ be in $I_{K}.$ It stands, that $\operatorname {con} _{L/K}(\alpha )\in I_{L}$ and therefore, we have in injective group homomorphism

$\operatorname {con} _{L/K}:I_{K}\hookrightarrow I_{L}.$

Since $\alpha$ is in $I_{L},$ in particular $\alpha$ is invertible, $N_{L/K}(\alpha )$ is invertible too, because $(N_{L/K}(\alpha ))^{-1}=N_{L/K}(\alpha ^{-1}).$ Therefore, it stands, that $N_{L/K}(\alpha )\in I_{K}.$ As a consequence, the restriction of the norm-function introduces the following function:

$N_{L/K}:I_{L}\rightarrow I_{K}.$

This function is continuous and fulfils the properties of the lemma about the properties from the trace and the norm.

Properties of the idele group

Lemma: $K^{\times }$ is a discrete subgroup of $I_{K}$

The group of units of the global field $K$ can be embedded diagonal in the idele group $I_{K,S}:$

${\begin{aligned}&K^{\times }\hookrightarrow I_{K,S},\\&a\mapsto (a,a,a,\dotsc ).\end{aligned}}$

Since $K^{\times }$ is a subset of $K_{v}^{\times }$ for all $v,$ the embedding is well-defined and injective.

Furthermore, it stands, that $K^{\times }$ is discrete and closed in $I_{K}.$ This statement can be proved with the same methods like the corresponding statement about the adele ring.

Corollary $A^{\times }$ is a discrete subgroup of $\mathbb {A} _{A}^{\times }.$

Definition: idele class group

In number theory, we can define the ideal class group for a given algebraic number field. In analogy to the ideal class group, we call the elements of $K^{\times }$ in $I_{K}$ principal ideles of $I_{K}.$ The quotient group $C_{K}:=I_{K}/K^{\times }$ is the so-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: Since $K^{\times }$ is closed in $I_{K},$ it follows, that $C_{K}$ is a locally compact, topological group and a Hausdorff space.

Let $L/K$ be a finite field extension of global fields. The embedding $I_{K}\hookrightarrow I_{L}$ induces an injective map on the idele class groups:

${\begin{aligned}C_{K}&\hookrightarrow C_{L},\\\alpha K^{\times }&\mapsto \alpha L^{\times }.\end{aligned}}$

This function is well-defined, because the injection $I_{K}\hookrightarrow I_{L}$ obviously maps $K^{\times }$ onto a subgroup of $L^{\times }.$

Theorem: the idele group is a locally compact, topological group

For each subset $S$ of the set of all places, $I_{K,S}$ is a locally compact, topological group.

Remark: In general, $I_{K,S}$ equipped with the subset topology is not a topological group, because the inversion function isn't continuous.

The local compactness follows from the descriptions of the idele group as a restricted product. The fact, that the idele group is a topological group follows from considerations about the group of units of a topological ring.

Since the idele group is a locally compact group, there exists a Haar measure $d^{\times }x$ on it. This can be normalised, so that $\textstyle \int _{I_{K,{\text{fin}}}}\mathbf {1} _{\widehat {\mathcal {O}}}\,d^{\times }x=1.$ This is the normalisation used for the finite places. In this equations, $I_{K,{\text{fin}}}$ is the finite idele group, meaning the group of units of the finite adele ring. For the infinite places, we use the multiplicative lebesgue measure $\textstyle {\frac {dx}{|x|}}.$

A neighbourhood system of $1$ in $\mathbb {A} _{K}^{\times }(P_{\infty })$ is a neighbourhood system of $1$ in $I_{K}.$ Alternatively, we can take all sets of the form:

$\prod _{v}U_{v},$

where $U_{v}$ is a neighbourhood of $1$ in $K_{v}^{\times }$ and $U_{v}={\mathcal {O}}_{v}^{\times }$ for almost all $v.$

Definition: absolute value on $I_{K}$ and the set of the $1$-idele of $K$

Let $K$ be a global field. We define an absolute value function on the idele group: For a given idele $\alpha =(\alpha _{v})_{v},$ we define:

$|\alpha |:=\prod _{v}|\alpha _{v}|_{v}.$

Since $\alpha \in I_{K},$ this product is finite and therefore well-defined. This definition can be extended onto the whole adele ring by allowing infinite products. This means, we consider convergence in $(\mathbb {R} ,|\cdot |_{\infty }).$ These infinite products are $0,$ so that the absolute value function is zero on $\mathbb {A} _{K}\setminus I_{K}.$ In the following, we will write $|\cdot |$ for this function on $\mathbb {A} _{K}$ respectively $I_{K}.$

It stands, that the absolute value function is a continuous group homomorphism, which means that the map $|\cdot |:I_{K}\rightarrow \mathbb {R} _{>0}$ is a continuous group homomorphism.

Proof: Let $\alpha$ and $\beta$ be in $I_{K}.$ It stands:

${\begin{aligned}|\alpha \cdot \beta |&=\prod _{v}|(\alpha \cdot \beta )_{v}|_{v}\\&=\prod _{v}|\alpha _{v}\cdot \beta _{v}|_{v}\\&=\prod _{v}(|\alpha _{v}|_{v}\cdot |\beta _{v}|_{v})\\&=\left(\prod _{v}|\alpha _{v}|_{v}\right)\cdot \left(\prod _{v}|\beta _{v}|_{v}\right)\\&=|\alpha |\cdot |\beta |,\end{aligned}}$

where we use that all products are finite. The map is continuous which can be seen using an argument dealing with sequences. This reduces the problem to the question, whether the absolute value function is continuous on the local fields $K_{v}.$ However, this is clear, because of the reverse triangle inequality.

We define the set of the $1$-idele, $\mathbb {A} _{K}^{1},$ as the following:

$\mathbb {A} _{K}^{1}:=\{x\in I_{K}:|x|=1\}=\ker(|\cdot |).$

It stands, that $\mathbb {A} _{K}^{1}$ is a subgroup of $I_{K}.$ In literature, the term $I_{K}^{1}$ is used as a synonym for the set of the $1$-Idele. We will use $\mathbb {A} _{K}^{1}$ in the following.

It stands, that $\mathbb {A} _{K}^{1}$ is a closed subset of $\mathbb {A} _{K},$ because $\mathbb {A} _{K}^{1}=|\cdot |^{-1}(\{1\}).$

The $\mathbb {A} _{K}$-topology on $\mathbb {A} _{K}^{1}$ equals the subset-topology of $I_{K}$ on $\mathbb {A} _{K}^{1}.$

Theorem: Artin's product formula

Let $K$ be a global field. The homomorphism $|\cdot |:z\mapsto |z|$ of $I_{K}$ to $\mathbb {R} _{>0}$ fulfils: $K^{\times }\subset \ker(|\cdot |).$ In other words, it stands, that $|k|=1$ for all $k\in K^{\times }.$ Artin's product formula says, that $K^{\times }$ is a subset of $\mathbb {A} _{K}^{1}.$

Proof: There are many proofs for the product formula. The one shown in the following is based on Neukirch 2007, p. 195. In the case of an algebraic number field, the main idea is to reduce the problem to the case $K=\mathbb {Q} .$ The case of a global function field can be proved similarly.

Let $a$ be in $K^{\times }.$ We have to show, that

$\prod _{v}|a|_{v}=1.$

It stands, that $v(a)=0$ and therefore $|a|_{v}=1$ for each $v\nmid \infty ,$ for which the corresponding prime ideal ${\mathfrak {p}}_{v}$ does not divide the principal ideal $(a).$ This is valid for almost all ${\mathfrak {p}}_{v}.$

It stands:

${\begin{aligned}\prod _{v}|a|_{v}&=\prod _{p\leq \infty }\prod _{v\mid p}|a|_{v}\\&=\prod _{p\leq \infty }\prod _{v\mid p}|N_{K_{v}/\mathbb {Q} _{p}}(a)|_{p}\\&=\prod _{p\leq \infty }|N_{K/\mathbb {Q} }(a)|_{p}.\end{aligned}}$

Note that in going from line 1 to line 2, we used the identity $|a|_{w}=|N_{L_{w}/K_{v}}(a)|_{v},$ where $v$ is a place of $K$ and $w$ is a place of $L,$ lying above $v.$ Going from line 2 to line 3, we use a property from the norm. We note, that the norm is in $\mathbb {Q} .$ Therefore, without loss of generality, we can assume that $a\in \mathbb {Q} .$ Then $a$ possesses a unique integer factorisation:

$a=\pm \prod _{p<\infty }p^{v_{p}},$

where $v_{p}\in \mathbb {Z}$ is $0$ for almost all $p.$ Due to Ostrowski's theorem every absolute values on $\mathbb {Q}$ is equivalent to either the usual real absolute value $|\cdot |_{\infty }$ or a $p$-adic absolute value, we can conclude, that

${\begin{aligned}|a|&=\left(\prod _{p<\infty }|a|_{p}\right)\cdot |a|_{\infty }\\&=\left(\prod _{p<\infty }p^{-v_{p}}\right)\cdot \left(\prod _{p<\infty }p^{v_{p}}\right)\\&=1,\end{aligned}}$

which is the desired result. In the mathematical literature many more proofs of the product formula can be found.

Theorem: characterisation of $\mathbb {A} _{\operatorname {End} (E)}^{\times }$

Let $E$ be a $n$-dimensional vector-space over $K.$ Define $A:=\operatorname {End} (E).$ Furthermore, let $a$ be in $\mathbb {A} _{A}.$ We obtain the following equivalent statements:

• $a\in \mathbb {A} _{A}^{\times },$
• $\det(a)\in \mathbb {A} _{K}^{\times },$
• $x\mapsto ax$ is an automorphism of $\mathbb {A} _{E}.$

If one of the three points above is true, we can conclude that $|a|=|\det(a)|.$ Moreover, it stands, that the maps $a\mapsto \det(a)$ and $a\mapsto |\det(a)|$ are homomorphism of $\mathbb {A} _{A}^{\times }$ to $\mathbb {A} _{K}^{\times }$ respectively to $\mathbb {R} _{>0}.$

Corollary: Let $A$ be a finite-dimensional algebra over $K$ und let $a$ be in $\mathbb {A} _{A}.$ We obtain the following equivalent statements:

• $a\in \mathbb {A} _{A}^{\times },$
• $N_{A/K}(a)\in \mathbb {A} _{K}^{\times },$
• $x\mapsto ax$ is an automorphism of the additive group $\mathbb {A} _{A}.$

If one of the three points above is true, we can conclude that $|a|=|N_{A/K}(a)|.$ Moreover, it stands, that the maps $a\mapsto N_{A/K}(a)$ and $a\mapsto |N_{A/K}(a)|$ are homomorphism of $\mathbb {A} _{A}^{\times }$ to $\mathbb {A} _{K}^{\times }$ respectively to $\mathbb {R} _{>0}.$ Based on this statement an alternative proof of the product formula is possible.

Theorem: $K^{\times }$ is a discrete and cocompact subgroup in the set of the $1$-idele

Prior to formulate the theorem, we require the following lemma:

Lemma: Let $K$ be a global field. There exists a constant $C,$ depending only on the global field $K,$ such that for every $\alpha =(\alpha _{v})_{v}\in \mathbb {A} _{K}$ with the property $\textstyle \prod _{v}|\alpha _{v}|_{v}>C,$ there exists a $\beta \in K^{\times },$ such that $|\beta _{v}|_{v}\leq |\alpha _{v}|_{v}$ for all $v.$

Corollary: Let $K$ be a global field. Let $v_{0}$ be a place of $K$ and let $\delta _{v}>0$ be given for all $v\neq v_{0}$ with the property $\delta _{v}=1$ for almost all $v.$ Then, there exists a $\beta \in K^{\times },$ so that $|\beta |\leq \delta _{v}$ for all $v\neq v_{0}.$

Proof: Let $C$ be the constant form of the prior lemma. Let $\pi _{v}$ be a uniformizing element of ${\mathcal {O}}_{v}.$ Define the adele $\alpha =(\alpha _{v})_{v}$ via $\alpha _{v}:=\pi _{v}^{k_{v}}$ with $k_{v}\in \mathbb {Z}$ minimal, so that $|\alpha _{v}|_{v}\leq \delta _{v}$ for all $v\neq v_{0}.$ It stands, that $k_{v}=0$ for almost all $v.$ Define $\alpha _{v_{0}}:=\pi _{v_{0}}^{k_{v_{0}}}$ with $k_{v_{0}}\in \mathbb {Z} ,$ so that $\textstyle \prod _{v}|\alpha _{v}|_{v}>C.$ This works, because $k_{v}=0$ for almost all $v.$ Using the lemma above, there exists a $\beta \in K^{\times },$ so that $|\beta |_{v}\leq |\alpha _{v}|_{v}\leq \delta _{v}$ for all $v\neq v_{0}.$

Now we are ready to formulate the theorem.

Theorem: Let $K$ be a global field, then $K^{\times }$ is discrete in $\mathbb {A} _{K}^{1}$ and the quotient $\mathbb {A} _{K}^{1}/K^{\times }$ is compact.

Proof: The fact that $K^{\times }$ is discrete in $I_{K}$ implies that $K^{\times }$ is also discrete in $\mathbb {A} _{K}^{1}.$

We have to show, that $\mathbb {A} _{K}^{1}/K^{\times }$ is compact. In the following, we will outline Cassels & Fröhlich 1967's idea of proof:

It is sufficient to show, that there exists a compact set $W\subset \mathbb {A} _{K},$ such that the natural projection $\pi :W\cap \mathbb {A} _{K}^{1}\rightarrow \mathbb {A} _{K}^{1}/K^{\times }$ is surjective, because $\pi$ is continuous. Let $\alpha \in \mathbb {A} _{K}$ with the property $\prod _{v}|\alpha _{v}|_{v}>C$ be given, where $C$ is the constant of the lemma above. Define

$W:=\{\xi =(\xi _{v})_{v}:|\xi _{v}|_{v}\leq |\alpha _{v}|_{v}\quad \forall v\}.$

Obviously, $W$ is compact. Let $\beta =(\beta _{v})_{v}$ be in $\mathbb {A} _{K}^{1}.$ We show, that there exists an $\eta \in K^{\times },$ so that $\eta \beta \in W.$ It stands, that

$\prod _{v}|\beta _{v}|_{v}=1,$

and therefore

$\prod _{v}|\beta _{v}^{-1}|_{v}=1.$

It follows, that

$\prod _{v}|\beta _{v}^{-1}\alpha _{v}|_{v}=\prod _{v}|\alpha _{v}|_{v}>C.$

Because of the lemma, there exists an $\eta \in K^{\times },$ such that $|\eta |_{v}\leq |\beta _{v}^{-1}\alpha _{v}|_{v}$ for all $v,$ and therefore $\eta \beta \in W.$ This ends the proof of the theorem.

Theorem: some isomorphisms in case $K=\mathbb {Q}$

In case $K=\mathbb {Q} ,$ there is a canonical isomorphism $\mathbb {A} _{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times }\cong {\widehat {\mathbb {Z} }}^{\times }.$ Furthermore, ${\widehat {\mathbb {Z} }}^{\times }\times \{1\}$ is a set of representatives of $\mathbb {A} _{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times },$ that means, that $({\widehat {\mathbb {Z} }}^{\times }\times \{1\})\mathbb {Q} ^{\times }=\mathbb {A} _{\mathbb {Q} }^{1}.$ Additionally, the absolute value function induces the following isomorphisms of topological groups:

${\begin{aligned}I_{\mathbb {Q} }&\cong \mathbb {A} _{\mathbb {Q} }^{1}\times (0,\infty ){\text{ and}}\\\mathbb {A} _{\mathbb {Q} }^{1}&\cong I_{\mathbb {Q} ,{\text{fin}}}\times \{\pm 1\}.\end{aligned}}$

Consequently, ${\widehat {\mathbb {Z} }}^{\times }\times (0,\infty )$ is a set of representatives of $I_{\mathbb {Q} }/\mathbb {Q} ^{\times }.$

Proof: Consider the map $\varphi :{\widehat {\mathbb {Z} }}^{\times }\rightarrow \mathbb {A} _{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times },$ via $(a_{p})_{p}\mapsto ((a_{p})_{p},1)\mathbb {Q} ^{\times }.$ This map is well-defined, since $|a_{p}|_{p}=1$ for all $p$ and therefore $\left(\prod _{p<\infty }|a_{p}|_{p}\right)\cdot 1=1.$ Obviously, this map is a continuous, group homomorphism. To show injectivity, let $((a_{p})_{p},1)\mathbb {Q} ^{\times }=((b_{p})_{p},1)\mathbb {Q} ^{\times }.$ As a result, it exists a $q\in \mathbb {Q} ^{\times },$ so that $((a_{p})_{p},1)q=((b_{p})_{p},1).$ By considering the infinite place, we receive $q=1$ and therefore $(a_{p})_{p}=(b_{p})_{p}.$ To show the surjectivity, let $((\beta _{p})_{p},\beta _{\infty })\mathbb {Q} ^{\times }$ be in $\mathbb {A} _{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times }.$ The absolute value of this element is $1$ and therefore $\textstyle |\beta _{\infty }|_{\infty }={\frac {1}{\prod _{p}|\beta _{p}|_{p}}}\in \mathbb {Q} .$ It follows, that $\beta _{\infty }\in \mathbb {Q} .$ It stands, that $\textstyle ((\beta _{p})_{p},\beta _{\infty })\mathbb {Q} ^{\times }=(({\frac {\beta _{p}}{\beta _{\infty }}})_{p},1)\mathbb {Q} ^{\times }$ and therefore the map $\varphi$ is surjective, since $\textstyle {\frac {|\beta _{p}|_{p}}{|\beta _{\infty }|_{p}}}=1$ für alle $p.$ The other isomorphisms are given by: $\psi :I_{\mathbb {Q} }\rightarrow \mathbb {A} _{\mathbb {Q} }^{1}\times (0,\infty ),$ via $a=(a_{p})_{p\leq \infty }\mapsto ((a_{p})_{p<\infty },a_{\infty }/|a|,|a|)$ and ${\widetilde {\psi }}:I_{\mathbb {Q} ,{\text{fin}}}\times \{\pm 1\}\rightarrow \mathbb {A} _{\mathbb {Q} }^{1},$ via $(a_{\text{fin}},\varepsilon )\mapsto (a_{\text{fin}},\varepsilon |a_{\text{fin}}|^{-1}).$

Theorem: relation between ideal class group and idele class group

For an algebraic number field $K,$ we define $C_{K,{\text{fin}}}:=I_{K,{\text{fin}}}/K^{\times }.$ It stands:

${\begin{aligned}J_{K}&\cong I_{K,{\text{fin}}}/{\widehat {\mathcal {O}}}^{\times }\qquad \quad {\text{and}}\\\operatorname {Cl} _{K}&\cong C_{K,{\text{fin}}}/{\widehat {\mathcal {O}}}^{\times }K^{\times }\qquad {\text{and}}\\\operatorname {Cl} _{K}&\cong C_{K}/({\widehat {\mathcal {O}}}^{\times }\times \prod _{v\mid \infty }K_{v}^{\times })K^{\times }.\end{aligned}}$

Here, $J_{K}$ is the group of fractional ideals of $K$ and $\operatorname {Cl} _{K}$ is the ideal class group of the Dedekind domain ${\mathcal {O}}:=\{\alpha \in K:\,\,{\text{ there exists a monic polynomial }}\,\,f\in \mathbb {Z} [x]:f(\alpha )=0\}.$ It stands, that ${\mathcal {O}}$ is the ring of integers of the algebraic number field $K.$ Per definition, it stands, that $\operatorname {Cl} _{K}=J_{K}/K^{\times }.$

Proof: In the following, we will use the fact, that for an algebraic number field $K,$ there is a one-to-one correspondence between the finite places of $K$ and the prime ideals of ${\mathcal {O}},$ which are different from $0:$

Let $v$ be a finite place of $K$ and let $|\cdot |_{v}$ be a representative of the equivalence class $v.$ Define

${\mathfrak {p}}_{v}:=\{x\in {\mathcal {O}}:|x|_{v}<1\}.$

Then is ${\mathfrak {p}}_{v}$ is a prime ideal in ${\mathcal {O}}.$ The map $v\mapsto {\mathfrak {p}}_{v}$ is a bijection between the set of the finite places of $K$ and the set of all prime ideals ${\mathfrak {p}}\neq 0$ of ${\mathcal {O}}.$ The inversion map is given by the following:

A given prime ideal ${\mathfrak {p}}$ is mapped to the valuation $v_{\mathfrak {p}},$ given by

${\begin{aligned}v_{\mathfrak {p}}(x)&:=\max\{k\in \mathbb {N} _{0}:x\in {\mathfrak {p}}^{k}\}\quad \forall x\in {\mathcal {O}}^{\times },\\v_{\mathfrak {p}}\left({\frac {x}{y}}\right)&:=v_{\mathfrak {p}}(x)-v_{\mathfrak {p}}(y)\quad \forall x,y\in {\mathcal {O}}^{\times }.\end{aligned}}$

Now, we can prove the theorem. The following map is well-defined:

${\begin{aligned}(\cdot ):\,&I_{K,{\text{fin}}}\rightarrow J_{K},\\&\alpha =(\alpha _{v})_{v}\mapsto \prod _{v\nmid \infty }{\mathfrak {p}}_{v}^{v(\alpha _{v})},\end{aligned}}$

where ${\mathfrak {p}}_{v}$ is the corresponding prime ideal to the place $v.$ The map $(\cdot )$ is obviously a surjective homomorphism. It stands, that $\ker((\cdot ))={\widehat {\mathcal {O}}}^{\times }.$ The first isomorphism of the theorem follows now using the fundamental theorem on homomorphism.

Now, we divide both sides from our map $(\cdot )$ by $K^{\times }.$ This is possible, because

${\begin{aligned}(\alpha )&=((\alpha ,\alpha ,\dotsc ))\\&=\prod _{v\nmid \infty }{\mathfrak {p}}_{v}^{v(\alpha )}\\&=(\alpha )\quad {\text{ for all }}\alpha \in K^{\times }.\end{aligned}}$

Please, note the abuse of notation: On the left hand side in line 1 of this chain of equations, $(\cdot )$ stands for the map defined above. Later, we use the embedding of $K^{\times }$ into $I_{K,{\text{fin}}}.$ In line 2, we use the definition of the map. Finally, we use the fact, that the ring of integers ${\mathcal {O}}$ is a Dedekind domain and therefore each ideal, in particular the principal ideal $(\alpha ),$ can be written as a product of prime ideals. In other words, the map $(\cdot )$ is a $K^{\times }$-equivariant group homomorphism. As a consequence, the map above induces a surjective homomorphism

${\begin{aligned}\varphi :\,&C_{K,{\text{fin}}}\rightarrow \operatorname {Cl} _{K},\\&\alpha K^{\times }\mapsto (\alpha )K^{\times }.\end{aligned}}$

To prove the second isomorphism, we have to show, that $\ker(\varphi )={\widehat {\mathcal {O}}}^{\times }K^{\times }.$ Consider $\xi =(\xi _{v})_{v}$ be in ${\widehat {\mathcal {O}}}^{\times }.$ Then $\textstyle \varphi (\xi K^{\times })=\prod _{v}{\mathfrak {p}}_{v}^{v(\xi _{v})}K^{\times }=K^{\times },$ because $v(\xi _{v})=0$ for all $v.$ On the other hand, consider $\xi K^{\times }\in C_{K,{\text{fin}}}$ with $\varphi (\xi K^{\times })={\mathcal {O}}K^{\times },$ which allows to write $\textstyle \prod _{v}{\mathfrak {p}}_{v}^{v(\xi _{v})}K^{\times }={\mathcal {O}}K^{\times }.$ As a consequence, there exists a representative, such that: $\textstyle \prod _{v}{\mathfrak {p}}_{v}^{v(\xi '_{v})}={\mathcal {O}}.$ Consequently, $\xi '\in {\widehat {\mathcal {O}}}^{\times }$ and therefore $\xi K^{\times }=\xi 'K^{\times }\in {\widehat {\mathcal {O}}}^{\times }K^{\times }.$ We have proved the second isomorphism of the theorem.

For the last isomorphism of the theorem, we note, that the map $\varphi$ induces a surjective group homomorphism

${\widetilde {\varphi }}:C_{K}\rightarrow \operatorname {Cl} _{K}$

with $\textstyle \ker({\widetilde {\varphi }})=({\widehat {\mathcal {O}}}^{\times }\times \prod _{v\mid \infty }K_{v}^{\times })K^{\times }.$ This ends the proof of the theorem.

Remark: The map $(\cdot )$ is continuous, if we install the following topology on the sets which are considered: On $I_{K,{\text{fin}}},$ we install the idele topology. On $J_{K},$ we install the discrete topology. Continuity follows, since we can show, that $(\{{\mathfrak {a}}\})^{-1}$ is open for each ${\mathfrak {a}}\in J_{K}.$ It stands, that $(\{{\mathfrak {a}}\})^{-1}=\alpha {\widehat {\mathcal {O}}}^{\times }$ is open, where $\alpha =(\alpha _{v})_{v}\in \mathbb {A} _{K,{\text{fin}}},$ so that $\textstyle {\mathfrak {a}}=\prod _{v}{\mathfrak {p}}_{v}^{v(\alpha _{v})}.$

Theorem: decomposition of $I_{K}$ and $C_{K}$

Let $K$ be a global field. If the characteristic of $K$ is greater than zero, it stands, that $I_{K}\cong \mathbb {A} _{K}^{1}\times \mathbb {Z} .$ If the characteristic of $K$ equals $0,$ then $I_{K}\cong M\times \mathbb {A} _{K}^{1},$ where $M$ is a closed subgroup of $I_{K},$ which is isomorph to $\mathbb {R} _{>0}.$ Furthermore, it stands, that

$C_{K}=I_{K}/K^{\times }\cong \mathbb {A} _{K}^{1}/K^{\times }\times N,$

where $N\cong \mathbb {Z} ,$ if $\operatorname {char} (K)>0$ or $N\cong \mathbb {R} _{>0},$ if $\operatorname {char} (K)=0.$

Proof: Let $p>1$ be the characteristic of $K.$ For each place $v$ of $K$ stands, that the characteristic of $K_{v}$ equals $p,$ so that for each $x\in K_{v}^{\times },$ the element $|x|_{v}$ is in the subgroup of $\mathbb {R} _{>0},$ generated by $p.$ It follows, that for each $z\in I_{K}$ the number $|z|$ is in the subgroup of $\mathbb {R} _{>0},$ generated by $p.$ It follows, that the image of the homomorphism $z\mapsto |z|$ is a discrete subgroup of $\mathbb {R} _{>0},$ generated by $p.$ Since this group isn't trivial, meaning $\operatorname {im} (|\cdot |)\neq \{1\},$ it is generated by a $Q=p^{N}$ for a $N\in \mathbb {N} .$ Choose $z_{1}\in I_{K},$ so that $|z_{1}|=Q,$ it follows, that $I_{K}$ is the direct product of $\mathbb {A} _{K}^{1}$ and the subgroup generated by $z_{1}.$ This subgroup is discrete and isomorphic to $\mathbb {Z} .$

If the characteristic of $K$ equals $0,$ we write $z(\lambda )$ for idele $(z_{v})_{v},$ fulfilling $z_{v}=1$ for all finite places $v$ and $z_{w}=\lambda$ for all infinite places $w$ of $K,$ where $\lambda \in \mathbb {R} _{>0}.$ It stands, that the map $\lambda \mapsto z(\lambda )$ is an isomorphism of $\mathbb {R} _{>0}$ in a closed subgroup $M$ of $I_{K}$ and it stands, that $I_{K}\cong M\times \mathbb {A} _{K}^{1}.$ The isomorphism is given by multiplication:

${\begin{aligned}\varphi :M\times \mathbb {A} _{K}^{1}&\rightarrow I_{K},\\((\alpha _{v})_{v},(\beta _{v})_{v})&\mapsto (\alpha _{v}\beta _{v})_{v}\end{aligned}}$

Obviously, $\varphi$ is a homomorphism. To show injectivity, let $(\alpha _{v}\beta _{v})_{v}=1.$ Since $\alpha _{v}=1$ for $v<\infty ,$ it stands that $\beta _{v}=1$ for $v<\infty .$ Moreover, it exists a $\lambda \in \mathbb {R} _{>0},$ so that $\alpha _{v}=\lambda$ for $v\mid \infty .$ Therefore, $\beta _{v}=\lambda ^{-1}$ for $v\mid \infty .$ In addition to that, since $\textstyle \prod _{v}|\beta _{v}|_{v}=1,$ it follows, that $\lambda ^{n}=1,$ where $n$ is the number of infinite places of $K.$ As a consequence, it stands, that $\lambda =1$ und therefore $\varphi$ is injectiv. To show surjectivity, let $\gamma =(\gamma _{v})_{v}$ be in $I_{K}.$ We define $\lambda :=|\gamma |^{1/n}$ and furthermore, we define $\alpha _{v}=1$ for $v<\infty$ and $\alpha _{v}=\lambda$ for $v\mid \infty .$ Define $\textstyle \beta ={\frac {\gamma }{\alpha }}.$ It stands, that $\textstyle |\beta |={\frac {|\gamma |}{|\alpha |}}={\frac {\lambda ^{n}}{\lambda ^{n}}}=1.$Therefore, $\varphi$ is surjective.

The other equations follow similarly.

Theorem: characterisation of the idele group

Let $K$ be an algebraic number field. There exists a finite subset $S$ of the set of all places, such that

$I_{K}=(I_{K,S}\times \prod _{v\notin S}{\mathcal {O}}_{v}^{\times })K^{\times }=\left(\prod _{v\in S}K_{v}^{\times }\times \prod _{v\notin S}{\mathcal {O}}_{v}^{\times }\right)K^{\times }.$

Proof: In this proof, we will use the fact, that the class number of an algebraic number field is finite. Let ${\mathfrak {a}}_{1},\dotsc ,{\mathfrak {a}}_{h}$ be the ideals, representing the $h$ classes in $\operatorname {Cl} _{K}.$ These ideals are generated of a finite number of prime ideals ${\mathfrak {p}}_{1},\dotsc ,{\mathfrak {p}}_{n}.$ Let $S$ be a finite set of places, which includes the infinite places of $K$ and those finite places corresponding to the prime ideals ${\mathfrak {p}}_{1},\dotsc ,{\mathfrak {p}}_{n}.$

We consider the isomorphism

$I_{K}/\left(\prod _{v\nmid \infty }{\mathcal {O}}_{v}^{\times }\times \prod _{v\mid \infty }K_{v}^{\times }\right)\cong J_{K},$

which is induced by $\textstyle (\alpha _{v})_{v}\mapsto \prod _{v<\infty }{\mathfrak {p}}_{v}^{v(\alpha _{v})}.$

In the following, we prove the statement for the finite places, because at the infinite places the statement is obvious. The inclusion ″$\supset$″ is obvious. Let $\alpha \in I_{K,{\text{fin}}}.$ The corresponding ideal $\textstyle (\alpha )=\prod _{v\nmid \infty }{\mathfrak {p}}_{v}^{v(\alpha _{v})}$ belongs to a class ${\mathfrak {a}}_{i}K^{\times },$ meaning $(\alpha )={\mathfrak {a}}_{i}(a)$ for a principal ideal $(a).$ The idele $\alpha '=\alpha a^{-1}$ maps to the ideal ${\mathfrak {a}}_{i}$ under the map $I_{K,{\text{fin}}}\rightarrow J_{K}.$ That means $\textstyle {\mathfrak {a}}_{i}=\prod _{v\nmid \infty }{\mathfrak {p}}_{v}^{v(\alpha '_{v})}.$ Since the prime ideals in ${\mathfrak {a}}_{i}$ are in $S,$ it follows $v(\alpha '_{v})=0$ for all $v\notin S,$ that means $\alpha '_{v}\in {\mathcal {O}}_{v}^{\times }$ for all $v\notin S.$ It follows, that $\alpha '=\alpha a^{-1}\in I_{K,S},$ therefore $\alpha \in I_{K,S}K^{\times }.$

Applications

Finiteness of the class number of an algebraic number field

In this section, we want to show that the class number of an algebraic number field is finite. Of course, there are many different proofs of this statement. In the proof of the characterisation of the idele group, we already used this fact.

Theorem: (finiteness of the class number of an algebraic number field) Let $K$ be an algebraic number field. It stands, that $|\operatorname {Cl} _{K}|<\infty .$

Proof: The map $\textstyle \mathbb {A} _{K}^{1}\rightarrow J_{K},((\alpha _{v})_{v<\infty },(\alpha _{v})_{v\mid \infty })\mapsto \prod _{v<\infty }{\mathfrak {p}}_{v}^{v(\alpha _{v})}$ is surjective and therefore $\operatorname {Cl} _{K}$ is the continuous image of the compact set $\mathbb {A} _{K}^{1}/K^{\times }.$ Thus, $\operatorname {Cl} _{K}$ is compact. In addition, $\operatorname {Cl} _{K}$ is discrete, thus it is 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.

Group of units and Dirichlet's unit theorem

Some definitions

Let $K$ be a global field. Let $P$ be a finite subset of the set of all places, containing $P_{\infty }.$ Define

${\begin{aligned}\Omega (P)&:=\prod _{v\in P}K_{v}^{\times }\times \prod _{v\notin P}{\mathcal {O}}_{v}^{\times }=(\mathbb {A} _{K}(P))^{\times },\\E(P)&:=K^{\times }\cap \Omega (P).\end{aligned}}$

It is valid, that $E(P)$ is a subgroup of $K^{\times },$ containing all elements $\xi \in K^{\times },$ which fulfil $v(\xi )=0$ for all $v\notin P.$ Since $K^{\times }$ is discrete in $I_{K},$ it follows, that $E(P)$ is a discrete subgroup of $\Omega (P)$ and with the same argument, $E(P)$ is discrete in $\Omega _{1}(P):=\Omega (P)\cap \mathbb {A} _{K}^{1}.$

An alternative definition of $E(P)$ is, that $E(P)=K(P)^{\times },$ where $K(P)$ is a subring of $K$ defined by $\textstyle K(P):=K\cap \left(\prod _{v\in P}K_{v}\times \prod _{v\notin P}{\mathcal {O}}_{v}\right).$ As a consequence, $K(P)$ contains all elements $\xi \in K,$ which fulfil $v(\xi )\geq 0$ for all $v\notin P.$

Let $0<c<C<\infty .$ It stands, that the set $\{\eta \in E(P):|\eta _{v}|_{v}=1\quad \forall v\notin P{\text{ and }}c\leq |\eta _{v}|_{v}\leq C\quad \forall v\in P\}$ is finite. In order to prove this statement, we define

$W:=\{(\alpha _{v})_{v}:|\alpha _{v}|_{v}=1\quad \forall v\notin P{\text{ and }}c\leq |\alpha _{v}|_{v}\leq C\quad \forall v\in P\}.$

It stands, that $W$ is compact and the set described above is the intersection of $W$ with the discrete subgroup $K^{\times }$ in $I_{K}.$ The finiteness follows from these arguments.

Define $F:=\{\xi \in K:|\xi |_{v}\leq 1\,\,{\text{ for all places }}\,\,v\,\,{\text{ of }}\,\,K\}=\{\xi \in K:|\xi |_{v}=1\,\,{\text{ for all places }}\,\,v\,\,{\text{ of }}\,\,K\}\cup \{0\},$ where the second equal sign is true because of Artin's product formula. Define $E:=F\setminus \{0\}.$ It stands

$E\subset E(P)$

for each finite subset of the set of all places of $K$ containing $P_{\infty }.$

Theorem: roots of unity of $K$

It stands, that $E$ is a finite, cyclic group, containing all roots of unity of $K.$ Furthermore, it is valid, that $E$ is the group of all roots of unity of $K.$

Proof: It stands, that $F=K\cap \{(x_{v})_{v}\in \mathbb {A} _{K}:|x_{v}|_{v}\leq 1\,\,{\text{ for all }}\,\,v\}.$ The last set is compact. Furthermore, $K$ is discrete in $\mathbb {A} _{K},$ thus $F$ is finite, because $F$ is a subset of a compact set and $F$ is discrete. Because of Artin's product formula, it stands for all $\xi \in E,$ that $|\xi |_{v}=1$ for all $v.$ It follows, that $E$ is a finite subgroup of $K^{\times }.$ Since $K$ is a field, $E$ is cyclic. It is obvious that each root of unity of $K$ is in $E,$ since roots of unity of $K$ all have absolute value $1$ and therefore have valuation $0.$ Suppose, that there exists a $\xi \in E,$ which isn't a root of unity of $K.$ It follows, that $\xi ^{n}\neq 1$ for all $n\in \mathbb {N} .$ This contradicts the finiteness of the group $E.$

Theorem: generalised form of Dirichlet's unit theorem

Let the situation be as above. It stands that $E(P)$ is the direct product of the group $E$ and a group being isomorphic to $\mathbb {Z} ^{s}.$ We note, that $s=0,$ if $P=\emptyset$ and that $s=\operatorname {card} (P)-1=|P|-1,$ if $P\neq \emptyset .$

Theorem: Dirichlet's unit theorem

Let $K$ be an algebraic number field. It stands

${\mathcal {O}}^{\times }\cong \mu (K)\times \mathbb {Z} ^{r+s-1},$

where $\mu (K)$ is the finite, cyclic group of all roots of unity of $K$ and $r$ is the number of real embeddings of $K$ and $s$ is the number of conjugate pairs of complex embeddings of $K.$ It stands, that $[K:\mathbb {Q} ]=r+2s.$

Remark: The theorem above is a generalisation of Dirichlet's unit theorem. For an algebraic number field $K,$ define $P=P_{\infty }$ and receive the Dirichlet's unit theorem. In literature, this theorem is also called "Theorem of the units″. Of course, Dirichlet's unit theorem is older than the theorems given above and can be proved on its own. With the help of the Dirichlet's unit theorem, we can prove the compactness of $\mathbb {A} _{K}^{1}/K^{\times }$ in an alternative way.

Proof of this remark:

We already know, that $E=\mu (K).$

Furthermore, it stands, that

${\begin{aligned}E(P_{\infty })&=K^{\times }\cap \left(\prod _{v\mid \infty }K_{v}^{\times }\times \prod _{v<\infty }{\mathcal {O}}_{v}^{\times }\right)\\&\cong K^{\times }\cap \left(\prod _{v<\infty }{\mathcal {O}}_{v}^{\times }\right)\\&\cong {\mathcal {O}}^{\times }.\end{aligned}}$

In addition to that, it stands that $|P_{\infty }|=r+s.$

Approximation theorems

Theorem: weak approximation theorem

Let $|\cdot |_{n},$ $1\leq n\leq N,$ be inequivalent, non-trivial valuations of the field $K.$ Let $K_{n}:=(K,|\cdot |_{n}).$ In particular, these are topological spaces. We embed $K$ diagonal in $\textstyle \prod _{n=1}^{N}K_{n}.$ It stands, that $K$ is everywhere dense in $\textstyle \prod _{n=1}^{N}K_{n}.$ In other words, for each $\varepsilon >0$ and for each $\textstyle (\alpha _{n})_{n}\in \prod _{n=1}^{N}K_{n},$ there exists a $\xi \in K,$ such that

$|\alpha _{n}-\xi |_{n}<\varepsilon \qquad \forall n=1,\dotsc ,N.$

Theorem: strong approximation theorem

Let $K$ be a global field. Let $v_{0}$ be a place of $K.$ Define

$V:={\widehat {\prod \limits _{v\neq v_{0}}}}^{{\mathcal {O}}_{v}}K_{v}.$

Then $K$ is dense in $V.$

Remark: The global field is discrete in its adele ring. To obtain this result, we had to include all places of the global field. The strong approximation theorem tells us that, if we omit one place (or more), the property of discreteness of $K$ is turned into a denseness of $K.$

″Local–global″ and Hasse principle

Local and global:

Let $L$ be a finite extension of the global field $K.$ We define $L/K$ as the global extension. Let $v$ be a place of $K$ and let $w$ be a place of $L,$ lying above $v.$ We define the (finite) extension $L_{w}/K_{v}$ as the local extension. Where do these names come from? In order to understand this, we consider the case of a global function field, for example $K=\mathbb {C} (t),$ although this isn't a global field. Let $L/K$ be a finite extension. The elements of $L$ are algebraic functions on a Riemann surface, a global object. On the other hand, if we consider the extension $L_{w}/K_{v},$ we change from studying these functions globally to a local one, which is equivalent to consider their power series.

Theorem: Minkowski–Hasse

A quadratic form on the global field $K$ is zero, if and only if, the quadratic form is zero in each completion $K_{v}.$

Remark: This is the Hasse principle for quadratic forms. For polynomials of degree larger than 2 the Hasse principle isn't valid in general.

Remark: The idea of the local–global principle is to solve a given problem of an algebraic number field $K$ by doing so in its completions $K_{v}$ and then concluding on a solution in $K.$

Characters on the adele ring

Definition: character group

Let $G$ be a locally compact, abelian group. Define the character group ${\widehat {G}}$ of $G$ as the set of all characters of $G,$ that means the set of all continuous group homomorphisms of $G$ to $\mathbb {T} :=\{z\in \mathbb {C} :|z|=1\}.$ We give ${\widehat {G}}$ the topology of the uniform convergence on compact subsets of $G.$ It can be shown, that ${\widehat {G}}$ is also a locally compact, abelian group.

Theorem: the adele ring is self-dual

Let $K$ be a global field. The adele ring is self-dual, that means, that

$\mathbb {A} _{K}\cong {\widehat {\mathbb {A} _{K}}}:=\{\chi :\mathbb {A} _{K}\rightarrow \mathbb {T} ,\chi {\text{ is a continuous group homomorphism}}\}.$

Proof: In a first step, we show that each $K_{v}$ is self-dual by fixing one character. We exemplify this for the case $K_{v}=\mathbb {R}$ by defining $e_{\infty }:\mathbb {R} \rightarrow \mathbb {T}$ via $e_{\infty }(t):=\exp(2\pi it).$ Now we consider the map $\varphi :\mathbb {R} \rightarrow {\widehat {\mathbb {R} }},$ $s\mapsto \varphi _{s},$ with $\varphi _{s}:\mathbb {R} \rightarrow \mathbb {T} ,t\mapsto \exp(2\pi ist)$ or in other words, $\varphi _{s}=e_{\infty }(\cdot s).$ It can be shown that $\varphi$ is an isomorphism which respects topologies. In a second step the problem for the adele ring is treated by reducing it to a problem in the local coordinates.

Theorem: Algebraic and continuous dual space of the adele ring

Let $K$ be a global field and let $\chi$ be a non-trivial character of $\mathbb {A} _{K},$ which is trivial on $K.$ Let $E$ be a finite-dimensional vector-space over $K.$ Let $E^{\star }$ be its algebraic dual space and let $\mathbb {A} _{E}^{\star }$ be the algebraic dual space of $\mathbb {A} _{E}.$ Furthermore, let $\mathbb {A} _{E}'$ be the topological dual of $\mathbb {A} _{E}.$ Then the formula $\langle e,e'\rangle =\chi ([e,e^{\star }])$ for all $e\in \mathbb {A} _{E}$ determines an isomorphism $e^{\star }\mapsto e'$ of $\mathbb {A} _{E}^{\star }$ onto $\mathbb {A} _{E}',$ where $e'\in \mathbb {A} _{E}'$ and $e^{\star }\in \mathbb {A} _{E}^{\star }.$ On this occasion $\langle {\cdot },{\cdot }\rangle$ and $[{\cdot },{\cdot }]$ are the bilinear pairings on $\mathbb {A} _{E}\times \mathbb {A} _{E}'$ and on $\mathbb {A} _{E}\times \mathbb {A} _{E}^{\star }.$ Moreover, if $e^{\star }\in \mathbb {A} _{E}^{\star }$ fulfils $\chi ([e,e^{\star }])=1$ for all $e\in E,$ then it stands, that $e^{\star }\in E^{\star }.$

With the help of the characters of $\mathbb {A} _{K},$ we can do Fourier analysis on the adele ring.

Tate's thesis

John Tate determines in his thesis "Fourier analysis in number fields and Heckes Zetafunctions" results about Dirichlet L-functions by 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 to study the zeta functions and the L-functions. We can define adelic forms of these functions and we can represent these functions as integrals over the adele ring, respectively the idele group, which respect to the corresponding Haar measures. Out of this, we can show functional equations and meromorphic continuations of these functions. For illustration purposes, we provide an example. For each complex number $s\in \mathbb {C}$ with $\Re (s)>1,$ it stands

$\int _{\widehat {\mathbb {Z} }}|x|^{s}d^{\times }x=\zeta (s),$

where $d^{\times }x$ is the normalised Haar measure on $I_{\mathbb {Q} ,{\text{fin}}}$ with $d^{\times }x({\widehat {\mathbb {Z} }}^{\times })=1,$ which is extended by zero on the finite adele ring. Note that it is unique. The equation above shows, that we can write the Riemann zeta function $\zeta$ as an integral over (a subset of) the adele ring.

Automorphic forms

We consider the case $K=\mathbb {Q} .$

In newer mathematical approaches, automorphic forms are described as being a function on the group $\mathrm {GL} _{2}(\mathbb {A} _{\mathbb {Q} })$ satisfying several additional conditions. For this purpose, we define $(\mathrm {GL} _{2}(\mathbb {A} _{\mathbb {Q} }))^{1}:=\{x\in \mathrm {GL} _{2}(\mathbb {A} _{\mathbb {Q} }):|\det(x)|=1\}$ and $Z_{\mathbb {R} }$ as the centre of the group $\mathrm {GL} _{2}(\mathbb {R} ).$ It stands, that $(\mathrm {GL} _{2}(\mathbb {Q} )\backslash (\mathrm {GL} _{2}(\mathbb {A} _{\mathbb {Q} }))^{1}\cong (\mathrm {GL} _{2}(\mathbb {Q} )Z_{\mathbb {R} })\backslash \mathrm {GL} _{2}(\mathbb {A} _{\mathbb {Q} }).$ We define an automorphic form as an element of the vector-space $L^{2}((\mathrm {GL} _{2}(\mathbb {Q} )Z_{\mathbb {R} })\backslash \mathrm {GL} _{2}(\mathbb {A} _{\mathbb {Q} })).$ For studying automorphic forms, it is important to know the representations of the group $\mathrm {GL} _{2}(\mathbb {A} _{\mathbb {Q} }),$ which are described in the tensor product theorem. It is also possible to study automorphic L-function, which can be described as an integral over the group $\mathrm {GL} _{2}(\mathbb {A} _{\mathbb {Q} }).$

Further applications

A generalisation of Artin reciprocity law leads to the connection of representations of $\mathrm {GL} _{2}(\mathbb {A} _{K})$ 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 number or function field. The Artin reciprocity law, which is a high level generalisation of the Gauss quadratic reciprocity law, states that the product vanishes on the multiplicative group of the number field. Thus, we obtain the global reciprocity map of the idele class group to the abelian part of the absolute Galois group of the field.

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

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• Weil, André (1967). Basic number theory. XVIII. Berlin; Heidelberg; New York: Springer. ISBN 978-3-662-00048-9. 294 pages.
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