In mathematics, the adele ring[a] (also adelic ring, ring of adeles or ring of adèles[1]) 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,[b] 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.

## Notation

Throughout this article, ${\displaystyle K}$ is a global field. That is, ${\displaystyle K}$ is a number field (a finite extension of ${\displaystyle \mathbb {Q} }$) or a global function field (a finite extension of ${\displaystyle \mathbb {F} _{p^{r}}(t)}$).

Let ${\displaystyle v}$ be a valuation of ${\displaystyle K}$ (we assume all valuations to be non-trivial). We write ${\displaystyle K_{v}}$ for the completion of ${\displaystyle K}$ with respect to ${\displaystyle v.}$ If ${\displaystyle v}$ is discrete we write ${\displaystyle {\mathcal {O}}_{v}}$ for the valuation ring of ${\displaystyle K_{v}}$ and ${\displaystyle {\mathfrak {m}}_{v}}$ for the maximal ideal of ${\displaystyle {\mathcal {O}}_{v}.}$ If this is a principal ideal we denote the uniformizing element by ${\displaystyle \pi _{v}.}$ A non-Archimedean valuation is written as ${\displaystyle v<\infty }$ or ${\displaystyle v\nmid \infty }$ and an Archimedean valuation as ${\displaystyle v|\infty .}$

By fixing a suitable constant ${\displaystyle C>1,}$ there is a one-to-one identification of valuations and absolute values. The valuation ${\displaystyle v}$ is assigned the absolute value ${\displaystyle |\cdot |_{v},}$ defined as:

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

Conversely, the absolute value ${\displaystyle |\cdot |}$ is assigned the valuation ${\displaystyle v_{|\cdot |},}$ defined as:

${\displaystyle \forall x\in K^{\times }:\quad v_{|\cdot |}(x):=-\log _{C}(|x|).}$

A place of ${\displaystyle K}$ is a representative of an equivalence class of valuations (or absolute values) of ${\displaystyle K.}$ Places corresponding to non-Archimedean valuations are called finite, whereas places corresponding to Archimedean valuations are called infinite. The set of infinite places of a global field is finite, we denote this set by ${\displaystyle P_{\infty }.}$

## 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 ${\displaystyle K}$ 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 ${\displaystyle K,}$ the valuations are zero for almost all places, i.e. for all places except a finite number. So, the global field can be embedded in the restricted product.
• The restricted product is a locally compact space, while the Cartesian product is not. Therefore, we can't apply harmonic analysis to the Cartesian product.

### Definition

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

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

This means, that the set of the finite adeles contains all ${\displaystyle \textstyle (x_{v})_{v}\in \prod _{v<\infty }K_{v},}$ so that ${\displaystyle x_{v}\in {\mathcal {O}}_{v}}$ for almost all ${\displaystyle v.}$ Addition and multiplication are defined component-wise. In this way ${\displaystyle \mathbb {A} _{K,{\text{fin}}}}$ 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:

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

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

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

${\displaystyle \mathbb {A} _{K}:=\mathbb {A} _{K,{\text{fin}}}\times \prod _{v|\infty }K_{v}={\widehat {\prod _{v<\infty }}}^{{\mathcal {O}}_{v}}K_{v}\times \prod _{v|\infty }K_{v}.}$

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 ${\displaystyle K.}$ In the following, we write

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

although this is generally not a restricted product.

Lemma. There is a natural diagonal embedding of ${\displaystyle K}$ into its adele ring ${\displaystyle \mathbb {A} _{K}}$ given by:
${\displaystyle {\begin{cases}K\to \mathbb {A} _{K}\\a\mapsto (a,a,a,\ldots )\end{cases}}}$

Proof. This embedding is well-defined, because for each ${\displaystyle a\in K,}$ we have ${\displaystyle a\in {\mathcal {O}}_{v}^{\times }}$ for almost all ${\displaystyle v.}$ The embedding is injective, because the embedding of ${\displaystyle K}$ in ${\displaystyle K_{v}}$ is injective for each ${\displaystyle v.}$

As a consequence, we can view ${\displaystyle K}$ as a subgroup of ${\displaystyle \mathbb {A} _{K}.}$ In the following, ${\displaystyle K}$ is a subring of its adele ring. The elements of ${\displaystyle K\subset \mathbb {A} _{K}}$ are the so-called principal adeles of ${\displaystyle \mathbb {A} _{K}.}$

Let ${\displaystyle S}$ a set of places of ${\displaystyle K.}$ Define the set of the ${\displaystyle S}$-adeles of ${\displaystyle K}$ as

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

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

Furthermore, define

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

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

### The adele ring of rationals

By Ostrowski's theorem we can identify all places of ${\displaystyle \mathbb {Q} }$ with ${\displaystyle \{p\in \mathbb {N} :p{\text{ prime}}\}\cup \{\infty \},}$ where we identify the prime number ${\displaystyle p}$ with the equivalence class of the ${\displaystyle p}$-adic absolute value and ${\displaystyle \infty }$ with the equivalence class of the absolute value ${\displaystyle |\cdot |_{\infty }}$ on ${\displaystyle \mathbb {Q} ,}$ defined as follows:

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

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

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

As a consequence, the rational adele ring is

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

We denote in short

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

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

We will illustrate the difference between restricted and unrestricted product topology using a sequence in ${\displaystyle \mathbb {A} _{\mathbb {Q} }}$:

Lemma. Consider the following sequence in ${\displaystyle \mathbb {A} _{\mathbb {Q} }}$:
{\displaystyle {\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}}}
In product topology this sequences converges to ${\displaystyle x=(1,1,\ldots ).}$ 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 ${\displaystyle a=(a_{p})_{p}\in \mathbb {A} _{\mathbb {Q} }}$ and for each restricted open rectangle ${\displaystyle \textstyle U=\prod _{p\in E}U_{p}\times \prod _{p\notin E}\mathbb {Z} _{p},}$ we have the result: ${\displaystyle \textstyle {\frac {1}{p}}-a_{p}\notin \mathbb {Z} _{p}}$ for ${\displaystyle a_{p}\in \mathbb {Z} _{p}}$ and therefore ${\displaystyle \textstyle {\frac {1}{p}}-a_{p}\notin \mathbb {Z} _{p}}$ for all ${\displaystyle p\notin F.}$ As a result, it stands, that ${\displaystyle (x_{n}-a)\notin U}$ for almost all ${\displaystyle n\in \mathbb {N} .}$ In this consideration, ${\displaystyle E}$ and ${\displaystyle F}$ 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).

### Alternative definition for number fields

Definition (profinite integers). Set

${\displaystyle {\widehat {\mathbb {Z} }}:=\varprojlim _{n}\mathbb {Z} /n\mathbb {Z} ,}$

that means ${\displaystyle {\widehat {\mathbb {Z} }}}$ is the profinite completion of the rings ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$ with the partial order ${\displaystyle n\geq m\Leftrightarrow m|n.}$ With the Chinese Remainder Theorem, it can be shown that the profinite integers are isomorphic to the product of the p-adic integers:

${\displaystyle {\widehat {\mathbb {Z} }}\cong \prod _{p{\text{ prime}}}\mathbb {Z} _{p}.}$
Lemma. Define ${\displaystyle \mathbb {A} _{\mathbb {Z} }:={\widehat {\mathbb {Z} }}\times \mathbb {R} .}$ Then we have an algebraic isomorphism ${\displaystyle \mathbb {A} _{\mathbb {Q} }\cong \mathbb {A} _{\mathbb {Z} }\otimes _{\mathbb {Z} }\mathbb {Q} .}$

Proof. We have:

${\displaystyle \mathbb {A} _{\mathbb {Z} }\otimes _{\mathbb {Z} }\mathbb {Q} =\left({\widehat {\mathbb {Z} }}\times \mathbb {R} \right)\otimes _{\mathbb {Z} }\mathbb {Q} \cong \left({\widehat {\mathbb {Z} }}\otimes _{\mathbb {Z} }\mathbb {Q} \right)\times (\mathbb {R} \otimes _{\mathbb {Z} }\mathbb {Q} )\cong \left({\widehat {\mathbb {Z} }}\otimes _{\mathbb {Z} }\mathbb {Q} \right)\times \mathbb {R} .}$

As a result, it is sufficient to show that ${\displaystyle \mathbb {A} _{\mathbb {Q} ,{\text{fin}}}={\widehat {\mathbb {Z} }}\otimes _{\mathbb {Z} }\mathbb {Q} .}$ We will use the universal property of the tensor product. Define a ${\displaystyle \mathbb {Z} }$-bilinear function

${\displaystyle {\begin{cases}\Psi :{\widehat {\mathbb {Z} }}\times \mathbb {Q} \to \mathbb {A} _{\mathbb {Q} ,{\text{fin}}}\\((a_{p})_{p},q)\mapsto (a_{p}\cdot q)_{p}\end{cases}}}$

This function is well-defined because only finitely many primes divide the denominator of ${\displaystyle q\in \mathbb {Q} .}$ Let ${\displaystyle Z}$ be another ${\displaystyle \mathbb {Z} }$-module together with a ${\displaystyle \mathbb {Z} }$-bilinear function ${\displaystyle \varphi :{\widehat {\mathbb {Z} }}\times \mathbb {Q} \to Z.}$ We have to show that there exists one and only one ${\displaystyle \mathbb {Z} }$-linear function ${\displaystyle \theta :\mathbb {A} _{\mathbb {Q} ,{\text{fin}}}\to Z}$ such that ${\displaystyle \theta \circ \Psi =\varphi .}$ We define ${\displaystyle \theta }$ as follows: for a given ${\displaystyle (u_{p})_{p}}$ there exist ${\displaystyle u\in \mathbb {N} }$ and ${\displaystyle (v_{p})_{p}\in {\widehat {\mathbb {Z} }},}$ such that ${\displaystyle u_{p}={\tfrac {1}{u}}\cdot v_{p}}$ for all ${\displaystyle p.}$ Define ${\displaystyle \theta ((u_{p})_{p}):=\varphi ((v_{p})_{p},{\tfrac {1}{u}}).}$ It can be shown that ${\displaystyle \theta }$ is well-defined, ${\displaystyle \mathbb {Z} }$-linear and satisfies ${\displaystyle \theta \circ \Psi =\varphi .}$ Furthermore, ${\displaystyle \theta }$ is unique with these properties.

With a similar proof we can show:

Lemma. For a number field ${\displaystyle K,\mathbb {A} _{K}=\mathbb {A} _{\mathbb {Q} }\otimes _{\mathbb {Q} }K.}$

Remark. Using ${\displaystyle \mathbb {A} _{\mathbb {Q} }\otimes _{\mathbb {Q} }K\cong \mathbb {A} _{\mathbb {Q} }\oplus \dots \oplus \mathbb {A} _{\mathbb {Q} },}$ where there are ${\displaystyle [K:\mathbb {Q} ]}$ summands, we give the right side the product topology and transport this topology via the isomorphism onto ${\displaystyle \mathbb {A} _{\mathbb {Q} }\otimes _{\mathbb {Q} }K.}$

### The adele ring of a finite extension

If ${\displaystyle L/K}$ be a finite extension then ${\displaystyle L}$ is a global field and thus ${\displaystyle \mathbb {A} _{L}}$ is defined. Let ${\displaystyle w}$ be a place of ${\displaystyle L}$ and ${\displaystyle v}$ a place of ${\displaystyle K.}$ We say ${\displaystyle w}$ lies above ${\displaystyle v,}$ denoted by ${\displaystyle w|v,}$ if the absolute value ${\displaystyle |\cdot |_{w}}$ restricted to ${\displaystyle K}$ is in the equivalence class of ${\displaystyle v.}$ Define

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

Note that both products are finite. It follows from the elementary properties of the restricted product that:

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

Remark: If ${\displaystyle w|v}$ we can embed ${\displaystyle K_{v}}$ in ${\displaystyle L_{w}.}$ Therefore, we can embed ${\displaystyle K_{v}}$ diagonally in ${\displaystyle L_{v}.}$ With this embedding ${\displaystyle L_{v}}$ is a commutative algebra over ${\displaystyle K_{v}}$ with degree

${\displaystyle \sum _{w|v}[L_{w}:K_{v}]=[L:K].}$

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

Lemma. If ${\displaystyle L/K}$ is a finite extension then ${\displaystyle \mathbb {A} _{L}\cong \mathbb {A} _{K}\otimes _{K}L}$ both algebraically and topologically.

With the help of this isomorphism, the inclusion ${\displaystyle \mathbb {A} _{K}\subset \mathbb {A} _{L}}$ is given via the function

${\displaystyle {\begin{cases}\mathbb {A} _{K}\hookrightarrow \mathbb {A} _{L}\\\alpha \mapsto \alpha \otimes _{K}1\end{cases}}}$

Furthermore, the principal adeles in ${\displaystyle \mathbb {A} _{K}}$ can be identified with a subgroup of principal adeles in ${\displaystyle \mathbb {A} _{L}}$ via the map

${\displaystyle {\begin{cases}K\hookrightarrow (K\otimes _{K}L)\cong L\\\alpha \mapsto 1\otimes _{K}\alpha \end{cases}}}$

Proof.[2] Let ${\displaystyle \omega _{1},\ldots ,\omega _{n}}$ be a basis of ${\displaystyle L}$ over ${\displaystyle K.}$ Then for almost all ${\displaystyle v,}$

${\displaystyle {\widetilde {{\mathcal {O}}_{v}}}\cong {\mathcal {O}}_{v}\omega _{1}\oplus \cdots \oplus {\mathcal {O}}_{v}\omega _{n}.}$

Furthermore, there are the following isomorphisms:

${\displaystyle K_{v}\omega _{1}\oplus \cdots \oplus K_{v}\omega _{n}\cong K_{v}\otimes _{K}L\cong L_{v}=\prod \nolimits _{w|v}L_{w}}$

For the second we used the map:

${\displaystyle {\begin{cases}K_{v}\otimes _{K}L\to L_{v}\\\alpha _{v}\otimes a\mapsto (\alpha _{v}\cdot (\tau _{w}(a)))_{w}\end{cases}}}$

in which ${\displaystyle \tau _{w}:L\to L_{w}}$ is the canonical embedding and ${\displaystyle w|v.}$ We take on both sides the restricted product with restriction condition ${\displaystyle {\widetilde {{\mathcal {O}}_{v}}}:}$

{\displaystyle {\begin{aligned}\mathbb {A} _{K}\otimes _{K}L&=\left({\widehat {\prod _{v}}}^{{\mathcal {O}}_{v}}K_{v}\right)\otimes _{K}L\\&\cong {\widehat {\prod _{v}}}^{({\mathcal {O}}_{v}\omega _{1}\oplus \cdots \oplus {\mathcal {O}}_{v}\omega _{n})}(K_{v}\omega _{1}\oplus \cdots \oplus K_{v}\omega _{n})\\&\cong {\widehat {\prod _{v}}}^{\widetilde {{\mathcal {O}}_{v}}}(K_{v}\otimes _{K}L)\\&\cong {\widehat {\prod _{v}}}^{\widetilde {{\mathcal {O}}_{v}}}L_{v}\\&=\mathbb {A} _{L}\end{aligned}}}
Corollary. As additive groups ${\displaystyle \mathbb {A} _{L}\cong \mathbb {A} _{K}\oplus \cdots \oplus \mathbb {A} _{K},}$ where the left side has ${\displaystyle [L:K]}$ summands.

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

### The adele ring of vector-spaces and algebras

Lemma. Suppose ${\displaystyle P\supset P_{\infty }}$ is a finite set of places of ${\displaystyle K}$ and define
${\displaystyle \mathbb {A} _{K}(P):=\prod _{v\in P}K_{v}\times \prod _{v\notin P}{\mathcal {O}}_{v}.}$
Define addition and multiplication component-wise and equip this ring with the product topology. Then ${\displaystyle \mathbb {A} _{K}(P)}$ is a locally compact, topological ring.

Remark. If ${\displaystyle P'}$ is another finite set of places of ${\displaystyle K}$ containing ${\displaystyle P}$ then ${\displaystyle \mathbb {A} _{K}(P)}$ is an open subring of ${\displaystyle \mathbb {A} _{K}(P').}$

Now, we are able to give an alternative characterization of the adele ring. The adele ring is the union of all sets ${\displaystyle \mathbb {A} _{K}(P)}$:

${\displaystyle \mathbb {A} _{K}=\bigcup _{P\supset P_{\infty },|P|<\infty }\mathbb {A} _{K}(P).}$

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

Fix a place ${\displaystyle v}$ of ${\displaystyle K.}$ Let ${\displaystyle P}$ be a finite set of places of ${\displaystyle K,}$ containing ${\displaystyle v}$ and ${\displaystyle P_{\infty }.}$ Define

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

Then:

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

Furthermore, define

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

where ${\displaystyle P}$ runs through all finite sets containing ${\displaystyle P_{\infty }\cup \{v\}.}$ Then:

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

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

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

#### The adele ring of a vector-space[3]

Let ${\displaystyle E}$ be a ${\displaystyle n}$-dimensional vector-space over ${\displaystyle K,}$ where ${\displaystyle n<\infty .}$ We fix a basis ${\displaystyle \omega _{1},\ldots ,\omega _{n}}$ of ${\displaystyle E}$ over ${\displaystyle K.}$ For each place ${\displaystyle v}$ of ${\displaystyle K,}$ we write ${\displaystyle E_{v}:=E\otimes _{K}K_{v}\cong K_{v}\omega _{1}\oplus \cdots \oplus K_{v}\omega _{n}}$ and ${\displaystyle {\widetilde {{\mathcal {O}}_{v}}}:={\mathcal {O}}_{v}\omega _{1}\oplus \cdots \oplus {\mathcal {O}}_{v}\omega _{n}.}$ We define the adele ring of ${\displaystyle E}$ as

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

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 ${\displaystyle \dim _{K}(E)=n<\infty .}$ We equip the adele ring ${\displaystyle \mathbb {A} _{E}}$ with the restricted product topology. Then ${\displaystyle \mathbb {A} _{E}=E\otimes _{K}\mathbb {A} _{K}}$ and we can embed ${\displaystyle E}$ in ${\displaystyle \mathbb {A} _{E}}$ naturally via the map ${\displaystyle e\mapsto e\otimes 1.}$

We give an alternative definition of the topology on the adele ring ${\displaystyle \mathbb {A} _{E}.}$ Consider all linear maps: ${\displaystyle E\to K.}$ Using the natural embeddings ${\displaystyle E\to \mathbb {A} _{E}}$ and ${\displaystyle K\to \mathbb {A} _{K},}$ we extend these linear maps to: ${\displaystyle \mathbb {A} _{E}\to \mathbb {A} _{K}.}$ The topology on ${\displaystyle \mathbb {A} _{E}}$ is the coarsest topology for which all these extensions are continuous.

We can define the topology in a different way. Take a basis ${\displaystyle \varepsilon }$ of ${\displaystyle E}$ over ${\displaystyle K.}$ This results in an isomorphism of ${\displaystyle E\cong K^{n}.}$ Therefore the basis ${\displaystyle \varepsilon }$ induces an isomorphism ${\displaystyle (\mathbb {A} _{K})^{n}\cong \mathbb {A} _{E}.}$ 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. More formally

{\displaystyle {\begin{aligned}\mathbb {A} _{E}&=E\otimes _{K}\mathbb {A} _{K}\\&\cong (K\otimes _{K}\mathbb {A} _{K})\oplus \cdots \oplus (K\otimes _{K}\mathbb {A} _{K})\\&\cong \mathbb {A} _{K}\oplus \cdots \oplus \mathbb {A} _{K}\end{aligned}}}

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

#### The adele ring of an algebra

Let ${\displaystyle A}$ be a finite-dimensional algebra over ${\displaystyle K.}$ In particular, ${\displaystyle A}$ is a finite-dimensional vector-space over ${\displaystyle K.}$ As a consequence, ${\displaystyle \mathbb {A} _{A}}$ is defined and ${\displaystyle \mathbb {A} _{A}\cong \mathbb {A} _{K}\otimes _{K}A.}$ Since we have a multiplication on ${\displaystyle \mathbb {A} _{K}}$ and ${\displaystyle A,}$ we can define a multiplication on ${\displaystyle \mathbb {A} _{A}}$ via:

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

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

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

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

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

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

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

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

One can show there is a finite set ${\displaystyle P_{0},}$ so that ${\displaystyle \mathbb {A} _{A}(P,\alpha )}$ is an open subring of ${\displaystyle \mathbb {A} _{A},}$ if ${\displaystyle P\supset P_{0}.}$ Furthermore ${\displaystyle \mathbb {A} _{A}}$ is the union of all these subrings and that for ${\displaystyle A=K,}$ the definition above is consistent with the definition of the adele ring.

### Trace and norm on the adele ring

Let ${\displaystyle L/K}$ be a finite extension. Since ${\displaystyle \mathbb {A} _{K}=\mathbb {A} _{K}\otimes _{K}K}$ and ${\displaystyle \mathbb {A} _{L}=\mathbb {A} _{K}\otimes _{K}L}$ from Lemma above we can interpret ${\displaystyle \mathbb {A} _{K}}$ as a closed subring of ${\displaystyle \mathbb {A} _{L}.}$ We write ${\displaystyle \operatorname {con} _{L/K}}$ for this embedding. Explicitly for all places ${\displaystyle w}$ of ${\displaystyle L}$ above ${\displaystyle v}$ and for any ${\displaystyle \alpha \in \mathbb {A} _{K},(\operatorname {con} _{L/K}(\alpha ))_{w}=\alpha _{v}\in K_{v}.}$

Let ${\displaystyle M/L/K}$ be a tower of global fields. Then:

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

Furthermore, restricted to the principal adeles ${\displaystyle \operatorname {con} }$ is the natural injection ${\displaystyle K\hookrightarrow L.}$

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

{\displaystyle {\begin{aligned}\alpha \omega _{1}&=\sum _{j=1}^{n}\alpha _{1j}\omega _{j}\\&\vdots \\\alpha \omega _{n}&=\sum _{j=1}^{n}\alpha _{nj}\omega _{j}\end{aligned}}}

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

{\displaystyle {\begin{aligned}\operatorname {Tr} _{L/K}(\alpha )&:=\operatorname {Tr} ((\alpha _{ij})_{i,j})=\sum _{i=1}^{n}\alpha _{ii}\\N_{L/K}(\alpha )&:=N((\alpha _{ij})_{i,j})=\det((\alpha _{ij})_{i,j})\end{aligned}}}

These are the trace and the determinant of the linear map

${\displaystyle {\begin{cases}\mathbb {A} _{L}\to \mathbb {A} _{L}\\x\mapsto \alpha x\end{cases}}}$

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

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

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

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

Moreover, it can be proven that:[4]

{\displaystyle {\begin{aligned}\operatorname {Tr} _{L/K}(\alpha )&=\left(\sum _{w|v}\operatorname {Tr} _{L_{w}/K_{v}}(\alpha _{w})\right)_{v}&&\forall \alpha \in \mathbb {A} _{L}\\N_{L/K}(\alpha )&=\left(\prod _{w|v}N_{L_{w}/K_{v}}(\alpha _{w})\right)_{v}&&\forall \alpha \in \mathbb {A} _{L}\end{aligned}}}

### Properties of the adele ring

Theorem. ${\displaystyle \mathbb {A} _{K,S}}$ is a topological ring for every subset ${\displaystyle S}$ of set of all places. Furthermore, ${\displaystyle (\mathbb {A} _{K,S},+)}$ is a locally compact group.

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

Proof.[5] ${\displaystyle \mathbb {A} _{K,S}}$ is locally compact, because ${\displaystyle {\mathcal {O}}_{v}}$ are compact and the adele ring is a restricted product. The continuity of the group operations follows from 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 ${\displaystyle E}$ over ${\displaystyle K}$ and an algebra ${\displaystyle A}$ over ${\displaystyle K.}$

Theorem. ${\displaystyle K\subset \mathbb {A} _{K}}$ is discrete and ${\displaystyle \mathbb {A} _{K}/K}$ is compact. In particular, ${\displaystyle K}$ is closed in ${\displaystyle \mathbb {A} _{K}.}$

Proof.[6] We prove the case ${\displaystyle K=\mathbb {Q} .}$ To show ${\displaystyle \mathbb {Q} \subset \mathbb {A} _{\mathbb {Q} }}$ is discrete it is sufficient to show that there exists a neighbourhood of ${\displaystyle 0}$ which contains no other rational number. The general case follows via translation. Define

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

${\displaystyle U}$ is an open neighbourhood of ${\displaystyle 0\in \mathbb {A} _{\mathbb {Q} }.}$ We claim ${\displaystyle U\cap \mathbb {Q} =\{0\}.}$ Let ${\displaystyle \beta \in U\cap \mathbb {Q} ,}$ then ${\displaystyle \beta \in \mathbb {Q} }$ and ${\displaystyle |\beta |_{p}\leq 1}$ for all ${\displaystyle p}$ and therefore ${\displaystyle \beta \in \mathbb {Z} .}$ Additionally, we have ${\displaystyle \beta \in (-1,1)}$ and therefore ${\displaystyle \beta =0.}$

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

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

We show that each element in ${\displaystyle \mathbb {A} _{\mathbb {Q} }/\mathbb {Q} }$ has a representative in ${\displaystyle W,}$ that is for each adele ${\displaystyle \alpha \in \mathbb {A} _{\mathbb {Q} },}$ there exists ${\displaystyle \beta \in \mathbb {Q} }$ such that ${\displaystyle \alpha -\beta \in W.}$ Take an arbitrary ${\displaystyle \alpha =(\alpha _{p})_{p}\in \mathbb {A} _{\mathbb {Q} },}$ and let ${\displaystyle p}$ be a prime for which ${\displaystyle |\alpha _{p}|>1.}$ Then there exists ${\displaystyle r_{p}=z_{p}/p^{x_{p}}}$ with ${\displaystyle z_{p}\in \mathbb {Z} ,x_{p}\in \mathbb {N} }$ and ${\displaystyle |\alpha _{p}-r_{p}|\leq 1.}$ Replace ${\displaystyle \alpha }$ with ${\displaystyle \alpha -r_{p}}$ and let ${\displaystyle q\neq p}$ be another prime. Then:

${\displaystyle \left|\alpha _{q}-r_{p}\right|_{q}\leq \max \left\{|a_{q}|_{q},|r_{p}|_{q}\right\}\leq \max \left\{|a_{q}|_{q},1\right\}\leq 1.}$

Next we claim:

${\displaystyle |\alpha _{q}-r_{p}|_{q}\leq 1\Longleftrightarrow |\alpha _{q}|_{q}\leq 1.}$

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 ${\displaystyle \alpha }$ are not in ${\displaystyle \mathbb {Z} _{p}}$ is reduced by 1. With iteration, we deduce that there exists ${\displaystyle r\in \mathbb {Q} }$ such that ${\displaystyle \alpha -r\in {\widehat {\mathbb {Z} }}\times \mathbb {R} .}$ Now we select ${\displaystyle s\in \mathbb {Z} }$ such that ${\displaystyle \alpha _{\infty }-r-s\in \left[-{\tfrac {1}{2}},{\tfrac {1}{2}}\right].}$ Since ${\displaystyle s\in \mathbb {Z} }$ it follows that ${\displaystyle \alpha -\beta \in W}$ for ${\displaystyle \beta :=r+s\in \mathbb {Q} .}$ Consider the continuous projection ${\displaystyle \pi :W\to \mathbb {A} _{\mathbb {Q} }/\mathbb {Q} .}$ Since it is surjective ${\displaystyle \mathbb {A} _{\mathbb {Q} }/\mathbb {Q} }$ is the continuous image of a compact set, and thus compact.

The last statement is a lemma about topological groups.

Corollary. Let ${\displaystyle E}$ be a finite-dimensional vector-space over ${\displaystyle K.}$ Then ${\displaystyle E}$ is discrete and cocompact in ${\displaystyle \mathbb {A} _{E}.}$
Theorem. We have the following:
• ${\displaystyle \mathbb {A} _{\mathbb {Q} }=\mathbb {Q} +\mathbb {A} _{\mathbb {Z} }.}$
• ${\displaystyle \mathbb {Z} =\mathbb {Q} \cap \mathbb {A} _{\mathbb {Z} }.}$
• ${\displaystyle \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} }$ is a divisible group.[7]
• ${\displaystyle \mathbb {Q} \subset \mathbb {A} _{\mathbb {Q} ,{\text{fin}}}}$ is dense.

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

In order to show ${\displaystyle \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} }$ is divisible we need to show that for any ${\displaystyle n\in \mathbb {N} }$ and ${\displaystyle y\in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} }$ the equation ${\displaystyle nx=y}$ has a solution ${\displaystyle x\in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} .}$ It is sufficient to show that ${\displaystyle \mathbb {A} _{\mathbb {Q} }}$ is divisible but this is true since ${\displaystyle \mathbb {A} _{\mathbb {Q} }}$ is a field with positive characteristic in each coordinate.

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

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

because ${\displaystyle (p^{m}\mathbb {Z} _{p})_{m\in \mathbb {N} }}$ is a neighbourhood system of ${\displaystyle 0}$ in ${\displaystyle \mathbb {Z} _{p}.}$ Using the Chinese Remainder Theorem there exists ${\displaystyle l\in \mathbb {Z} }$ such that ${\displaystyle l\equiv a_{p}{\bmod {p}}^{l_{p}}.}$ Since powers of distinct primes are coprime ${\displaystyle l\in V}$ follows.

Remark. ${\displaystyle \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} }$ is not uniquely divisible. Let ${\displaystyle y=(0,0,\ldots )+\mathbb {Z} \in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} }$ and ${\displaystyle n\geq 2}$ be given. Then ${\displaystyle x_{1}=(0,0,\ldots )+\mathbb {Z} \in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} }$ and ${\displaystyle x_{2}=(1/n,1/n,\ldots )+\mathbb {Z} \in \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} }$ (${\displaystyle x_{2}}$ is well-defined, because only finitely many primes divide ${\displaystyle n}$), both satisfy the equation ${\displaystyle nx=y}$ and clearly ${\displaystyle x_{1}\neq x_{2}.}$ In this case, being uniquely divisible is equivalent to being torsion-free, which is not true for ${\displaystyle \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} }$ since ${\displaystyle n\cdot x_{2}=0,}$ but ${\displaystyle x_{2}\neq 0}$ and ${\displaystyle n\neq 0.}$

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

### Haar measure on the adele ring

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

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

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

We construct this measure by defining it on simple sets ${\displaystyle \textstyle \prod _{v}A_{v},}$ where ${\displaystyle A_{v}\subset K_{v}}$ is open and ${\displaystyle A_{v}={\mathcal {O}}_{v}}$ for almost all ${\displaystyle v.}$ Since the simple sets generate the entire Borel ${\displaystyle \sigma }$-algebra, the measure can be defined on the entire ${\displaystyle \sigma }$-algebra.[8]

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

## The idele group

### Definition

Definition. We define the idele group of ${\displaystyle K}$ as the group of units of the adele ring of ${\displaystyle K,}$ that is ${\displaystyle I_{K}:=\mathbb {A} _{K}^{\times }.}$ The elements of the idele group are called the ideles of ${\displaystyle K.}$

Remark. We would like to equip ${\displaystyle I_{K}}$ with a topology so that it becomes a topological group. The subset topology inherited from ${\displaystyle \mathbb {A} _{K}}$ is not a suitable candidate since the group of units of a topological ring equipped with subset topology may not be a topological group. For example the inverse map in ${\displaystyle \mathbb {A} _{\mathbb {Q} }}$ is not continuous. The sequence

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

converges to ${\displaystyle 1\in \mathbb {A} _{\mathbb {Q} }.}$ To see this let ${\displaystyle U}$ be neighbourhood of ${\displaystyle 0,}$ without loss of generality we can assume:

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

Since ${\displaystyle (x_{n})_{p}-1\in \mathbb {Z} _{p}}$ for all ${\displaystyle p,}$ ${\displaystyle x_{n}-1\in U}$ for ${\displaystyle n}$ large enough. However as we saw above the inverse of this sequence does not converge in ${\displaystyle \mathbb {A} _{\mathbb {Q} }.}$

Lemma. Let ${\displaystyle R}$ be a topological ring. Define:
${\displaystyle {\begin{cases}\iota :R^{\times }\to R\times R\\x\mapsto (x,x^{-1})\end{cases}}}$
Equipped with the topology induced from the product on topology on ${\displaystyle R\times R}$ and ${\displaystyle \iota ,R^{\times }}$ is a topological group and the inclusion map ${\displaystyle R^{\times }\subset R}$ is continuous. It is the coarsest topology, emerging from the topology on ${\displaystyle R,}$ that makes ${\displaystyle R^{\times }}$ a topological group.

Proof. Since ${\displaystyle R}$ is a topological ring, it is sufficient to show that the inverse map is continuous. Let ${\displaystyle U\subset R^{\times }}$ be open, then ${\displaystyle U\times U^{-1}\subset R\times R}$ is open. We have to show ${\displaystyle U^{-1}\subset R^{\times }}$ is open or equivalently, that ${\displaystyle U^{-1}\times (U^{-1})^{-1}=U^{-1}\times U\subset R\times R}$ is open. But this is the condition above.

We equip the idele group with the topology defined in the Lemma making it a topological group.

Lemma. Define ${\displaystyle I_{K,S}:=\mathbb {A} _{K,S}^{\times },I_{K}^{S}:=(\mathbb {A} _{K}^{S})^{\times }.}$ Then the following identities of topological groups hold:
{\displaystyle {\begin{aligned}I_{K,S}&={\widehat {\prod _{v\in S}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times }\\I_{K}^{S}&={\widehat {\prod _{v\notin S}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times }\\I_{K}&={\widehat {\prod _{v}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times }\end{aligned}}}
The restricted product has the restricted product topology, which is generated by restricted open rectangles of the form
${\displaystyle \prod _{v\in E}U_{v}\times \prod _{v\notin E}{\mathcal {O}}_{v}^{\times },}$
where ${\displaystyle E}$ is a finite subset of the set of all places and ${\displaystyle U_{v}\subset K_{v}^{\times }}$ are open sets.

Proof. We prove the identity for ${\displaystyle I_{K},}$ the other two follow similarly. First we show the two sets are equal:

{\displaystyle {\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 _{v}}}^{{\mathcal {O}}_{v}^{\times }}K_{v}^{\times }.\end{aligned}}}

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

Now, we can 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 ${\displaystyle U\subset I_{K},}$ which is open in the topology of the idele group, meaning ${\displaystyle U\times U^{-1}\subset \mathbb {A} _{K}\times \mathbb {A} _{K}}$ is open, it stands that for each ${\displaystyle u\in U}$ there exists an open restricted rectangle, which is a subset of ${\displaystyle U}$ and contains ${\displaystyle u.}$ Therefore, ${\displaystyle U}$ is the union of all these restricted open rectangles and therefore is open in the restricted product topology.

Lemma. For each set of places, ${\displaystyle S,I_{K,S}}$ is a locally compact topological group.

Proof. The local compactness follows from the descriptions of the idele group as a restricted product. That ${\displaystyle I_{K,S}}$ is a topological group follows from the above discussion about the group of units of a topological ring.

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

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

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

Define

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

and let ${\displaystyle {\widehat {\mathcal {O}}}^{\times }}$ be its group of units. Then:

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

Since the idele group is a locally compact, there exists a Haar measure ${\displaystyle d^{\times }x}$ on it. This can be normalised, so that

${\displaystyle \int _{I_{K,{\text{fin}}}}\mathbf {1} _{\widehat {\mathcal {O}}}\,d^{\times }x=1.}$

This is the normalisation used for the finite places. In this equations, ${\displaystyle I_{K,{\text{fin}}}}$ 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 ${\displaystyle {\tfrac {dx}{|x|}}.}$

### The idele group of a finite extension

Lemma. Let ${\displaystyle L/K}$ be a finite extension. Define
{\displaystyle {\begin{aligned}L_{v}^{\times }&:=\prod _{w|v}L_{w}^{\times },\\{\widetilde {{\mathcal {O}}_{v}}}^{\times }&:=\prod _{w|v}{\mathcal {O}}_{w}^{\times }.\end{aligned}}}
Note, that both products are finite. Then:
${\displaystyle I_{L}={\widehat {\prod _{v}}}^{{\widetilde {{\mathcal {O}}_{v}}}^{\times }}L_{v}^{\times }.}$
Lemma. There is a canonical embedding of ${\displaystyle I_{K}}$ in ${\displaystyle I_{L}.}$

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

### The case of vector-spaces and algebras

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

Definition. Let ${\displaystyle E}$ be a finite-dimensional vector space over ${\displaystyle K.}$ Define:

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

This is an algebra over ${\displaystyle K}$ and ${\displaystyle \operatorname {End} (E)^{\times }=\operatorname {Aut} (E),}$ where the inverse of a linear map exists if and only if the determinant is non-zero. ${\displaystyle \operatorname {Aut} (E)}$ is an open subset of ${\displaystyle \operatorname {End} (E).}$ ${\displaystyle \operatorname {End} (E)\setminus \operatorname {Aut} (E)=\det \nolimits ^{-1}(\{0\}).K}$ is a topological field, therefore ${\displaystyle \{0\}}$ is closed, since ${\displaystyle \det }$ is continuous, ${\displaystyle \operatorname {Aut} (E)}$ is open.

#### The idele group of an algebra

Let ${\displaystyle A}$ be a finite-dimensional algebra over ${\displaystyle K.}$ Since ${\displaystyle \mathbb {A} _{A}^{\times }}$ is not a topological group with the subset-topology in general, we equip ${\displaystyle \mathbb {A} _{A}^{\times }}$ with the topology similar to ${\displaystyle I_{K}}$ above and call ${\displaystyle \mathbb {A} _{A}^{\times }}$ the idele group. The elements of the idele group are called idele of ${\displaystyle A.}$

Proposition. Let ${\displaystyle \alpha }$ be a finite subset of ${\displaystyle A,}$ containing a basis of ${\displaystyle A}$ over ${\displaystyle K.}$ For each finite place ${\displaystyle v}$ of ${\displaystyle K,}$ let ${\displaystyle \alpha _{v}}$ be the ${\displaystyle {\mathcal {O}}_{v}}$-module generated by ${\displaystyle \alpha }$ in ${\displaystyle A_{v}.}$ There exists a finite set of places ${\displaystyle P_{0}}$ containing ${\displaystyle P_{\infty }}$ such that for all ${\displaystyle v\notin P_{0},}$${\displaystyle \alpha _{v}}$ is a compact subring of ${\displaystyle A_{v}.}$ Furthermore, ${\displaystyle \alpha _{v}}$ contains ${\displaystyle A_{v}^{\times }.}$ For each ${\displaystyle v,A_{v}^{\times }}$ is an open subset of ${\displaystyle A_{v}}$ and the map ${\displaystyle x\mapsto x^{-1}}$ is continuous on ${\displaystyle A_{v}^{\times }.}$ As a consequence, the function ${\displaystyle x\mapsto (x,x^{-1})}$ maps ${\displaystyle A_{v}^{\times }}$ homeomorphically on its image in ${\displaystyle A_{v}\times A_{v}.}$ For each ${\displaystyle v\notin P_{0},}$ the ${\displaystyle \alpha _{v}^{\times }}$ are the elements of ${\displaystyle A_{v}^{\times },}$ mapping in ${\displaystyle \alpha _{v}\times \alpha _{v}}$ with the function above. Therefore, ${\displaystyle \alpha _{v}^{\times }}$ is an open and compact subgroup of ${\displaystyle A_{v}^{\times }.}$[9]

#### Alternative characterisation of the idele group

Proposition. Let ${\displaystyle P\supset P_{\infty }}$ be a finite set of places. Then
${\displaystyle \mathbb {A} _{A}(P,\alpha )^{\times }:=\prod _{v\in P}A_{v}^{\times }\times \prod _{v\notin P}\alpha _{v}^{\times }}$
is an open subgroup of ${\displaystyle \mathbb {A} _{A}^{\times },}$ where ${\displaystyle \mathbb {A} _{A}^{\times }}$ is the union of all ${\displaystyle \mathbb {A} _{A}(P,\alpha )^{\times }.}$[10]
Corollary. In the special case when ${\displaystyle A=K}$ for each finite set of places ${\displaystyle P\supset P_{\infty },}$
${\displaystyle \mathbb {A} _{K}(P)^{\times }=\prod _{v\in P}K_{v}^{\times }\times \prod _{v\notin P}{\mathcal {O}}_{v}^{\times }}$
is an open subgroup of ${\displaystyle \mathbb {A} _{K}^{\times }=I_{K}.}$ Furthermore, ${\displaystyle I_{K}}$ is the union of all ${\displaystyle \mathbb {A} _{K}(P)^{\times }.}$

### 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 ${\displaystyle \alpha \in I_{K}.}$ Then ${\displaystyle \operatorname {con} _{L/K}(\alpha )\in I_{L}}$ and therefore, we have in injective group homomorphism

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

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

${\displaystyle N_{L/K}:I_{L}\to I_{K}.}$

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

### The Idele class group

The group of units of ${\displaystyle K}$ can be embedded diagonally in the idele group ${\displaystyle I_{K,S}:}$

${\displaystyle {\begin{cases}K^{\times }\to I_{K,S}\\a\mapsto (a,a,a,\ldots )\end{cases}}}$

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

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

Defenition. In analogy to the ideal class group, the elements of ${\displaystyle K^{\times }}$ in ${\displaystyle I_{K}}$ are called principal ideles of ${\displaystyle I_{K}.}$ The quotient group ${\displaystyle C_{K}:=I_{K}/K^{\times }}$ is called idele class group of ${\displaystyle K.}$ This group is related to the ideal class group and is a central object in class field theory.

Remark. ${\displaystyle K^{\times }}$ is closed in ${\displaystyle I_{K},}$ therefore ${\displaystyle C_{K}}$ is a locally compact topological group and a Hausdorff space.

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

${\displaystyle {\begin{cases}C_{K}\hookrightarrow C_{L}\\\alpha K^{\times }\mapsto \alpha L^{\times }\end{cases}}}$

This function is well-defined, because the injection ${\displaystyle I_{K}\hookrightarrow I_{L}}$ obviously maps ${\displaystyle K^{\times }}$ onto a subgroup of ${\displaystyle L^{\times }.}$[11]

### Properties of the idele group

#### Absolute value on ${\displaystyle I_{K}}$ and ${\displaystyle 1}$-idele

For a given idele ${\displaystyle \alpha =(\alpha _{v})_{v},}$ we define its absolute value as:

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

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

Theorem. ${\displaystyle |\cdot |:I_{K}\to \mathbb {R} _{+}}$ is a continuous group homomorphism.

Proof. Let ${\displaystyle \alpha ,\beta \in I_{K}.}$

{\displaystyle {\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 ${\displaystyle K_{v}.}$ However, this is clear, because of the reverse triangle inequality.

We define the set of ${\displaystyle 1}$-idele as:

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

${\displaystyle I_{K}^{1}}$ is a subgroup of ${\displaystyle I_{K}.}$ Since ${\displaystyle I_{K}^{1}=|\cdot |^{-1}(\{1\}),}$ it is a closed subset of ${\displaystyle \mathbb {A} _{K}.}$ Finally the ${\displaystyle \mathbb {A} _{K}}$-topology on ${\displaystyle I_{K}^{1}}$ equals the subset-topology of ${\displaystyle I_{K}}$ on ${\displaystyle I_{K}^{1}.}$[12][13]

Artin's Product Formula. ${\displaystyle |k|=1}$ for all ${\displaystyle k\in K^{\times }.}$

Proof.[14] We prove the formula for number fields, the case of a global function field can be proved similarly. Let ${\displaystyle K}$ be a number field and ${\displaystyle a\in K^{\times }.}$ We have to show:

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

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

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

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

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

{\displaystyle {\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}}}
Lemma. There exists a constant ${\displaystyle C,}$ depending only on ${\displaystyle K,}$ such that for every ${\displaystyle \alpha =(\alpha _{v})_{v}\in \mathbb {A} _{K}}$ satisfying ${\displaystyle \textstyle \prod _{v}|\alpha _{v}|_{v}>C,}$ there exists a ${\displaystyle \beta \in K^{\times },}$ such that ${\displaystyle |\beta _{v}|_{v}\leq |\alpha _{v}|_{v}}$ for all ${\displaystyle v.}$[15]
Corollary. Let ${\displaystyle v_{0}}$ be a place of ${\displaystyle K}$ and let ${\displaystyle \delta _{v}>0}$ be given for all ${\displaystyle v\neq v_{0}}$ with the property ${\displaystyle \delta _{v}=1}$ for almost all ${\displaystyle v.}$ Then, there exists a ${\displaystyle \beta \in K^{\times },}$ so that ${\displaystyle |\beta |\leq \delta _{v}}$ for all ${\displaystyle v\neq v_{0}.}$

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

Theorem. ${\displaystyle K^{\times }}$ is discrete and cocompact in ${\displaystyle I_{K}^{1}.}$

Proof.[16] Since ${\displaystyle K^{\times }}$ is discrete in ${\displaystyle I_{K}}$ it is also discrete in ${\displaystyle I_{K}^{1}.}$ To prove ${\displaystyle I_{K}^{1}/K^{\times }}$ is compact it is sufficient to show there exists a compact set ${\displaystyle W\subset \mathbb {A} _{K},}$ such that the natural projection ${\displaystyle \pi :W\cap I_{K}^{1}\to I_{K}^{1}/K^{\times }}$ is surjective, because ${\displaystyle \pi }$ is continuous. Let ${\displaystyle \alpha \in \mathbb {A} _{K}}$ with the property ${\displaystyle \textstyle \prod _{v}|\alpha _{v}|_{v}>C}$ be given, where ${\displaystyle C}$ is the constant of the lemma above. Define

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

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

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

and therefore

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

It follows that

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

By the lemma, there exists an ${\displaystyle \eta \in K^{\times },}$ such that ${\displaystyle |\eta |_{v}\leq |\beta _{v}^{-1}\alpha _{v}|_{v}}$ for all ${\displaystyle v,}$ and therefore ${\displaystyle \eta \beta \in W.}$

Theorem. There is a canonical isomorphism ${\displaystyle I_{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times }\cong {\widehat {\mathbb {Z} }}^{\times }.}$ Furthermore, ${\displaystyle {\widehat {\mathbb {Z} }}^{\times }\times \{1\}}$ is a set of representatives of ${\displaystyle I_{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times },}$ in other words:
${\displaystyle \left({\widehat {\mathbb {Z} }}^{\times }\times \{1\}\right)\mathbb {Q} ^{\times }=I_{\mathbb {Q} }^{1}.}$
Additionally, the absolute value function induces the following isomorphisms of topological groups:
{\displaystyle {\begin{aligned}I_{\mathbb {Q} }&\cong I_{\mathbb {Q} }^{1}\times (0,\infty )\\I_{\mathbb {Q} }^{1}&\cong I_{\mathbb {Q} ,{\text{fin}}}\times \{\pm 1\}.\end{aligned}}}
Consequently, ${\displaystyle {\widehat {\mathbb {Z} }}^{\times }\times (0,\infty )}$ is a set of representatives of ${\displaystyle I_{\mathbb {Q} }/\mathbb {Q} ^{\times }.}$[17]

Proof. Consider the map

${\displaystyle {\begin{cases}\varphi :{\widehat {\mathbb {Z} }}^{\times }\to I_{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times }\\(a_{p})_{p}\mapsto ((a_{p})_{p},1)\mathbb {Q} ^{\times }\end{cases}}}$

This map is well-defined, since ${\displaystyle |a_{p}|_{p}=1}$ for all ${\displaystyle p}$ and therefore ${\displaystyle \textstyle \left(\prod _{p<\infty }|a_{p}|_{p}\right)\cdot 1=1.}$ Obviously, this map is a continuous, group homomorphism. To show injectivity, let ${\displaystyle ((a_{p})_{p},1)\mathbb {Q} ^{\times }=((b_{p})_{p},1)\mathbb {Q} ^{\times }.}$ As a result, it exists a ${\displaystyle q\in \mathbb {Q} ^{\times },}$ so that ${\displaystyle ((a_{p})_{p},1)q=((b_{p})_{p},1).}$ By considering the infinite place, we receive ${\displaystyle q=1}$ and therefore ${\displaystyle (a_{p})_{p}=(b_{p})_{p}.}$ To show the surjectivity, let ${\displaystyle ((\beta _{p})_{p},\beta _{\infty })\mathbb {Q} ^{\times }\in I_{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times }.}$ The absolute value of this element is ${\displaystyle 1}$ and therefore ${\displaystyle \textstyle |\beta _{\infty }|_{\infty }={\frac {1}{\prod _{p}|\beta _{p}|_{p}}}\in \mathbb {Q} .}$ It follows, that ${\displaystyle \beta _{\infty }\in \mathbb {Q} .}$ It stands, that