Adele ring

The ring of adeles solves the technical problem of "doing analysis on the rational numbers ${\displaystyle \mathbf {Q} }$." The classical solution was to pass to the standard metric completion ${\displaystyle \mathbf {R} }$ and use analytic techniques there. But, as was learned later on, there are many more absolute values other than the Euclidean distance, one for each prime number ${\displaystyle p\in \mathbf {Z} }$, as was classified by Ostrowski. The Euclidean absolute value, denoted ${\displaystyle |\cdot |_{\infty }}$, is only one among many others, ${\displaystyle |\cdot |_{p}}$, but the ring of adeles makes it possible to comprehend and use all of the valuations at once. This has the advantage of enabling analytic techniques while also retaining information about the primes, since their structure is embedded by the restricted infinite product.

The purpose of the adele ring is to look at all completions of ${\displaystyle K}$ at once. The adele ring is defined with the restricted product, rather than the Cartesian product. There are two reasons for this:

• For each element of ${\displaystyle K}$ 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, there cannot be any application of harmonic analysis to the Cartesian product. This is because local compactness ensures the existence (and uniqueness) of Haar measure, a crucial tool in analysis on groups in general.

The restricted infinite product is a required technical condition for giving the number field ${\displaystyle \mathbf {Q} }$ a lattice structure inside of ${\displaystyle \mathbf {A} _{\mathbf {Q} }}$, making it possible to build a theory of Fourier analysis (cf. Harmonic analysis) in the adelic setting. This is analogous to the situation in algebraic number theory where the ring of integers of an algebraic number field embeds

${\displaystyle {\mathcal {O}}_{K}\hookrightarrow K}$

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

${\displaystyle \mathbf {A} _{\mathbf {Z} }=\mathbf {R} \times {\hat {\mathbf {Z} }}=\mathbf {R} \times \prod _{p}\mathbf {Z} _{p},}$

then the ring of adeles can be equivalently defined as

${\displaystyle {\begin{aligned}\mathbf {A} _{\mathbf {Q} }&=\mathbf {Q} \otimes _{\mathbf {Z} }\mathbf {A} _{\mathbf {Z} }\\&=\mathbf {Q} \otimes _{\mathbf {Z} }\left(\mathbf {R} \times \prod _{p}\mathbf {Z} _{p}\right).\end{aligned}}}$

The restricted product structure becomes transparent after looking at explicit elements in this ring. The image of an element ${\displaystyle b/c\otimes (r,(a_{p}))\in \mathbf {A} _{\mathbf {Q} }}$ inside of the unrestricted product ${\textstyle \mathbf {R} \times \prod _{p}\mathbf {Q} _{p}}$ is the element

${\displaystyle \left({\frac {br}{c}},\left({\frac {ba_{p}}{c}}\right)\right).}$

The factor ${\displaystyle ba_{p}/c}$ lies in ${\displaystyle \mathbf {Z} _{p}}$ whenever ${\displaystyle p}$ is not a prime factor of ${\displaystyle c}$, which is the case for all but finitely many primes ${\displaystyle p}$.[3]

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

The rationals ${\displaystyle K={\mathbf {Q}}}$ have a valuation for every prime number ${\displaystyle p}$, with ${\displaystyle (K_{\nu },{\mathcal {O}}_{\nu })=(\mathbf {Q} _{p},\mathbf {Z} _{p})}$, and one infinite valuation ∞ with ${\displaystyle \mathbf {Q} _{\infty }=\mathbf {R} }$. Thus an element of

${\displaystyle \mathbf {A} _{\mathbf {Q} }\ =\ \mathbf {R} \times \prod _{p}(\mathbf {Q} _{p},\mathbf {Z} _{p})}$

is a real number along with a p-adic rational for each ${\displaystyle p}$ of which all but finitely many are p-adic integers.

Secondly, take the function field ${\displaystyle K=\mathbf {F} _{q}(\mathbf {P} ^{1})=\mathbf {F} _{q}(t)}$ of the projective line over a finite field. Its valuations correspond to points ${\displaystyle x}$ of ${\displaystyle X=\mathbf {P} ^{1}}$, i.e. maps over ${\displaystyle {\text{Spec}}\mathbf {F} _{q}}$

${\displaystyle x\$ :\ {\text{Spec}}\mathbf {F} _{q^{n}}\ \longrightarrow \ \mathbf {P} ^{1}.}

For instance, there are ${\displaystyle q+1}$ points of the form ${\displaystyle {\text{Spec}}\mathbf {F} _{q}\ \longrightarrow \ \mathbf {P} ^{1}}$. In this case ${\displaystyle {\mathcal {O}}_{\nu }={\widehat {\mathcal {O}}}_{X,x}}$ is the completed stalk of the structure sheaf at ${\displaystyle x}$ (i.e. functions on a formal neighbourhood of ${\displaystyle x}$) and ${\displaystyle K_{\nu }=K_{X,x}}$ is its fraction field. Thus

${\displaystyle \mathbf {A} _{\mathbf {F} _{q}(\mathbf {P} ^{1})}\ =\ \prod _{x\in X}({\mathcal {K}}_{X,x},{\widehat {\mathcal {O}}}_{X,x}).}$

The same holds for any smooth proper curve ${\displaystyle X/\mathbf {F_{\mathit {q}}} }$ over a finite field, the restricted product being over all points of ${\displaystyle x\in X}$.

The Artin reciprocity law says that for a global field ${\displaystyle K}$,

${\displaystyle {\widehat {C_{K}}}={\widehat {\mathbf {A} _{K}^{\times }/K^{\times }}}\ \simeq \ {\text{Gal}}(K^{\text{ab}}/K)}$

where ${\displaystyle K^{ab}}$ is the maximal abelian algebraic extension of ${\displaystyle K}$ and ${\displaystyle {\widehat {(\dots )}}}$ means the profinite completion of the group.

If ${\displaystyle X/\mathbf {F_{\mathit {q}}} }$ is a smooth proper curve then its Picard group is[4]

${\displaystyle {\text{Pic}}(X)\ =\ K^{\times }\backslash \mathbf {A} _{X}^{\times }/\mathbf {O} _{X}^{\times }}$

and its divisor group is ${\displaystyle {\text{Div}}(X)=\mathbf {A} _{X}^{\times }/\mathbf {O} _{X}^{\times }}$. Similarly, if ${\displaystyle G}$ is a semisimple algebraic group (e.g. ${\textstyle SL_{n}}$, it also holds for ${\displaystyle GL_{n}}$) then Weil uniformisation says that[5]

${\displaystyle {\text{Bun}}_{G}(X)\ =\ G(K)\backslash G(\mathbf {A} _{X})/G(\mathbf {O} _{X}).}$

Applying this to ${\displaystyle G=\mathbf {G} _{m}}$ gives the result on the Picard group.

There is a topology on ${\displaystyle \mathbf {A} _{K}}$ for which the quotient ${\displaystyle \mathbf {A} _{K}/K}$ is compact, allowing one to do harmonic analysis on it. John Tate in his thesis "Fourier analysis in number fields and Hecke Zeta functions"[6] proved results about Dirichlet L-functions using Fourier analysis on the adele ring and the idele group. Therefore, the adele ring and the idele group have been applied to study the Riemann zeta function and more general zeta functions and the L-functions.

If ${\displaystyle X}$ is a smooth proper curve over the complex numbers, one can define the adeles of its function field ${\displaystyle \mathbf {C} (X)}$ exactly as the finite fields case. John Tate proved[7] that Serre duality on ${\displaystyle X}$

${\displaystyle H^{1}(X,{\mathcal {L}})\ \simeq \ H^{0}(X,\Omega _{X}\otimes {\mathcal {L}}^{-1})^{*}}$

can be deduced by working with this adele ring ${\displaystyle \mathbf {A} _{\mathbf {C} (X)}}$. Here L is a line bundle on ${\displaystyle X}$.

Throughout this article, ${\displaystyle K}$ is a global field, meaning it is either a number field (a finite extension of ${\displaystyle \mathbb {Q} }$) or a global function field (a finite extension of ${\displaystyle \mathbb {F} _{p^{r}}(t)}$ for ${\displaystyle p}$ prime and ${\displaystyle r\in \mathbb {N} }$). By definition a finite extension of a global field is itself a global field.

For a valuation ${\displaystyle v}$ of ${\displaystyle K}$ it can be written ${\displaystyle K_{v}}$ for the completion of ${\displaystyle K}$ with respect to ${\displaystyle v.}$ If ${\displaystyle v}$ is discrete it can be written ${\displaystyle O_{v}}$ for the valuation ring of ${\displaystyle K_{v}}$ and ${\displaystyle {\mathfrak {m}}_{v}}$ for the maximal ideal of ${\displaystyle O_{v}.}$ If this is a principal ideal denoting the uniformising 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 ${\displaystyle v|\infty .}$ Then assume all valuations to be non-trivial.

There is a one-to-one identification of valuations and absolute values. Fix a constant ${\displaystyle C>1,}$ 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. Infinite places of a global field form a finite set, which is denoted by ${\displaystyle P_{\infty }.}$

Define ${\displaystyle \textstyle {\widehat {O}}:=\prod _{v<\infty }O_{v}}$ and let ${\displaystyle {\widehat {O}}^{\times }}$ be its group of units. Then ${\displaystyle \textstyle {\widehat {O}}^{\times }=\prod _{v<\infty }O_{v}^{\times }.}$

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

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

(Note that both products are finite.)

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

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

By Ostrowski's theorem the places of ${\displaystyle \mathbb {Q} }$ are ${\displaystyle \{p\in \mathbb {N} :p{\text{ prime}}\}\cup \{\infty \},}$ it is possible to identify a prime ${\displaystyle p}$ with the equivalence class of the ${\displaystyle p}$-adic absolute value and ${\displaystyle \infty }$ with the equivalence class of the absolute value ${\displaystyle |\cdot |_{\infty }}$ defined as:

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

The completion of ${\displaystyle \mathbb {Q} }$ with respect to the place ${\displaystyle p}$ is ${\displaystyle \mathbb {Q} _{p}}$ with valuation ring ${\displaystyle \mathbb {Z} _{p}.}$ For the place ${\displaystyle \infty }$ the completion is ${\displaystyle \mathbb {R} .}$ Thus:

${\displaystyle {\begin{aligned}\mathbb {A} _{\mathbb {Q} ,{\text{fin}}}&={\prod _{p<\infty }}^{'}\mathbb {Q} _{p}\\\mathbb {A} _{\mathbb {Q} }&=\left({\prod _{p<\infty }}^{'}\mathbb {Q} _{p}\right)\times \mathbb {R} \end{aligned}}}$

Or for short

${\displaystyle \mathbb {A} _{\mathbb {Q} }={\prod _{p\leq \infty }}^{'}\mathbb {Q} _{p},\qquad \mathbb {Q} _{\infty }:=\mathbb {R} .}$

the difference between restricted and unrestricted product topology can be illustrated using a sequence in ${\displaystyle \mathbb {A} _{\mathbb {Q} }}$:

Lemma. Consider the following sequence in ${\displaystyle \mathbb {A} _{\mathbb {Q} }}$:

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

In the product topology this converges to ${\displaystyle (1,1,\ldots )}$, but it does not converge at all in the restricted product topology.

Proof. In product topology convergence corresponds to the convergence in each coordinate, which is trivial because the sequences become stationary. The sequence doesn't converge in restricted product topology. For each adele ${\displaystyle a=(a_{p})_{p}\in \mathbb {A} _{\mathbb {Q} }}$ and for each restricted open rectangle ${\displaystyle \textstyle U=\prod _{p\in E}U_{p}\times \prod _{p\notin E}\mathbb {Z} _{p},}$ it has: ${\displaystyle {\tfrac {1}{p}}-a_{p}\notin \mathbb {Z} _{p}}$ for ${\displaystyle a_{p}\in \mathbb {Z} _{p}}$ and therefore ${\displaystyle {\tfrac {1}{p}}-a_{p}\notin \mathbb {Z} _{p}}$ for all ${\displaystyle p\notin F.}$ As a result ${\displaystyle x_{n}-a\notin U}$ 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.

Definition (profinite integers). The profinite integers are defined as the profinite completion of the rings ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$ with the partial order ${\displaystyle n\geq m\Leftrightarrow m|n,}$ i.e.,

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

Lemma. ${\displaystyle \textstyle {\widehat {\mathbb {Z} }}\cong \prod _{p}\mathbb {Z} _{p}.}$

Proof. This follows from the Chinese Remainder Theorem.

Lemma. ${\displaystyle \mathbb {A} _{\mathbb {Q} ,{\text{fin}}}={\widehat {\mathbb {Z} }}\otimes _{\mathbb {Z} }\mathbb {Q} .}$

Proof. 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}}}\\\left((a_{p})_{p},q\right)\mapsto (a_{p}q)_{p}\end{cases}}}

This is well-defined because for a given ${\displaystyle q={\tfrac {m}{n}}\in \mathbb {Q} }$ with ${\displaystyle m,n}$ co-prime there are only finitely many primes dividing ${\displaystyle n.}$ Let ${\displaystyle M}$ be another ${\displaystyle \mathbb {Z} }$-module with a ${\displaystyle \mathbb {Z} }$-bilinear map ${\displaystyle \Phi$ :{\widehat {\mathbb {Z} }}\times \mathbb {Q} \to M.} It must be the case that ${\displaystyle \Phi }$ factors through ${\displaystyle \Psi }$ uniquely, i.e., there exists a unique ${\displaystyle \mathbb {Z} }$-linear map ${\displaystyle {\tilde {\Phi }}:\mathbb {A} _{\mathbb {Q} ,{\text{fin}}}\to M}$ such that ${\displaystyle \Phi ={\tilde {\Phi }}\circ \Psi .}$ ${\displaystyle {\tilde {\Phi }}}$ can be defined 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 ${\displaystyle {\tilde {\Phi }}((u_{p})_{p}):=\Phi ((v_{p})_{p},{\tfrac {1}{u}}).}$ One can show ${\displaystyle {\tilde {\Phi }}}$ is well-defined, ${\displaystyle \mathbb {Z} }$-linear, satisfies ${\displaystyle \Phi ={\tilde {\Phi }}\circ \Psi }$ and is unique with these properties.

Corollary. Define ${\displaystyle \mathbb {A} _{\mathbb {Z} }:={\widehat {\mathbb {Z} }}\times \mathbb {R} .}$ This results in an algebraic isomorphism ${\displaystyle \mathbb {A} _{\mathbb {Q} }\cong \mathbb {A} _{\mathbb {Z} }\otimes _{\mathbb {Z} }\mathbb {Q} .}$

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

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, give the right side receives the product topology and transport this topology via the isomorphism onto ${\displaystyle \mathbb {A} _{\mathbb {Q} }\otimes _{\mathbb {Q} }K.}$

If ${\displaystyle L/K}$ be a finite extension, then ${\displaystyle L}$ is a global field. Thus ${\displaystyle \mathbb {A} _{L}}$ is defined, and ${\displaystyle \textstyle \mathbb {A} _{L}={\prod _{v}}^{'}L_{v}.}$ ${\displaystyle \mathbb {A} _{K}}$ can be identified with a subgroup of ${\displaystyle \mathbb {A} _{L}.}$ Map ${\displaystyle a=(a_{v})_{v}\in \mathbb {A} _{K}}$ to ${\displaystyle a'=(a'_{w})_{w}\in \mathbb {A} _{L}}$ where ${\displaystyle a'_{w}=a_{v}\in K_{v}\subset L_{w}}$ for ${\displaystyle w|v.}$ Then ${\displaystyle a=(a_{w})_{w}\in \mathbb {A} _{L}}$ is in the subgroup ${\displaystyle \mathbb {A} _{K},}$ if ${\displaystyle a_{w}\in K_{v}}$ for ${\displaystyle w|v}$ and ${\displaystyle a_{w}=a_{w'}}$ for all ${\displaystyle w,w'}$ lying above 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 by

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

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

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

Proof.[8] Let ${\displaystyle \omega _{1},\ldots ,\omega _{n}}$ be a basis of ${\displaystyle L}$ over ${\displaystyle K.}$ Then for almost all ${\displaystyle v,}$

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

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 use 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 ${\displaystyle w|v.}$ The restricted product is taken on both sides with respect to ${\displaystyle {\widetilde {O_{v}}}:}$

${\displaystyle {\begin{aligned}\mathbb {A} _{K}\otimes _{K}L&=\left({\prod _{v}}^{'}K_{v}\right)\otimes _{K}L\\&\cong {\prod _{v}}^{'}(K_{v}\omega _{1}\oplus \cdots \oplus K_{v}\omega _{n})\\&\cong {\prod _{v}}^{'}(K_{v}\otimes _{K}L)\\&\cong {\prod _{v}}^{'}L_{v}\\&=\mathbb {A} _{L}\end{aligned}}}$

Corollary. As additive groups ${\displaystyle \mathbb {A} _{L}\cong \mathbb {A} _{K}\oplus \cdots \oplus \mathbb {A} _{K},}$ where the right 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 ${\displaystyle K}$ is considered as a subset of ${\displaystyle \mathbb {A} _{K}.}$

Lemma. Suppose ${\displaystyle P\supset P_{\infty }}$ is a finite set of places of ${\displaystyle K}$ and define

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

Equip ${\displaystyle \mathbb {A} _{K}(P)}$ with the product topology and define addition and multiplication component-wise. 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, an alternative characterisation of the adele ring can be presented. 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

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

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 ${\displaystyle \mathbb {A} _{K}\twoheadrightarrow K_{v}.}$

Let ${\displaystyle E}$ be a finite dimensional vector-space over ${\displaystyle K}$ and ${\displaystyle \{\omega _{1},\ldots ,\omega _{n}\}}$ a basis for ${\displaystyle E}$ over ${\displaystyle K.}$ For each place ${\displaystyle v}$ of ${\displaystyle K}$:

${\displaystyle {\begin{aligned}E_{v}&:=E\otimes _{K}K_{v}\cong K_{v}\omega _{1}\oplus \cdots \oplus K_{v}\omega _{n}\\{\widetilde {O_{v}}}&:=O_{v}\omega _{1}\oplus \cdots \oplus O_{v}\omega _{n}\end{aligned}}}$

The adele ring of ${\displaystyle E}$ is defined as

${\displaystyle \mathbb {A} _{E}:={\prod _{v}}^{'}E_{v}.}$

This definition is based on the alternative description of the adele ring as a tensor product equipped with the same topology that was defined when giving an alternate definition of adele ring for number fields. Next, ${\displaystyle \mathbb {A} _{E}}$ is equipped with the restricted product topology. Then ${\displaystyle \mathbb {A} _{E}=E\otimes _{K}\mathbb {A} _{K}}$ and ${\displaystyle E}$ is embedded in ${\displaystyle \mathbb {A} _{E}}$ naturally via the map ${\displaystyle e\mapsto e\otimes 1.}$

An alternative definition of the topology on ${\displaystyle \mathbb {A} _{E}}$ can be provided. Consider all linear maps: ${\displaystyle E\to K.}$ Using the natural embeddings ${\displaystyle E\to \mathbb {A} _{E}}$ and ${\displaystyle K\to \mathbb {A} _{K},}$ 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.

The topology can be defined in a different way. Fixing a basis for ${\displaystyle E}$ over ${\displaystyle K}$ results in an isomorphism ${\displaystyle E\cong K^{n}.}$ Therefore fixing a basis induces an isomorphism ${\displaystyle (\mathbb {A} _{K})^{n}\cong \mathbb {A} _{E}.}$ The left-hand side is supplied with the product topology and transport this topology with the isomorphism onto the right-hand side. The topology doesn't depend on the choice of the basis, because another basis defines a second isomorphism. By composing both isomorphisms, a linear homeomorphism which transfers the two topologies into each other is obtained. More formally

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

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 ${\displaystyle L/K.}$

[9]

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 there is multiplication on ${\displaystyle \mathbb {A} _{K}}$ and ${\displaystyle A,}$ a multiplication on ${\displaystyle \mathbb {A} _{A}}$ can be defined via:

${\displaystyle \forall \alpha ,\beta \in \mathbb {A} _{K}{\text{ and }}\forall a,b\in A:\qquad (\alpha \otimes _{K}a)\cdot (\beta \otimes _{K}b):=(\alpha \beta )\otimes _{K}(ab).}$

As a consequence, ${\displaystyle \mathbb {A} _{A}}$ is an algebra with a unit over ${\displaystyle \mathbb {A} _{K}.}$ Let ${\displaystyle {\mathcal {B}}}$ be a finite subset of ${\displaystyle A,}$ containing a basis for ${\displaystyle A}$ over ${\displaystyle K.}$ For any finite place ${\displaystyle v}$ , ${\displaystyle M_{v}}$ is defined as the ${\displaystyle O_{v}}$-module generated by ${\displaystyle {\mathcal {B}}}$ in ${\displaystyle A_{v}.}$ For each finite set of places, ${\displaystyle P\supset P_{\infty },}$ define

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

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 for ${\displaystyle A=K,}$ the definition above is consistent with the definition of 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 the Lemma above, ${\displaystyle \mathbb {A} _{K}}$ can be interpreted as a closed subring of ${\displaystyle \mathbb {A} _{L}.}$ For this embedding, write ${\displaystyle \operatorname {con} _{L/K}}$. 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 ${\displaystyle K\to L.}$

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. 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, 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

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