아델 반지

수학에서 지구영역의 아델 링(역시 아델릭 링, 아델링 링 또는 아델링 링^[1])은 대수적 수 이론의 한 분야인 계급장 이론의 중심 대상이다.글로벌 분야의 모든 완성품 중 제약제품으로, 자기이중 위상학적 고리의 한 예다.

아델의 고리는 2차적 상호주의 및 유한 분야에 대한 다른 상호주의 법칙의 광대한 일반화인 아르틴 상호주의 법칙을 우아하게 묘사할 수 있게 한다.또한 $유한장$에 대한 대수곡선의 G ${\displaystyle G}$ -번들$$(bundles)을 환원군 $G{\displaystyle }$ ${\displaystyle G}$에 대한 아델(adel)의 관점에서 설명할 수 있는 것이 Weil로부터의 고전적인 정리다$.$

정의

$K{\displaystyle }$ ${\displaystyle K}$을(를) 전역 필드($Q{\displaystyle \mathbf{}}$ ${\$의 유한 확장 또는$$ 유한 필드 위에 있는 곡선 X/F의_q 함수 필드)로$$ 한다.$K{\displaystyle }$ ${\displaystyle K}$의 아델 링이 서브링이다$$.

${\displaystyle \mathbf {A} _{K}\ =\\prod(K_{\nu }},{\mathcal {O}_{\nu }}\subsetecq \prod K_{\nu }}}}}}}}\subsetecq \prod K_{prod K_{\nu }}}}}}$

consisting of the tuples ${\displaystyle (a_{\nu })}$ where ${\displaystyle a_{\nu }}$ lies in the subring ${\displaystyle {\mathcal {O}}_{\nu }\subset K_{\nu }}$ for all but finitely many places ${\displaystyle \nu }$. Here the index ${\displaystyle \nu }$ ranges over all글로벌 필드 $K{\displaystyle }$ ${\displaystyle K$ $K{\displaystyle {}_{}}$ ${\displaystyle {\nu }_{}}$ ${\$은(는) 해당 평가에서 완료되고 $O{\displaystyle {\mathcal{}}_{}}$ ${\displaystyle {\nu }_{}}$ ${\displaystyle{O}_{\nu}}$ 평가 링에서는 완료된다$$.

동기

아델의 링은 합리적인 숫자 $Q{\displaystyle \mathbf{}}$ ${\$에 대해 "실행"하는 기술적 문제를 해결한다$.$ 이전에 사람들이 사용한 "클래식" 솔루션은 $표준{\displaystyle \mathbf{}}$ 메트릭 완성 R ${\\$에 전달하고 거기서$$ 분석 기법을 사용하는 것이었다.그러나 나중에 알게 된 바와 같이 유클리드 거리 외에 더 많은 절대값이 존재하는데, 오스트로우스키가 분류한 황금수 $p{\displaystyle }$ ${\displaystyle \in }$ ${\displaystyle Z\mathbf{}}$ ${\$에 각각 하나씩 있다$.$$⋅{\$ $\}\}$로 표시된 유클리드 절대값은 여러 가지 중 하나일 뿐이므로${\displaystyle }$ad ${\displaystyle {}_{}}$ ${\$아델의 고리는 절충을 가능하게 하고 모든 가치를 한 번에 사용할 수 있게 한다.이것은 분석 기법에 접근하는 동시에 제한된 무한 제품에 의해 구조물이 내장되기 때문에 프리타임에 대한 정보를 보유할 수 있는 장점이 있다.

왜 제한된 제품인가?

제한된 무한 생산물은 숫자 $필드{\displaystyle \mathbf{}}$ Q$${\$displaystyle \mathbf {Q}$을(를$)$ $A{\displaystyle {\mathbf{}}_{}}$ ${\displaystyle {\mathbf{}}_{}}${\$displaystyle \mathbf {A} _{\mathbf {Q}}}}$의 내부에 격자 구조를 부여하기 위해 필요한 기술 조건으로서$,$ 아델릭한 설정에서 푸리에 분석 이론을 구축할 수 있다.이것은 대수적 수장의 정수 링이 내장되어 있는 대수적 수 이론의 상황과 완전히 유사하다.

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

격자로푸리에 분석의 새로운 이론의 힘으로 테이트는 L 기능의 특별한 등급을 증명할 수 있었고 복잡한 평면에서는 데데킨드 제타 기능이 메로모르픽이었다.이러한 기술적 조건이 유지되는 또 다른 자연적인 이유는 아델의 링을 링의 텐서 제품으로 구성함으로써 직접 볼 수 있다.적분 아델 $A{\displaystyle {\mathbf{}}_{}}$ ${\displaystyle {\mathbf{}}_{}}$ ${\$의 링을 링으로$$ 정의하면

${\displaystyle \mathbf {A} _{\mathbf {Z} }=\mathbf {R} \times {\hatbf {Z}}}}}} \time \prod _{p}\mathbf {Z} _{p}}}}}}}}}}} }}}}}}}}}}}}}}}}}$

그러면 아델의 링은 다음과 같이 동등하게 정의될 수 있다.

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

제한된 제품 구조는 이 링의 명시적 요소를 보고 투명해진다.If we take a rational number ${\displaystyle b/c\in \mathbf {Q} }$ we find ${\displaystyle c=p_{1}^{k_{1}}\cdots p_{n}^{k_{n}}}$. For any tuple ${\displaystyle (r,(a_{p}))\in \mathbf {R} \times {\hat {\mathbf {Z} }}}$ we have the follo날개 계열의 등가.

${\displaystyle {\c}\frac {b}{c}\right)\otimes(r, (a_{p})&=b\otimes \leftences\frac {c},{1}{c}{c}(a_{p}\right)\\\\\&=b\otimes \left\frac {r}{c}},\left\frac {a_{p}}\{p}\오른쪽)\\\end{aigned}}}}$

Then, for any ${\displaystyle p\notin \{p_{1},\ldots ,p_{n}\}}$ we still have ${\displaystyle a_{p}/c\in \mathbf {Z} _{p}}$ for ${\displaystyle a_{p}\in \mathbf {Z} _{p}}$, but for ${\displaystyle p\in \{p_{1},\ldots ,p_{$$n{\displaystyle }$$}\}$${\displaystyle p_{}}$ /${\displaystyle {}_{}c}$ ${\displaystyle \in }$ ${\displaystyle {}_{}}$ p ${\$ p ${\displaystyle p}$의 역동력이 있기 때문에$.$이것은 이 새로운 아델의 링에 있는 어떤 요소도 $단지{\displaystyle {\mathbf{}}_{}}$ 미세하게 많은 $장소{\displaystyle }$ p$$ ${\displaystyle$$\mathbf{Q} _{$p$}$의 요소를 가질 수 있다는 것을 보여준다$.$

이름의 유래

지역 계급장 이론에서는 지역 영역의 단위 집단이 중심적인 역할을 한다.글로벌 클래스 필드 이론에서 이데클 클래스 그룹이 이 역할을 맡는다.용어 "이상"(프랑스어: idelle)은 프랑스의 수학자 Claude Chevalley(1909–1984)의 발명으로 "이상적 요소"(약칭: id.el.)를 의미한다."에델"이라는 용어는 첨가된 이념을 의미한다.

아델 반지의 아이디어는 $K{\displaystyle }$ ${\displaystyle K}$의 모든 보완을 한꺼번에$$ 살펴보는 것이다.얼핏 보면 까르테시안 제품이 좋은 후보가 될 수 있다.그러나 아델 링은 제한된 제품으로 정의된다.이에 대한 두 가지 이유가 있다.

• $K{\displaystyle }$ ${\displaystyle K}$의 각 요소에 대해 거의$$ 모든 장소, 즉 한정된 숫자를 제외한 모든 장소에 대해 0이 평가된다.그래서 글로벌 분야는 제한 제품에 내장할 수 있다.
• 제한된 제품은 국소적으로 좁은 공간인 반면, 데카르트 제품은 그렇지 않다.따라서 우리는 카르테시안 제품에 조화 분석을 적용할 수 없다.

예

합리적인 숫자에 대한 아델의 고리

이성애자 K=Q는 (K_ν,O_ν)=(Q_p,Z_p)로 모든 소수 p에 대한 평가와_∞ Q=R로 하나의 무한 평가 ∞을 가진다.따라서 의 한 요소

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

p-adic 합리성과 함께 실제 숫자로 p-adic 정수를 제외한 모든 p-adic 정수를 의미한다.

투영 라인의 함수 필드용 아델 링

두 번째로 유한한 필드 위에 투사선의 함수 필드 K=F_q(P¹)=F_q(t)를 취한다.그것의 가치는 X=P의¹ X 포인트, 즉 Spec F에_q 대한 지도에 해당한다.

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

예를 들어, SpecF_q → P¹ 형식의 q+1 지점이 있다. 이 경우 O==은_νX,x x(즉, x의 공식적인 인접 지역에 대한 함수)에서 구조물의 피복의 완성된 줄기가 되며 K=K는_νX,x 그 분수장이 된다.그러므로,

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

제한 제품이 x∈X의 모든 포인트에 걸쳐 있는 유한한 영역에 걸친 매끄러운 적절한 곡선 X/F에도_q 동일한 조건이 적용된다.

관련 개념

아델 링에 있는 유닛들의 그룹을 이데르 그룹이라고 부른다.

${\displaystyle I_{K}\\\\\\\mathbf {A} _{K}^{\times }}$

부분군 KII에^×_K 의한 공칭의 인수를 이데클 클래스 그룹이라고 한다.

${\displaystyle C_{K}\ =\ I_{K}/K^{\times }}$

일체형 아델은 서브링이다.

${\displaystyle \mathbf {O} _{K}\\\prod O_{\nu }\subseteq \mathbf {A} _{K}.}$

적용들

아르틴 상호주의 표현

아르틴 상호주의 법은 글로벌 분야 K에 대해

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

여기서 K는^ab K의 최대 아벨리아 대수적 확장이며${\displaystyle \stackrel{}{}}$(${\displaystyle \stackrel{}{}}$… )${\displaystyle \stackrel{}{}}$^ ${\${\$widehat {(\dots )}}}$은 그룹의 확실한 완성을 의미한다$$.

곡선 피카르 그룹의 아델릭 공식화

X/F가_q 매끄러운 적절한 곡선일 경우 해당^[2] Picard 그룹은

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

그리고 그것의 구분 그룹은 Div(X)=A_K^×/O이다_K^×.마찬가지로, G가 반이행 대수 그룹(예: SL_n, GL도_n 보유)이라면 Weil 균일화는 다음과^[3] 같이 말한다.

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

이것을 G=G에_m 적용하면 피카르 그룹에 대한 결과가 나온다.

테이트의 논문

A에는_K 지수 A_K/K가 콤팩트한 토폴로지가 있어 조화분석을 할 수 있다.존 테이트는 논문 '숫자 분야 푸리에 분석과 헤케스 제타 기능'^[4]에서 아델 링과 이데르 그룹에 푸리에 분석을 이용한 디리클레 L 기능 결과를 입증했다.따라서 아델 링과 이데르 그룹이 적용되어 리만 제타 기능과 보다 일반적인 제타 기능과 L-기능을 연구하게 되었다.

부드러운 곡선에서 Serre 이중성 입증

X가 복잡한 숫자에 대한 매끄러운 적절한 곡선인 경우, 함수 필드 C(X)의 아델을 유한장 케이스와 정확하게 정의할 수 있다.존 테이트는 세레의 이중성이 X에 있다는 것을^[5] 증명했다.

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

아델 링 A로_C(X) 작업하면 추론할 수 있어여기 L은 X에 줄다발이다.

표기법 및 기본정의

전역 필드

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 ${$$\displaystyle r\in \mathb {N}$}.$$정의에 따르면 글로벌 영역의 한정된 확장은 그 자체로 글로벌 영역이다.

가치평가

$K{\displaystyle }$ ${\displaystyle K}$의 평가 $v{\displaystyle }$ ${\$$displaystyle$ $v}$에 대해서는 $v{\displaystyle .}$ ${\displaystyle v.}$ v ${\displaystyle v}$에 $대해$$$ K v$${\$displaystyle$ v$}$의 $완료$에 ${\displaystyle {대해}_{}}$ K$$v {\displaystystyleyled인 경우$$, $K{\displaystyle {}_{}}$$v}$를 ${\displaystyle {작성}_{}}$한다.Ov.{\displaystyle O_{v}의 최대 이상을 위해\displaystyle K_{v}}과 결제 v{\displaystyle{\mathfrak{m}}_{v}}.}우리는 π v.{\displaystyle \pi_{v}.}한non-Archimedean 평가 v<>로 쓰여 있는uniformizing 요소를 의미한다 만약 이것이 교장 이상, ∞{\displaystyle v<,\infty}또는.v${\displaystyle \mathrm{\nmid }}$ ${\displaystyle v\nmid \inft }$ 및$$ Archimedious as $v{\displaystyle }$ ${\displaystyle \infty .}$ ${\displaystyle$ v $\inft.}$우리는$$ 모든 평가액이 비경쟁적이라고 가정한다.

평가와 절대값을 일대일로 파악하는 것이 있다.상수 > ,${\displaystyle C>1,}$ 평가 $v{\displaystyle }$ ${\displaystyle v}$에 다음과 같이 정의된$$ 절대값 ${\displaystyle ∆}$ ${\displaystyle {}_{}}$ ${\displaystyle \cdot _{v}$이 할당됨$$:

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

반대로 절대값${\displaystyle }$⋅ $displaystyle \cdot }$에는 다음과 같이 정의된$$ $가치평가{\displaystyle {}_{}}$ v ${\displaystyle {\cdot }_{}}$, ${\displaystyle v_{ \cdot }}}$이(가) 할당된다$$.

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

$K{\displaystyle }$ ${\displaystyle K}$의 장소는 $K{\displaystyle .}$ ${\displaystyle K.}$ 비 아르키메데스적 평가(또는$$ 절대값)에 해당하는 곳을 유한이라고 하고, 아르키메데스적 평가(Archimedeal 평가)에 해당하는 곳을 무한이라고 한다$$.글로벌 필드의 무한한 장소 집합은 유한하며, $우리$는 이 집합을 P$∞.{\displaystyle P_{\inflit }}$에 의해 나타낸다.

^^= < O ${\$ {$O}:=\prod _${v$<\\infit }O_{v}}}}$를 $정의$하고 O ${\displaystyle {^}^{}}$ $×$ {\$displaystystyle${\O$}^{\time}}}}}}}}}}}}}}}}}}}}}}}}}}}$을 그 단위 그룹으로$$ 두십시오.그러면 ^= <<O .${\$

유한확장

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 ${\displaystyle K.}$ We say ${\displaystyle w}$ lies above ${\displaystyle v,}$ denoted by $$${\displaystyle w}$ ${\displaystyle v}$, ${\displaystyle$w ${\displaystyle {}_{}}$$,}$ $K$로 제한된$$ 절대값${\displaystyle }$⋅ w ${\$$}$이 $v{\displaystyle }$. ${\displaystyle v.}$의 동등성 클래스에 있는$$ 경우$$ 정의$$

${\displaystyle {\reasoned}L_{v}&:=\prod _{w},\\widetilde {O_{v}}}&:=\prod _{w v}O_{w}.\end{정렬}}}$

두 제품 모두 유한하다는 점에 유의하십시오.

${\displaystyle w}$ v ${\displaystyle$w$v}$이면 L ${\displaystyle w_{}}$ . ${\displaystyle L_{w}$에 $K{\displaystyle {}_{}}$ ${\displaystyle v_{}}$ ${\$를 삽입할 수 있다.$}}$ 따라서$$ ${\displaystyle L_{}}$ v${\displaystyle {}_{}}$. ${\displaystyle L_{v}}$에 대각선으로$$ K ${\displaystyle v_{}}$ ${\$displaystyle L_{v}}을(를$$) 삽입할 수 있다. 이 $내장$ ${\displaystyle L_{}}$ v ${\$는 $K{\displaystyle {}_{}}$ ${\displaystyle v_{}}$ ${\$에 대한 정류 대수학이다$$.

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

아델 반지

글로벌 필드 $K{\displaystyle }$, ${\displaystyle$ K$,}$의 유한한 아델 집합은 ${\displaystyle O_{}}$ $v{\displaystyle {}_{}}$: ${\displaystyle O_{$$K,{\text{$$v$}}}}에 대해$$ ${\displaystyle A_{}}$${\displaystyle {}_{}}$, ${\displaystyle {\text{fin}}_{}}$,$${\$displaystyle \mathb$ {$A}$ $_{v}}}}}}}}$을 나타내는$$ 제한된 제품으로 정의된다$.$

${\displaystyle \mathb{A} _{K,{\text{fin}}}}}={\prod_{v<\infline}}}={v}^{{v}=\왼쪽.(x_{v}_{v}\in \prod _{v}\{v}\right x_{v}\in O_{v}{v}{\text{}in }} 거의 모든 }v\right\}.}$

제한된 개방 직사각형에 의해 생성되는 토폴로지인 제한된 제품 토폴로지를 갖추고 있으며, 이 토폴로지는 다음과 같은 형태를 가지고 있다.

${\displaystyle U=\prod _{v\in E}U_{v}\time \prod _{v\notin E}O_{v}\subset {\prod }{v<\infty }^{'}K_{v}}}}}$

여기서 $E{\displaystyle }$ ${\displaystyle E}$은(완료) 장소의 유한 집합이며 $U{\displaystyle {}_{}}$ v${\displaystyle {}_{}{v}_{}}$ ${\displaystyle {}_{}}$ v ${\$은(는) 열려 있다$$.성분별 덧셈과 $곱셈{\displaystyle {\mathbb{}}_{}}$ A ${\displaystyle {}_{}}$가 있는${\displaystyle {}_{}}$핀 ${\$도 링이다$$.

글로벌 필드 $K{\displaystyle }$ ${\displaystyle K}$의 아델 링은 ${\displaystyle {\text{무한}}_{}}$의 장소에서$$ ${\displaystyle$$\mathb {A} _{K,{\text{fin}}}$의 제품으로 정의된다$$.무한 장소의 수는 유한하며, 완성도는 $R{\displaystyle \mathbb{}}$ ${\$ 또는$$ $C{\displaystyle \mathbb{}.}$ ${\displaystyle \mathb {C}$이다$.$ 요약하면$$:

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

성분으로 정의되는 덧셈과 곱셈으로 아델 링은 반지 입니다.아델 반지의 원소를 $K{\displaystyle .}$ ${\displaystyle$ K$.}$라고 하는데$$, 다음과 같이 쓴다.

${\displaystyle \mathb {A} _{K}={\prod_{v}^{'}K_{v}}}$

비록 이것이 일반적으로 제한된 제품은 아니지만.

비고. 글로벌 기능 분야는 무한한 공간이 없기 때문에 한정된 아델 링은 아델 링과 같다.

보조정리. 대각선 지도가 $주는$ ${\displaystyle {}_{}}$${\displaystyle }$$K{\displaystyle {\mathbb{}}_{}}$ ${\$ {$A$${\displaystyle \}}$$_{K}$에 ${\$$displaystyle a\mapsto(a,a,\dots)$가 $자연$스럽게 내장되어 있다$.$$}$

증명. 만약 ${\displaystyle \in }$ ${\displaystyle K,}$ ${\displaystyle a\in K,}$이(가) 거의 모든 $v{\displaystyle .}$ ${\displaystyle$ $a_{v}^{\time }}}$에 대해$$ ${\displaystyle \in }$ ${\displaystyle O_{}^{}}$ ${\displaystyle {}_{}^{}}$ ${\displaystyle {}_{}^{}}$ ${\$ $v.}$이(가$$)가 잘 정의되어 있음을 보여준다.$또한{\displaystyle {}_{}}$ ${\displaystyle K_{}}$ v ${\$에 K$${\$displaystyle K}$을(를) 내장하는 것이 $모든{\displaystyle }$v .$${\$displaystyle$ v$.}$에 대해 주입하기$$ 때문에 주입이 가능하다.

비고. 대각선 지도 아래 $이미지$와$$함께 K {\$displaystyle K}$을(를) 식별함으로써 우리는 이를 $A{\displaystyle {\mathbb{}}_{}}$ ${\displaystyle {}_{}}$${\displaystyle {}_{}}$.${\displaystyle \mathb$ ${$$A} _{K}$의 하위 링으로 간주한다$.$$}$

정의.$S{\displaystyle }$ ${\$$displaystyle S$$}$을(를) $K{\displaystyle }$. ${\displaystyle$ $K$$}$의 위치 집합으로 $설정$하십시오$.$

${\displaystyle \mathb {A} _{K,S}={\prod_{v\in S}^{'}K_{v}}}$

더 나아가 우리가 정의한다면

${\displaystyle \mathb{A}_{K}^{S}={\prod_{v\notin S}^{'}K_{v}}}}}}$

we have: $A{\displaystyle {\mathbb{}}_{}}$ ${\displaystyle {}_{}{}_{}}$= ${\displaystyle A_{}}$K${\displaystyle {}_{}}$ S × A ${\displaystyle {}_{}^{}}$ $.$ {\$displaystyle \mathb {$A$} _{$K}=\$mathb$ {$A} _{K,S$}\time \$mathb {A} _${K$}^{K}^}$$}$

이성애자들의 아델 반지

By Ostrowski's theorem the places of ${\displaystyle \mathbb {Q} }$ are ${\displaystyle \{p\in \mathbb {N} :p{\text{ prime}}\}\cup \{\infty \},}$ where we identify a prime ${\displaystyle p}$ with the equivalence class of the ${\displaystyle p}$-adic absolute value and ${\dis$절대값 $⋅{\$의 동등 등급이 다음과$$ 같이 정의된$$ $Playstyle \inflt }:$

${\displaystyle \forall x\in \mathb {Q} :\quad x _{\infit }={\begin}x&x\geq 0\-x&x<0\case}}}}}}$

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 ${\displaystyle \mathbb {R} .}$따라서 다음과 같다.

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

아니면 간단히 말하면

${\displaystyle \mathb {A} _{\mathb {Q}={\prod _{p\leq \infit }^{}}\mathb {Q} _{p},\qqqad \mathb {Q} _{\infort }=\mathb {R}.}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

A ${\displaystyle {\mathbb{}}_{}}$ ${\$ {$A} _{\mathb$ {$Q}}}}$의 시퀀스를 사용하여 제한 제품과 제한되지 않은 제품 토폴로지의 차이를 설명하겠다$.$

보조정리. A ${\displaystyle {\mathbb{}}_{}}$ ${\$의 다음 순서를 고려하십시오$.$

${\displaystyle {\displaysty}x_{1}=\좌측값\frac {1}{2}},1,1,\ldots \오른쪽)\\x_{2}&=\왼쪽(1,{\frac {1}{3},1,\ldots \오른쪽)\x_{3}&=\좌측(1,1,{\frac {1}{5},1,\ldots \오른쪽)\\x_{4}&=\left(1,1,1,{\frac {1}{7},1,\ldots \right)\\&\vdots \end{aigned}}$

제품 위상에서는 (${\displaystyle 1},{\displaystyle }$ ${\displaystyle 1}$,${\displaystyle }$…${\displaystyle )}$로 수렴한다$.$ {\$디스플레이$ 스타일(1$,1,\ldots).$${}$ 제한된$$ 제품 토폴로지에 수렴하지 않는다.

증명. 제품 위상 수렴에서 각 좌표의 수렴에 해당하며, 이는 시퀀스가 정지하기 때문에 사소한 것이다.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$$E}\mathbb {Z} _{p},}$ we have: ${\displaystyle {\tfrac {1}{p}}-a_{p}\notin \mathbb {Z} _{p}}$ for ${\displaystyle a_{p}\in \mathbb {Z} _{p}}$ and therefore ${\displaystyle {\tfrac {1}{p}}-a_{p}\notin \mathbb {Z} _{p}}$ for all ${\dis$$Playstyle p\notin F$${\displaystyle {}_{}}$ ${\displaystyle \mathrm{결과적}}$으로 ${\displaystyle {거의}_{}}$ $모든{\displaystyle }$ n ${\displaystyle \in }$ ${\displaystyle N\mathbb{}}$${\displaystyle .}$${\displaystyle x_{n}-a\notin$ U$}$이$($가$$) 모든 위치 집합의 유한 하위 집합으로 간주된다$$. 이 고려에서$$ $E{\displaystyle }$ ${\displaystystyle E}$ $및$ F$${\$displaystystystystysty F}$

숫자 필드에 대한 대체 정의

정의(확실한 정수).우리는 무한정 정수를 $부분{\displaystyle \mathbb{}}$ 순서 n${\displaystyle n}$ m${\displaystyle }$${\displaystyle n}$ ${\displaystyle m\mathbb{},}${\$displaystyle \mathb {Z}$ /$n\mathb {Z}$의 확실한$$완료로 정의한다. 즉, 부분$$ 순서 n $n$ m ≥ ${\displaystyle m,}$ ${\displaystystyle$ n$\geq$ m$\leftrighrow m,}.$

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

보조정리. ${\displaystyle \textstyle {\widehat {\mathb {Z}}}}}}\cong \prod_{p}\mathb {Z} _{p}.}$

증명. 이것은 중국 잔존 정리(Chinese Legacy Organization에서 따온 것이다.

보조정리. ${\displaystyle \mathb {A} _{\mathb {Q},{\text{fin}}={\widehat {\\mathb {Z}}}}}}\mathb {Q}}}}$

증명. 우리는 텐서 제품의 보편적 특성을 사용할 것이다.$Z{\displaystyle \mathbb{}}$ ${\$ - 이선$$ 함수 정의

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

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}}$${\displaystyle \mathbb {Z} }$-bilinear map ${\displaystyle \Phi :{\widehat {\mathbb {Z} }}\times \mathbb {Q} \to M.}$ We have to show ${\displaystyle \Phi }$ factors through ${\displaystyle \Psi }$ uniquely, i.e., there exists a unique ${\displaystyle \mathbb {Z} }$-linear map ${\displaystyle \tilde{\mathrm{\Phi }}}$${\displaystyle {\tilde {\Phi }}:\mathbb {A} _{\mathbb {Q} ,{\text{fin}}}\to M}$ such that ${\displaystyle \Phi ={\tilde {\Phi }}\circ \Psi .}$ We define ${\displaystyle {\tilde {\Phi }}}$ 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{\mathrm{\Phi }}((u_p{)}_p):=\mathrm{\Phi }((v_p}$${\displaystyle {\tilde {\Phi }}}}(u_{p}_{p}):$$=\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.

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

증명 ${\displaystyle \mathbb {A} _{\mathbb {Z} }\otimes _{\mathbb {Z} }\mathbb {Q} =\left({\widehat {\mathbb {Z} }}\times \mathbb {R} \right)\otimes _{\mathbb {Z} }\mathbb {Q} \cong \left({\widehat {\mathbb {Z} }}\otimes _{\mathbb {Z} }\mathbb {Q} \right)\times (\mathbb {R} \otimes _{\mathbb {Z} }\mathbb {Q} )\cong \left({\widehat {\mathbb {Z} }}\otimes_{\mathb {Z} \mathb {Q}\오른쪽)\time \mathb {R} =\mathb {A}{\mathb {Q}, {\text{fin}}\time \mathb {R} =\mathb {A} _{\mathb {Q}}}}}}}}}}}}}}}}}$

lema. 숫자 필드 $K$의 경우${\displaystyle ,}$ ${\displaystyle {}_{}}$ K${\displaystyle {}_{}}$= ${\displaystyle {\mathbb{}}_{}}$ ${\displaystyle {\mathbb{}}_{}}$ ${\displaystyle {\otimes }_{}{}_{}}$ ${\displaystyle {Q\mathbb{}}_{}}$ ${\displaystyle K_{}.}$ ${\displaystyle$ K$,\mathb {A} _{K}=\mathb {Q}$$_{\mathb${Q}\$otimes _{\mathb {Q}}}}}K.$$}$

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 및 이 토폴로지를 A ${\displaystyle {\mathbb{}}_{}}$ ${\displaystyle {\otimes }_{}{}_{}}$ ${\displaystyle Q_{}}$ K${\displaystyle }$. ${\displaystyle \mathb {A} _{\mathb {Q}\otimes _{\mathb {Q}}}K$로 전송하십시오$.$$}$

유한확장의 아델 링

If ${\displaystyle L/K}$ be a finite extension then ${\displaystyle L}$ is a global field and thus ${\displaystyle \mathbb {A} _{L}}$ is defined and ${\displaystyle \textstyle \mathbb {A} _{L}={\prod _{v}}^{'}L_{v}.}$ We claim ${\displaystyle \math$$bb {A} _{K}}$ can be identified with a subgroup of ${\displaystyle \mathbb {A} _{L}.}$ Map ${\displaystyle a=(a_{v})_{v}\in \mathbb {A} _{K}}$ to ${\displaystyle a'=(a'_{w})_{w}\in \mathbb {A} _{L}}$ where ${\displaystyle a_w^{\prime }=a_v\in K_v\subset L_{}}$${\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 wv}$${\displaystyle$ w $v}$ 및$$ ${\displaystyle a_{}}$= 모든 $w$에 대한$$ ${\displaystyle {w{}^{}}_{}}$ ${\displaystyle {\prime }^{}}$ ${\$$w'}}},$$K{\displaystyle }$. ${\displaystyle$ $K$$.}$의 동일한 $위치{\displaystyle }$ v ${\displaystyle v}$ 위에 놓여$$ 있음

보조정리. $L{\displaystyle }$ /${\displaystyle K}$ ${\displaystyle$ L/$K}$이(가) 유한한 확장인$$ 경우 $A$ ${\displaystyle {}_{}}$ ${\displaystyle {}_{}}$ ${\displaystyle {}_{}}$ ${\displaystyle K_{}{}_{}{}_{}}$ ${\displaystyle \mathb {A} _{L}\cong \mathb {A} _{K}\otimes _{K}L}}.$

이러한 이형성의 도움으로 A ${\displaystyle {}_{}}$ ${\displaystyle {}_{}}$ A ${\displaystyle {}_{}}$ ${\$이(가) 포함되어 있다$$.

${\displaystyle {\begin}\mathb {A} \mathb {A} \{L}\\\\alpha \mapsto \alpha \otimes \{K}1\case}}}}}$

또한 A ${\displaystyle {}_{}}$ ${\$의 주요 아델은 지도를 통해$$ ${\displaystyle {}_{}}$ L ${\$의 주요 아델의 하위 그룹으로 식별할 수 있다$$.

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

증명.^[6] $\Omega {\displaystyle {}_{}}$ ${\displaystyle {1\mathrm{}}_{}}$,${\displaystyle }$… ,${\displaystyle {}_{}}$ ${\displaystyle n_{}}$ ${\$을(를) K${\displaystyle .}${\$displaystyle K.}$ $위$에 L ${\displaystyle L}$을(를) 기본으로 한다. 그러면$$ 거의 $모든{\displaystyle }$ v${\displaystyle }$, ${\displaystyleyle$, $v,}$

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

게다가 다음과 같은 이형성이 있다.

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

두번째로 우리는 지도를 사용했다:

${\displaystyle {\begin}K_{v}\time _{K}L\to L_{v}\\\alpha _{v}\nalpha _\cdot(\tau _{w})({w}a)_{w}end{case}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}?$

여기서 $\tau {\displaystyle {}_{}}$ ${\displaystyle w_{}}$: ${\displaystyle L}$→ L ${\displaystyle w_{}}$ ${\$$L\to L_{w}}$은(는) 표준 임베딩이며 w ${\displaystyle v.}$ ${\displaystyle$w$v.}$ ${\displaystyle \stackrel{}{{Ov}_{}}}$~${\displaystyle }$: ${\displaystyle {\widetilde{O_{v}}}}}}}}$에 대한 제한 제품을 양쪽에 맡는다$$.

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

코롤러리.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 we consider ${\displaystyle K}$ as a subset of ${\displaystyle \mathbb {A} _{K$$}.}$

벡터 스페이스와 알헤브라의 아델 링

보조정리. $P{\displaystyle }$ ${\displaystyle \supset }$ $∞{\$이(가) $K{\displaystyle }$ {\$displaystyle K}$의 유한 집합이라고 가정하고$$ 정의한다.

${\displaystyle \mathb {A} _{K}(P):=\prod _{v\in P}K_{v}\time \prod _{v\notin P}O_{v}}}$

$제품{\displaystyle {\mathbb{}}_{}}$ 토폴로지를 A$$ ${\displaystyle {}_{}}$( ${\displaystyle P}$) ${\displaystyle \mathb {A} _{K}(P)$에 장착하고 추가 및 곱셈 구성요소를 정의하십시오.그러면 $A{\displaystyle {\mathbb{}}_{}}$ ${\displaystyle {}_{}}$( ${\displaystyle P}$) ${\displaystyle \mathb {A} _{K}(P)}$은(는) 로컬 컴팩트 위상학적 링이다.

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

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

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

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 ${\displaystyle K,}$ we write:

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

We define the adele ring of ${\displaystyle E}$ 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 we defined when giving an alternate definition of adele ring for number fields. We equip ${\displaystyle \mathbb {A} _{E}}$ 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 ${\displaystyle \mathbb {A} _{E}.}$ Consider all linear maps: ${\displaystyle E\to K.}$ Using the natural embeddings ${\displaystyle E\to \mathbb {A} _{E}}$ and ${\displaystyle K\to \mathbb {A} _{K},}$ 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. 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 ${\displaystyle (\mathbb {A} _{K})^{n}\cong \mathbb {A} _{E}.}$ 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 which 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 ${\displaystyle L/K.}$

^[7]

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:

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

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 ${\displaystyle v}$ we define ${\displaystyle M_{v}}$ as the ${\displaystyle O_{v}}$-module generated by ${\displaystyle {\mathcal {B}}}$ in ${\displaystyle A_{v}.}$ For each finite set of places, ${\displaystyle P\supset P_{\infty },}$ we 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.

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 ${\displaystyle K\to L.}$

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. 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:^[8]

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

Properties of the adele ring

Theorem.^[9] For every set of places ${\displaystyle S,\mathbb {A} _{K,S}}$ is a locally compact topological ring.

Remark. The result above also holds for the adele ring of vector-spaces and algebras over ${\displaystyle K.}$

Theorem.^[10] ${\displaystyle K}$ is discrete and cocompact in ${\displaystyle \mathbb {A} _{K}.}$ In particular, ${\displaystyle K}$ is closed in ${\displaystyle \mathbb {A} _{K}.}$

Proof. 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 the existence of a neighbourhood of ${\displaystyle 0}$ which contains no other rational number. The general case follows via translation. Define

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

${\displaystyle U}$ 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 compactness, define:

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

We show each element in ${\displaystyle \mathbb {A} _{\mathbb {Q} }/\mathbb {Q} }$ has a representative in ${\displaystyle W,}$ that is for each ${\displaystyle \alpha \in \mathbb {A} _{\mathbb {Q} },}$ there exists ${\displaystyle \beta \in \mathbb {Q} }$ such that ${\displaystyle \alpha -\beta \in W.}$ Let ${\displaystyle \alpha =(\alpha _{p})_{p}\in \mathbb {A} _{\mathbb {Q} },}$ be arbitrary and ${\displaystyle p}$ 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 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 ${\displaystyle \alpha _{\infty }-r-s\in \left[-{\tfrac {1}{2}},{\tfrac {1}{2}}\right].}$ Then ${\displaystyle \alpha -(r+s)\in W.}$ The continuous projection ${\displaystyle \pi :W\to \mathbb {A} _{\mathbb {Q} }/\mathbb {Q} }$ is surjective, therefore ${\displaystyle \mathbb {A} _{\mathbb {Q} }/\mathbb {Q} ,}$ as the continuous image of a compact set, is compact.

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.^[11]
• ${\displaystyle \mathbb {Q} \subset \mathbb {A} _{\mathbb {Q} ,{\text{fin}}}}$ is dense.

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

By definition ${\displaystyle \mathbb {A} _{\mathbb {Q} }/\mathbb {Z} }$ is divisible if 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 ${\displaystyle \mathbb {A} _{\mathbb {Q} }}$ 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 ${\displaystyle \mathbb {Z} _{p}.}$ By 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

${\displaystyle {\begin{aligned}x_{1}&=(0,0,\ldots )+\mathbb {Z} \\x_{2}&=\left({\tfrac {1}{n}},{\tfrac {1}{n}},\ldots \right)+\mathbb {Z} \end{aligned}}}$

both satisfy the equation ${\displaystyle nx=y}$ and clearly ${\displaystyle x_{1}\neq x_{2}}$ (${\displaystyle x_{2}}$ is well-defined, because only finitely many primes divide ${\displaystyle n}$). 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 ${\displaystyle nx_{2}=0,}$ 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

Definition. A function ${\displaystyle f:\mathbb {A} _{K}\to \mathbb {C} }$ is called simple if ${\displaystyle \textstyle f=\prod _{v}f_{v},}$ where ${\displaystyle f_{v}:K_{v}\to \mathbb {C} }$ are measurable and ${\displaystyle f_{v}=\mathbf {1} _{O_{v}}}$ for almost all ${\displaystyle v.}$

Theorem.^[12] Since ${\displaystyle \mathbb {A} _{K}}$ is a locally compact group with addition, there is an additive Haar measure ${\displaystyle dx}$ on ${\displaystyle \mathbb {A} _{K}.}$ This measure can be normalized such that every integrable simple function ${\displaystyle \textstyle f=\prod _{v}f_{v}}$ satisfies:

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

where for ${\displaystyle v<\infty ,dx_{v}}$ is the measure on ${\displaystyle K_{v}}$ such that ${\displaystyle O_{v}}$ has unit measure and ${\displaystyle dx_{\infty }}$ is the Lebesgue measure. The product is finite, i.e. almost all factors are equal to one.

The idele group

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.

Definition. For ${\displaystyle S}$ a subset of places of ${\displaystyle K}$ set: ${\displaystyle I_{K,S}:=\mathbb {A} _{K,S}^{\times },I_{K}^{S}:=(\mathbb {A} _{K}^{S})^{\times }.}$

Lemma. The following identities of topological groups hold:

${\displaystyle {\begin{aligned}I_{K,S}&={\prod _{v\in S}}^{'}K_{v}^{\times }\\I_{K}^{S}&={\prod _{v\notin S}}^{'}K_{v}^{\times }\\I_{K}&={\prod _{v}}^{'}K_{v}^{\times }\end{aligned}}}$

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

${\displaystyle \prod _{v\in E}U_{v}\times \prod _{v\notin E}O_{v}^{\times },}$

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:

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

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 ${\displaystyle x_{v}\in O_{v}}$ for almost all ${\displaystyle v}$ and ${\displaystyle x_{v}^{-1}\in O_{v}}$ for almost all ${\displaystyle v.}$ Therefore, ${\displaystyle x_{v}\in O_{v}^{\times }}$ 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, so 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 description of ${\displaystyle I_{K,S}}$ as a restricted product. It being a topological group follows from the above discussion on the group of units of a topological ring.

A neighbourhood system of ${\displaystyle 1\in \mathbb {A} _{K}(P_{\infty })^{\times }}$ 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 ${\displaystyle U_{v}=O_{v}^{\times }}$ for almost all ${\displaystyle v.}$

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

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

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. Then:

${\displaystyle I_{L}={\prod _{v}}^{'}L_{v}^{\times }.}$

where the restricted product is with respect to ${\displaystyle {\widetilde {O_{v}}}^{\times }.}$

Lemma. There is a canonical embedding of ${\displaystyle I_{K}}$ in ${\displaystyle I_{L}.}$

Proof. We map ${\displaystyle a=(a_{v})_{v}\in I_{K}}$ to ${\displaystyle a'=(a'_{w})_{w}\in I_{L}}$ 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

^[13]

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 ${\displaystyle O_{v}}$-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 ${\displaystyle x\mapsto (x,x^{-1})}$ 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 ${\displaystyle A_{v}^{\times }.}$^[14]

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 ${\displaystyle \mathbb {A} _{A}(P,\alpha )^{\times }.}$^[15]

Corollary. In the special case of ${\displaystyle A=K}$ for each finite set of places ${\displaystyle P\supset P_{\infty },}$

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

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 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 ${\displaystyle \alpha \in I_{L},}$ it 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 a continuous function:

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

The Idele class group

Lemma. There is natural embedding of ${\displaystyle K^{\times }}$ into ${\displaystyle I_{K,S}}$ given by the diagonal map: ${\displaystyle a\mapsto (a,a,a,\ldots ).}$

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

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.

Lemma.^[16] Let ${\displaystyle L/K}$ be a finite extension. The embedding ${\displaystyle I_{K}\to I_{L}}$ induces an injective map:

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

Properties of the idele group

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

Definition. For ${\displaystyle \alpha =(\alpha _{v})_{v}\in I_{K}}$ define: ${\displaystyle \textstyle \alpha :=\prod _{v} \alpha _{v} _{v}.}$ Since ${\displaystyle \alpha }$ is an idele this product is finite and therefore well-defined.

Remark. The definition can be extended to ${\displaystyle \mathbb {A} _{K}}$ by allowing infinite products. However these infinite products vanish and so ${\displaystyle \cdot }$ vanishes on ${\displaystyle \mathbb {A} _{K}\setminus I_{K}.}$ We will use ${\displaystyle \cdot }$ to denote both the function on ${\displaystyle I_{K}}$ and ${\displaystyle \mathbb {A} _{K}.}$

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

Definition. 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 ${\displaystyle I_{K}^{1}.}$^[17][18]

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

Proof.^[19] We prove the formula for number fields, the case of global function fields can be proved similarly. Let ${\displaystyle K}$ be a number field and ${\displaystyle a\in K^{\times }.}$ We have to show:

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

For a finite place ${\displaystyle v}$ for which the corresponding prime ideal ${\displaystyle {\mathfrak {p}}_{v}}$ does not divide ${\displaystyle (a)}$ we have ${\displaystyle v(a)=0}$ and therefore ${\displaystyle a _{v}=1.}$ This is valid for almost all ${\displaystyle {\mathfrak {p}}_{v}.}$ We have:

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

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 of the norm. We note the norm is in ${\displaystyle \mathbb {Q} }$ so without loss of generality we can assume ${\displaystyle a\in \mathbb {Q} .}$ 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 ${\displaystyle p.}$ By Ostrowski's theorem all absolute values on ${\displaystyle \mathbb {Q} }$ are equivalent to the real absolute value ${\displaystyle \cdot _{\infty }}$ or a ${\displaystyle p}$-adic absolute value. Therefore:

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

Lemma.^[20] 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 ${\displaystyle \beta \in K^{\times }}$ such that ${\displaystyle \beta _{v} _{v}\leq \alpha _{v} _{v}}$ for all ${\displaystyle v.}$

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 ${\displaystyle \beta \in K^{\times },}$ 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 ${\displaystyle O_{v}.}$ 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 ${\displaystyle v.}$ By the Lemma there exists ${\displaystyle \beta \in K^{\times },}$ 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.^[21] Since ${\displaystyle K^{\times }}$ is discrete in ${\displaystyle I_{K}}$ it is also discrete in ${\displaystyle I_{K}^{1}.}$ To prove the compactness of ${\displaystyle I_{K}^{1}/K^{\times }}$ let ${\displaystyle C}$ is the constant of the Lemma and suppose ${\displaystyle \alpha \in \mathbb {A} _{K}}$ satisfying ${\displaystyle \textstyle \prod _{v} \alpha _{v} _{v}>C}$ is given. Define:

${\displaystyle W_{\alpha }:=\left\{\xi =(\xi _{v})_{v}\in \mathbb {A} _{K} \xi _{v} _{v}\leq \alpha _{v} _{v}{\text{ for all }}v\right\}.}$

Clearly ${\displaystyle W_{\alpha }}$ is compact. We claim the natural projection ${\displaystyle W_{\alpha }\cap I_{K}^{1}\to I_{K}^{1}/K^{\times }}$ is surjective. Let ${\displaystyle \beta =(\beta _{v})_{v}\in I_{K}^{1}}$ be arbitrary, then:

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

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 ${\displaystyle \eta \in K^{\times }}$ such that ${\displaystyle \eta _{v}\leq \beta _{v}^{-1}\alpha _{v} _{v}}$ for all ${\displaystyle v,}$ and therefore ${\displaystyle \eta \beta \in W_{\alpha }}$ proving the surjectivity of the natural projection. Since it is also continuous the compactness follows.

Theorem.^[22] There is a canonical isomorphism ${\displaystyle I_{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times }\cong {\widehat {\mathbb {Z} }}^{\times }.}$ Furthermore, ${\displaystyle {\widehat {\mathbb {Z} }}^{\times }\times \{1\}\subset I_{\mathbb {Q} }^{1}}$ is a set of representatives for ${\displaystyle I_{\mathbb {Q} }^{1}/\mathbb {Q} ^{\times }}$ and ${\displaystyle {\widehat {\mathbb {Z} }}^{\times }\times (0,\infty )\subset I_{\mathbb {Q} }}$ is a set of representatives for ${\displaystyle I_{\mathbb {Q} }/\mathbb {Q} ^{\times }.}$

Proof. Consider the map

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

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 ${\displaystyle \phi }$ is a continuous group homomorphism. Now suppose ${\displaystyle ((a_{p})_{p},1)\mathbb {Q} ^{\times }=((b_{p})_{p},1)\mathbb {Q} ^{\times }.}$ Then there exists ${\displaystyle q\in \mathbb {Q} ^{\times }}$ such that ${\displaystyle ((a_{p})_{p},1)q=((b_{p})_{p},1).}$ By considering the infinite place we see ${\displaystyle q=1}$ proving injectivity. To show 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

${\displaystyle \beta _{\infty } _{\infty }={\frac {1}{\prod _{p} \beta _{p} _{p}}}\in \mathbb {Q} .}$

Hence ${\displaystyle \beta _{\infty }\in \mathbb {Q} }$ and we have:

${\displaystyle ((\beta _{p})_{p},\beta _{\infty })\mathbb {Q} ^{\times }=\left(\left({\frac {\beta _{p}}{\beta _{\infty }}}\right)_{p},1\right)\mathbb {Q} ^{\times }.}$

Since

${\displaystyle \forall p:\qquad \left {\frac {\beta _{p}}{\beta _{\infty }}}\right _{p}=1,}$

we conclude ${\displaystyle \phi }$ is surjective.

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

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

Proof. The isomorphisms are given by:

${\displaystyle {\begin{cases}\psi :I_{\mathbb {Q} }\to I_{\mathbb {Q} }^{1}\times (0,\infty )\\a=(a_{\text{fin}},a_{\infty })\mapsto \left(a_{\text{fin}},{\frac {a_{\infty }}{ a }}, a \right)\end{cases}}\qquad {\text{and}}\qquad {\begin{cases}{\widetilde {\psi }}:I_{\mathbb {Q} ,{\text{fin}}}\times \{\pm 1\}\to I_{\mathbb {Q} }^{1}\\(a_{\text{fin}},\varepsilon )\mapsto \left(a_{\text{fin}},{\frac {\varepsilon }{ a_{\text{fin}} }}\right)\end{cases}}}$

Relation between ideal class group and idele class group

Theorem. Let ${\displaystyle K}$ be a number field with ring of integers ${\displaystyle O,}$ group of fractional ideals ${\displaystyle J_{K},}$ and ideal class group ${\displaystyle \operatorname {Cl} _{K}=J_{K}/K^{\times }.}$ We have the following isomorphisms

${\displaystyle {\begin{aligned}J_{K}&\cong I_{K,{\text{fin}}}/{\widehat {O}}^{\times }\\\operatorname {Cl} _{K}&\cong C_{K,{\text{fin}}}/{\widehat {O}}^{\times }K^{\times }\\\operatorname {Cl} _{K}&\cong C_{K}/\left({\widehat {O}}^{\times }\times \prod _{v \infty }K_{v}^{\times }\right)K^{\times }\end{aligned}}}$

where we have defined ${\displaystyle C_{K,{\text{fin}}}:=I_{K,{\text{fin}}}/K^{\times }.}$

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

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

Then ${\displaystyle {\mathfrak {p}}_{v}}$ is a prime ideal in ${\displaystyle O.}$ The map ${\displaystyle v\mapsto {\mathfrak {p}}_{v}}$ is a bijection between finite places of ${\displaystyle K}$ and non-zero prime ideals of ${\displaystyle O.}$ The inverse is given as follows: a prime ideal ${\displaystyle {\mathfrak {p}}}$ is mapped to the valuation ${\displaystyle v_{\mathfrak {p}},}$ given by

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

The following map is well-defined:

${\displaystyle {\begin{cases}(\cdot ):I_{K,{\text{fin}}}\to J_{K}\\\alpha =(\alpha _{v})_{v}\mapsto \prod _{v<\infty }{\mathfrak {p}}_{v}^{v(\alpha _{v})},\end{cases}}}$

The map ${\displaystyle (\cdot )}$ is obviously a surjective homomorphism and ${\displaystyle \ker((\cdot ))={\widehat {O}}^{\times }.}$ The first isomorphism follows from fundamental theorem on homomorphism. Now, we divide both sides by ${\displaystyle K^{\times }.}$ This is possible, because

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

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

${\displaystyle {\begin{cases}\phi :C_{K,{\text{fin}}}\to \operatorname {Cl} _{K}\\\alpha K^{\times }\mapsto (\alpha )K^{\times }\end{cases}}}$

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

For the last isomorphism note that ${\displaystyle \phi }$ induces a surjective group homomorphism ${\displaystyle {\widetilde {\phi }}:C_{K}\to \operatorname {Cl} _{K}}$ with

${\displaystyle \ker({\widetilde {\phi }})=\left({\widehat {O}}^{\times }\times \prod _{v \infty }K_{v}^{\times }\right)K^{\times }.}$

Remark. Consider ${\displaystyle I_{K,{\text{fin}}}}$ with the idele topology and equip ${\displaystyle J_{K},}$ with the discrete topology. Since ${\displaystyle (\{{\mathfrak {a}}\})^{-1}}$ is open for each ${\displaystyle {\mathfrak {a}}\in J_{K},(\cdot )}$ is continuous. It stands, that ${\displaystyle (\{{\mathfrak {a}}\})^{-1}=\alpha {\widehat {O}}^{\times }}$ is open, where ${\displaystyle \alpha =(\alpha _{v})_{v}\in \mathbb {A} _{K,{\text{fin}}},}$ so that ${\displaystyle \textstyle {\mathfrak {a}}=\prod _{v}{\mathfrak {p}}_{v}^{v(\alpha _{v})}.}$

Decomposition of ${\displaystyle I_{K}}$ and ${\displaystyle C_{K}}$

Theorem.

${\displaystyle {\begin{aligned}I_{K}&\cong I_{K}^{1}\times M:\quad {\begin{cases}M\subset I_{K}{\text{ discrete and }}M\cong \mathbb {Z} &\operatorname {char} (K)>0\\M\subset I_{K}{\text{ closed and }}M\cong \mathbb {R} _{+}&\operatorname {char} (K)=0\end{cases}}\\C_{K}&\cong I_{K}^{1}/K^{\times }\times N:\quad {\begin{cases}N=\mathbb {Z} &\operatorname {char} (K)>0\\N=\mathbb {R} _{+}&\operatorname {char} (K)=0\end{cases}}\end{aligned}}}$

Proof. ${\displaystyle \operatorname {char} (K)=p>0.}$ For each place ${\displaystyle v}$ of ${\displaystyle K,\operatorname {char} (K_{v})=p,}$ so that for all ${\displaystyle x\in K_{v}^{\times },}$ ${\displaystyle x _{v}}$ belongs to the subgroup of ${\displaystyle \mathbb {R} _{+},}$ generated by ${\displaystyle p.}$ Therefore for each ${\displaystyle z\in I_{K},}$ ${\displaystyle z }$ is in the subgroup of ${\displaystyle \mathbb {R} _{+},}$ generated by ${\displaystyle p.}$ Therefore the image of the homomorphism ${\displaystyle z\mapsto z }$ is a discrete subgroup of ${\displaystyle \mathbb {R} _{+},}$ generated by ${\displaystyle p.}$ Since this group is non-trivial, it is generated by ${\displaystyle Q=p^{m}}$ for some ${\displaystyle m\in \mathbb {N} .}$ Choose ${\displaystyle z_{1}\in I_{K},}$ so that ${\displaystyle z_{1} =Q,}$ then ${\displaystyle I_{K}}$ is the direct product of ${\displaystyle I_{K}^{1}}$ and the subgroup generated by ${\displaystyle z_{1}.}$ This subgroup is discrete and isomorphic to ${\displaystyle \mathbb {Z} .}$

${\displaystyle \operatorname {char} (K)=0.}$ For ${\displaystyle \lambda \in \mathbb {R} _{+}}$ define:

${\displaystyle z(\lambda )=(z_{v})_{v},\quad z_{v}={\begin{cases}1&v\notin P_{\infty }\\\lambda &v\in P_{\infty }\end{cases}}}$

The map ${\displaystyle \lambda \mapsto z(\lambda )}$ is an isomorphism of ${\displaystyle \mathbb {R} _{+}}$ in a closed subgroup ${\displaystyle M}$ of ${\displaystyle I_{K}}$ and ${\displaystyle I_{K}\cong M\times I_{K}^{1}.}$ The isomorphism is given by multiplication:

${\displaystyle {\begin{cases}\phi :M\times I_{K}^{1}\to I_{K},\\((\alpha _{v})_{v},(\beta _{v})_{v})\mapsto (\alpha _{v}\beta _{v})_{v}\end{cases}}}$

Obviously, ${\displaystyle \phi }$ is a homomorphism. To show it is injective, let ${\displaystyle (\alpha _{v}\beta _{v})_{v}=1.}$ Since ${\displaystyle \alpha _{v}=1}$ for ${\displaystyle v<\infty ,}$ it stands that ${\displaystyle \beta _{v}=1}$ for ${\displaystyle v<\infty .}$ Moreover, it exists a ${\displaystyle \lambda \in \mathbb {R} _{+},}$ so that ${\displaystyle \alpha _{v}=\lambda }$ for ${\displaystyle v \infty .}$ Therefore, ${\displaystyle \beta _{v}=\lambda ^{-1}}$ for ${\displaystyle v \infty .}$ Moreover ${\displaystyle \textstyle \prod _{v} \beta _{v} _{v}=1,}$ implies ${\displaystyle \lambda ^{n}=1,}$ where ${\displaystyle n}$ is the number of infinite places of ${\displaystyle K.}$ As a consequence ${\displaystyle \lambda =1}$ and therefore ${\displaystyle \phi }$ is injective. To show surjectivity, let ${\displaystyle \gamma =(\gamma _{v})_{v}\in I_{K}.}$ We define ${\displaystyle \lambda := \gamma ^{\frac {1}{n}}}$ and furthermore, we define ${\displaystyle \alpha _{v}=1}$ for ${\displaystyle v<\infty }$ and ${\displaystyle \alpha _{v}=\lambda }$ for ${\displaystyle v \infty .}$ Define ${\displaystyle \textstyle \beta ={\frac {\gamma }{\alpha }}.}$ It stands, that ${\displaystyle \textstyle \beta ={\frac { \gamma }{ \alpha }}={\frac {\lambda ^{n}}{\lambda ^{n}}}=1.}$ Therefore, ${\displaystyle \phi }$ is surjective.

The other equations follow similarly.

Characterisation of the idele group

Theorem.^[24] Let ${\displaystyle K}$ be a number field. There exists a finite set of places ${\displaystyle S,}$ such that:

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

Proof. The class number of a number field is finite so let ${\displaystyle {\mathfrak {a}}_{1},\ldots ,{\mathfrak {a}}_{h}}$ be the ideals, representing the classes in ${\displaystyle \operatorname {Cl} _{K}.}$ These ideals are generated by a finite number of prime ideals ${\displaystyle {\mathfrak {p}}_{1},\ldots ,{\mathfrak {p}}_{n}.}$ Let ${\displaystyle S}$ be a finite set of places containing ${\displaystyle P_{\infty }}$ and the finite places corresponding to ${\displaystyle {\mathfrak {p}}_{1},\ldots ,{\mathfrak {p}}_{n}.}$ Consider the isomorphism:

${\displaystyle I_{K}/\left(\prod _{v<\infty }O_{v}^{\times }\times \prod _{v \infty }K_{v}^{\times }\right)\cong J_{K},}$

induced by

${\displaystyle (\alpha _{v})_{v}\mapsto \prod _{v<\infty }{\mathfrak {p}}_{v}^{v(\alpha _{v})}.}$

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

Applications

Finiteness of the class number of a number field

In the previous section we used the fact that the class number of a number field is finite. Here we would like to prove this statement:

Theorem (finiteness of the class number of a number field). Let ${\displaystyle K}$ be a number field. Then ${\displaystyle \operatorname {Cl} _{K} <\infty .}$

Proof. The map

${\displaystyle {\begin{cases}I_{K}^{1}\to J_{K}\\\left((\alpha _{v})_{v<\infty },(\alpha _{v})_{v \infty }\right)\mapsto \prod _{v<\infty }{\mathfrak {p}}_{v}^{v(\alpha _{v})}\end{cases}}}$

is surjective and therefore ${\displaystyle \operatorname {Cl} _{K}}$ is the continuous image of the compact set ${\displaystyle I_{K}^{1}/K^{\times }.}$ Thus, ${\displaystyle \operatorname {Cl} _{K}}$ is compact. In addition it is discrete and so finite.

Remark. There is a similar result for the case of a global function field. In this case, the so-called divisor group is defined. It can be shown, that the quotient of the set of all divisors of degree ${\displaystyle 0}$ by the set of the principal divisors is a finite group.^[25]

Group of units and Dirichlet's unit theorem

Let ${\displaystyle P\supset P_{\infty }}$ be a finite set of places. Define

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

Then ${\displaystyle E(P)}$ is a subgroup of ${\displaystyle K^{\times },}$ containing all elements ${\displaystyle \xi \in K^{\times }}$ satisfying ${\displaystyle v(\xi )=0}$ for all ${\displaystyle v\notin P.}$ Since ${\displaystyle K^{\times }}$ is discrete in ${\displaystyle I_{K},}$ ${\displaystyle E(P)}$ is a discrete subgroup of ${\displaystyle \Omega (P)}$ and with the same argument, ${\displaystyle E(P)}$ is discrete in ${\displaystyle \Omega _{1}(P):=\Omega (P)\cap I_{K}^{1}.}$

An alternative definition is: ${\displaystyle E(P)=K(P)^{\times },}$ where ${\displaystyle K(P)}$ is a subring of ${\displaystyle K}$ defined by

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

As a consequence, ${\displaystyle K(P)}$ contains all elements ${\displaystyle \xi \in K,}$ which fulfil ${\displaystyle v(\xi )\geq 0}$ for all ${\displaystyle v\notin P.}$

Lemma 1. Let ${\displaystyle 0<c\leq C<\infty .}$ The following set is finite:

${\displaystyle \left\{\eta \in E(P):\left.{\begin{cases} \eta _{v} _{v}=1&\forall v\notin P\\c\leq \eta _{v} _{v}\leq C&\forall v\in P\end{cases}}\right\}\right\}.}$

Proof. Define

${\displaystyle W:=\left\{(\alpha _{v})_{v}:\left.{\begin{cases} \alpha _{v} _{v}=1&\forall v\notin P\\c\leq \alpha _{v} _{v}\leq C&\forall v\in P\end{cases}}\right\}\right\}.}$

${\displaystyle W}$ is compact and the set described above is the intersection of ${\displaystyle W}$ with the discrete subgroup ${\displaystyle K^{\times }}$ in ${\displaystyle I_{K}}$ and therefore finite.

Lemma 2. Let ${\displaystyle E}$ be set of all ${\displaystyle \xi \in K}$ such that ${\displaystyle \xi _{v}=1}$ for all ${\displaystyle v.}$ Then ${\displaystyle E=\mu (K),}$ the group of all roots of unity of ${\displaystyle K.}$ In particular it is finite and cyclic.

Proof. All roots of unity of ${\displaystyle K}$ have absolute value ${\displaystyle 1}$ so ${\displaystyle \mu (K)\subset E.}$ For converse note that Lemma 1 with ${\displaystyle c=C=1}$ and any ${\displaystyle P}$ implies ${\displaystyle E}$ is finite. Moreover ${\displaystyle E\subset E(P)}$ for each finite set of places ${\displaystyle P\supset P_{\infty }.}$ Finally Suppose there exists ${\displaystyle \xi \in E,}$ which is not a root of unity of ${\displaystyle K.}$ Then ${\displaystyle \xi ^{n}\neq 1}$ for all ${\displaystyle n\in \mathbb {N} }$ contradicting the finiteness of ${\displaystyle E.}$

Unit Theorem. ${\displaystyle E(P)}$ is the direct product of ${\displaystyle E}$ and a group isomorphic to ${\displaystyle \mathbb {Z} ^{s},}$ where ${\displaystyle s=0,}$ if ${\displaystyle P=\emptyset }$ and ${\displaystyle s= P -1,}$ if ${\displaystyle P\neq \emptyset .}$^[26]

Dirichlet's Unit Theorem. Let ${\displaystyle K}$ be a number field. Then ${\displaystyle O^{\times }\cong \mu (K)\times \mathbb {Z} ^{r+s-1},}$ where ${\displaystyle \mu (K)}$ is the finite cyclic group of all roots of unity of ${\displaystyle K,r}$ is the number of real embeddings of ${\displaystyle K}$ and ${\displaystyle s}$ is the number of conjugate pairs of complex embeddings of ${\displaystyle K.}$ It stands, that ${\displaystyle [K:\mathbb {Q} ]=r+2s.}$

Remark. The Unit Theorem is a generalisation of Dirichlet's Unit Theorem. To see this let ${\displaystyle K}$ be a number field. We already know that ${\displaystyle E=\mu (K),}$ set ${\displaystyle P=P_{\infty }}$ and note ${\displaystyle P_{\infty } =r+s.}$ Then we have:

${\displaystyle {\begin{aligned}E\times \mathbb {Z} ^{r+s-1}=E(P_{\infty })&=K^{\times }\cap \left(\prod _{v \infty }K_{v}^{\times }\times \prod _{v<\infty }O_{v}^{\times }\right)\\&\cong K^{\times }\cap \left(\prod _{v<\infty }O_{v}^{\times }\right)\\&\cong O^{\times }\end{aligned}}}$

Approximation theorems

Weak Approximation Theorem.^[27] Let ${\displaystyle \cdot _{1},\ldots , \cdot _{N},}$ be inequivalent valuations of ${\displaystyle K.}$ Let ${\displaystyle K_{n}}$ be the completion of ${\displaystyle K}$ with respect to ${\displaystyle \cdot _{n}.}$ Embed ${\displaystyle K}$ diagonally in ${\displaystyle K_{1}\times \cdots \times K_{N}.}$ Then ${\displaystyle K}$ is everywhere dense in ${\displaystyle K_{1}\times \cdots \times K_{N}.}$ In other words, for each ${\displaystyle \varepsilon >0}$ and for each ${\displaystyle (\alpha _{1},\ldots ,\alpha _{N})\in K_{1}\times \cdots \times K_{N},}$ there exists ${\displaystyle \xi \in K,}$ such that:

${\displaystyle \forall n\in \{1,\ldots ,N\}:\quad \alpha _{n}-\xi _{n}<\varepsilon .}$

Strong Approximation Theorem.^[28] Let ${\displaystyle v_{0}}$ be a place of ${\displaystyle K.}$ Define

${\displaystyle V:={\prod _{v\neq v_{0}}}^{'}K_{v}.}$

Then ${\displaystyle K}$ is dense in ${\displaystyle V.}$

Remark. The global field is discrete in its adele ring. The strong approximation theorem tells us that, if we omit one place (or more), the property of discreteness of ${\displaystyle K}$ is turned into a denseness of ${\displaystyle K.}$

Hasse principle

Hasse-Minkowski Theorem. A quadratic form on ${\displaystyle K}$ is zero, if and only if, the quadratic form is zero in each completion ${\displaystyle K_{v}.}$

Remark. This is the Hasse principle for quadratic forms. For polynomials of degree larger than 2 the Hasse principle isn't valid in general. The idea of the Hasse principle (also known as local–global principle) is to solve a given problem of a number field ${\displaystyle K}$ by doing so in its completions ${\displaystyle K_{v}}$ and then concluding on a solution in ${\displaystyle K.}$

Characters on the adele ring

Definition. Let ${\displaystyle G}$ be a locally compact abelian group. The character group of ${\displaystyle G}$ is the set of all characters of ${\displaystyle G}$ and is denoted by ${\displaystyle {\widehat {G}}.}$ Equivalently ${\displaystyle {\widehat {G}}}$ is the set of all continuous group homomorphisms from ${\displaystyle G}$ to ${\displaystyle \mathbb {T} :=\{z\in \mathbb {C} : z =1\}.}$ We equip ${\displaystyle {\widehat {G}}}$ with the topology of uniform convergence on compact subsets of ${\displaystyle G.}$ One can show that ${\displaystyle {\widehat {G}}}$ is also a locally compact abelian group.

Theorem. The adele ring is self-dual: ${\displaystyle \mathbb {A} _{K}\cong {\widehat {\mathbb {A} _{K}}}.}$

Proof. By reduction to local coordinates it is sufficient to show each ${\displaystyle K_{v}}$ is self-dual. This can done by using a fixed character of ${\displaystyle K_{v}.}$ We illustrate this idea by showing ${\displaystyle \mathbb {R} }$ is self-dual. Define:

${\displaystyle {\begin{cases}e_{\infty }:\mathbb {R} \to \mathbb {T} \\e_{\infty }(t):=\exp(2\pi it)\end{cases}}}$

Then the following map is an isomorphism which respects topologies:

${\displaystyle {\begin{cases}\phi :\mathbb {R} \to {\widehat {\mathbb {R} }}\\s\mapsto {\begin{cases}\phi _{s}:\mathbb {R} \to \mathbb {T} \\\phi _{s}(t):=e_{\infty }(ts)\end{cases}}\end{cases}}}$

Theorem (algebraic and continuous duals of the adele ring).^[29] Let ${\displaystyle \chi }$ be a non-trivial character of ${\displaystyle \mathbb {A} _{K},}$ which is trivial on ${\displaystyle K.}$ Let ${\displaystyle E}$ be a finite-dimensional vector-space over ${\displaystyle K.}$ Let ${\displaystyle E^{\star }}$ and ${\displaystyle \mathbb {A} _{E}^{\star }}$ be the algebraic duals of ${\displaystyle E}$ and ${\displaystyle \mathbb {A} _{E}.}$ Denote the topological dual of ${\displaystyle \mathbb {A} _{E}}$ by ${\displaystyle \mathbb {A} _{E}'}$ and use ${\displaystyle \langle \cdot ,\cdot \rangle }$ and ${\displaystyle [{\cdot },{\cdot }]}$ to indicate the natural bilinear pairings on ${\displaystyle \mathbb {A} _{E}\times \mathbb {A} _{E}'}$ and ${\displaystyle \mathbb {A} _{E}\times \mathbb {A} _{E}^{\star }.}$ Then the formula ${\displaystyle \langle e,e'\rangle =\chi ([e,e^{\star }])}$ for all ${\displaystyle e\in \mathbb {A} _{E}}$ determines an isomorphism ${\displaystyle e^{\star }\mapsto e'}$ of ${\displaystyle \mathbb {A} _{E}^{\star }}$ onto ${\displaystyle \mathbb {A} _{E}',}$ where ${\displaystyle e'\in \mathbb {A} _{E}'}$ and ${\displaystyle e^{\star }\in \mathbb {A} _{E}^{\star }.}$ Moreover, if ${\displaystyle e^{\star }\in \mathbb {A} _{E}^{\star }}$ fulfils ${\displaystyle \chi ([e,e^{\star }])=1}$ for all ${\displaystyle e\in E,}$ then ${\displaystyle e^{\star }\in E^{\star }.}$

Tate's thesis

With the help of the characters of ${\displaystyle \mathbb {A} _{K},}$ we can do Fourier analysis on the adele ring.^[30] John Tate in his thesis "Fourier analysis in number fields and Heckes Zeta functions"^[4] 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. We can define adelic forms of these functions and we can represent them as integrals over the adele ring or the idele group, with respect to corresponding Haar measures. We can show functional equations and meromorphic continuations of these functions. For example, for all ${\displaystyle s\in \mathbb {C} }$ with ${\displaystyle \Re (s)>1,}$

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

where ${\displaystyle d^{\times }x}$ is the unique Haar measure on ${\displaystyle I_{\mathbb {Q} ,{\text{fin}}}}$ normalized such that ${\displaystyle {\widehat {\mathbb {Z} }}^{\times }}$ has volume one and is extended by zero to the finite adele ring. As a result the Riemann zeta function can be written as an integral over (a subset of) the adele ring.^[31]

Automorphic forms

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

${\displaystyle {\begin{aligned}I_{\mathbb {Q} }&=\operatorname {GL} (1,\mathbb {A} _{\mathbb {Q} })\\I_{\mathbb {Q} }^{1}&=(\operatorname {GL} (1,\mathbb {A} _{\mathbb {Q} }))^{1}:=\{x\in \operatorname {GL} (1,\mathbb {A} _{\mathbb {Q} }): x =1\}\\\mathbb {Q} ^{\times }&=\operatorname {GL} (1,\mathbb {Q} )\end{aligned}}}$

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

${\displaystyle {\begin{aligned}I_{\mathbb {Q} }&\leftrightsquigarrow \operatorname {GL} (2,\mathbb {A} _{\mathbb {Q} })\\I_{\mathbb {Q} }^{1}&\leftrightsquigarrow (\operatorname {GL} (2,\mathbb {A} _{\mathbb {Q} }))^{1}:=\{x\in \operatorname {GL} (2,\mathbb {A} _{\mathbb {Q} }): \det(x) =1\}\\\mathbb {Q} &\leftrightsquigarrow \operatorname {GL} (2,\mathbb {Q} )\end{aligned}}}$

And finally

${\displaystyle \mathbb {Q} ^{\times }\backslash I_{\mathbb {Q} }^{1}\cong \mathbb {Q} ^{\times }\backslash I_{\mathbb {Q} }\leftrightsquigarrow (\operatorname {GL} (2,\mathbb {Q} )\backslash (\operatorname {GL} (2,\mathbb {A} _{\mathbb {Q} }))^{1}\cong (\operatorname {GL} (2,\mathbb {Q} )Z_{\mathbb {R} })\backslash \operatorname {GL} (2,\mathbb {A} _{\mathbb {Q} }),}$

where ${\displaystyle Z_{\mathbb {R} }}$ is the centre of ${\displaystyle \operatorname {GL} (2,\mathbb {R} ).}$ Then we define an automorphic form as an element of ${\displaystyle L^{2}((\operatorname {GL} (2,\mathbb {Q} )Z_{\mathbb {R} })\backslash \operatorname {GL} (2,\mathbb {A} _{\mathbb {Q} })).}$ In other words an automorphic form is a functions on ${\displaystyle \operatorname {GL} (2,\mathbb {A} _{\mathbb {Q} })}$ satisfying certain algebraic and analytic conditions. For studying automorphic forms, it is important to know the representations of the group ${\displaystyle \operatorname {GL} (2,\mathbb {A} _{\mathbb {Q} }).}$ It is also possible to study automorphic L-functions, which can be described as integrals over ${\displaystyle \operatorname {GL} (2,\mathbb {A} _{\mathbb {Q} }).}$^[32]

We could generalize even further by replacing ${\displaystyle \mathbb {Q} }$ with a number field and ${\displaystyle \operatorname {GL} (2)}$ with an arbitrary reductive algebraic group.

Further applications

A generalisation of Artin reciprocity law leads to the connection of representations of ${\displaystyle \operatorname {GL} (2,\mathbb {A} _{K})}$ and of Galois representations of ${\displaystyle K}$ (Langlands program).

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

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

Notes

• What problem do the adeles solve?
• Some good books on adeles

References

Sources

• Cassels, John; Fröhlich, Albrecht (1967). Algebraic number theory: proceedings of an instructional conference, organized by the London Mathematical Society, (a NATO Advanced Study Institute). Vol. XVIII. London: Academic Press. ISBN 978-0-12-163251-9. 366 pages.
• Neukirch, Jürgen (2007). Algebraische Zahlentheorie, unveränd. nachdruck der 1. aufl. edn (in German). Vol. XIII. Berlin: Springer. ISBN 9783540375470. 595 pages.
• Weil, André (1967). Basic number theory. Vol. XVIII. Berlin; Heidelberg; New York: Springer. ISBN 978-3-662-00048-9. 294 pages.
• Deitmar, Anton (2010). Automorphe Formen (in German). Vol. VIII. Berlin; Heidelberg (u.a.): Springer. ISBN 978-3-642-12389-4. 250 pages.
• Lang, Serge (1994). Algebraic number theory, Graduate Texts in Mathematics 110 (2nd ed.). New York: Springer-Verlag. ISBN 978-0-387-94225-4.