%&latex % January 12, 2000 \documentclass[11pt]{amsart} \usepackage{amsmath,amssymb,amscd} \usepackage[mathscr]{eucal} \newcommand{\Aut}{{\mathrm{Aut} }} \newcommand{\End}{{\mathrm{End} }} \newcommand{\GL}{{\mathbf{GL} }} \newcommand{\Gal}{{\mathrm{Gal}\, }} \newcommand{\G}{{\mathbf G}_{m}} \newcommand{\Hom}{{\mathrm{Hom} }} \newcommand{\pr}{{\mathfrak p}} \newcommand{\Tr}{{\mathrm{Tr}\, }} \newcommand{\Q}{\mathbf Q} \newcommand{\Z}{\mathbf Z} \newcommand{\R}{\mathbf R} \newcommand{\C}{\mathbf C} \newcommand{\F}[1]{\mathbf{F}_{#1}} \newcommand{\Fbar}[1]{\overline{\mathbf{F}}_{#1}} \newcommand{\fr}[1]{\,{\rm Frob}_{\, #1}} \newcommand{\q}[1]{\mathbf{Q}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\msr}[1]{\mathscr{#1}} \newcommand{\mbf}[1]{\mathbf{#1}} \newcommand{\mfr}[1]{\mathfrak{#1}} \newcommand{\mr}[1]{\mathrm{#1}} \newcommand{\ok}[1]{O_{#1}} \newcommand{\re}[1]{{\rm Re} (#1)} \newcommand{\im}[1]{{\rm Im} (#1)} \theoremstyle{plain} \newtheorem{thm}{Theorem}[section] \newtheorem{lem}[thm]{Lemma} \newtheorem{pro}[thm]{Proposition} \newtheorem{cor}[thm]{Corollary} \theoremstyle{definition} \newtheorem{Def}[thm]{Definition} %\theoremstyle{remark} \newtheorem{rem}[thm]{Remark} \newtheorem{exa}[thm]{Example} \begin{document} \title{Lecture {\rm 23}} \author{Mustafa Arslan} \address{Louisiana State University \\ Department of Mathematics \\ Baton Rouge, LA 70803, USA} \date{\today} \email{arslan@math.lsu.edu} \maketitle %\section{Lecture ?} \newcommand{\ov}{{\mathcal O}_v} In this section $K$ denotes a number field. Also we assume that all valuations are normalized. Last time we defined the restricted topological product\[\prod_\lambda(X_\lambda,U_\lambda)\] as a subspace of the product space $\prod X_\lambda$ such that the coordinates of the each element are restricted to open sets $U_\lambda$ for almost all $\lambda$. We introduced the {\em ring of adeles} as the restricted topological product of the family $(K_v,\ov)$ where $\ov$ is the ring associated with the valuation $v$ if it is non-archimedean, $K_v$ is just the same as $K$ indexed with $v$ (so in this lecture it does not denote the residue field). We can view the restricted topological product as the union \[X=\bigcup_{\begin{array}{c}{\ }^{S\subset\Lambda}\\ {\ }^{|S|<\infty}\end{array}} X_S\] of product spaces \[X_S=\left(\prod_{\lambda\in S}X_\lambda\right)\times\left(\prod_{\lambda\not\in S}U_\lambda\right)\] \begin{lem} Suppose that each $X_\lambda$ is locally compact and $U_\lambda$ are all compact. Then $X$ is locally compact. \end{lem} \begin{proof} It is enough to prove that each $X_S$ is locally compact. Since finite product of locally compact spaces is locally compact, the product $\prod_{\lambda\in S}X_\lambda$ is locally compact and since any product of compact spaces is compact (Tychonoff's theorem) $\prod_{\lambda\not\in S}U_\lambda$ is compact and hence $X_S$ is locally compact. \end{proof} When $K$ is a locally compact field, this lemma says that the ring of adeles is a locally compact topological ring since $\ov$ is compact for all $v$. Let us denote by $M_K$ the set of all valuations on $K$ and by $S_\infty$ the set of all archimedean places on $K$. As we wrote above, for $S$ a finite subset of $M_K$ we write \[ A_{K,S}=\left(\prod_{v\in S}K_v\right)\times\left(\prod_{v\not\in S}\ov\right).\] \begin{thm} \begin{itemize} \item[(a)] The natural map $K\rightarrow A_K$ defined by \[x\mapsto(\ldots,x,\ldots)\] embeds $K$ as a discrete subgroup of $A_K$. \item[(b)]$K+A_{K,S_\infty} = A_K$. \item[(c)]$A_K/K$ is compact. \end{itemize} \end{thm} \begin{proof} \begin{itemize} \item[(a)] It is enough to show that $0$ is an isolated point in the image of $K$. To say that an element $x\neq$ is close to $0$ in the adele topology means \begin{enumerate} \item $|x|_v\leq 1$ for all places $v$, \item $|x|_v\approx 0$ for all archimedean valuation $v$. \end{enumerate} By product formula \[\prod |x|_v =1\] whereas this is impossible when (1) and (2) hold. \item[(b)] $A_K = K+A_{K,S_\infty}$ means for every adele $\alpha\in A_K$ there exists $x\in K$ such that \[\alpha_v-x\in\ov\] for every non-archimedean $v$. Since $\alpha_v$ is in $\ov$ for almost all non-archimedean valuation $v$, there is an integer $m$ such that $v(m\alpha_v)\geq 0$ for all non-archimedean $v$. Let \[S=\{\pr_1,\ldots,\pr_s\}\] be that the set of all primes of $K$ dividing $m$ By Chinese remainder theorem, the map \[{\mathcal O}_K\rightarrow\left({\mathcal O}_K\big/\pr_1^\nu\right)\times\cdots\times\left({\mathcal O}_K\big/\pr_s^\nu\right)\]is onto. Hence there exists an element $y$ in ${\mathcal O}$ such that \[\begin{array}{c} y\equiv m\alpha_{\pr_1}\mod{\pr_1^\nu}\\ \vdots\ \ \ \\ y\equiv m\alpha_{\pr_s}\mod{\pr_s^\nu}\\ \end{array}\] We will show that for sufficiently big $\nu$, $\alpha -\frac{y}{m}\in A_ {K,S_\infty}$ or equivalently \begin{equation}\label{ARS3eqn1} v\left(\alpha_v-\frac{y}{m}\right)\geq 0 \end{equation} for all non-archimedean $v$ and this will finish the proof. Since in a number field, all non-archimedean valuations are places at a prime we will show \ref{ARS3eqn1} for $v = v_\pr$. If $\pr$ is a prime different from all $\pr_j$ then $v_\pr(m)=0$ and \begin{eqnarray*} v_\pr\left(\alpha_\pr-\frac{y}{m}\right)&\geq& \inf\{v_\pr(\alpha_\pr),v_\pr(y)-v_\pr(m)\}\\ &\geq&\inf\{v_\pr(\alpha_\pr),v_\pr(y)\} \\ &\geq& 0. \end{eqnarray*} On the other hand if $\pr$ is in $S$ say $\pr =\pr_j$ then \begin{eqnarray*} v_{\pr_j}\left(\alpha_{\pr_j}-\frac{y}{m}\right) &=&v_{\pr_j}\left(\frac{m\alpha_{\pr_j}-y}{m}\right)\\ &=&v_{\pr_j}(m\alpha_{\pr_j}-y)-v_{\pr_j}(m)\\ &\geq& \nu -v_{\pr_j}(m)\\ &\geq& 0. \end{eqnarray*} \item[(c)] >From the basic theory we know that the image $L=\lambda({\mathcal O}_K)$ of ${\mathcal O}_K$ under the embedding \[K\stackrel{\lambda}{\hookrightarrow}\prod_{v\in S_\infty} K_v \cong \R^N\] is a lattice (more specifically it is a free abelian group of rank $N$ such that $\R^N/L$ is compact). Let $\alpha$ be an adele. By part (b) we can translate $\alpha$ by an element $ x$ of $K$ so that $\beta =\alpha+x$ lies in $A_{K,S_\infty}$ that is, \[\beta =\alpha+x\in A_{K,S_\infty}=\left(\prod_{v\in S_\infty}K_v\right)\times \left(\prod_{v\not\in S_\infty}\ov\right).\] We can further translate $\beta$ by an element $z$ of ${\mathcal O}_K$ so that the archimedean component of lies of $\beta+z$ lies in a compact fundamental region $W$ for $L$ in $\R^N$. In other words we can translate an arbitrary adele $\alpha$ by an element in $K$ so that it lands in the compact set\footnote{The compactness of \[W\times \left(\prod_{v\not\in S_\infty}\ov\right)\] is a consequence of Tychonoff's theorem, since all components of this product are compact topological spaces.} \[W\times\left(\prod_{v\not\in S_\infty}\ov\right).\] This shows that $A_K/K$ is compact \end{itemize} \end{proof} \end{document}