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\begin{document}

\title{Lecture 22}

\author{Uroyo\'an R. Walker}

\address{Louisiana State University,
Department of Mathematics, Baton Rouge, LA 70803, USA}

\date{March 10, 2000}


\email{walker@math.lsu.edu} \maketitle

\section{Local Fields}
\setcounter{subsection}{7}
\subsection{Absolute Values of Algebraic Number Fields (continued)}
\begin{thm} \label{8.4}For a number field $k$, and $0\neq x\in k$,
$\displaystyle{\prod _{v\in M_k}\norm x_v}=1.$
\end{thm}
\begin{rem} $\norm x_v=1$ for all but finitely many
$v$'s.  This follows since for $(x)=\mfr p_1^{n_1}\cdots\mfr
p_r^{n_r}$ in $k$, $\norm x_{\mfr p}=1$ if $\mfr p\not\in\set{\mfr
p_1,\ldots ,\,\mfr p_r}$.
\end{rem}\proof Let's start by considering the special case when
$k=\Q$.  Note that the LHS is multiplicative in $x$. Hence it will
suffice to prove the identity when $x=\ell$, a prime number.
Recall that by Ostrowski's Theorem $\displaystyle{M_\Q
=\set{\infty ,\, 2,\, 3,\, 5,\ldots ,\, p,\ldots }}$.  Now,
$$\norm \ell_v=\left\{\begin{array}{ll}\ell & \mbox{if } v=\infty
\\ \frac{1}{\ell} & \mbox{if } v=\ell \\ 1 & \mbox{if }
v\neq\ell \end{array}\right.$$ \noindent This follows since $\norm
x_{\mfr p} = \left(\frac{1}{q}\right)^{v_{\mfr p}(x)}$ where $q=\#
\ok k/\mfr p$.  Now, let $k/\Q$ be any number field, and $0\neq
x\in k$, recall that $N(x)\in\Q ^\times$ so
\begin{eqnarray*}1 &=& \prod _{v\in M_\Q}{\norm {N(x)}_v}\\ &=&
\prod _{v\in M_\Q}\left( \prod _{w\mid v}{\abs
x_w^{n(w)}}\right)\quad\mbox{where }n(w)=[k_w\colon\Q
_v]\enskip\mbox{by Corollary }8.2\\ &=& \prod _{w\in M_k}{\abs
x_w^{n(w)}} \end{eqnarray*}\noindent Hence we need only show that
$\abs x_w^{n(w)}=\norm x_w$.  But for $w\mid v$, if we set
$e=e(w|v)$ and $f=f(w|v)=[k_w\colon k_v]$, by definition, $\norm
x_w=q_w^{-w(x)}$, where $q_w=q_v^f$.  We also have $\abs
x_w=q_v^{-\frac{1}{e}w(x)}$.  To see this let $\pi$ be the
uniformizer down below, hence lifting it to the level of $k_w$ we
must have $\pi =u\Pi ^e$, for some unit $u$ and $e$ the
ramification index. What do we have? $$\abs
y_w=q_v^{-Cw(y)}\enskip\mbox{for some } C\enskip\mbox{and }\norm
x_v=q_v^{-v(x)}$$ Recall that $\Vert\cdot\Vert$ denotes the
normalized absolute value.  We must have $\norm\pi _v=\abs\pi _w$.
But $$\norm\pi _v=q_v^{-v(\pi )}=q_v^{-1}$$ and $$\abs\pi _w
=q_v^{-Cw(\pi )}=q_v^{-Ce}$$ Thus, $C=1/e$. Hence $$\abs
x_w^{n(w)}=\abs x_w^{ef}=\left(
q_v^{-\frac{1}{e}w(x)}\right)^{ef}=q_v^{-fw(x)}=q_w^{-w(x)}=\norm
x_w$$ \endproof \begin{pro}\label{8.1} Let $L/k$ be a Galois
extension of number fields with $\displaystyle{\gal=\Gal(L/k)}$.
Let $v$ be any place of $k$. Then $\gal$ acts transitively on the
places of $L$ lying over $v$.\end{pro}\proof First note that
$\gal$ acts on these places. We define the action of $\gal$ on the
places of $L$ lying over $v$ as follows, for $w\mid v$ and
$\sigma\in\gal$, we define $\sigma\ast w$ by $\abs x_{\sigma\ast
w}=\abs {\sigma^{-1}x}_w$. Now, the various places are given by
embeddings $$\begin{array}{rll}L & & \bar k_v
\\ \mid & & \mid \\ k & \hookrightarrow & k_v
\end{array}$$ Write $L=k(\alpha )$ then the diagram above is
$$\begin{array}{rll}L=k(\alpha ) & \hookrightarrow & \bar k_v \\
\mid & & \mid \\ k & \hookrightarrow & k_v \end{array}$$ Take
$f(x)$ to be the minimal polynomial for $\alpha$ over $k[X]$, then
$$f(x)=\varphi _1(x)\cdots\varphi _g(x)$$ over $ k_v[X]$, and if
$\beta$ is a root of (say) $\varphi _1(x)$ we get
$$\begin{array}{llll}L & \hookrightarrow & \bar k_v & \\ \mid & &
\mid & \diagdown \\ \mid & & \mid & k_v[\beta ] \\ k &
\hookrightarrow & k_v & \diagup
\end{array}$$ And this proves the proposition.\endproof \noindent
In general we have various $w_i$ lying over a given $v$
$$\begin{array}{ccl}w_1,\ldots ,w_g & L & \\ \mid & \mid & n=\sum
_{i=1}^g{e(w_i|v)f(w_i|v)}\\ v & k & \end{array}$$ In the Galois
case all the $e$'s and the $f$'s over a given $v$ are equal.  In
particular, $n=efg$.\vskip.2cm\noindent We move on to adeles and
ideles.
\section{Adeles and Ideles}\subsection{Restricted Topological
Product}\begin{Def}\label{1.1}Let
$\set{X_\lambda}_{\lambda\in\Lambda}$ be a family of topological
spaces, and for almost all, {\it i.e.\/}, all but finitely many,
$\lambda$ we are given an open subset $U_\lambda\subset
X_\lambda$.  Let $X=\set{(x_\lambda )\in\prod
_{\lambda\in\Lambda}{X_\lambda}\colon x_\lambda\in
U_\lambda\,\mbox{for almost all }\lambda\in\Lambda}$.  We make $X$
into a topological space where a basis of open sets are $W=\prod
_{\lambda\in\Lambda}{W_\lambda}$, where $W_\lambda\subset
X_\lambda$ is open for all $\lambda$ and equals $U_\lambda$ for
almost all $\lambda$.  This is called the {\it restricted
topological product\/} of $\set{X_\lambda}_{\lambda\in\Lambda}$
with respect to $\set{U_\lambda}_{\lambda\in\Lambda}$, sometimes
the restricted topological product is denoted $\displaystyle{\prod
_{\lambda\in\Lambda}{\left( X_\lambda ,U_\lambda\right)}}$. Notice
that $X$ is a subset of the Cartesian product of the
$X_\lambda$'s.\end{Def}\noindent One of the main examples that we
will be interested in is \begin{exa}Let $k$ be a number field,
$\Lambda =M_k,\, X_\lambda =k_v$ and $U_\lambda=\ok v\subset k_v$.
Note that the latter only makes sense for $v$ a non-archimedean
valuation.  Then $A_k=\displaystyle{\prod _{v\in M_k}{\left( k_v,
\ok v\right)}}$ is called the ring of adeles. \end{exa}\noindent
Now, what is an adele? In an extension of degree $n=r_1+2r_2$ an
adele $\alpha$ has the form $$\alpha =(\underbrace{\alpha
_1,\ldots ,\alpha _{r_1},\,\alpha _{r_1+1},\ldots ,\,\alpha
_{r_1+r_2}}_{ {}_{\mbox{archimedean}}},\underbrace{\,\ldots
,\,\overbrace{\alpha _{\mfr p}}^{\in k_{\mfr p}}
,\ldots}_{{}_{\mbox{non-archimedean}}})$$ Here $\alpha _{\mfr
p}\in k_{\mfr p}$ and moreover $\alpha _{\mfr p}\in\ok {\mfr p}$
for almost all $\mfr p$.  This could be reformulated by saying
that for $\alpha =\alpha _v$ we have $\abs {\alpha _v}_v\leqslant
1$ for almost all $v$.\\ \noindent That $A_k$ is a topological
ring is almost clear.  We add and multiply vectors componentwise
and it can be shown that addition and multiplication are
continuous maps in this topology.
\begin{lem}\label{1.1}Let $S\subset\Lambda$ be a finite subset.
Define $$X_S=\left(\prod _{\lambda\in
S}{X_\lambda}\right)\left(\prod _{\lambda\not\in
S}{U_\lambda}\right)$$ Then $X_S$ is open in this topology and
$X_S$ as a subset of $X$ is endowed with the product topology.
\end{lem}\proof It's clear that $X_S$ is open.  In the product
topology the basic open sets are $\prod {W_\lambda}$ where
$W_\lambda$ is open and $W_\lambda$ equals the whole thing for
almost all $\lambda$.  In particular, in the product topology of
the $X_S$'s, $W_\lambda =U_\lambda$, for almost all
$\lambda$.\endproof\noindent Note that $X=\bigcup _S{X_S}$.
\end{document}
¬