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\title{Math 129: Algebraic Number Theory\\
  {\bf\Large Lecture 21: Global Fields and Adeles}}

\author{William Stein (based closely on Cassels's {\em Global Fields}
  article in Cassels-Fr\"ohlich)}

\date{Tuesday, April 27, 2004}
\begin{document}
\maketitle 

\setcounter{section}{11}
\section{Global Fields}
\begin{definition}[Global Field]
  A {\em global field} is a number field or a finite separable
  extension of $\F(t)$, where $\F$ is a finite field, and $t$ is
  transcendental over $\F$.
\end{definition}

Below we will focus attention on number fields leaving the function
field case to the reader.   

The following lemma essentially says that the denominator of an
element of a global field is only ``nontrivial'' at a finite number of
valuations.
\begin{lemma}\label{lem:absbig}
Let $a\in K$ be a nonzero element of a global field $K$.  Then
there are only finitely many inequivalent valuations $\absspc$
of $K$ for which 
$$
  \abs{a} > 1.
$$
\end{lemma}
\begin{proof}
  If $K=\Q$ or $\F(t)$ then the lemma follows by Ostrowski's
  classification of all the valuations on~$K$.\edit{ref} For example,
  when $a=\frac{n}{d}\in\Q$, with $n,d\in \Z$, then the valuations
  where we could have $\abs{a}>1$ are the archimedean one, or the
  $p$-adic valuations $\absspc_p$ for which $p\mid d$.

Suppose now that $K$ is a finite extension of $\Q$, so
$a$ satisfies a monic polynomial
$$
  a^n + c_{n-1} a^{n-1} + \cdots + c_0 = 0,
$$
for some $n$ and $c_0,\ldots, c_{n-1}\in\Q$.
If $\absspc$ is a non-archimedean valuation on $K$, we have
\begin{align*}
  \abs{a}^n &= \abs{-(c_{n-1} a^{n-1} + \cdots + c_0)} \\
      &\leq \max(1,\abs{a}^{n-1})\cdot \max(\abs{c_0},\ldots,\abs{c_{n-1}}).
\end{align*}
Raising each side to the power $1/(n-1)$, we have
whenever $\abs{a}\geq 1$ that
$$
   \abs{a} \leq \abs{a}^{n/(n-1)}
     \leq \max(\abs{c_0},\ldots,\abs{c_{n-1}})^{1/(n-1)},
$$
so in all cases we have
\begin{equation}\label{eqn:maxabs}
   \abs{a} \leq \max(1, \abs{c_0},
   \ldots,\abs{c_{n-1}})^{1/(n-1)}.
\end{equation}
We know the lemma for $\Q$, so there are only finitely
many valuations $\absspc$ on $\Q$ such that the right
hand side of (\ref{eqn:maxabs}) is bigger than~$1$.  
Since each valuation of $\Q$ has finitely many extensions
to~$K$, and there are only finitely many archimedean valuations, it follows that there are only finitely many valuations
on~$K$ such that $\abs{a}>1$.  
\end{proof}

Any valuation on a global field is either archimedean, or discrete
non-archimedean with finite residue class field, since this is true of
$\Q$ and $\F(t)$ and is a property preserved by extending a valuation
to a finite extension of the base field.  Hence it makes sense to talk
of normalized valuations.  Recall that the normalized $p$-adic
valuation on $\Q$ is $\abs{x}_p = p^{-\ord_p(x)}$, and if~$v$ is a
valuation on a number field~$K$ equivalent to an extension of
$\absspc_p$, then the normalization of $v$ is the composite of the
sequence of maps
$$
  K\hra K_v \xra{\Norm} \Q_p \xra{\absspc_p} \R,
$$
where $K_v$ is the completion of $K$ at $v$.

\begin{example}
Let $K=\Q(\sqrt{2})$, and let $p=2$.  Because $\sqrt{2}\not\in\Q_2$, there is
exactly one extension of $\absspc_2$ to~$K$, and
it sends $a=1/\sqrt{2}$ to
$$
  \abs{\Norm_{\Q_2(\sqrt{2})/\Q_2}(1/\sqrt{2})}^{1/2}_2 = \sqrt{2}.
$$
Thus the normalized valuation of $a$ is $2$.

There are two extensions of $\absspc_7$ to $\Q(\sqrt{2})$,
since $\Q(\sqrt{2})\tensor_\Q \Q_7 \isom \Q_7 \oplus \Q_7$,
as $x^2-2 = (x-3)(x-4)\pmod{7}$.  The image of $\sqrt{2}$
under each embedding into $\Q_7$ is a unit in $\Z_7$, so
the normalized valuation of $a=1/\sqrt{2}$ is, in both
cases, equal to $1$.  More generally, for any valuation
of $K$ of characteristic an odd prime $p$, the
normalized valuation of $a$ is $1$.

Since $K=\Q(\sqrt{2})\hra \R$ in two ways, there are exactly
two normalized archimedean valuations on $K$, and
both of their values on $a$ equal $1/\sqrt{2}$.
Notice that the product of the absolute values of $a$
with respect to all normalized valuations is
$$
   2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \cdot 1
   \cdot 1 \cdot 1 \cdots  = 1.
$$
This ``product formula'' holds in much more generality, as
we will now see.
\end{example}

\begin{theorem}[Product Formula]\label{thm:product_formula}
Let $a\in K$ be a nonzero element of a global field~$K$.
Let $\absspc_v$ run through the normalized valuations
of $K$.  Then $\abs{a}_v=1$ for almost all $v$, and 
$$
 \prod_{\text{\rm all }v} \abs{a}_v = 1\qquad{\text{\rm (the product formula).}}
$$
\end{theorem}
%We will later give a less computational proof of this.\edit{ref}
\begin{proof}
By Lemma~\ref{lem:absbig}, we have $\abs{a}_v\leq 1$
for almost all~$v$.  Likewise, $1/\abs{a}_v = \abs{1/a}_v\leq 1$
for almost all~$v$, so $\abs{a}_v = 1$ for almost all~$v$.

Let $w$ run through all normalized valuations of $\Q$ (or of $\F(t)$),
and write $v\mid w$ if the restriction of $v$ to $\Q$ is equivalent to $w$.
Then (by the previous section\edit{ref}), 
$$
 \prod_{v} \abs{a}_v = \prod_w \left(\prod_{v\mid w} \abs{a}_v\right)
     = \prod_w \abs{\Norm_{K/\Q}(a)}_w,
$$
so it suffices to prove the theorem for $K=\Q$.

By multiplicativity of valuations, if the theorem is true for $b$ and
$c$ then it is true for the product $b c$ and quotient $b/c$ (when
$c\neq 0$). The theorem is clearly true for $-1$, which has valuation
$1$ at all valuations.  Thus to prove the theorem for $\Q$ it suffices
to prove it when $a=p$ is a prime number.  Then we have
$\abs{p}_\infty = p$, $\abs{p}_p = 1/p$, and for primes $q\neq p$ that
$\abs{p}_q = 1$.  Thus
$$\prod_v \abs{p}_v = p \cdot \frac{1}{p} \cdot 1 \cdot 1 \cdot 1 \cdots = 1,$$
as claimed.
\end{proof}


If $v$ is a valuation on a field $K$, recall that 
we let $K_v$ denote the completion of $K$ with respect to $v$. Also when
$v$ is non-archimedean, let
$$
  \O_v = \O_{K,v} = \{x \in K_v : \abs{x} \leq 1\}
$$
be the ring of integers of the completion. 

\begin{definition}[Almost All]
We say a condition holds for {\em almost all} elements
of a set if it holds for all but finitely many elements.
\end{definition}

We will use the following lemma later\edit{ref} to
prove that formation of the adeles of a global field 
is compatible with base change. 
\begin{lemma} 
Let $\omega_1,\ldots, \omega_n$ be a basis for $L/K$,
where $L$ is a finite separable extension of the global field
$K$ of degree~$n$. 
Then for almost all normalized non-archimedean valuations $v$ on $K$ we
have
\begin{equation}\label{eqn:sum_int}
   \omega_1 \O_{v} \oplus \cdots \oplus \omega_n \O_{v}
      = \O_{w_1} \oplus \cdots \oplus \O_{w_g}
      \subset K_v\tensor_K L,
\end{equation}
where $w_1,\ldots, w_g$ are the extensions of $v$
to $L$.   Here we have identified $a\in L$ with
its canonical image in $K_v\tensor_K L$, and the direct 
sum on the left is the sum taken inside the tensor
product (so directness means that the intersections are
trivial). 
\end{lemma}
\begin{proof}
  The proof proceeds in two steps.  First we deduce easily from
  Lemma~\ref{lem:absbig} that for almost all $v$ the left hand side 
of (\ref{eqn:sum_int}) is
  contained in the right hand side.  Then we use a trick involving
  discriminants to show the opposite inclusion for all but finitely
  many primes.
  
  Since $\O_v\subset \O_{w_i}$ for all $i$, the left hand side of
  (\ref{eqn:sum_int}) is contained in the right hand side if
  $\abs{\omega_i}_{w_j}\leq 1$ for $1\leq i\leq n$ and $1\leq j\leq
  g$.  Thus by Lemma~\ref{lem:absbig}, for all but finitely many~$v$
  the left hand side of (\ref{eqn:sum_int}) is contained in the right
  hand side.  We have just eliminated the finitely many primes
  corresponding to ``denominators'' of some $\omega_i$, and now only
  consider~$v$ such that $\omega_1,\ldots, \omega_n \in \O_{w}$ for
  all $w\mid v$.
  
  For any elements $a_1,\ldots, a_n \in K_v\tensor_K L$, consider the
  discriminant
  $$
  D(a_1,\ldots, a_n) = \det(\Tr(a_i a_j)) \in K_v,
  $$
  where the trace is induced from the $L/K$ trace.
  Since each $\omega_i$ is in each $\O_w$, for $w\mid v$, the
  traces lie in $\O_v$, so
  $$d = D(\omega_1,\ldots, \omega_n)\in \O_v.$$
  Also note that $d\in
  K$ since each $\omega_i$ is in $L$.  Now suppose that
  $$
  \alpha = \sum_{i=1}^n a_i \omega_i \in \O_{w_1} \oplus \cdots
  \oplus \O_{w_g},
  $$
  with $a_i \in K_v$.  Then by properties of determinants for any
  $m$ with $1\leq m\leq n$, we have
  \begin{equation}\label{eqn:discsquare} 
  D(\omega_1,\ldots, \omega_{m-1}, \alpha, \omega_{m+1}, \ldots, \omega_n)
    = a_m^2 D(\omega_1,\ldots, \omega_n).
  \end{equation}
  The left hand side of (\ref{eqn:discsquare}) is in $\O_v$, so the
  right hand side is well, i.e.,
  $$
  a_m^2 \cdot d \in \O_v, \qquad\text{(for }m=1,\ldots, n\text{)},
  $$
  where $d\in K$. Since $\omega_1,\ldots, \omega_n$ are a basis for
  $L$ over $K$ and the trace pairing is nondegenerate, we have $d\neq
  0$, so by Theorem~\ref{thm:product_formula} we have $\abs{d}_v=1$
  for all but finitely many~$v$.  Then for all but finitely many~$v$
  we have that $a_m^2\in \O_v$.  For these $v$, that $a_m^2\in\O_v$
  implies $a_m\in \O_v$ since $a_m\in K_v$, i.e., $\alpha$ is in the
  left hand side of (\ref{eqn:sum_int}).
\end{proof}
\begin{example}
Let $K=\Q$ and $L=\Q(\sqrt{2})$.  Let $\omega_1 = 1/3$ and $\omega_2 = 2\sqrt{2}$.  In the first stage of the above proof we would eliminate
$\absspc_3$ because $\omega_2$ is not integral at $3$.  The discriminant
is
$$
 d = D\left(\frac{1}{3}, 2\sqrt{2}\right)
   =\det \mtwo{\frac{2}{9}}{0}{0}{16} = \frac{32}{9}.
$$
As explained in the second part of the proof, as long as $v\neq 2, 3$,
we have equality of the left and right hand sides in (\ref{eqn:sum_int}).
\end{example}


\setcounter{section}{12}
\section{Restricted Topological Products}
\newcommand{\bx}{\mathbf{x}}
\newcommand{\by}{\mathbf{y}}

In this section we describe a topological tool, which we need in order
to define adeles (see Definition~\ref{def:adele}).

\begin{definition}[Restricted Topological Products]
  Let $X_\lambda$, for $\lambda\in\Lambda$, be a family of topological
  spaces, and for almost all~$\lambda$ let $Y_{\lambda}\subset
  X_{\lambda}$ be an open subset of $X_{\lambda}$.  Consider the space
  $X$ whose elements are sequences $\bx = \{x_\lambda\}_{\lambda\in
    \Lambda}$, where $x_\lambda\in X_\lambda$ for every $\lambda$, and
  $x_\lambda\in Y_{\lambda}$ for almost all~$\lambda$.  We give $X$ a
  topology by taking as a basis of open sets the sets $\prod
  U_{\lambda}$, where $U_{\lambda}\subset X_{\lambda}$ is open for all
  $\lambda$, and $U_{\lambda} = Y_{\lambda}$ for almost all $\lambda$.
  We call~$X$ with this topology the {\em restricted topological
    product} of the $X_{\lambda}$ with respect to the $Y_{\lambda}$.
\end{definition}


\begin{corollary}\label{lem:xs} 
  Let $S$ be a finite subset of $\Lambda$, and let $X_S$ be the set of
  $\bx\in X$ with $x_\lambda\in Y_\lambda$ for all $\lambda\not\in S$,
  i.e.,
  $$
  X_S = \prod_{\lambda \in S} X_{\lambda} \times
  \prod_{\lambda\not\in S} Y_{\lambda} \subset X.
  $$
  Then $X_S$ is an open subset of $X$, and the topology induced on
  $X_S$ as a subset of $X$ is the same as the product topology.
\end{corollary}

The restricted topological product depends on the totality of the
$Y_{\lambda}$, but not on the individual $Y_{\lambda}$:
\begin{lemma}
  Let $Y_{\lambda}'\subset X_{\lambda}$ be open subsets, and suppose
  that $Y_{\lambda} = Y_{\lambda}'$ for almost all~$\lambda$.  Then
  the restricted topological product of the $X_\lambda$ with respect
  to the $Y_{\lambda}'$ is canonically isomorphic to the restricted
  topological product with respect to the $Y_{\lambda}$.
\end{lemma}

\begin{lemma}
  Suppose that the $X_\lambda$ are locally compact and that the
  $Y_\lambda$ are compact.  Then the restricted topological
product $X$ of the $X_\lambda$ is locally compact.
\end{lemma}
\begin{proof}
  For any finite subset $S$ of $\Lambda$, the open subset $X_S\subset
  X$ is locally compact, because by Lemma~\ref{lem:xs} it is a product
  of finitely many locally compact sets with an infinite product of
  compact sets.  (Here we are using Tychonoff's theorem from
  topology, which asserts that an arbitrary product of compact
  topological spaces is compact (see Munkres's {\em Topology, a first course}, chapter
  5).) Since $X=\cup_{S} X_S$, and the $X_S$ are open in $X$, the
  result follows.
\end{proof}

\edit{where will this be used later?}
\begin{definition}[Product Measure]
  For all $\lambda\in\Lambda$, suppose $\mu_\lambda$ is a measure on
  $X_\lambda$ with $\mu_\lambda(Y_\lambda) = 1$ when $Y_\lambda$ is
  defined.  We define the {\em product measure} $\mu$ on $X$ to be
  that for which a basis of measurable sets is $$\prod_\lambda
  M_\lambda$$
  where each $M_\lambda\subset X_\lambda$ has finite
  $\mu_\lambda$-measure and
  $M_\lambda=Y_\lambda$ for almost all $\lambda$, and where
  $$
  \mu\left(\prod_\lambda M_\lambda\right) = \prod_\lambda
  \mu_\lambda(M_\lambda).
  $$
\end{definition}

%% \begin{corollary}
%%   The restriction of $\mu$ to $X_S$ is the ordinary product
%%   measure.
%% \end{corollary}

\setcounter{section}{13}

\section{The Adele Ring}
Let $K$ be a global field.  For each normalization $\absspc_v$ of $K$,
let $K_v$ denote the completion of $K$.  If $\absspc_v$ is
non-archimedean, let $\O_v$ denote the ring of integers of $K_v$.
\renewcommand{\AA}{\mathbb{A}}

\begin{definition}[Adele Ring]\label{def:adele}
  The {\em adele ring} $\AA_K$ of $K$ is the topological ring whose
  underlying topological space is the restricted topological product
  of the $K_v$ with respect to the $\O_v$, and where addition and
  multiplication are defined componentwise:
\begin{equation}\label{eqn:adelearith} 
(\bx \by)_v = \bx_v \by_v \qquad
(\bx + \by)_v = \bx_v + \by_v\qquad
\text{for }\bx, \by\in\AA_K.
\end{equation}
\end{definition}
It is readily verified that (i) this definition makes sense, i.e., if
$\bx, \by\in \AA_K$, then $\bx\by$ and $\bx+\by$, whose components are
given by (\ref{eqn:adelearith}), are also in $\AA_K$, and (ii) that
addition and multiplication are continuous in the $\AA_K$-topology, so
$\AA_K$ is a topological ring, as asserted.  Also, $\AA_K$ is locally
compact because the $K_v$ are locally compact, and the $\O_v$ are
compact.\edit{ref}

There is a natural continuous ring inclusion
\begin{equation}\label{eqn:princ_inc} 
K\hra \AA_K
\end{equation}
that sends $x\in K$ to the adele every one of whose components is $x$.
This is an adele because $x\in \O_v$ for almost all $v$, by
Lemma~\ref{lem:absbig}.  The map is injective because each map $K\to
K_v$ is an inclusion.
 
\begin{definition}[Principal Adeles]
  The image of (\ref{eqn:princ_inc}) is the ring of {\em principal
    adeles}.
\end{definition}
It will cause no trouble to identify $K$ with the principal adeles, so
we shall speak of~$K$ as a subring of $\AA_K$.

Formation of the adeles is compatibility with base change, in the
following sense.
\begin{lemma}
  Suppose $L$ is a finite (separable) extension of the global field
  $K$.  Then
  $$
  \AA_K \tensor_K L \isom \AA_L
  $$
  both algebraically and topologically.  Under this isomorphism,
  $L\isom K\tensor_K L \subset \AA_K \tensor_K L$ maps isomorphically onto
  $L\subset \AA_L$.
\end{lemma}

\begin{center}
*\qquad * \qquad *
\end{center}

Next time we will start by proving the above lemma.  Here are some
further highlights of what will come in the final three lectures.
\begin{theorem}
  The global field $K$ is discrete in $\AA_K$ and the quotient
  $\AA_K^+/K^+$ of additive groups is compact in the quotient
  topology.
\end{theorem}

If we remove even one valuation, then the situation changes
dramatically:
\begin{theorem}[Strong Approximation]
  Let $v_0$ be any valuation of the global field~$K$.  Define $\AA_K'$
  to be the restricted topological product of the $K_v$ with respect
  to the $\O_v$, where $v$ runs through all normalized valuations
  $v\neq v_0$.  Then $K$ is dense in $\AA_K'$.
\end{theorem}

\newcommand{\II}{\mathbb{I}}

\begin{definition}[Idele Group]
  The idele group $\II_K$ of $K$ is the group $\AA_K^*$ of invertible
  elements of the adele ring $\AA_K$.
%with the topology just defined.
\end{definition}

The subgroup $\II_K^1$ is the subgroup of ideles $\bx=\{x_v\}$ such
that $\ds\prod_v \abs{x_v}_v=1$.  Note that $K^*\subset \II_K^1$.

\begin{theorem} The quotient
  $\II_K^1/K^*$ with the quotient topology is compact.
\end{theorem}

\begin{theorem} The
  ideal class group of $K$ (with the discrete topology) is canonically
  isomorphic to $\II_K^1/K^*$.  
\end{theorem}
Since a discrete compact group is finite, this proves
that the ideal class group of a global field is finite.

I think we will likely stop here, and not do sections 18 and 19.  This
will be a nice conclusion, because you'll finish the class having
learned the basic theorems and objects of algebraic number theory,
both from the classical and adelic points of view, and will have seen
a nontrivial result proved from both directions.  You have also
learned something about the structure of local fields (completions of
global fields).
\end{document}

