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% Class notes on March 20, 2000 Lecture Notes no3 prepared for Dr. Hoffman's 
%Number Theory Class.

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\begin{document}

\title{Lecture 26}

\author{Cem Guneri}

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

\date{\today}

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

\begin{pro}Let $K$ be a number field. There exists a constant $C>0$, depending 
only on $K$, such that the following holds:

If $\alpha \in A_{K}$ is given with 
\[\prod_{v\in M_{K}} |\alpha_{v}|_{v} > C\]
then there exists a principal adele $0\neq \beta \in K \subset A_{K}$ such that 
$|\beta|_{v}\leq |\alpha_{v}|_{v}$ for every $v$ in $M_{K}$.
\end{pro}

\proof We will use some facts about Haar measure $\mu$ on $A_{K}$. Therefore even 
though Haar measures will be discussed in more detail later on, we will just 
mention some basic facts that will serve our purposes at this time.

$\mu$ is a translation invariant volume measure. A very important fact that we 
need is that every locally compact group has a Haar measure. Therefore $A_{K}$ 
has one, too. Furthermore $\mu$ on $A_{K}$ induces a measure on $A_{K}/K$ and 
since $A_{K}/K$ is compact let $\mu(A_{K}/K)=c_{0} < \infty$.

Now define the following subset of $A_{K}$:
\[W=\{\gamma \in A_{K}; \hspace{2pt} v:arch. \Rightarrow |\gamma_{v}|_{v} \leq 
\frac{1}{10}, \hspace{2pt} v:nonarch. \Rightarrow |\gamma_{v}|_{v} \leq 1\} \]

Then archimedean components of $W$ are closed and bounded (hence compact) sets 
and nonarchimedean ones are exactly $O_{\mfr{p}}$ for every prime $\mfr{p}$. Then by 
Tychonoff's Theorem $W$ is compact and hence it has a finite measure. 
Let $\mu(W)=c_{1} < \infty$. Our claim is to show that $C=\frac{c_{0}}{c_{1}}$ 
will do it.

Let $\alpha\in A_{K}$ be an arbitrary element such that $\prod |\alpha_{v}|_{v} > C$. 
Consider the set 
\[V=\{\zeta\in A_{K}; \hspace{2pt} v:arch. \Rightarrow |\zeta_{v}|_{v} \leq \frac{1}{10}
|\alpha_{v}|_{v}, \hspace{2pt} v:nonarch. \Rightarrow |\zeta_{v}|_{v} \leq |\alpha_{v}|_{v}\} \]

Then 
\[\mu(V)=\bigl(\prod |\alpha_{v}|_{v}\bigr) \mu(W) > Cc_{1}=c_{0}\]

But $V \subset A_{K}$ and if we consider the natural map from $A_{K}$ to $A_{K}/K$, 
there is a continuous map $V\longrightarrow A_{K}/K$. $\mu(A_{K}/K)=c_{0}$ and 
$\mu(V)>c_{0}$ imply that there exist two distinct elements, $\zeta_{1}\neq 
\zeta_{2}$, in $V$ which map to the same element in $A_{K}/K$. Let $\beta=
\zeta_{1}- \zeta_{2} \in K$. (Observe that if $\beta$ is zero the proof is 
obviously over) The $v$-value of $\beta$ is:
\[|\beta|_{v}=|\zeta_{1_v} - \zeta_{2_v}|_{v} \]

If $v$ is archimedean then $|\beta|_{v}\leq \frac{1}{5}|\alpha_{v}|_{v}$ and if 
$v$ is nonarchimedean then $|\beta|_{v}\leq max(|\zeta_{1_v}|_{v}, 
|\zeta_{2_v}|_{v})\leq |\alpha_{v}|_{v}$. Hence we found an element $\beta\in
K^*$ so that $|\beta|_{v}\leq|\alpha_{v}|_{v}$ for every $v\in M_{K}$. \endproof 

\begin{cor}Let $v_{0}$ be a normalized valuation on a number field $K$ and suppose 
$\delta_{v}>0$ is given for every $v_{0}\neq v\in M_{K}$ with $\delta_{v}=1$ for 
almost all $v$. Then there exists $\beta\in K^*$ such that $|\beta|_{v}\leq 
\delta_{v}$ for every $v\neq v_{0}$.
\end{cor}

\proof Let $K_{v}$ denote the completion of $K$ at $v$ and for every $v\in M_{K}$ 
except $v_{0}$ choose an element $\alpha_{v}\in K_{v}$ such that
\[0< |\alpha_{v}|_{v}\leq \delta_{v} \hspace{10pt} if \hspace{2pt} \delta_{v}\neq 1\]
\[|\alpha_{v}|_{v}=1 \hspace{18pt} if \hspace{2pt} \delta_{v}=1 \]

At this point the product of all $|\alpha_{v}|_{v}$ (excluding $v_{0}$ of course) 
is actually a finite product since $|\alpha_{v}|_{v}=1$ for infinitely many $v$. 
Now choos $\alpha_{v_0}\in K_{v_0}$ such that product over all $v\in M_{K}$ is greater 
than the constant $C$ assigned to the number field $K$. (i.e; the constant in 
Prop 3.2) This way we obtain $\alpha=(\alpha_{v})$ in $A_{K}$ such that
\[\prod_{v\in M_{K}} |\alpha_{v}|_{v} > C \]

Then by Prop 3.2 there exists $\beta\in K^*\subset A_{K}$ such that $|\beta|_{v}
\leq|\alpha_{v}|_{v}$ for all $v$. But this clearly finishes our proof.
\endproof

\begin{thm}$K^*\hookrightarrow J_{K}^1$ is a discrete subgroup and $J_{K}^1/K^*$ 
is compact.
\end{thm}

\proof We know that $K^*$ is embedded in $J_{K}^1$ due to the product formula. 
The fact that it is a discrete subgroup comes from the fact that $K$ in $A_{K}$ 
is a discrete subgroup. Because the idele topology of $J_{K}^1$ is the same as 
the subspace topology it inherits from $A_{K}$ and hence if $K\subset A_{K}$ is 
a discrete subgroup then $K^* \hookrightarrow J_{K}^1 \subset A_{K}$ is a discrete 
subgroup. Hence we only need to show the compactness of $J_{K}^1/K^*$. 

For this we will come up with a compact set in $J_{K}^1$ whose image under the 
natural map $J_{K}^1 \longrightarrow J_{K}^1/K^*$ is onto. Also observe that 
finding an $A_{K}$-compact set $W$ such that $W\cap J_{K}^1$ maps onto $J_{K}^1/K^*$ 
is enough again since the topology of $J_{K}^1$ is the $A_{K}$ subspace topology. 

Now let $C$ be the constant in Prop 3.2. that is associated to the number field $K$ 
and choose $\alpha \in A_{K}$ such that $\prod |\alpha_{v}|_{v} > C$. Define 
\[W=\{\zeta \in A_{K}; \hspace{2pt} |\zeta_{v}|_{v} \leq |\alpha_{v}|_{v},
\hspace{2pt} \forall v\in M_{K}\} \]

Observe that since $\prod |\alpha_{v}|_{v} > C$, $|\alpha_{v}|_{v}=1$ for almost 
all $v\in M_{K}$. (Otherwise the infinite product would approach to zero since 
infinitely many terms in the product would have a value less than one) So the 
set $W$ is exactly the same as the $W$ we used in the proof of Prop 3.2. and hence 
it is a compact subset of $A_{K}$. 

The last thing to show is that $W\cap J_{K}^1$ maps onto $J_{K}^1/K^*$. Take an 
arbitrary element $\beta$ in $J_{K}^1$. (i.e; $\prod |\beta_{v}|_{v} =1$) Then we 
have the following:
\[\prod_{v\in M_{K}} |\beta_{v}^{-1} \alpha_{v}|_{v}=\bigl(\prod_{v\in M_{K}} 
|\beta_{v}|_{v} \bigr)^{-1} \bigl(\prod_{v\in M_{K}} |\alpha_{v}|_{v} \bigr ) > C \]

Therefore, by Prop 3.2., there exists $\gamma \in K^*$ such that $|\gamma|_{v} 
\leq |\beta_{v}^{-1} \alpha_{v}|_{v}$ for every $v\in M_{K}$. Then $|\gamma \beta_
{v}|_{v} \leq |\alpha_{v}|_{v}$ for every $v$ and hence $\gamma \beta \in A_{K}$ 
is actually in $W$. This element is also in $J_{K}^1$ since $\beta$ was chosen 
from $J_{K}^1$ and $\gamma\in K^* \hookrightarrow J_{K}^1$. Therefore for any 
$\beta \in J_{K}^1$, there exists $\gamma\in K^*$ such that $\gamma \beta \in 
W\cap J_{K}^1$. Finally observe that the image of $\gamma \beta$ under the natural 
map is in the same class as $\beta$ in $J_{K}^1/K^*$ and we are done.
\endproof

\section{Some Classical Theorems Revisited}
\begin{thm} $Cl(K)=I_{K}/P_{K}$ is finite for a number field $K$.
\end{thm}

\proof Consider the map $J_{K}\longrightarrow I_{K}$ 
\[ \alpha \longmapsto \prod_{v:nonarch} \mfr{p}_{v}^{ord_v(\alpha_v)} \]
which was defined earlier and shown to be continuous and onto. In fact the 
restriction of this map to $J_{K}^1$ is, naturally, continuous and can be 
shown that it is also onto. (If you take a fractional ideal $\prod \mfr{p}^{n_v}$ 
where $n_{v}\in \Z$, then choose an idele $\alpha$ such that $ord_v(\alpha_v)=n_v$ 
for these nonarchemedian places, order at all other nonarchemedian places is one 
and orders at archemedian places are chosen so that $\prod |\alpha_{v}|_{v} =1$) 
So we know that $J_{K}^1$ maps onto $I_{K}$ and then obviously $J_{K}^1/K^*$ maps 
onto $I_{K}/P_{K}$. But we showed that $J_{K}^1/K^*$ is compact and hence 
$I_{K}/P_{K}$ must be compact as well. But if a space is both discrete and compact, 
then it is finite.
\endproof


\end{document}