\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{\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 9}

\maketitle 
Today we discuss some general facts about absolute values.

\begin{lem} \label{L:AV} Let $|\quad|_{1}$, $\cdots , |\quad|_{J}$ be pairwise inequivalent nontrivial absolute values on K.  Then there exists an $a \in K$ such that 
$$
|a|_{1} > 1  \quad \text{,} \quad |a|_{j} < 1 \quad \text{for} \quad j = 2, \cdots, J.
$$
\end{lem}

\begin{proof}  We proceed by induction on J.  We start with the case $J = 2$.  Recall two absolute values $|\quad|_{1}$ and $|\quad|_{2}$ are equivalent if and only if $\{x : |x|_1 < 1\} = \{x : |x|_2 < 1\}$.  Since $|\quad|_1$ is not equivalent to $|\quad|_2$, there exists $b \in K$ such that $|b|_1 < 1$ and $|b|_2 \geq 1$.  By the same argument, there exists a $c \in K$ such that $|c|_1 \geq 1$ and $|c|_2 < 1$.  Let $a = cb^{-1}$. Then $|a|_1 = |cb^{-1}| = \frac{|c|_1}{|b|_2} > 1$ and $|a|_2 = \frac{|c|_2}{|b|_2} < 1$.  Now consider the case where $J > 2$.  By induction, there exists $b \in k$ such that $|b|_1 > 1$, $|b|_{j} < 1$ for $2 \leq j \leq J - 1$.  By the $J = 2$ case for the absolute values $|\quad|_1$ and $|\quad|_J$, there exists $c \in K$ such that $|c|_1 > 1$, $|c|_{J} < 1$.  We consider three cases. \\
Case 1: If $|b|_J < 1$, take $a = b$. \\
Case 2 : If $|b|_J = 1$, take $a =b^{n}c$ for $n >> 0$.
For $j=1$, we have $|a|_1  = |b|_{1}^{n}|c|_1 > 1$.  Also for $j = J$, $|a|_J = |b|_{J}^{n}|c|_{J} = |c|_{J} < 1$.  Now for $j = 2, \cdots, J$,  $|a|_j = |b|_j^{n}|c|_{J} < 1$  as $|b|_j^{n} < 1$ for large n. \\
Case 3: If $|b|_{J} > 1$, take $a = (\frac{b^{n}}{1 + b^n})c$ for $n >> 0$.
For $j = 1$, we have 
$$
|a|_1  = \Big|\frac{b^n}{1 + b^n}c\Big|_1 = \Big|\frac{1}{1 + (\frac{1}{b^n})}\Big|_1|c|_1 > 1
$$
as  $\Big|\frac{1}{1 + (\frac{1}{b^n})}\Big|_1 \to 1$ for large n and $|c|_1 > 1$.

For $j = J$, we have
$$
|a|_J = \Big|\frac{b^n}{1 + b^n}c\Big|_J = \Big|\frac{1}{1 + (\frac{1}{b^n})}\Big|_J|c|_J < 1
$$
as $\Big|\frac{1}{1 + (\frac{1}{b^n})}\Big|_J \to 1$ for large n and $|c|_1 < 1$.

For $j = 2, \cdots , J$, 
$$
|a|_j = \Big|\frac{b^n}{1 + b^n}c\Big|_j = \Big|\frac{1}{1 + (\frac{1}{b^n})}\Big|_j|c|_j \to 0 < 1
$$
as $\Big|\frac{1}{1 + (\frac{1}{b^n})}\Big|_j \to 0$ for large n. 
\end{proof}

We will now use the previous lemma for the following "Weak Approximation Theorem" by Artin and Whaples.


\begin{thm} \label{T:WAT} Let $|\quad|_1, \cdots, |\quad|_J$ be pairwise inequivalent nontrivial absolute values on K.  Let $x_1, \cdots, x_J \in K$.  Then for any $\epsilon > 0$, there exists $x \in K$ such that 
$$
|x - x_j|_j < \epsilon
$$
for all $j = 1, \cdots, J$.
\end{thm}


Note that the above theorem is true for any field K.  Later we will prove a "Strong Approximation Theorem" for number fields.


\begin{proof} By the previous lemma, there exists $c_j \in K$ such that $|c_j|_j > 1$ and $|c_j|_i < 1$ for $i \neq j$.  We claim that 
$$
x = \sum_{j=1}^{J} \Big(\frac{c^n_j}{1 + c^n_j}\Big)x_j
$$
works for $n >> 0$.
From the triangle inequality, we have
$$
|x - x_j| _{j}\leq \Big|\frac{x_j}{1 + c^n_j}\Big|_j + \sum_{i \neq j}\Big|\frac{c^n_i}{1 + c^n_i}x_i\Big|_j.
$$
Thus for n sufficiently large, we obtain for any $\epsilon > 0$, $|x - x_j|_j < \epsilon$.
\end{proof}


\begin{thm} \label{T:Ost} The only nontrivial absolute values on $\Bbb Q$ are equivalent to $|\quad|_p$ or $|\quad|_{\infty}$.
\end{thm}

\begin{proof} Without loss of generality, assume $|\quad|$ satisfies the triangle inequality.  Let $1 < a \in \Bbb Z$.  Then every positive $b \in \Bbb Z$ can be expanded in an a-adic expansion. That is, 
$$
b = b_m a^m + \cdots + b_1 a + b_0 \quad \text{,} \quad b_j \in \{0, 1, \cdots, a - 1\} \quad \text{,} \quad b_m \neq 0.
$$
Note that $b \geq a^m$ implies $\log b \geq \log a^m$ and so $m \leq \frac{\log b}{\log a}$ since $b_m \neq 0$.  Also we have 
$$
\begin{aligned}
|b|
& \leq \sum_{j = 0}^{M} |b_j| |a|^j \leq M\sum_{j=0}^{M} |a|^j \quad \mbox{where $M = max \{|1|, \cdots, |a-1|\}$} \\  
& \leq M(m + 1) \max_{0 \leq j \leq M} (|a|^j) \leq M(m + 1) \max(1, |a|^m).
\end{aligned}
$$

So $|b| \leq M(m + 1) \max(1, |a|^{\frac{\log b}{\log a}})$.  Now let $b = c^n$.  Then we have $|c|^n \leq M(m + 1) \max(1, |a|^{\frac{n\log c}{log a}})$.  Taking $n^{\rm th}$ roots and letting $n \to \infty$, we get $|c| \leq \max(1, |a|^{\frac{\log c}{log a}})$.

 
\end{proof} 
  
We will finish the proof in lecture 10.

\end{document}