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\begin{document}
\selectlanguage{english}%
\global\long\def\zz{\mathbb{Z}}
\global\long\def\im{\operatorname{im}}
\global\long\def\re{\operatorname{re}}
\global\long\def\rr{\mathbb{R}}
\global\long\def\cc{\mathbb{C}}
\global\long\def\vv{\mathbb{V}}
\global\long\def\SL{\operatorname{SL}}
\global\long\def\ord{\operatorname{ord}}
\global\long\def\Spec{\operatorname{Spec}}
\global\long\def\qq{\mathbb{Q}}
\global\long\def\pp{\mathbb{P}}
\global\long\def\Hom{\operatorname{Hom}}
\global\long\def\id{\operatorname{id}}
\global\long\def\gcd{\operatorname{gcd}}
\global\long\def\lcm{\operatorname{lcm}}
\global\long\def\Sym{\operatorname{Sym}}
\global\long\def\ff{\mathbb{F}}
\global\long\def\nn{\mathbb{N}}
\global\long\def\hh{\mathbb{H}}
\global\long\def\Ann{\operatorname{Ann}}
\global\long\def\End{\operatorname{End}}
\global\long\def\sgn{\operatorname{sgn}}
 \global\long\def\rk{\operatorname{rk}}
\global\long\def\Br{\operatorname{Br}}
 \global\long\def\Hasse{\operatorname{Hasse}}
\global\long\def\GL{\operatorname{GL}}
\global\long\def\Coll{\operatorname{Col}}
 \global\long\def\Tr{\operatorname{Tr}}
\global\long\def\aa{\mathbb{A}}
\global\long\def\ld{\operatorname{in}_{<}}
\global\long\def\ob{\operatorname{Ob}}
\global\long\def\mor{\operatorname{mor}}
\global\long\def\ext{\operatorname{Ext}}
\global\long\def\tor{\operatorname{Tor}}
\global\long\def\cok{\operatorname{coker}}
\global\long\def\ilim{\varprojlim}


\selectlanguage{british}%

\title{Local Fields}


\author{Tim Browning\\
Notes by Florian Bouyer}

\maketitle
\begin{center}
Copyright (C) Bouyer 2013. 
\par\end{center}

\begin{center}
Permission is granted to copy, distribute and/or modify this document 
\par\end{center}

\begin{center}
under the terms of the GNU Free Documentation License, Version 1.3 
\par\end{center}

\begin{center}
or any later version published by the Free Software Foundation; 
\par\end{center}

\begin{center}
with no Invariant Sections, no Front-Cover Texts, and no Back-Cover
Texts. 
\par\end{center}

\begin{center}
A copy of the license can be found at \href{http://www.gnu.org/licenses/fdl.html}{http://www.gnu.org/licenses/fdl.html}
\par\end{center}

\tableofcontents{}

\newpage{}


\section{Foundations}


\subsection*{Absolute Values}

Let $K$ be a field.
\begin{defn}
An \emph{absolute value} on $K$ is a map $|\cdot|:K\to\mathbb{R}_{>0}$
\begin{enumerate}
\item $|x|=0\iff x=0$
\item $|xy|=|x|\cdot|y|\forall x,y\in K$
\item $|x+y|\leq|x|+|y|$ (the $\triangle$ inequality)
\end{enumerate}
\end{defn}
\begin{defn*}
An absolute value on $K$ is called \emph{non-archimedean} if also
\begin{enumerate}[resume]
\item $|x+y|\leq\max\{|x|,|y|\}$ (the ultrametric inequality)
\end{enumerate}
Otherwise we say the absolute value is \emph{archimedean}\end{defn*}
\begin{example*}
~{}
\begin{enumerate}
\item $K=\qq$ and $|\cdot|$ to be the usual absolute value given by inclusion
$\qq\hookrightarrow\rr$. This is an archimedean absolute value.
\item Take $|x|=\begin{cases}
1 & x\neq0\\
0 & x=0
\end{cases}$. This is non-archimedean absolute value. ``The trivial absolute
value''
\item $K=\qq$ and $p$ a prime. For $x\in\qq^{*}$ the $p$-adic valuation
is $\nu_{p}(x)=r$ if $x=p^{r}\frac{u}{v}$ for $u,v\in\zz,r\in\zz$
and $p\nmid uv$. We extend to all of $\qq$ by setting $\nu_{p}(0)=+\infty$


\textbf{Check: }$\nu_{p}(xy)=\nu_{p}(x)+\nu_{p}(y)$ and $\nu_{p}(x+y)\geq\min\{\nu_{p}(x),\nu_{p}(y)\}$
$(*)$


Define the $p$-adic absolute value on $\qq$ to be $|\cdot|_{p}:\qq\to\rr_{\geq0}$
by $|x|_{p}=\begin{cases}
p^{\nu_{p}(x)} & x\neq0\\
0 & x=0
\end{cases}$. This satisfies the axioms of being a non-archimedean absolute value
(using $(*)$)
\begin{note*}
$|p^{n}|_{p}=p^{-n}$ so $p^{n}\to0$ as $n\to\infty$.
\end{note*}
\item Let $K$ be any field. Put $F=K(T)=\left\{ \frac{P(T)}{Q(T)}:P,Q\in K[T],Q\neq0\right\} $.
Define the valuation 
\[
\nu_{\infty}\left(\frac{P(T)}{Q(T)}\right)=\begin{cases}
\deg Q-\deg P & \frac{P}{Q}\neq0\\
+\infty & \frac{P}{Q}=0
\end{cases}
\]
Check this satisfies $(*)$. If $c>1$, then we get a non-archimedean
absolute value on $F$ given by $|F(T)|_{\infty}:=c^{-\nu_{\infty}(f(T))}$. 

\begin{note*}
If $K=\ff_{q}$ then convenient to take $c=q$.
\end{note*}
\end{enumerate}
\end{example*}
\begin{lem}
Let $|\cdot|$ be an absolute value on a field $K$. Then
\begin{enumerate}
\item $|1|=1$
\item $x\in K$ such that $x^{n}=1$, then $|x|=1$.
\item $x\in K$, then $|-x|=|x|$
\item $K$ is a finite field then the absolute value has to be the trivial
absolute value
\end{enumerate}
\end{lem}
\begin{proof}
~{}
\begin{enumerate}
\item Note that $x\neq0\Rightarrow x>0$. We have $|1|=|1^{2}|=|1|\cdot|1|$.
So 1.~holds.
\item Note that $1=|1|=|x^{n}|=|x|^{n}\Rightarrow|x|=1$
\item Note that $-x=-1\cdot x$
\item Follows from 2.~since any non-zero element $x$ of a finite field
satisfies $x^{n}=1$ for some $n$. 
\end{enumerate}
\end{proof}
The following result gives a criterion for checking whether an absolute
value is non-archimedean. 
\begin{lem}
\label{lem:1.3}Let $|\cdot|$ be an absolute value on a field $K$.
Then $|\cdot|$ is non-archimedean if and only if $|e|\leq1$ for
all $e$ in the additive ring generated by $1$.\end{lem}
\begin{proof}
``$\Rightarrow$'' Since $|n|=|-n|$ we may as well assume that
$n\geq1$. Then $|n|=|\underset{n\mathrm{\, times}}{\underbrace{1+\dots+1}|}\leq|1|=1$

``$\Leftarrow$'' Suppose $|e|\leq1$ for all elements $e$ in the
additive ring generated by $1$. Let $x,y\in K$, then 
\begin{eqnarray*}
|x+y|^{m} & = & \left|\sum_{j=0}^{m}{m \choose j}x^{j}y^{m-j}\right|\\
 & \leq & \sum_{j=0}^{m}\left|{m \choose j}\right||x|^{j}|y|^{m-j}\\
 & \leq & \sum_{j=0}^{m}|x|^{j}|y|^{m-j}\qquad\mathrm{by\, assumption\,}\left|{m \choose j}\right|\leq1\\
 & \leq & \max(\{|x|,|y|\}^{m}
\end{eqnarray*}
 Take $m$th root and let $m\to\infty$ (since $(m+1)^{1/m}\to1$
as $m\to\infty$) \end{proof}
\begin{cor}
If $\mathrm{char}(K)\neq0$ then all absolute values are non-archimedean\end{cor}
\begin{proof}
The ring in Lemma 1.3 is a finite field. Then apply Lemma 1.2 part
4.~\end{proof}
\begin{cor}
Suppose $F\subset K$ is a subfield of $K$ and $|\cdot|$ is an absolute
value on $K$. Then $|\cdot|$ is non-archimedean on $K$ if and only
if $|\cdot|$ is non-archimedean on $F$
\end{cor}

\subsection*{Topology}

Let $K$ be a field with absolute value $|\cdot|$ on $K$. Then we
get a metric on $K$ induced by $|\cdot|$. Call it $d:K\times K\to R_{\geq0}$
defined by $d(x,y)\mapsto|x-y|$. 
\begin{xca*}
Check this is a metric.
\end{xca*}
The notion of distance on fields with non-archimedean values is weird. 
\begin{lem}
Let $K$ be a field with non-archimedean absolute value. If $x,y\in K$
with $|x|\neq|y|$, then $|x+y|=\max\{|x|,|y|\}$\end{lem}
\begin{proof}
Without loss of generality assume $|x|>|y|$. Then $|x+y|\leq\max\{|x|,|y|\}=|x|$
and $|x|=|x+y-y|\leq\max\{|x+y|,|y|\}$. Hence $|x|\leq|x+y|\leq|x|$.\end{proof}
\begin{defn}
Let $K$ be a field with absolute value $|\cdot|$. Let $a\in K$
and $r\in\rr_{\geq0}$. The \emph{open ball }of radius $r$ and centre
$a$ is $B(a,r)=\{x\in K:|x-a|<r\}$. The \emph{closed ball} of radius
$r$ and centre $a$ is $\overline{B}(a,r)=\{x\in K:|x-a|\leq r\}$.

A set $U\subset K$ is \emph{open} if and only if $\forall x\in U$
there exists an open ball around $x$ contained in $U$. A set is
\emph{closed} if and only if its complement in $K$ is open\end{defn}
\begin{lem}
Let $K$ be a field with non-archimedean absolute value $|\cdot|$.
Then
\begin{enumerate}
\item $b\in B(a,r)\Rightarrow B(a,r)=B(b,r)$
\item $b\in\overline{B}(a,r)\Rightarrow\overline{B}(a,r)=\overline{B}(b,r)$
\item $B(a,r)\cap B(a',r')\neq0\iff B(a,r)\subset B(a',r')$ or $B(a,r)\supset B(a',r')$
\item $\overline{B}(a,r)\cap\overline{B}(a',r')\neq0\iff\overline{B}(a,r)\subset\overline{B}(a',r')$
or $\overline{B}(a,r)\supset\overline{B}(a',r')$
\item $B(a,r)$ is both open and closed
\item $\overline{B}(a,r)$ is both open and closed.
\end{enumerate}
\end{lem}
\begin{proof}
We prove 1.~3.~5.~only.

1.~$b\in B(a,r)$ and $c\in B(b,r)$. $|c-a|\leq\max\{|c-b|,|b-a|\}<r$,
i.e., $B(b,r)\subset B(a,r)$. Reverse inclusion follows from symmetry
since $a\in B(b,r)$.

3.~Follows form 1\@.~

5.~$b\in B(a,r)$ implies $B(b,r)\subset B(a,r)$, so any open ball
is open. To show that it is closed, note that $b\notin B(a,r)\Rightarrow a\notin B(b,r)$.
So neither ball is contained in the other and they are disjoint. Hence
$B(b,r)\subset K\setminus B(a,r)$ and the complement of $B(a,r)$
in $K$ is open. \end{proof}
\begin{rem*}
Recall that a set $S$ is said to be \emph{disconnected} if there
exists open sets $U,V$ such that
\begin{itemize}
\item $U\cap V=\emptyset$, 
\item $S\subset U\cup V$
\item $S\cap U\neq\emptyset$ and $S\cap V\neq\emptyset$
\end{itemize}
Otherwise $S$ is \emph{connected. }If $x\in K$ then the \emph{connected
component} of $x$ is the union of all connected sets containing it. \end{rem*}
\begin{example*}
$K=\rr$ with usual absolute value, then connected component of any
$x\in\rr$ is $\rr$. \end{example*}
\begin{xca*}
If $|\cdot|$ is a non-archimedean absolute value on a filed $K$,
then the connected component of any $x\in K$ is $\{x\}$, i.e., $K$
is \emph{totally disconnected }topological space.
\end{xca*}

\subsection*{Equivalence}
\begin{defn}
Two absolute values $|\cdot|_{1}$ and $|\cdot|_{2}$ on a field $K$
are \emph{equivalent} if they induce the same topology on $K$. (i.e.,
every set which is open with respect to $|\cdot|_{1}$ is open with
respect to $|\cdot|_{2}$)

Given an absolute value $|\cdot|$ on a field $K$, a sequence $\{a_{n}\}_{n}$
in $K$ \emph{converges} to $a$ in the induced topology if and only
if $\forall\epsilon>0\exists N\in\nn$ such that for $n>N$, $|a_{n}-a|<\epsilon$.
Equivalently, for all open sets $U$ containing $a$, there exists
$N$ such that $a_{n}\in U$ for $n>N$
\end{defn}
Thus the notion of convergence depends on the topology induced by
the absolute value. 
\begin{lem}
\label{lem:1.10}Let $|\cdot|_{1},|\cdot|_{2}$ be absolute values
on field $K$, with $|\cdot|_{1}$ non-trivial. Then the following
are equivalent
\begin{enumerate}
\item $|\cdot|_{1}$ , $|\cdot|_{2}$ are equivalent
\item $\forall x\in K$, $|x|_{1}<1\iff|x|_{2}<1$
\item $\exists\alpha>0$ such that $\forall x\in K$, $|x|_{1}=|x|_{2}^{\alpha}$. 
\end{enumerate}
\end{lem}
\begin{proof}
~{}
\begin{elabeling}{00.00.0000}
\item [{$3.\,\Rightarrow1.\,$}] Then $|x-a|_{2}<r\iff|x-a|_{1}<r^{\alpha}$.
So any open ball with respect to $|\cdot|_{2}$ is an open ball with
respect to $|\cdot|_{1}$. Hence the topology must be the same and
the absolute value are equivalent.
\item [{$1.\,\Rightarrow2.\,$}] $|x|_{1}<1\iff x^{n}\to0$ as $n\to\infty$
with respect to $|\cdot|_{1}$ $\overset{1.}{\iff}$$x^{n}\to0$ as
$n\to\infty$ with respect $|\cdot|_{2}\iff|x|_{2}<1$
\item [{$2.\,\Rightarrow3.\,$}] Now $|x|_{1}>1\iff|x^{-1}|<1\iff|x^{-1}|_{2}<1\iff|x|_{2}>1$.
Also $|x|_{1}=1\iff|x|_{2}=1$. Now pick (and fix) $a\in K^{*}$ such
that $|a|_{1}<1$ (which is possible since $|\cdot|_{1}$ is non-trivial).
Then also $|a|_{2}<1$. Let $\alpha=\frac{\log|a|_{1}}{\log|a|_{2}}>0$.
Choose $b\in k^{*}$

\begin{enumerate}
\item $\left|b\right|_{1}=1$ then $\left|b\right|_{2}=1$ and $1=1^{\alpha}$
\item $\left|b\right|_{1}<1$ by assumption $\left|b\right|_{2}<1$. Define
$\beta_{i}=\frac{\log\left|a\right|_{i}}{\log\left|b\right|_{i}}$
for $o=1,2$. We show that $\beta_{1}=\beta_{2}$ which implies $\frac{\log\left|b\right|_{1}}{\log\left|b\right|_{2}}=\frac{\log\left|a\right|_{1}}{\log\left|a\right|_{2}}=\alpha$.


Suppose that $\beta_{1}>\beta_{2}$, then $\exists\frac{m}{n}\in\qq$
such that $\beta_{2}\leq\frac{m}{n}<\beta_{1}$. Set $x=a^{n}b^{-m}\in k$,
then $\log\left|x\right|_{i}=n\log\left|a\right|_{i}-m\log\left|b\right|_{i}=\underset{<0}{\underbrace{n\log\left|b\right|_{i}}}\underset{\begin{cases}
>0 & i=1\\
<0 & i=2
\end{cases}}{\underbrace{\left(\beta_{i}-\frac{m}{n}\right)}}$, hence we have a contradiction with $\left|x\right|_{1}<1$ and $\left|x\right|_{2}>1$.
Similarly if $\beta_{2}>\beta_{1}$. Hence $\beta_{1}=\beta_{2}$

\item If $\left|b\right|_{1}>1$, $\left|b\right|_{2}>1$, replace $b$
by $b^{-1}$ and get $\left|b^{-1}\right|_{1}<1$ and $\left|b^{-1}\right|_{2}<1$
\end{enumerate}
\end{elabeling}
\end{proof}
How independent inequivalent absolute value are
\begin{lem}
\label{lem:Inequivalent}Let $\left|\,\right|_{1},\dots,\left|\,\right|_{J}$
be non trivial inequivalent absolute values on $K$. Then there exists
$x\in K$ such that $\left|x\right|_{1}>1$ and $\left|x\right|_{j}<1$
for $2\leq j\leq J$.\end{lem}
\begin{proof}
By induction on $J$.
\begin{elabeling}{00.00.0000}
\item [{$J=2$}] Since $\left|\,\right|_{1},\left|\,\right|$ are non-trivial
and non-equivalent, by the previous lemma there exists $y\in K$ such
that $\left|y\right|_{1}<1$ and $\left|y\right|_{2}\geq1$, and $z\in K$
such that $\left|z\right|_{1}\geq1$ and $\left|z\right|_{2}<1$.
Let $x=zy^{-1}$, then $\left|x\right|_{1}=\left|z\right|_{1}\left|y\right|_{1}^{-1}>1$
and $\left|x\right|_{2}=\left|z\right|_{2}\left|y\right|_{2}^{-1}<1$
\item [{$J>2$}] By induction, there exists $y,z\in K$ such that $\left|y\right|_{1}>1$,
$\left|y\right|_{j}<1$ for $2\leq j<J$ and $\left|z\right|_{1}<1$,
$\left|z\right|_{j}>1$ for $2\leq j<J$. Consider $\left|y\right|_{J}$
and we have different cases:

\begin{enumerate}
\item $\left|y\right|_{J}<1$ so take $x=y$
\item $\left|y\right|_{J}=1$ so take $x=y^{n}z$ for large enough $n$
\item $\left|y\right|_{J}>1$, then $\left|\frac{y^{n}}{1+y^{n}}\right|_{j}=\left|\frac{1}{1+y^{-n}}\right|_{j}\underset{n\to\infty}{\to}\begin{cases}
1 & j=2,\dots,J\\
0 & \mathrm{else}
\end{cases}$. So Let $x=\left(\frac{y^{n}}{1+y^{n}}\right)z$ for large enough
$n$
\end{enumerate}
\end{elabeling}
\end{proof}
\begin{thm}[Weak Approximation]
 \label{thm:Weak_approx}Let $\left|\,\right|_{1},\dots,\left|\,\right|_{J}$
be non trivial inequivalent absolute values on $K$. Let $b_{j}\in K$
for $j=1,\dots,J$ and let $\epsilon>0$. Then there exists $x\in K$
such that $\left|x-b_{j}\right|_{j}<\epsilon$ for all $j=1,\dots,J$.\end{thm}
\begin{proof}
By Lemma \ref{lem:Inequivalent}, there exists $x_{j}\in K$ such
that $\left|x_{j}\right|_{j}>1$ but $\left|x_{j}\right|_{i}<1$ for
$i\neq j$. Consider $\left|\frac{x_{j}^{n}}{1+x_{j}^{n}}\right|_{j}\underset{n\to\infty}{\to}\begin{cases}
1 & i=j\\
0 & \mathrm{else}
\end{cases}$. Take $w_{n}=\sum_{j=1}^{J}b_{j}\left(\frac{x_{j}^{n}}{1+x_{j}^{n}}\right)\underset{n\to\infty}{\to}b_{j}$,
so take $x=w_{n}$ for $n$ large enough.\end{proof}
\begin{rem*}
This is clearly related to the Chinese Remainder Theorem. Let $p_{1},\dots,p_{j}$
be distinct primes and $m_{j}\in\nn,b_{j}\in\zz$. Then there exists
$x\in\zz$ such that $x\equiv b_{j}\mod p_{j}^{m_{j}}$. Using the
Theorem above, $\left|x-b_{j}\right|_{j}<p_{j}^{-m_{j}}$ with $p_{j}$-adic
absolute value
\end{rem*}

\subsection{Completion}
\begin{defn}
~{}
\begin{enumerate}
\item A sequence $\{x_{n}\}$ is a field $K$ is called \emph{Cauchy} if
$\forall\epsilon>0,\exists N>0$ such that $\forall m,n>N$, $\left|x_{m}-x_{n}\right|<\epsilon$
\item $(K,\left|\cdot\right|)$ is \emph{complete} if every Cauchy sequence
is convergent
\item A subset $S\subset K$ is dense if $\forall x\in K$, $\forall\epsilon>0$,
$B(x,\epsilon)\cap S\neq0$. That is, $\forall x\in K$, there exists
a sequence $\{x_{n}\}\in S$ such that $\left\{ x_{n}\right\} \to x$. 
\item A field $\left(\widehat{K},||\,||\right)$ is a \emph{completion }of
$\left(K,\left|\,\right|\right)$ if 

\begin{enumerate}
\item There exists an embedding $\iota:K\to\widehat{K}$ which respect absolute
values
\item $\im(K)$ is dense in $\widehat{K}$
\item $\left(\widehat{K},||\,||\right)$ is complete
\end{enumerate}
\end{enumerate}
\end{defn}
\begin{thm}
\label{thm:1.14}Let $(K,\left|\,\right|)$ be a field. Then there
exists a completion $(\widehat{K},||\,||)$ of $K$ and it is unique
as any two completions are canonically isomorphic. That is if $(\widehat{K}_{j},||\,||_{j})$
for $j=1,2$ then there exists a unique isomorphism of $\widehat{K}_{1}\cong\widehat{K}_{j}$
which is the identity of $K$ and preserves $||\,||_{1}=||\,||_{2}$\end{thm}
\begin{proof}
~{}
\begin{elabeling}{00.00.0000}
\item [{Existence~of~Completion}] Let $\mathcal{K}$ be the set of all
Cauchy Sequences in $K$. This is a ring as $\left\{ a_{n}\right\} +\left\{ b_{n}\right\} =\left\{ a_{n}+b_{n}\right\} $,
$\left\{ a_{n}\right\} \times\left\{ b_{n}\right\} =\left\{ a_{n}b_{n}\right\} $
and $\id=\{1\}$. Define $||\,||:\mathcal{K}\to\rr_{>0}$ by $\left\{ a_{n}\right\} \to\lim_{n\to\infty}|a_{n}|$
($\rr$ is complete). Let $\mathcal{N\subset K}$ be the subset of
all null sequences ($||a_{n}||=0$). Then $\mathcal{N}$ is a maximal
ideal (Exercise). Hence $\mathcal{K}/\mathcal{N}$ is a field $\widehat{K}$.
We have $||\,||$ (not an absolute value since $||a_{n}||=0$ for
non zero elements) only depends on $\mathcal{K}/\mathcal{N}$. We
get a well defined functions $||\,||:\widehat{K}\to\rr_{>0}$. This
is an absolute value. Define $\iota:K\to\widehat{K}$ by $a\mapsto\left\{ a\right\} \mod\mathcal{N}$.
Then $\iota(K)$ is dense and $(\widehat{K},||\,||)$ is complete.
\item [{Uniqueness}] Suppose $(\widehat{K'},||\,||')$ is complete and
is a completion, $\iota':K\to\widehat{K}'$ satisfy the embedding
properties above.

\begin{claim*}
$\iota'$ extends uniquely to an embedding $\lambda:\widehat{K}\to\widehat{K}'$
such that 
\[
\xymatrix{K\ar[r]^{\iota'}\ar[rd]^{\iota} & \widehat{K}'\\
 & \widehat{K}\ar[u]^{\lambda}
}
\]
Let $x\in\widehat{K}$ and $\{x_{n}\}$ is a sequence in $K$ such
that $\{\iota(x_{n})\}$ converges to $x$ (dense). Define $\lambda(x)=\lim_{n\to\infty}\{\iota'(x_{n})\}$.
Construct $\lambda':\widehat{K}'\to\widehat{K}$ in the same way
\end{claim*}
\end{elabeling}
\end{proof}
\begin{cor}
Let $K$ be a field and $|\,|_{j}$ ($j\leq J$) be non-trivial and
inequivalent absolute values on $K$. Let $\widehat{K}_{j}$ be the
respective completions, let $\Delta:K\hookrightarrow\prod_{j}\widehat{K}_{j}$
defined by $x\mapsto(\iota_{j}(x))$. Then $\Delta(K)$ is dense,
i.e., its closure $\overline{\Delta(K)}$ is $\prod_{j}K_{j}$.\end{cor}
\begin{rem*}
We have $\qq\hookrightarrow\rr$ but $\qq\hookrightarrow\rr\times\rr$
is not dense.\end{rem*}
\begin{proof}
Let $\alpha_{j}\in\widehat{K}_{j}$, for $1\leq j\leq J$, then $\forall\epsilon>0$
there exists $a_{j}\in K$ such that $|a_{j}-\alpha_{j}|<\epsilon$
for $1\leq j\leq J$. By Theorem \ref{thm:Weak_approx} there exists
$b\in K$ such that $|b-a_{j}|_{j}<\epsilon$. Then $|b-\alpha_{j}|_{j}<2\epsilon$
so arbitrary closed to $\alpha_{j}$, hence dense.

\newpage{}
\end{proof}

\section{The $p$-adic}
\begin{thm}[Ostrowski ]
 Every non trivial absolute value on $\qq$ is equivalent to $|\,|_{v}$
where $v=p$ a prime or $v=\infty$.\end{thm}
\begin{proof}
Let $|\,|$ be an absolute value on $\qq$ and $a>1$, $b>0$ be integers.
Let $t=\max\{\left|0\right|,|1|,\dots,\left|a-1\right|\}$, $b=b_{m}a^{m}+\dots+b_{1}a+b_{0}$
with $b_{i}\in\{0,\dots,a-1\}$, $b_{m}\neq0$ and $m\leq\frac{\log b}{\log a}$.
Then $\left|b\right|\leq\sum_{j=0}^{m}\left|b_{j}a^{j}\right|\leq(m+1)t\max\{1,|a|^{m}\}\leq(\log b/\log a+1)t\max\{1,|a|^{m}\}$.
Replace $b$ by $b^{n}$ and take $n$th root,
\[
|b|\leq\underset{\underset{n\to\infty}{\to}1}{\underbrace{\left(n\frac{\log b}{\log a}+1\right)^{1/n}}}t^{1/n}\max\{1,\left|a\right|\}^{\log b/\log a}
\]
Take the limit as $n\to\infty$, then $\left|b\right|\leq\max\{1,|a|\}^{\log b/\log a}\:(*)$.
We have two cases
\begin{enumerate}
\item $|\,|$ is archimedean, then there exists $|b|>1$ for some $b$ by
Lemma \ref{lem:1.3}. So apply $(*)$, then $|a|>1$ for all $a>1$,
so $|b|\leq|a|^{\log b/\log b}$. Reversing $a$ and $b$ we get $\left|a\right|\leq\left|b\right|^{\log a/\log b}.$
Hence $\left|a\right|^{1/\log a}=\left|b\right|^{\log b},$so $\frac{\log\left|a\right|}{\log a}=\frac{\log\left|b\right|}{\log b}=\alpha>0$,
and it is independent of $a$ and $b$. Hence $|a|=a^{\alpha}=\left|a\right|_{\infty}^{\alpha}$
for all $a\in\nn$. But $|\pm1|=1$, hence $|a|=|a|_{\infty}^{\alpha}$
for all $a\in\zz$. Let $q=\frac{a}{b}$, hence true for all $q\in\qq$
\item $|\,|$ is non-archimedean. Then there exists $a\in\nn$ such that
$|a|<1$. Let $b$ be the such least integer.

\begin{claim*}
$b=p$ a prime number

We prove this by contradiction. Suppose $b$ is not a prime, $b=uv$.
Now $|uv|<1$, but as $b$ is the least such number, we have $|u|=|v|=1$,
hence $|b|=1$ a contradiction.

\end{claim*}

So $b$ is a prime, let $b=p$.
\begin{claim*}
$p|a$ if and only if $|a|<1$.
\begin{elabeling}{00.00.0000}
\item [{$\Rightarrow$:}] Let $a=up$, then $|a|=|u||p|$, hence $|u|<1$
and $|p|<1$. 
\item [{$\Leftarrow$:}] Suppose that if $p\nmid a$ then $a=up+r$ where
$r<p$. By minimality of $p$, $|r|=1$, $|up|<1$, hence $|a|=\max\{|up|,|r|\}=1$
\end{elabeling}
\end{claim*}

So let $\alpha==\frac{\log\left|p\right|}{\log p}$, $|p|=|p|_{p}^{\alpha}$.
For all $a\in\zz$ we have $a=p^{r}a'$ where $p\nmid a'$, hence
$|a'|=|a'|_{p}=1$. Therefore, $|a|=|p^{r}a'|=|p|_{p}^{r\alpha}=|a|_{p}^{\alpha}$.
And $|q|=|q|_{p}^{\alpha}$ for all $q\in\qq$

\end{enumerate}
\end{proof}
\begin{defn}
~{}
\begin{enumerate}
\item The \emph{field of $p$-adic numbers} $\qq_{p}$ is the completion
of $\qq$ with respect to $|\,|_{p}$. $(\qq_{p},|\,|)$ is a non
archimedean complete field.
\item The \emph{ring of }$p$\emph{-adic integers} $\zz_{p}$ is $\zz_{p}=\{x\in\qq_{p}||x|_{p}\leq1\}=\overline{B}(0,1)$
(check it is a ring, by using non archimedean properties)
\end{enumerate}
\end{defn}
\begin{lem}
$\zz$ is dense in $\zz_{p}$\end{lem}
\begin{proof}
$\qq$ is dense in $\qq_{p}$, $\zz_{p}$ is open in $\qq_{p}$ so
$\qq\cap\zz_{p}$ is dense in $\zz_{p}$. Now $\qq\cap\zz_{p}=\left\{ \frac{a}{b}\in\qq|p\nmid b\right\} $.
Let $\frac{a}{b}\in\qq$ be such that $p\nmid b$ for $n\geq1$ pick
$y_{n}\in\zz$ such that $by_{n}\equiv1\mod p^{n}$ ($b$ is a unit
in $\zz_{p}$). Then $by_{n}\to1$ as $n\to\infty$. Hence $\zz$
is dense in $\qq\cap\zz_{p}$, hence dense in $\zz_{p}$
\end{proof}
What do elements of $\qq_{p}$ look like?

Let $x\in\zz_{p}$, let $n\in\nn,$ then by density there exists $q=\frac{a}{b}\in\qq$
such that $\left|x-\frac{a}{b}\right|_{p}\leq p^{-n}$. But then $\left|\frac{a}{b}\right|_{p}\leq\max\{\left|x\right|_{p},\left|x-\frac{a}{b}\right|_{p}\}$.
Hence $p\nmid b$ and there exists $b'\in\zz$ such that $bb'\equiv1\mod p^{n}$.
But then $\left|\frac{a}{b}-ab'\right|_{p}=\left|\frac{a}{b}(1-bb')\right|_{p}\leq p^{-n}$.
Hence $\left|x-ab'\right|_{p}\leq\max\left\{ \left|x-\frac{a}{b}\right|_{p},\left|\frac{a}{b}-ab'\right|_{p}\right\} \leq p^{-n}$.
Now let $\alpha\in\{0,\dots,p^{n}-1\}$ be the unique integer such
that $ab'\equiv\alpha\mod p^{n}$.

\uline{Conclusion}: $\forall x\in\zz_{p},\forall n\in\nn$, $\exists\alpha\in\{0,\dots,p^{n}-1\}$
such that $x\equiv\alpha\mod p^{n}$
\begin{lem}
For all $n\in\nn$ there exists an exact sequence of rings $0\to\zz_{p}\overset{p^{n}}{\to}\zz_{p}\overset{\phi_{n}}{\to}\zz/(p^{n}\zz)\to0$\end{lem}
\begin{proof}
Note that $\ker p^{n}=\{z\in\zz_{p}|p^{n}z=0\}=\{0\}$ (take absolute
value on both side). We have that $\phi_{n}$ is surjective since
$\{0,\dots,p^{n}-1\}\subset\zz_{p}$.

We show that $\im(p^{n})=\ker(\phi_{n})$. Suppose that $x\in\im(p^{n})$,
then $x=p^{n}y$ for some $y\in\zz_{p},$ then $\left|p^{n}y-0\right|_{p}\leq p^{-n}$.
Thence $\phi_{n}(x)=0$ and $x\in\ker\phi_{n}$.

Conversely, let $x\in\ker(\phi_{n})$. Then $|x|_{p}=|x-0|_{p}\leq p^{-n}$,
hence $|p^{-n}x|\leq1$ so $x=p^{n}\underset{\in\zz_{p}}{\underbrace{p^{-n}x}}\in\im(p^{n})$
\end{proof}
Hence $\zz_{p}/(p^{n}\zz_{p})\cong\zz/(p^{n}\zz)$. We will see in
a more general context that elements of $\qq_{p}$ can be uniquely
written as a Laurent series expansion in $p$. Later we will consider
the extensions of $\qq_{p}$.

In a Global setting: $[k:\qq]<\infty$, $\mathcal{O}_{K}$ is the
integral closure of $\zz$ in $k$. In a local setting: $[k:\qq_{p}]<\infty$,
$\mathcal{O}_{k}$ is the integral closure of $\zz_{p}$ in $k$.
But in the global setting $\mathcal{O}_{k}$ is not necessarily a
Unique Factorisation Domain while in a local setting it always is.

\newpage{}


\section{Non-archimedean Local Fields}

We will examine a general theory of fields which are complete with
respect to a non archimedean absolute value.
\begin{thm}[Ostrowski ]
 Let $K$ be a field complete with respect to a archimedean absolute
value. Then $K\cong\rr$ or $K\cong\cc$ and the absolute value are
equivalent to the usual absolute value.\end{thm}
\begin{proof}
See Chapter 3 of Cassels \emph{Local Fields}
\end{proof}
\uline{Basics} Let $K$ be a field with a non-trivial non-archimedean
absolute value $|\,|$.
\begin{itemize}
\item $\mathcal{O}_{K}=\{x\in K|\left|x\right|\leq1\}$, the\emph{ ring
of integers} of $K$
\item $\mathcal{P}_{K}=\{x\in K|\left|x\right|<1\}$
\end{itemize}
Check that $\mathcal{O}_{K}$ is an integral domain and that $\mathcal{P}_{K}$
is maximal. (If $J\supsetneq\mathcal{P}_{K}$ is an ideal of $\mathcal{O}_{K}$
then there exists $x\in J$ such that $|x|=1$. Then $|x^{-1}|=1$
and $|1=xx^{-1}|\in J$)
\begin{itemize}
\item $\mathcal{U}_{K}=\{x\in K|\left|x\right|=1\}$, the \emph{group of
units} in $K$
\item $k_{K}=\mathcal{O}_{K}/\mathcal{P}_{K}$, the \emph{residue field}
of $K$
\end{itemize}
The characteristic of $k_{K}$ is the residual characteristic.
\begin{note*}
In general $\mbox{char}k_{K}\neq\mbox{char}K$.\end{note*}
\begin{itemize}
\item $\Gamma_{K}=\{\left|x\right||x\in K^{*}\}$, the \emph{value group}
of $|\,|$ on $K$.
\end{itemize}
This is a multiplicative subgroup of $\rr_{>0}$.
\begin{defn}
A non-archimedean absolute value is \emph{discrete} if $\Gamma_{K}$
is discrete. (i.e., $\Gamma_{K}\cong\zz$)\end{defn}
\begin{lem}
A non-archimedean absolute value is discrete if and only if the maximal
ideal is principal.\end{lem}
\begin{proof}
By problem A.6 $\Gamma_{K}$ is discrete if and only if $\Gamma_{K}$
is cyclic.
\begin{elabeling}{00.00.0000}
\item [{$\Leftarrow$:}] Suppose that $\mathcal{P}_{K}$ is principal,
say $\left\langle \pi\right\rangle $. Let $\gamma=\left|\pi\right|<1$.
Hence for all $x\in\mathcal{P}_{K}$, we have $\left|x\right|\leq\gamma$
(since $x=\pi y$ with $y\in\mathcal{O}_{K}$). So for all $x\in K$,
there exists $n\in\zz$ such that $\gamma^{n}\leq\left|x\right|<\gamma^{n-1}$.
Dividing through by $\gamma^{n-1}$ we get that $\gamma\leq\left|x\pi^{1-n}\right|<1$,
whence $x\pi^{1-n}\in\mathcal{P}_{K}$. So $\gamma\leq\left|x\pi^{1-n}\right|\leq\gamma$,
thus $\left|x\pi^{1-n}\right|=\gamma$. So $\left|x\right|=\gamma^{n}$,
hence $\Gamma_{K}$ is cyclic generated by $\gamma$.
\item [{$\Rightarrow$:}] Suppose that $\Gamma_{K}$ is cyclic with generator
$\gamma<1$ say. Let $\pi\in K$ be such that $\left|\pi\right|=\gamma$.
Clearly $\left\langle \pi\right\rangle \subset\mathcal{P}_{K}$. Conversely,
for $x\in\mathcal{P}_{K}$, then $|x|=\gamma^{n}$ for some $n\geq1$
since $\Gamma_{K}$ is cyclic. So $|x\pi^{-1}|=\gamma^{n-1}\leq1$,
i.e. $x\pi^{-1}\in\mathcal{O}_{K}$ and $x\in\left\langle \pi\right\rangle $
\end{elabeling}
\end{proof}
From now on $\left|\,\right|$ is a discrete non-archimedean absolute
value on a field $K$. So by the previous lemma $\mathcal{P}_{K}=\left\langle \pi\right\rangle .$We
call $\pi$ the \emph{uniformiser} for the absolute value. Any $x\in K^{*}$
can be written as 
\[
x=\pi^{n}\epsilon
\]
 with $n\in\zz$ and $\epsilon\in\mathcal{U}_{K}$. We write $V_{K}(x)=n\in\zz$
for the \emph{order} of $x$. This gives a valuation $V_{K}:K\to\zz\cup\left\{ \infty\right\} $
by setting $V_{K}(0)=\infty$.
\begin{lem}
Let $0\neq I\subset\mathcal{O}_{K}$ be an integral ideal. Then $I=\mathcal{P}_{K}^{n}:=\{x_{1}\dots x_{n}|x_{i}\in\mathcal{P}_{K}\}$
for some $n\in\nn$. \end{lem}
\begin{proof}
The subset $\left\{ \left|x\right||x\in I\right\} \subset\Gamma_{K}$
is bounded and so it attains its maximal at $x_{0}=\pi^{n}\epsilon$,say
($\Gamma_{K}$ is discrete). Then $I=\left\langle x_{0}\right\rangle =\mathcal{P}_{K}^{n}$
\end{proof}
This implies that $\mathcal{P}_{K}$ is the unique non-zero prime
ideal in $\mathcal{O}_{K}$. Furthermore, $\mathcal{O}_{K}$ is a
PID and a local ring (with a unique maximal ideal)

Let $\overline{K}$ be the completion of $K$ with respect to the
absolute value $\left|\,\right|$. Let $\mathcal{O}_{\overline{K}},\mathcal{P}_{\overline{K}}$
be the ring of integers and the maximal ideal of $\overline{K}$ respectively.
then $\mathcal{O}_{K}=\mathcal{O}_{\overline{K}}\cap K$ and $\mathcal{P}_{K}=\mathcal{P}_{\overline{K}}\cap K$.
There is an inclusion map $\mathcal{O}_{K}\hookrightarrow\mathcal{O}_{\overline{K}}$
and so a map $\mathcal{O}_{K}\to k_{\overline{K}}:=\mathcal{O}_{\overline{K}}/\mathcal{P}_{\overline{K}}$
defined by $x\mapsto x+\mathcal{P}_{\overline{K}}$. The kernel of
this map is $\mathcal{P}_{K}$, so it induces a natural map $k_{K}\overset{\sim}{\to}k_{\overline{K}}$.
\begin{claim*}
This map is an isomorphism. It suffices to show that it is surjective\end{claim*}
\begin{proof}
Let $\overline{x}\in\mathcal{O}_{\overline{K}}$. By density of $K$
in $\overline{K}$, there exists $x\in K$ such that $\left|x-\overline{x}\right|<1$.
Then $x-\overline{x}\in\mathcal{P}_{\overline{K}}$ and $\left|x\right|\leq\max\left\{ \underset{<1}{\left|x-\overline{x}\right|},\underset{\leq1}{\left|\overline{x}\right|}\right\} \leq1$.
Thus $x\in K\cap\mathcal{O}_{\overline{K}}=\mathcal{O}_{K}$\end{proof}
\begin{defn}
A \emph{non-archimedean local field }is a field which is complete
with respect to a non-trivial discrete non-archimedean absolute value
such that the residue class $k_{K}$ is finite.\end{defn}
\begin{example*}
\begin{tabular}{|c|c|c|}
\hline 
$K$ & $\qq_{p}$ & $\ff_{q}\left(\left(T\right)\right)$\tabularnewline
\hline 
\hline 
Completion of  & $\qq$ & $\ff_{q}(T)$\tabularnewline
\hline 
$\mathcal{O}_{K}$ & $\zz_{p}$ & $\ff_{q}\left[\left[T\right]\right]$\tabularnewline
\hline 
$\mathcal{P}_{K}$ & $p\zz_{p}$ & $(T)$\tabularnewline
\hline 
$k_{K}$ & $\cong\ff_{p}$ & $\ff_{q}$\tabularnewline
\hline 
\end{tabular}
\end{example*}
From now on, $K$ is a non-archimedean local field.

Say that an infinite sum $\sum_{n=0}^{\infty}x_{n}$, $x_{n}\in K$,
converges to $s$ if $s=\lim_{N\to\infty}\sum_{n=0}^{N}x_{n}$
\begin{lem}
\label{lem:Converges iff}$\sum_{n=0}^{\infty}x_{n}$ converges if
and only if $x_{n}\to\infty$ as $n\to\infty$ \end{lem}
\begin{proof}
Exercise\end{proof}
\begin{lem}
\label{lem:3.7}Let $\pi$ be a uniformiser of $K$ and let $\mathcal{A}\subset\mathcal{O}_{K}$
be set of representative of $\mathcal{O}_{K}/\mathcal{P}_{K}$. Then
$\mathcal{O}_{K}=\left\{ \sum_{n=0}^{\infty}x_{n}\pi^{n}:x_{n}\in\mathcal{A}\right\} $\end{lem}
\begin{proof}
By Lemma \ref{lem:Converges iff} we have $\sum_{n=0}^{\infty}$ converges
and lies in $\mathcal{O}_{K}$. Conversely, if $x\in\mathcal{O}_{K}$,
then there exists a unique $x_{0}\in\mathcal{A}$ such that $|x-x_{0}|<1$.
Hence $x=x_{0}+\pi y_{1}$ for some $y_{1}\in\mathcal{O}_{K}$. Continue
inductively with $y_{1}$ etc
\end{proof}
Suppose $x\in K^{*}$, Then $\pi^{-N}x\in\mathcal{O}_{K}$ for some
$N\in\zz$. Apply Lemma \ref{lem:3.7} to get $K^{*}=\left\{ \sum_{n=N}^{\infty}x_{n}\pi^{n}:x_{n}\in\mathcal{A},N\in\zz,x_{N}\neq0\right\} $

Let us return topology. A subset $V\subset K$ is said to be \emph{compact}
if whenever we have a family $U_{\lambda}$ ($\lambda\in\Lambda$)
of open sets of $K$ such that $V\subset\cup_{\lambda\in\Lambda}U_{\lambda}$,
then there exists a finite subset $\Lambda_{0}\subset\Lambda$ such
that $V\subset\cup_{\lambda\in\Lambda_{0}}U_{\lambda}$. We say that
$K$ \emph{locally compact} if every point of $K$ has a compact neighbourhood.
(i.e., $\forall x\in K$ there exists $V_{x}\subset K$ which is compact
and contains $B(x,r)$ for some $r>0$)
\begin{lem}
\label{lem:3.8}Let $K$ be a non-archimedean local field. Then $\mathcal{O}_{K}$
is compact, and hence $K$ is locally compact.\end{lem}
\begin{proof}
First we prove that $\mathcal{O}_{K}$ is compact. Let $U_{\lambda}$
($\lambda\in\Lambda$) be open sets covering $\mathcal{O}_{K}$. Suppose
that there does not exists a finite subcovering. Now $\mathcal{O}_{K}=\cup_{x\in\mathcal{A}}(x+\pi\mathcal{O}_{K})$
where $\mathcal{A}$ is set of representation for (finite field) $\mathcal{O}_{K}/\mathcal{P}_{K}$.
Then there exists $x_{0}\in\mathcal{A}$ such that $x_{0}+\pi\mathcal{O}_{K}$
is not covered by finitely many $U_{\lambda}$. Similarly there exists
$x_{1}\in\mathcal{A}$ such that $x_{0}+x_{1}\pi+\pi^{2}\mathcal{O}_{K}$
is not finitely covered and so on. Let $\overline{x}=x_{0}+x_{1}\pi+x_{2}\pi^{2}+\dots\in\mathcal{O}_{K}$.
There exists $\lambda_{0}\in\Lambda$ such that $\overline{x}\in U_{\lambda_{0}}$.
Since $U_{\lambda_{0}}$ is open it follows that $\overline{x}+\pi^{n}\mathcal{O}_{K}\in U_{\lambda_{0}}$
for some $N$, which is a contradiction.

Next we prove that $K$ is locally compact. Put $V_{x}=\overline{B}(x,1)=x+\mathcal{O}_{K}$. \end{proof}
\begin{rem*}
In fact: $F$ locally compact with respect non-archimedean absolute
value $\iff$ $F$ non-archimedean local field.
\end{rem*}

\subsection{Hensel's Lemma}
\begin{thm}
\label{thm:Hensel}Let $K$ be a non-archimedean local field and $f\in\mathcal{O}_{K}[X]$.
Suppose $x_{0}\in\mathcal{O}_{K}$ satisfies $\left|f(x_{0})\right|<\left|f'(x_{0})\right|^{2}$.
Then there exists a unique $x\in\mathcal{O}_{K}$ such that $f(x)=0$,
$\left|x-x_{0}\right|\leq\frac{\left|f(x_{0})\right|}{\left|f'(x_{0})\right|}$. \end{thm}
\begin{proof}
Define $f_{j}\in\mathcal{O}_{K}[X]$ via 
\begin{equation}
f(X+Y)=f(X)+f_{1}(X)Y+f_{2}(X)Y^{2}+\dots\label{eq:3.1}
\end{equation}
In particular $f_{1}(X)+f'(X)$. Defined $y_{0}\in\mathcal{O}_{K}$
by $f(x_{0})+y_{0}f'(x_{0})=0$. Then 
\begin{eqnarray*}
\left|f(x_{0}+y_{0})\right| & \leq & \max_{j\geq2}\left|f_{j}(x_{0})y_{0}^{j}\right|\quad\mathrm{By\,(3.1)}\\
 & \leq & \max_{j\geq2}\left|y_{0}^{j}\right|\\
 & \leq & \left|y_{0}\right|^{2}\\
 & = & \left|\frac{f(x_{0})}{f'(x_{0})}\right|^{2}\\
 & < & \left|f(x_{0})\right|
\end{eqnarray*}
 Similarly $\left|f_{1}(x_{0}+y_{0})-f_{1}(x_{0})\right|\leq\left|y_{0}\right|<\left|f_{1}(x_{0})\right|$.
Then $\left|f_{1}(x_{0}+y_{0})\right|=\left|f_{1}(x_{0})\right|$.
Put $x_{1}=x_{0}+y_{0}$. Then $\left|f(x_{1})\right|\leq\frac{\left|f(x_{0})\right|^{2}}{\left|f_{1}(x_{0})\right|^{2}}$,
$\left|f_{1}(x_{1})\right|=\left|f_{1}(x_{0})\right|$ and $|x_{1}-x_{0}|=\frac{\left|f(x_{0})\right|}{\left|f'(x_{0})\right|}$.
So repeat the process and obtain a sequence of $x_{n+1}=x_{n}+y_{n}$
such that $\left|f_{1}(x_{n})\right|=\left|f_{1}(x_{0})\right|$ and
$\left|f(x_{n+1})\right|\leq\frac{\left|f(x_{n})\right|^{2}}{\left|f_{1}(x_{n})\right|^{2}}=\frac{\left|f(x_{n})\right|^{2}}{\left|f_{1}(x_{0})\right|^{2}}$.
So $f(x_{n})\to0$ as $n\to\infty$. Finally $\left|x_{n+1}-x_{n}\right|=\left|y_{n}\right|=\frac{\left|f(x_{n})\right|}{\left|f_{1}(x_{n})\right|}\to0$
as $n\to\infty$. So $\{x_{n}\}$ is Cauchy and it has a limit as
required. 

Now suppose that we have another solution $x+z$ with $z\neq0$ and
$\left|z\right|\leq\frac{\left|f(x_{0})\right|}{\left|f_{1}(x_{0})\right|}<\left|f_{1}(x_{0})\right|=\left|f_{1}(x)\right|$.
Then, putting $X=x$ and $Y=z$ in \eqref{3.1}, we get $0=f(x+z)-f(x)=xf_{1}(x)+z^{2}f_{2}(x)+\dots$.
But $\left|zf_{1}(x)\right|>\left|z^{j}\right|\geq\left|z^{j}f_{j}(x)\right|$
for all $j\geq2$. Which gives a contradiction. \end{proof}
\begin{example*}
~{}
\begin{enumerate}
\item Squares in $\qq_{p}$. 

\begin{itemize}
\item $p\neq2$. Suppose that $y\in\zz_{p}^{*}$. If there exists $x_{0}\in\zz_{p}$
such that $\left|x_{0}^{2}-y\right|<1$ then there exists $x\in\zz_{p}$
such that $x^{2}=y$ . (Take $f(X)=X^{2}-y$, so $\left|f(x_{0})\right|<1$
but $\left|f'(x_{0})\right|=\left|2x_{0}\right|=1$). 

\begin{thm*}
Any $z\in\zz$ with $p\nmid z$, is a square in $\zz_{p}\iff\left(\frac{z}{p}\right)=+1$\end{thm*}
\begin{claim*}
$\qq_{p}^{*}/(\qq_{p}^{*})^{2}$ has $4$ elements represented by
$1,c,p,cp$ where $c\in\{1,\dots,p-1\}$ is a quadratic non-residue. \end{claim*}
\begin{cor*}
It follows that $\qq_{p}$ has exactly $3$ quadratic extensions. \end{cor*}
\begin{proof}[Proof of the claim]
 Suppose $x\in\qq_{p}^{*}$. We may assume $x=u$ or $pu$ for $u\in\zz_{p}$
(on multiplying $x$ by a power of $p^{2}\in\qq^{*2}$). Let $\alpha\in\{1,\dots,p-1\}$
be such that $u\equiv\alpha\mod p\zz_{p}$ (i.e. $u=\alpha+v$ for
some $v\in p\zz_{p}$). Then $u=\alpha(1+\alpha^{-1}v)$ and $1+\alpha^{-1}v\equiv1\mod p\zz_{p}$
which is a square. Thus we may assume $u=\alpha$. But $\left(\frac{u}{p}\right)=1\Rightarrow u\in\qq_{p}^{*2}$,
otherwise $uc$ is in $\qq_{p}^{*2}$. 
\end{proof}
\item $p=2$. See exercise B.1
\end{itemize}
\item Since residue field $k_{K}$is finite, it follows that $k_{K}^{*}$
is cyclic group of order $q-1$ where $q=p^{r}$ for some prime $p$.
Now show there exists an alternative set of representative for $k_{K}=\mathcal{O}_{K}/\mathcal{P}_{K}$,
besides $\{0,\dots,q-1\}$.


Note $p\cdot1\in\mathcal{O}_{K}$ and so $q-1\in\mathcal{O}_{K}^{*}$.
For each $\alpha\in k_{K}^{*}$, let $x_{0}\in\mathcal{O}_{K}^{*}$
such that $x_{0}\equiv\alpha\mod p$ and consider $f(x)=x^{q-1}-1$.
Then $\left|f(x_{0})\right|<1$, $\left|f'(x_{0})\right|=\left|q-1\right|\cdot\left|x_{0}\right|^{q-2}=1$.
Hence by Theorem (\ref{thm:Hensel}), there exists a unique \emph{Teichmuller
representative} $\widehat{\alpha}\in\mathcal{O}_{K}^{*}$ of $\alpha$
such that $f(\widehat{\alpha})=0$ and $\widehat{\alpha}\equiv\alpha\mod p$.
We can take $\{0\}\cup\{\widehat{\alpha}:\alpha\in k_{K}^{*}\}$ as
a set of representative for $k_{K}$. 


Define \emph{principal congruence subgroup} $\mathcal{U}_{K}^{n}=\{u\in\mathcal{U}_{K}=\mathcal{O}_{K}^{*}:u-1\in\mathcal{P}_{K}^{n}\}=1+\mathcal{P}_{K}^{n}$.
Then $\mathcal{U}_{K}$ and $\mathcal{U}_{K}^{n}$ are open and closed
and compact in $K^{*}$ (with induced topology). 


We have isomorphism of topological groups:
\begin{itemize}
\item $K^{*}/\mathcal{U}_{K}\to\zz$ defined by $x\mathcal{U}_{K}\mapsto V_{K}(x)$. 
\item $\mathcal{U}_{K}/\mathcal{U}_{K}^{1}\to k_{K}^{*}$ defined by $\xi^{v}\mathcal{U}_{K}^{1}\mapsto g^{v}$
where $\xi$ is a primitive $(q-1)$th root of unity in $K$ and $g$
is a generator for $k_{K}^{*}$. 
\end{itemize}

Hence any $x\in K^{*}$ can be uniquely written as $\pi^{u}\xi^{v}\epsilon$
for $\epsilon\in\mathcal{U}_{K}^{1}$, i.e., $K^{*}\cong\zz\times\zz/(q-1)\zz\times\mathcal{U}_{K}^{1}$. 

\end{enumerate}
\end{example*}
\newpage{}


\section{Extensions of local fields}

We consider field extensions of non-archimedean local fields. We would
like to show that these extension are non-archimedean local fields.


\subsection{Normed vector spaces}

Let $K$ be a non-archimedean local field
\begin{defn}
Let $V$ be a vector space over $K$. A function $\Vert\;\Vert:V\to\rr_{\geq0}$
is a \emph{norm} if
\begin{enumerate}
\item $\Vert x\Vert=0$ if and only if $x=0$
\item $\Vert x+y\Vert\leq\Vert x\Vert+\Vert y\Vert$
\item $\Vert\lambda x\Vert=|\lambda|\cdot\Vert x\Vert$ for all $\lambda\in K$
\end{enumerate}
\end{defn}
\begin{note*}
The norm induces a metric $d(x,y)=\Vert x-y\Vert$ on $V$, which
gives a topology. \end{note*}
\begin{defn}
Two norms $\Vert\;\Vert_{1}$ and $\Vert\,\,\Vert{}_{2}$ one a vector
$V$ are \emph{equivalent} if $\exists c_{1,}c_{2}>0$ such that $c_{1}\Vert x\Vert_{2}\leq\Vert x\Vert_{1}\leq c_{2}\Vert x\Vert_{2}$
$\forall x\in V$\end{defn}
\begin{xca*}
Show that $\Vert\,\,\Vert_{1},\Vert\:\Vert_{2}$ are equivalent if
and only if they induce the same topology on $V$.\end{xca*}
\begin{lem}
\label{lem:4.3}Let $V$ be a finite dimensional vector space over
$K$. Then any $2$ norms on $V$ are equivalent. Moreover, $V$ is
complete with respect to the induced metric.\end{lem}
\begin{proof}
We proof by induction on $n=\dim_{K}V$
\begin{elabeling}{00.00.0000}
\item [{$n=1$}] Trivial
\item [{$n>1$}] Let $e_{1},\dots,e_{n}$ be a basis for $V$ over $K$.
Put $a=a_{1}e_{1}+\dots+a_{n}e_{n}$ ($a_{j}\in K$) and define $\Vert a\Vert_{0}:=\max_{j}|a_{j}|$.
Check that $\Vert\,\,\Vert_{0}$ is a norm and that $V$ is complete
with respect to it. It will suffice to show any norm $\Vert\,\,\Vert$
on $V$ is equivalent to $\Vert\,\,\Vert_{0}$. Firstly $\Vert a\Vert\leq\sum_{j}|a_{j}|\cdot\Vert e_{j}\Vert\leq c_{2}\Vert a\Vert_{0}$
with $c_{2}=\sum\Vert e_{j}\Vert$. 


We now need to show $\exists c>0$ such that $\Vert a\Vert_{0}\leq c\Vert a\Vert$
for all $a\in V$~~~ $(*)$. If not, $\forall\epsilon>0$, there
exists $b=b_{\epsilon}\in V$ such that $\Vert b\Vert\leq\epsilon\Vert b\Vert_{0}$.
Assume without loss of generality that $\Vert b\Vert_{0}=|b_{n}|$.
Replacing $b$ by $b_{n}^{-1}b$ we have $b=c+e_{n}$ where $c\in\left\langle e_{1},\dots,e_{n-1}\right\rangle _{K}$. 


Summary: $(*)$ false, implies we can find a sequence $c^{(m)}\in W=\left\langle e_{1},\dots,e_{n-1}\right\rangle _{K}$
such that $\Vert c^{(m)}+e_{n}\Vert\to0$ as $m\to\infty$. But then
$\Vert c^{(m)}-c^{(l)}\Vert\to0$. So now use induction hypothesis.
Since $\dim W=n-1$, it is complete under $\Vert\;\Vert$. Thus there
exists $c^{*}\in W$ such that $\Vert c^{(m)}-c\Vert=0$. Hence $\Vert c^{*}+e_{n}\Vert=\lim_{m\to\infty}\Vert c^{(m)}+e_{n}\Vert=0$.
Therefore $c^{*}+e_{n}=0$, which is impossible. Hence $(*)$ hold
and so $\Vert\:\Vert$ and $\Vert\,\,\Vert_{0}$ are equivalent.

\end{elabeling}
\end{proof}
\begin{cor}
\label{cor:4.4}$V$ finite dimensional normed vector space over $K$.
Then $V$ is locally compact. (i.e., $v\in V$ has a compact neighbourhood)\end{cor}
\begin{proof}
By Lemma \ref{lem:4.3} we can assume $\Vert\,\,\Vert$ is $\Vert\,\,\Vert_{0}$,
with respect to some fixed basis $e_{1},\dots,e_{n}$. Now imitate
the proof of Lemma \ref{lem:3.8} to show that $\{v\in V:\Vert v\Vert_{0}\leq1\}$
is compact.
\end{proof}

\subsection{Extension of Absolute Values}

Let $K$ be a non-archimedean local field and $L\supset K$ an extension.
We say that an absolute value $\Vert\,\,\Vert$ $L$ \emph{extends
to the absolute value} $|\,\,|$ on $K$ if $\Vert\lambda\Vert=|\lambda|$
for all $\lambda\in K$
\begin{thm}
\label{thm:4.5}Let $L\supset K$ be a finite extension. Then there
exists a unique extension $\Vert\,\,\Vert$ of $|\,\,|$ to $L$.
Moreover, $L$, $\Vert\,\,\Vert$ is a non-archimedean local field. \end{thm}
\begin{proof}
~{}
\begin{elabeling}{00.00.0000}
\item [{Uniqueness:}] Suppose $\Vert\,\Vert_{1},\Vert\,\Vert_{2}$ extend
$|\,\,|$ to $L$. Then, regarding $L$ as a finite dimensional vector
space over $K$, Lemma \ref{lem:4.3}implies $\Vert\,\,\Vert_{1}$
and $\Vert\,\,\Vert_{2}$ are equivalent and some define the same
topology on $L$. But then they are equivalent as absolute values
and so by Lemma \ref{lem:1.10}, there exists $\alpha$ such that
$\Vert x\Vert_{1}=\Vert x\Vert_{2}^{\alpha}\forall x\in L$. But the
two absolute values are equal on $K$, so that $\alpha=1$. 
\item [{Second\_part}] Apply \ref{cor:4.4} and converse of Lemma \ref{lem:3.8}
\item [{Existence}] We will show that the extension of $\Vert\,\,\Vert$
of $|\,\,|$ to $L$ is given by $\Vert x\Vert=|N_{L/K}(x)|^{1/n}$
for $x\in L$, where $n=[L:K]$. Here $N_{L/K}:L\to K$ is the norm
map. (Recall: Thinking of $L$ as a vector space over $K$, multiplication
by $\alpha\in L$ gives a linear map $m_{\alpha}:K\to L$, with matrix
$A_{\alpha}\in M_{n}(K)$. Put $N_{L/K}(\alpha):=\det A_{\alpha}$).
For $x\in K$, $\Vert x\Vert=|x^{n}|^{1/n}=|x|$. So $\Vert\,\,\Vert$
does extend $|\,\,|$. 


For $x\in L^{*}$, the linear map $m_{x}:L\to L$ is invertible with
inverse $m_{x^{-1}}$. Thus the matrix $A_{x}$ is invertible, and
$\det A_{x}\neq0$. Hence $\Vert x\Vert\neq0$. Multiplicativity follows
from the multiplicativity of the norm map. 


Remains to prove the ultrametric inequality. Suffices to show $\Vert x\Vert\leq1$,
then $\Vert1+x\Vert\leq1.$ (Then, assuming $\Vert x\Vert\leq\Vert y\Vert$
then $\Vert x+y\Vert=\Vert y\Vert\cdot\Vert\frac{x}{y}+1\Vert\leq\Vert y\Vert$).
Suppose $\Vert x\Vert\leq1$. Let $\chi(x)$ be the characteristic
polynomial of the linear map $m_{x}:L\to L$. Let $f(X)=X^{r}+f_{r-1}X^{r-1}+\dots+f_{0}$
be the minimal polynomial of $x$. Here $r$ is the degree of $x$
over $K$. Then $\chi(X)=f(X)^{n/r}$ (where $n/r$ is the degree
of $L$ over $K(x)$, a proof of this can be found in Cassels book
Lemma B.3). Then $|f_{0}^{\mathrm{power}}|=|N_{L/K}(x)|\leq1,$ hence
$|f_{0}|\leq1$. Since $f$ is irreducible and monic it follows from
consideration of Newton polygon associated to it that $|f_{i}|\leq1$
(See Cassels chapter 4). Hence $f\in\mathcal{O}_{K}[X]$ and also
$\chi\in\mathcal{O}_{K}[X]$. Now $N_{L/K}(1+x)=\det(I_{n}+A_{x})=(-1)^{n}\chi(-1)$.
So $\Vert1+x\Vert=|\chi(-1)|^{1/n}\leq1$. This completes the proof

\end{elabeling}
\end{proof}
Since the absolute value on $L$ is unique, we will usually write
it as $|\,\,|$ instead of $\Vert\,\,\Vert$. 
\begin{cor}
$|\,\,|_{p}$ on $\qq_{p}$ extends uniquely to an absolute value
on algebraic closure $\overline{\qq_{p}}$.\end{cor}
\begin{proof}
$x\in\overline{\qq_{p}}$ then $x\in K$ for some finite extension
$K/\qq_{p}$. Let $|\,\,|=|\,\,|_{K}$ where $|\,\,|$ is the unique
absolute value on $K$ extending $|\,\,|_{p}$. This is independent
of choice of $K$ by Theorem \ref{thm:4.5}
\end{proof}

\subsection{Ramification}

Suppose $L/K$ is a finite extension of non-archimedean local fields
of degree $n=[L:K]$.
\begin{lem}
\label{lem:4.7}There exists a natural injection $k_{K}\to k_{L}$
such that $k_{L}$ is an extension of $k_{K}$ of degree $f=f(L/K):=[k_{L}:k_{K}]\leq n$. \end{lem}
\begin{proof}
There is certainly an inclusion $\mathcal{O}_{K}\hookrightarrow\mathcal{O}_{L}$.
But $\mathcal{P}_{K}=\mathcal{O}_{K}\cap\mathcal{P}_{L}$, this induces
the injection $k_{K}\to k_{L}$.

Let $\alpha_{1},\dots,\alpha_{n+1}\in k_{L}$. Show that there are
linearly dependent over $k_{K}$. Then we will have shown that $f=\dim_{k_{K}}k_{L}\leq n$.
Let $\widehat{\alpha_{1}},\dots,\widehat{\alpha_{n+1}}\in\mathcal{O}_{L}$
such that $\alpha_{i}=\widehat{\alpha}_{i}+\mathcal{P}_{L}$ for $1\leq i\leq n+1$.
Since $\dim_{K}L=n$, there are linearly dependent over $K$, i.e.,
there exists $\lambda_{i}\in K$ not all zeroes such that $\sum_{i=1}^{n+1}\lambda_{i}\widehat{\alpha_{i}}=0$.
Without loss of generality, we assume that $\lambda_{n+1}\neq0$.
Let $\mu_{i}\in k_{K}$ be the reduction of $\lambda_{i}\lambda_{n+1}^{-1}$
modulo $\mathcal{P}_{L}$. Then $\sum_{i=1}^{n}\mu_{i}\alpha_{i}+\alpha_{n+1}=0$,
as required.\end{proof}
\begin{defn}
If $f=f(L/K)=n$, we say $L/K$ is \emph{unramified}.

If $f=f(L/K)=1$, we say $L/K$ is \emph{totally ramified}. 

If $f=f(L/K)<n$, we say $L/K$ is \emph{ramified}. \end{defn}
\begin{rem*}
If $K\subset L\subset E$ is a tower of extensions then $f(E/K)=f(E/L)\cdot f(L/K)$. 
\end{rem*}
We shall see that unramified extensions are easy to characterise. 
\begin{thm}
\label{thm:4.9} Let $\alpha\in k_{L}=\mathcal{O}_{L}/\mathcal{P}_{L}$.
Then there exists $\widehat{\alpha}\in\mathcal{O}_{L}$ such that
$\widehat{\alpha}+\mathcal{P}_{L}=\alpha$ and $[K(\widehat{\alpha}):K]=[k_{K}(\alpha):k_{K}]$. 

Furthermore, the field $K(\widehat{\alpha})$ depends on $\alpha$. \end{thm}
\begin{rem*}
The extension $K(\widehat{\alpha})/K$ is unramified. \end{rem*}
\begin{proof}
Let $\phi\in k_{x}[x]$ be the minimal polynomial of $\alpha$. Let
$\Phi\in K[X]$ be any \emph{lift} of $\phi$ (i.e., $\deg\phi=\deg\Phi$
and $\phi=\overline{\Phi}$, meaning coefficients of $\Phi$ are reduced
modulo $\mathcal{P}_{K}$). Let $\widehat{\alpha_{0}}\in\mathcal{O}_{L}$
be an element of the residue class of $\alpha$. Then $\overline{\Phi}(\widehat{\alpha_{0}})=\phi(\alpha)=0$
and $\overline{\Phi}'(\widehat{\alpha}_{0})=\phi'(\alpha)\neq0$ (since
$k_{K}$ is a finite field so it is perfect). Thus $\left|\Phi(\widehat{\alpha_{0}})\right|<1$
and $\left|\Phi'(\widehat{\alpha_{0}})\right|=1$. Hence by Hensel's
lemma, with $K(\widehat{\alpha}_{0})$ as the ground field, implies
there exists $\widehat{\alpha}\in K(\widehat{\alpha}_{0})\subset L$
such that $\Phi(\widehat{\alpha})=0$, $\left|\widehat{\alpha}-\widehat{\alpha_{0}}\right|<1$.
Hence $\widehat{\alpha}$ in residue class of $\alpha$ and $[K(\widehat{\alpha}):K]=[k_{K}(\alpha):k_{K}]$
since $\Phi$ is irreducible. 

Now suppose that $\widehat{\alpha}'$ is also in the residue class
of $\alpha$ and satisfies $[K(\widehat{\alpha}'):K]=[k_{K}(\alpha):k_{K}]$.
Then the above argument implies $\widehat{\alpha}\in K(\widehat{\alpha}')$
and so $K(\widehat{\alpha})=K(\widehat{\alpha}')$. But we must have
equality since the degrees are the same.\end{proof}
\begin{cor}
\label{cor:Cor 4.10}There exists a bijection between intermediate
fields $E$ (with $K\subset E\subset L$) which are unramified and
the fields $k$ with $k_{K}\subset k\subset k_{L}$, given by $E\to k_{E}=E\cap\mathcal{O}_{L}/E\cap\mathcal{P}_{L}$\end{cor}
\begin{proof}
The previous theorem gives one direction.

Let $k$ be an intermediate field $k_{K}\subset k\subset k_{L}$.
Let $q=\#k$. Then $k=k_{K}(\alpha)$ for some $(q-1)$th root of
unity $\alpha$. Then apply Theorem \ref{thm:4.9}\end{proof}
\begin{cor}
\label{cor:4.11}Let $K$ be a non-archimedean local field. For all
$n\in\nn$ there exists a unique (up to isomorphism) unramified extension
of degree $n$. It is the splitting field over $K$ of $X^{q}-X$
where $q=q_{K}^{n}$, with $q_{K}=\#k_{K}$\end{cor}
\begin{proof}
Let $L/K$ be unramified extension of degree $n$. Then $k_{L}$ has
$q=q_{K}^{n}$ elements. Then $L$ contains a full set of $(q-1)$th
roots of unity (By example $2$ after Hensel's lemma). In particular
$X^{q}-X$ splits in $L$ and so $L$ contains its splitting field,
say $F$. However $q_{L}=q_{F}=q$ and so by Corollary \ref{cor:Cor 4.10}
we must have $F=L$\end{proof}
\begin{cor}
\label{cor:4.12}Let $f\in\mathcal{O}_{K}[X]$ be monic of degree
$n$ and reduction $\overline{f}\mod\mathcal{P}_{K}$ is irreducible.
Then
\begin{enumerate}
\item if $L=K(\alpha)$ and $\alpha$ has minimal polynomial $f$, then
$L/K$ is unramified
\item The splitting field of $f$ over $K$ is unramified and has degree
$n$. 
\end{enumerate}
\end{cor}
\begin{proof}
~{}
\begin{enumerate}
\item Note that $k_{L}\supset k_{K}(\overline{\alpha})$, where $\overline{\alpha}$
is the reduction of $\alpha\mod\mathcal{P}_{L}$. Moreover, $k_{K}(\overline{\alpha})$
has degree $n$ over $k_{K}$. Hence $f(L/K)\geq n$. But we also
have $f(L/K)\leq n=[L:K]$ by Lemma \ref{lem:4.7}. 
\item Let $L$ be the splitting field of $f$ over $K$ and let $\alpha,\beta$
be roots of $f$ in $L$. Then part 1.~implies that $K(\alpha)$
and $K(\beta)$ are both unramified extensions of degree $n$. Then
Corollary \ref{cor:Cor 4.10} implies they are equal, therefore $L=K(\alpha)$.
\end{enumerate}
\end{proof}
\begin{summary*}
Unramified extensions of $K$ are obtained by adjoining a root of
unity of order coprime to residual characteristic of $K$. 
\end{summary*}
Now let us look at ramified extensions.

Suppose $L/K$ is a finite extension of non-archimedean local fields.
Consider the relationship between value groups $\Gamma_{L}=\{\left|x\right|:x\in L^{*}\}$
is a discrete (cyclic) subgroup of $\rr_{>0}$. 
\begin{defn}
The \emph{ramification index} of $L/K$ is $e=e(L/K)=[\Gamma_{L}:\Gamma_{K}]$. 
\end{defn}
If $\pi_{L},\pi_{K}$ are uniformisers for $L$ and $K$ respectively.
Recall $\left|\pi_{K}\right|<1$ is a generator for $\Gamma_{K}$
and similarly $\pi_{L}$ for $\Gamma_{L}$. Then $\left|\pi_{K}\right|=\left|\pi_{L}\right|^{e}$.
This implies $e(E/K)=e(E/L)e(L/K)$ for any tower $K\subset L\subset E$. 
\begin{thm}
\label{thm:4.14}$L/K$ be a finite extensions of non-archimedean
local fields of degree $n$. Then $n=ef$.
\end{thm}
It follows from this that $L/K$ is unramified if and only if $e(L/K)=1$

$L/K$ is totally ramified if and only if $e(L/K)=n$

It is ramified if and only if $e(L/K)>1$
\begin{proof}
Let $\pi_{L}$ be a uniformiser of $L$ and let $\widehat{\alpha}_{1},\dots,\widehat{\alpha}_{f}$
be any lift to $\mathcal{O}_{L}$ of a basis for $k_{L}/k_{K}$. (As
in Theorem \ref{thm:4.9}) We will prove that $\mathcal{B}=\left\{ \widehat{\alpha}_{i}\pi_{L}^{j}:1\leq i\leq f,0\leq j\leq e-1\right\} $
is a basis for $L/K$. (In fact we will prove that it is an $\mathcal{O}_{K}$
basis for $\mathcal{O}_{L}$.) 

Firs suppose that $\mathcal{B}$ is not linearly independent over
$K$. Then there exists $a_{ij}\in K$, not all zeroes, such that
\[
\sum_{i,j}a_{ij}\widehat{\alpha}_{i}\pi_{L}^{j}=0\,\,\,(*)
\]
Without loss of generality, assume that $\max_{i,j}|a_{ij}|=1$. Hence,
there exists integers, $I,J$ such that $\left|a_{ij}\right|\leq\left|\pi_{K}\right|$
for $1\leq i\leq f,j<J$, $\left|a_{IJ}\right|=1$. If we reduce $\sum_{i}a_{iJ}\widehat{\alpha}_{i}$
module $\mathcal{P}_{L}$, then we get a non-zero coefficient $\overline{a}_{IJ}$.
Since $\widehat{\alpha}_{i}\mod\mathcal{P}_{L}$ are linearly independent
over $k_{K},$ this reduction is non-zero. Thus $\left|\sum_{i}a_{iJ}\widehat{\alpha}_{i}\right|=1$.
We now get 
\[
\left|\sum_{i}a_{ij}\widehat{\alpha}_{i}\pi_{L}^{j}\right|\begin{cases}
\leq\left|\pi_{K}\right|=\left|\pi_{L}\right|^{e} & j<J\\
=\left|\pi_{L}\right|^{J} & j=J\\
\leq\left|\pi_{L}\right|^{J+1} & j>J
\end{cases}
\]
Recalling $J\leq e-1$, one term in $(*)$ has to be bigger than all
the others. Contradiction

Now let $x\in L$. We claim that $x$ is in the $K$-span of $\mathcal{B}$.
Multiplying by a suitable power of $\pi_{K}$, we reduce to the case
$x\in\mathcal{O}_{L}$. (If $\pi_{K}^{n}x=\sum_{ij}a_{ij}\widehat{\alpha}_{i}\pi_{L}^{j}$
with $a_{ij}\in K$, then putting $b_{ij}=\pi_{k}^{-n}a_{ij}$ gives
$x=\sum b_{ij}\widehat{\alpha}_{i}\pi_{L}^{j}$). Since $\alpha_{i}\equiv\widehat{\alpha}_{i}\mod\mathcal{P}_{L}$
form a basis for $k_{L}/k_{K}$, there exists $c_{i0}\in k_{K}$ such
that $\overline{x}=\sum_{i}c_{i0}\alpha_{i}$. Choose any lifts $\widehat{c_{i0}}\in\mathcal{O}_{K}$,
we have $x-\sum_{i}\widehat{c}_{i0}\widehat{\alpha}_{i}=\pi_{L}x_{1}\in\mathcal{P}_{L}$
for some $x_{1}\in\mathcal{O}_{L}$. Now repeat process on $x_{1}$,
and so on, until we have obtained $\widehat{c}_{ij}\in\mathcal{O}_{K}$
such that 
\[
x-\sum_{j=0}^{e-1}\sum_{i}\widehat{c}_{ij}\widehat{\alpha}_{i}\pi_{L}^{j}=\pi_{L}^{e}x_{e}
\]
 for some $x_{e}\in\mathcal{O}_{L}$. Now $\left|\pi_{L}\right|^{e}=\left|\pi_{K}\right|$
and so $\pi_{L}^{e}x_{e}=\pi_{K}x^{(1)}$ for some $x^{(1)}\in\mathcal{O}_{L}$.
Now we start again with $x^{(1)}$ instead of $x$. Carrying on in
this way, we find a succession of linear combinations 
\[
c_{r}=\sum_{j=0}^{e-1}\sum_{i}\widehat{c}_{ij}^{(r)}\widehat{\alpha}_{i}\pi_{L}^{j}
\]
of elements of $\mathcal{B}$ with coefficients in $\mathcal{O}_{K}$
such that $x-c_{0}-c_{1}\pi_{K}-\dots-c_{s}\pi_{K}^{s}\in\pi_{K}^{s+1}\mathcal{O}_{L}$,
$\forall s\geq0$. Now let $s\to\infty$ and, using completeness,
put 
\[
a_{ij}=\sum_{r=0}^{\infty}\widehat{c}_{ij}^{(r)}\pi_{K}^{r}
\]
 Then $x=\sum_{i,j}a_{ij}\widehat{\alpha}_{i}\pi_{L}^{j}$ as required. 
\end{proof}
A polynomial $f(X)=f_{n}X^{n}+f_{n-1}X^{n-1}+\dots+f_{0}\in\mathcal{O}_{K}[X]$
is said to be \emph{Eisenstein }if 
\[
\left|f_{n}\right|=1,\left|f_{j}\right|<1\,\forall0\leq j<n,\left|f_{0}\right|=\left|\pi_{K}\right|\quad(\dagger)
\]


Aside on irreducibility: Let $f=f_{0}+f_{1}X+\dots+f_{n}X^{n}\in K[X]$
with $f_{0}\neq0,f_{n}\neq0$. The \emph{Newton polygon} of $f$ is
the convex hull in $\rr^{2}$ of the points $p(j)=(j,\log\left|f_{j}\right|)$
for $f_{j}\neq0$. It consist of line segments $\sigma_{s}$ for $1\leq s\leq r$,
where $\sigma_{s}$ joins $P(m_{s-1})$ to $P(m_{s})$ and $0=m_{0}<m_{1}<\dots<m_{r}=n$.
The slope of $\sigma_{s}$ is $\gamma_{s}=\left(\log\left|f_{m_{s}}\right|-\log\left|f_{m_{s-1}}\right|\right)/(m_{s}-m_{s-1})$.
We say $f$ is of type $(l_{1},\gamma_{1},\dots,l_{r},\gamma_{r})$
where $l_{s}=m_{s}-m_{s-1}$. If $r=1$ then $f$ is said to be pure. 
\begin{fact*}
\emph{(Cassels Local Field pg 100):} $f$ of type $(l_{1},\gamma_{1},\dots,l_{r},\gamma_{r})$
then $f(X)=g_{1}(X)\dots g_{r}(X)$ where $g_{s}$ is pure of type
$(l_{s},\gamma_{s})$. 
\end{fact*}
Totally ramified extensions are quite easy to classify. 
\begin{thm}
\label{thm:4.15}Let $L/K$ be a finite extension of non-archimedean
local fields. Then $L/K$ is totally ramified if and only if $L=K(\beta)$,
where $\beta$ is the root of an Eisenstein polynomial. \end{thm}
\begin{proof}
~{}
\begin{elabeling}{00.00.0000}
\item [{$\Rightarrow$}] $L/K$ totally ramified of degree $n$, let $\beta=\pi_{L}$
be a uniformiser for $L$. Then $1,\pi_{L},\dots,\pi_{L}^{n-1}$ are
linearly independent over $K$ (as in the proof of Theorem \ref{thm:4.14}).
Hence there exists an equation $\beta^{n}+f_{n-1}\beta^{n-1}+\dots+f_{0}=0$
with $f_{j}\in K$. Two of the summands must have the same absolute
value and this must be the first and the last. (Suppose $|f_{k}\beta^{k}|=|f_{l}\beta^{l}|$
for some $n\geq k>l\geq0$, then there exists $a_{k},a_{l}\in\zz$
such that $\left|\pi_{L}^{k-l}\right|=\left|\beta\right|^{k-l}=\left|\pi_{K}\right|^{a_{k}-a_{l}}=\left|\pi_{L}\right|^{n(a_{k}-a_{l})}$,
hence $l=0$ and $k=n$). Therefore $\left|f_{0}\right|=\left|\pi_{L}\right|^{n}=\left|\pi_{K}\right|$
and $\left|f_{j}\right|<1$ for all $j$. Hence a polynomial in the
equation is Eisenstein.
\item [{$\Leftarrow$}] Suppose $f_{n}\beta^{n}+\dots+f_{0}=0$, where
$f_{j}\in K$ such that $\left|f_{n}\right|=1$, $\left|f_{j}\right|<1$
and $\left|f_{0}\right|=\left|\pi_{K}\right|.$Then $\left|\beta^{n}\right|<1$
and so $\left|\beta\right|<1$. Hence the last term in the sum is
bigger than all the others. except possibly for the first. Since the
sum is zero, they must be equal (in absolute value) and so $\left|\beta\right|^{n}=\left|f_{0}\right|=\left|\pi_{K}\right|$.
Hence $\beta=\pi_{L}^{g}y$ for some unit $y$, then $\left|\pi_{L}\right|^{gn}=\left|\pi_{K}\right|$,
whence $e(L/K)=gn\geq n$. But we must have equality by Theorem \ref{thm:4.14},
since $n=[L:K]$. Hence $L/K$ is totally ramified.
\end{elabeling}
\end{proof}
\begin{example*}
Let $f(X)=X^{4}-2X^{3}+17X^{2}+22X+66$. We are going to look at the
splitting field $E$ over $\qq_{p}$ for various prime $p$. In each
case, we want to calculate:
\begin{itemize}
\item The degree $[E:\qq_{p}]$?
\item Residue class degree $f(E/\qq_{p})$?
\item Ramification index $e(E/\qq_{p})$?
\item (If possible) maximal unramified subextension $L$? i.e., $E\supset L\supset\qq_{p}$
with $L/\qq_{p}$ unramified and hence $E/L$ totally ramified.\end{itemize}
\begin{elabeling}{00.00.0000}
\item [{$p=2$}] Newton polygon. Note $\log\left|2^{a}b\right|_{2}=\log2^{-1}$
\[
\xyR{0.5pc}\xyC{1pc}\xymatrix{ &  &  & \underset{0}{\bullet}\ar@{-}[rr] &  & \underset{0}{\bullet}\\
 & \underset{-\log2}{\bullet}\ar@{-}[rru] & \underset{-\log2}{\bullet} &  & \underset{-\log2}{\bullet}\\
\ar@{-}[rrrrrr] & \underset{0}{\mid} & \underset{1}{\mid} & \underset{2}{\mid} & \underset{3}{\mid} & \underset{4}{\mid} & \,
}
\]
$l_{1}=2-0=2$, $\gamma_{1}=\log2/2$, $l_{2}=4-2=2$, $\gamma_{2}=0$.
So the type is $(2,\frac{1}{2}\log2,2,0)$ and factorises as a product
of $2$ quadratic. Trial an error over $\zz$ gives $f(X)=\underset{:=g(X)}{\underbrace{(X^{2}-2X+6)}}\cdot\underset{:=h(X)}{\underbrace{(X^{2}-11)}}$
and $g,h$ irreducible over $\qq_{2}$ (Eisenstein criterion for $g$
and $11\not\equiv1\mod8$ so apply Exercise B.1). Let $\alpha$be
a root of $g$ and $\beta$ a root of $h$ in $E$. Then by Theorem
\ref{thm:4.15}, we have that $\qq_{2}(\alpha)/\qq_{2}$ is totally
ramified. Since $\beta-1$ satisfies $0=h(X+1)=X^{2}+2X-10$ which
is Eisenstein, we also have $\qq_{2}(\beta)/\qq_{2}$ is totally ramified.
Note that $[\qq_{2}(\alpha):\qq_{2}]=[\qq_{2}(\beta):\qq_{2}]=2$. 


Next $\gamma=\alpha-1$ satisfies $0=g(X+1)=X^{2}-7$, so $\gamma^{2}=7$.
Let $\delta=\beta\gamma$, then $\delta^{2}=7\cdot11$. We claim that
$\qq_{2}(\delta)/\qq_{2}$ is unramified of degree $2$. Then $[E:\qq_{2}]=4$,
$e(E/\qq_{2})=f(E/\qq_{2})=2$ and $L=\qq_{2}(\delta)$. To show that
$\qq_{2}(\delta)/\qq_{2}$ is unramified, we need to show (by Corollary
\ref{cor:4.11}) that $\qq_{2}(\delta)$ is obtained from $\qq_{2}$
by adjoining a root of $X^{2}+X+1$ (i.e., a primitive $(2^{2}-1)$th
root of unity). Do this by applying Hensel to $(\delta-1)/2$. 

\item [{$p$~odd}] Use the fact that $E=\qq_{p}(\gamma,\beta)$ where
$\gamma$ and $\beta$ are as above ($\gamma^{2}=7$, $\beta^{2}=11$)

\begin{elabeling}{00.00.0000}
\item [{$p=3$}] Since $\left(\frac{7}{3}\right)=1$, then $\gamma\in\qq_{3}$,
while $\left(\frac{11}{3}\right)=-1$, so $X^{2}-11$ is irreducible
in $\ff_{3}[X].$ Hence it follows that $E=\qq_{3}(\beta)$ and it
is unramified of degree $2$ over $\qq_{3}$ (by Corollary \ref{cor:4.12})
\item [{$p=19$}] Since $\left(\frac{7}{19}\right)=\left(\frac{11}{19}\right)=1$
so $E=\qq_{19}$ 
\end{elabeling}

(all primes behave like $3,5,11,13$ or $19$)

\end{elabeling}
\end{example*}
\newpage{}


\section{Algebraic Closure}

Recall that a field $K$ is called \emph{algebraically closed} if
every polynomial with coefficients in $K$ has a root in $K$.
\begin{defn}
An extension $\overline{K}\supset K$ is the \emph{algebraic closure}
of $K$ if
\begin{enumerate}
\item $\overline{K}$ is algebraically closes
\item Any $\alpha\in\overline{K}$ is algebraic over $K$. 
\end{enumerate}
\end{defn}
For example, $\cc$ is the algebraic closure of $\rr$, $[\cc:\rr]=2$.
If $\overline{\qq_{p}}$ is the closure of $\qq_{p}$ then $[\overline{\qq_{p}}:\qq_{p}]=\infty$.
(Note $\overline{\qq_{p}}$ must contain roots of $X^{n}-p$ for all
$n\in\nn$)
\begin{thm}
\label{thm:5.2}Let $K$ be a field. Then there exists an algebraic
closure $\overline{K}$ of $K$ and it is unique up to isomorphism. \end{thm}
\begin{proof}
(Sketch) Let $\Lambda$ be the set of all irreducible polynomials
over $K$ of degree $\geq2$. Let $\Xi=\{X_{f}:f\in\Lambda\}$ be
a family of indeterminate indexed by $\Lambda$. Put $R=K[\Xi]$.
Consider the ideal $I=\{f(X_{f}):f\in\Lambda\}$. This is a proper
ideal: if not we would have an equation $1=u_{1}f_{1}(X_{f_{1}})+\dots+u_{n}f_{n}(X_{f_{n}})$
for some $u_{j}\in R.$ Let $E/K$ formed by adjoining roots $\alpha_{1},\dots,\alpha_{n}$
of $f_{1},\dots,f_{n}$ respectively. Then we deduce that $1=0$ a
contradiction. 

Since $I$ is proper, it is contained in a maximal ideal $m$ of $R$.
Then $\overline{K}=R/m$ is a field and the homomorphism $K\to K[\Xi]\twoheadrightarrow\overline{K}$
is an embedding of $K\to\overline{K}$. We claim $\overline{K}$ is
an algebraic closure of $K$. If $f$ is an irreducible polynomial
in $K[X]$, then $\alpha=X_{f}+m$ is an root of $f$ in $\overline{K}$
(since $f(X_{f})\in I\subset m$). Moreover, each $X_{f}+m$ is algebraic
over $\overline{K}$ and $\overline{K}$ is generated by them.

Uniqueness: Essentially follows form uniqueness of splitting fields
of polynomials over $K$.
\end{proof}
From now on $K$ is a non-archimedean local field with algebraic closure
$\overline{K}$. Recall that the absolute value on $K$ extends uniquely
to $\overline{K}$. ($\forall\alpha\in\overline{K},$there exists
$K\subset L\subset\overline{K}$ such that $L=K(\alpha)$, then $\left|\alpha\right|=\left|N_{L/K}(\alpha)\right|^{1/[L:K]}$). 

We want to know if it is possible for $\overline{K}$ to be a non-archimedean
local field:
\begin{elabeling}{00.00.0000}
\item [{$\Gamma_{\overline{K}}$}] Let us ask is the value group $\Gamma_{\overline{K}}$
is discrete? Suppose $\Gamma_{K}=\{\left|x\right|:x\in K\}$ is generated
by $g<1$. Suppose $r\in\Gamma_{\overline{K}}$. Then $r=\left|\alpha\right|$
for some $\alpha\in\overline{K}$. Let $L/K$ of degree $n$ such
that $\alpha\in L$. Then $\left|\alpha\right|=g^{m/n}$ for some
$m\in\zz$. Hence $\Gamma_{\overline{K}}\subset\left\{ g^{m/n}:\frac{m}{n}\in\qq\right\} $.
In fact we have equality. Let $L\subset\overline{K}$ be an extension
obtained by adjoining a root $\alpha$ of the Eisenstein polynomial
$X^{n}-\pi_{K}X-\pi_{K}$. Then $\alpha$ is the uniformiser for $L$
and $L/K$ is totally ramified of degree $n$. Hence $\left|\alpha\right|=g^{1/n}$
and so $\left|\alpha^{m}\right|=g^{m/n}$. This shows that $\Gamma_{\overline{K}}$
is \uline{not} discrete. 
\item [{$k_{\overline{K}}$}] Consider the residue field $k_{\overline{K}}.$
Let $\alpha\in k_{\overline{K}}=\mathcal{O}_{\overline{K}}/\mathcal{P}_{\overline{K}}$
and let $\widehat{\alpha}$ be a lift of $\alpha$ to $\mathcal{O}_{\overline{K}}$.
Then $\widehat{\alpha}\in\overline{K}$ and so there exists a minimal
polynomial $\Phi\in\mathcal{O}_{K}[X]$ of $\widehat{\alpha}$ over
$K$. Let $\phi\in k_{K}[X]$ denote the reduction modulo $\mathcal{P}_{K}$
of $\Phi$. Then it follows $\phi(\alpha)=0$ and so $\alpha$ is
algebraic over $k_{K}$. Thus $k_{\overline{k}}\subset\overline{k_{K}}$.
In fact we have equality here. Suppose $\phi\in k_{K}[X]$ is irreducible
and let $\Phi$ be a lift of $\phi$. Then $\Phi$ has a root $\alpha\in\mathcal{O}_{\overline{K}}$
(since $\overline{K}$ is algebraic closed). Then $\overline{\alpha}=\alpha+\mathcal{P}_{\overline{K}}$
is a root of $\phi$ in $k_{\overline{K}}$. Hence $k_{\overline{K}}=\overline{k_{K}}$. 
\end{elabeling}
Suppose $L/K$ is Galois. 
\begin{xca*}
If $\left|\,\,\right|$ is an absolute value on $L$ which extends
the absolute value on $K$, then so $\Vert x\Vert=\left|\sigma(x)\right|$
for all $\sigma\in\mbox{Gal}(L/K)$. By uniqueness, we have $\left|x\right|=\left|\sigma(x)\right|$
for all $x\in L\forall\sigma\in\mbox{Gal}(L/K)$. \end{xca*}
\begin{thm}[Krasner's Lemma ]
\label{thm:Krasmer's Lemma} $K$ field of characteristic $0$, which
is complete with respect to a non-archimedean absolute value $\left|\,\,\right|$.
Let $a,b\in\overline{K}$ and suppose that $\left|b-a\right|<\left|a-a_{i}\right|$for
all $2\leq i\leq n$ where $a_{1}=a,a_{2},\dots,a_{n}$ are roots
of the minimal polynomial of $a$ in $K[X]$. Then $K(a)\subset K(b)$. \end{thm}
\begin{proof}
Put $L=K(b)$. Suppose for contradiction that $a\notin L$. Let $f\in L[X]$
be minimal polynomial of $a$ over $L$. Let $E$ be the splitting
field $f$ over $L$. Then $E/L$ is Galois and since $a\notin L$,
there exists $\sigma\in\mbox{Gal}(E/L)$ which does not fix $a$.
Then $\sigma(a)=a_{i}$ for some $i>1$. $\left|a-a_{i}\right|\leq\max\{\left|a-b\right|,\underset{=\left|\sigma(b-a)\right|=\left|b-a\right|}{\underbrace{\left|b-a_{i}\right|}}\}=\left|a-b\right|$,
which is a contradiction. 
\end{proof}

\subsection*{Incompleteness}

$K=\qq_{p}$
\begin{thm}
$\overline{\qq}_{p}$ is not complete with respect to $\left|\,\,\right|_{p}$\end{thm}
\begin{proof}
We need to find a Cauchy sequence $\left\{ \alpha_{n}\right\} $ in
$\overline{\qq}_{p}$ which does not converge. For $i\geq0$, let
$\zeta_{i}$ be a root of unity of order $p^{(i+1)!}-1$. Put $F_{i}=\qq_{p}(\zeta_{i})$,
then
\begin{itemize}
\item $F_{i}$ is the splitting field of $X^{p^{(1+i)!}}-X$ over $\qq_{p}$.
Thus it is an unramified extension of $\qq_{p}$ of degree $(i+1)!$
and it is Galois (Corollary \ref{cor:4.11})
\item $\zeta_{i-1}\subset F_{i}$ since $[F_{i}:F_{i-1}]=i+1$ and moreover
$p^{i!}-1|p^{(i+1)!}-1$. 
\end{itemize}
Consider the sequence $\alpha_{n}=\sum_{i=1}^{n}\zeta_{i}p^{i}$.
Since $\left|\alpha_{m}-\alpha_{n}\right|_{p}=\left(\frac{1}{p}\right)^{\min\{m,n\}}$,
so this is certainly Cauchy. We claim it does not have a limit in
$\overline{\qq}_{p}$. Suppose that it does have a limit, $\alpha=\sum_{n=0}^{\infty}\zeta_{i}p^{^{i}}\in\overline{\qq}_{p}$.
Let $d$ be the degree of the minimal polynomial $m_{\alpha}$ of
$\alpha$ over $\qq_{p}$. Now $F_{d}/F_{d-1}$ is Galois of degree
$d+1$. Hence there exists $\sigma_{1},\dots,\sigma_{d+1}\in\mbox{Gal}(F_{d}/F_{d-1})$
such that the images of $\zeta_{d}$ are all distinct. Note that $\left|\sigma_{i}(\alpha-\alpha_{d})\right|_{p}=\left|\alpha-\alpha_{d}\right|_{p}\leq p^{-(d+1)}$.
Also for $i\neq j$, we have 
\begin{eqnarray*}
\sigma_{i}(\alpha_{d})-\sigma_{j}(\alpha_{d}) & = & \sum_{k=0}^{d-1}\zeta_{k}p^{k}+\sigma_{i}(\zeta_{d})p^{d}-\left(\sum_{k=0}^{d-1}\zeta_{k}p^{k}+\sigma_{j}(\zeta_{d})p^{d}\right)\\
 & = & p^{d}(\sigma_{i}(\zeta_{d})-\sigma_{j}(\zeta_{d})).
\end{eqnarray*}
Hence for $i\neq j$, we have $\left|\sigma_{i}(\alpha_{d})-\sigma_{j}(\alpha_{d})\right|_{p}=p^{-d}$
(since $\sigma_{i}(\zeta_{d})$ and $\sigma_{j}(\zeta_{d})$ are distinct
and $(p^{(d+1)!}-1)$th root of unity). We conclude that 
\begin{eqnarray*}
\left|\sigma_{i}(\alpha)-\sigma_{j}(\alpha)\right|_{p} & = & \left|\sigma_{i}(\alpha-\alpha_{d})+\sigma_{i}(\alpha_{d})-\sigma_{j}(\alpha_{d})-\sigma_{j}(\alpha-\alpha_{d})\right|_{p}\\
 & = & p^{-d}
\end{eqnarray*}
This implies that $\sigma_{i}(\alpha)\neq\sigma_{j}(\alpha)$ for
all $i\neq j$. But then $\sigma_{1}(\alpha),\dots,\sigma_{d+1}(\alpha)$
are distinct conjugates of $\alpha$. This is a contradiction to the
fact that the degree of $m_{\alpha}$ is $d$. \end{proof}
\begin{note*}
Our sequence $\left\{ \alpha_{n}\right\} $ was actually in $\qq_{p}^{\mathrm{un}}:=\qq_{p}\left(\cup_{(n,p)=1}\mu_{n}\right)$,
which we've shown is not complete.
\end{note*}
We let $\cc_{p}$ denote the completion of $\overline{\qq}_{p}$ (as
in Theorem \ref{thm:1.14})
\begin{thm}
$\cc_{p}$ is algebraic closed. \end{thm}
\begin{proof}
The proof is based on the 
\begin{lem*}
Let $\mbox{char}(K)=0$ and $K$ complete with respect to a non-archimedean
value. Let $f=X^{n}+a_{n-1}X^{n-1}+\dots+a_{0}\in K[X]$. Assume $f$
is irreducible over $K$. Then there exists $\delta>0$ such that
for all $g=X^{n}+b_{n-1}X^{n-1}+\dots+b_{0}\in K[X]$ with $|a_{i}-b_{i}|<\delta$
($0\leq i\leq n-1$), $g$ is irreducible. \end{lem*}
\begin{proof}
Let $\lambda_{1},\dots,\lambda_{n}$ be the roots of $f$ in $\overline{K}$
and similarly let $\mu_{1},\dots,\mu_{n}$ be the roots of $g$ in
$\overline{K}$. Put $C=\max\{1,\left|a_{i}\right|\}$. Define $r=\min_{i\neq j}\left|\lambda_{i}-\lambda_{j}\right|$,
$R(f,g)=\prod_{i,,j}(\lambda_{i}-\mu_{j})=\prod_{i}g(\lambda_{i})=\prod_{j}f(\mu_{j})\cdot(-1)^{n}$
(the resultant).
\begin{elabeling}{00.00.0000}
\item [{Step~1}] If $0<\delta<C$ then for all $g$ with $\left|a_{i}-b_{i}\right|<\delta$,
every root $\mu_{j}$ over $g$ has $\left|\mu_{j}\right|\leq C$. 


Suppose for contradiction, we have $\left|\mu\right|>C$. Then for
$0\leq i\leq n-1$, $\left|b_{i}\mu^{i}\right|\leq C\left|\mu\right|^{i}<\left|\mu\right|^{i+1}\leq\left|\mu\right|^{n}$.
This is a contradiction.

\item [{Step~2}] For all $\epsilon>0$, there exists $\delta>0$ such
that if $\left|a_{i}-b_{i}\right|<\delta$ for all $i$ then $\left|R(f,g)\right|<\epsilon$


If $\left|a_{i}-b_{i}\right|<\delta<C$ for all $i$ then 
\begin{eqnarray*}
\left|f(\mu_{j})\right| & = & \left|f(\mu_{j})-g(\mu_{j})\right|\\
 & = & \left|\sum_{i=0}^{n-1}(a_{i}-b_{i})\mu_{j}^{i}\right|\\
 & \leq & \max_{i}\left|a_{i}-b_{i}\right|\cdot\max\{1,\left|\mu_{j}\right|^{n}\}\\
 & < & \delta C^{n}
\end{eqnarray*}
 by step $1$. Thus for all $\delta<\min\{C,\epsilon^{1/n}C^{-n}\}$,
we have $\left|R(f,g)\right|=\prod_{j}\left|f(\mu_{j})\right|<\delta^{n}C^{n^{2}}<\epsilon$.

\item [{Step~3}] If $\left|R(f,g)\right|<r^{n^{2}}$ then $g$ is irreducible
over $K$. 


The condition means at least one of the factors $\left|\lambda_{I}-\mu_{J}\right|<r=\min_{i\neq j}\left|\lambda_{i}-\lambda_{j}\right|$.
Then by Krasner's lemma (Theorem \ref{thm:Krasmer's Lemma}) , we
have $K(\lambda_{I})\subset K(\mu_{J})$, hence $K(\mu_{J})$ has
degree $n$ and so $g$ is irreducible.

\end{elabeling}
\end{proof}
We apply the sublemma with $K=\cc_{p}$. Let $f\in\cc_{p}[X]$ be
irreducible, which is monic. Let $\delta>0$ be as in the sublemma.
Since $\overline{\qq}_{p}$ is dense in $\cc_{p}$, there exists a
monic polynomial $g\in\overline{\qq}_{p}[X]$ satisfying the hypothesis
of the sublemma. Thus $g$ is irreducible of degree $n$ in $\cc_{p}[X]$,
so also in $\overline{\qq}_{p}[X]$. But since $\overline{\qq}_{p}$
is algebraic closed, so $\deg g=1$
\end{proof}

\section{Algebraic Number Fields}

Let $K/\qq$ be a number field. A \emph{place} is an equivalence class
of non-trivial absolute values on $k$, denote the completion of $k$
at $\mathcal{P}$ by $k_{\mathfrak{p}}$. If $\mathcal{P}$ is non-archimedean,
then absolute values in $\qq\subset K$ are equivalent to $p$-adic
absolute value $\left|\,\,\right|_{p}$, we write $\mathfrak{p}|p$.
Then $k_{\mathfrak{p}}$is an extension of $\qq_{p}$ (and so is a
non-archimedean local field). Let $q_{\mathfrak{p}}$ be the cardinality
of residue field of $k_{\mathfrak{p}}$
\begin{defn}
The \emph{renormalised absolute value }$\left|\,\,\right|_{\mathfrak{p}}$
on $k_{\mathfrak{p}}$ is determined by $\left|\pi_{\mathfrak{p}}\right|_{\mathfrak{p}}=q_{\mathfrak{p}}^{-1}$
where $\pi_{\mathfrak{p}}$ is a uniformiser. By problem C.1, we have
$\left|\alpha\right|_{\mathfrak{p}}=\left|\alpha\right|_{p}^{[k_{\mathfrak{p}}:\qq_{p}]}$
for all $\alpha\in k_{\mathfrak{p}}$
\end{defn}
If $\mathfrak{r}$ is an archimedean place, the relevant completion
$k_{\mathfrak{r}}$ is either $\rr$ ($\mathfrak{r}$ is a real place)
or $\cc$ ($\mathfrak{r}$ is a complex place) The renormalised absolute
value is 
\[
\left|\,\,\right|_{\mathfrak{r}}=\begin{cases}
\left|\,\,\right|_{\infty} & \mathfrak{r}\mathrm{\, real}\\
\left|\,\,\right|_{\infty}^{2} & \mathfrak{r}\,\mathrm{complex}
\end{cases}
\]
An archimedean place $\mathfrak{r}$ is an extension of an archimedean
place $\infty$ on $\qq$, write $\mathfrak{r}|\infty$
\begin{lem}
\label{lem:6.2}Let $\alpha\in k^{*}$. Then $\left|\alpha\right|_{\mathfrak{p}}=1$
for all but finitely many places $\mathfrak{p}$\end{lem}
\begin{proof}
Let $f=X^{n}+a_{n-1}X^{n-1}+\dots+a_{0}\in\qq[X]$ be a minimal polynomial
of $\alpha$ over $\qq$. Then $a_{j}\in\zz_{p}$ ($0\leq j\leq n-1$)
for almost all primes $p$. Hence $\left|\alpha\right|_{\mathfrak{p}}\leq1$
for almost all $\mathfrak{p}$ (not $a_{j}\in\zz_{p}$ implies $\left|\alpha\right|_{\mathfrak{p}}\leq1\forall\mathfrak{p}|p$)

Similarly $\left|\alpha^{-1}\right|_{\mathfrak{p}}\leq1$ for almost
all $\mathfrak{p}$\end{proof}
\begin{thm}[Product Formula]
 Let $\alpha\in k^{*}$. Then 
\[
\prod_{\mathfrak{p}\mathrm{\, archimedean\,\&\, non-archimedean}}\left|\alpha\right|_{\mathfrak{p}}=1
\]
\end{thm}
\begin{proof}
By standard field theory we have $k\otimes_{\qq}\qq_{p}=\oplus_{\mathfrak{p}|p}k_{\mathfrak{p}}$
and $\sum_{\mathfrak{p}|p}[k_{\mathfrak{p}}:\qq_{p}]=[k:\qq]$

Similarly $k\otimes_{\qq}\rr=\oplus_{\mathfrak{r}|\infty}k_{\mathfrak{r}}$
and $\sum_{\mathfrak{r}|\infty}[k_{\mathfrak{r}}:\rr]=[k:\qq]$

Hence for all $w\in\{p,\infty\}$ (with $\qq_{\infty}:=\rr$)

\begin{eqnarray*}
\prod_{\mathfrak{p}|w}\left|\alpha\right|_{\mathfrak{p}} & = & \prod_{\mathfrak{p}|w}\left|\alpha\right|_{w}^{[k_{\mathfrak{p}}:\qq_{w}]}\\
 & = & \left|N_{k/\qq}(\alpha)\right|_{w}
\end{eqnarray*}
 since $N_{k/\qq}=\prod_{\mathfrak{p}|w}N_{k_{\mathfrak{p}}/\qq_{w}}$
(c.f. Theorem \ref{thm:4.5}). This reduces the statement to of the
theorem to the case $k=\qq$. Apply Problem A.2\end{proof}
\begin{thm}[Strong Approximation]
 Let $S$ be a finite set of non-archimedean places of a number field
$k$. Let $\epsilon>0$. Let $\alpha_{\mathfrak{p}}\in k_{\mathfrak{p}}$
for $\mathfrak{p}\in S$. Then there exists $\alpha\in k$ such that
\begin{enumerate}
\item $\left|\alpha-\alpha_{\mathfrak{p}}\right|_{\mathfrak{p}}<\epsilon$
for all $\mathfrak{p}\in S$
\item $\left|\alpha\right|_{\mathfrak{p}}\leq1$, $\mathfrak{p}\notin S$,
$\mathfrak{p}$ non-archimedean
\end{enumerate}
\end{thm}
Note: If $\alpha_{\mathcal{P}}\in\mathcal{O}_{\mathcal{P}}$ (for
$\mathfrak{p}\in S)$, then 2.~can be replaces by $\alpha\in\mathcal{O}$.
\begin{proof}
Let $S_{0}$ be the set of rational primes $p$ such that $\mathfrak{p}|p$
for some $\mathfrak{p}\in S$. Without loss of generality we assume
$S$ contains \uline{all} $\mathfrak{p}$ extending $p\in S_{0}$
(put $\alpha_{\mathcal{\mathfrak{p}}}=0$ for $\mathfrak{p}$ not
in original $S$). By the Weak Approximation Theorem (Theorem \ref{thm:Weak_approx})
there exists $\beta\in k$ such that $|\beta-\alpha_{\mathcal{\mathfrak{p}}}|_{\mathfrak{p}}<\epsilon$
(for $\mathfrak{p}\in S)$ . Lemma \ref{lem:6.2} implies the set
$R$ of non-archimedean places $\mathfrak{p}\notin S$ for $\left|\beta\right|_{\mathcal{\mathfrak{p}}}>1$
is finite. Let $R_{0}$ be the set of rational primes $p$ such that
$\mathfrak{p}|p$ for some $\mathfrak{p}\in R$. Then $R_{0}\cap S_{0}=\emptyset$.

Let $\eta>0$. By the Chinese Remainder Theorem we can find $l\in\zz$
such that $\left|l-1\right|_{p}<\eta$ for $p\in S_{0}$ and $\left|l\right|_{p}<\eta$
for $p\in R_{0}$. Check that $\alpha=l\beta$ satisfies the conclusion
of the theorem.
\end{proof}

\section{Diaphantine Equations}


\subsection{Quadratic forms}

Let $K$ be a field of characteristic not $2$, $Q=\sum a_{ij}x_{i}x_{j}\in K[x_{1},\dots,x_{n}]$
is a \emph{quadratic form }of rank $n$, We say $Q$ is \emph{soluble
}if there exists $x\in K^{n}\setminus\{0\}$ such that $Q(x)=0$
\begin{lem}
\label{lem:7.1}Suppose $[K:\qq_{p}]<\infty$, $p\neq2$. Assume without
loss of generality that $Q=\sum a_{i}x_{i}^{2}$, then $Q$ is soluble
if either
\begin{enumerate}
\item $n\geq3$ and $a_{i}\in\mathcal{O}_{K}^{*}$ for all $i$
\item $n\geq5$
\end{enumerate}
\end{lem}
\begin{proof}
{}
\begin{enumerate}
\item Without loss of generality, assume $Q=ax^{2}+by^{2}-z^{2}$ for $a,b\in\mathcal{O}_{K}^{*}$.
Let $k=k_{K}$ and assume $q=\#k$. The maps $x\to\overline{a}x^{2}$
and $y\to1-\overline{b}y^{2}$ have images of size $\frac{q+1}{2}$
in $k$. Thus the images overlap and there exists $x,y\in\mathcal{O}_{K}$
such that $ax^{2}+by^{2}\equiv1\mod\pi_{K}$. By Hensel's lemma, $Q$
is soluble
\item On multiplying by the square of the uniformiser we may assume $v_{K}(a_{i})\in\{0,1\}$.
As $n\geq5$, without loss of generality, $v_{K}(a_{1})=v_{K}(a_{2})=v_{K}(a_{3})$.
If $v_{K}(a_{1})=v_{K}(a_{2})=v_{K}(a_{3})=0$, then apply part $1.$~.Otherwise
if $v_{K}(a_{1})=v_{K}(a_{2})=v_{K}(a_{3})=1$, then divide through
by uniformiser and apply part~1.~
\end{enumerate}
\end{proof}
\begin{note*}
Part 2.~is still true when $p=2$: quinary quadratic forms are isotropic
over any $p$-adic field. 
\end{note*}
On the arXiv, there is a recent paper by Bhargava, Cremona, Fisher
which looks at the density of isotropic quadratic forms in $4$ variables
(roughly $97\%$).
\begin{thm}[Hasse-Minkowski Theorem]
\label{thm:7.2} Q is a quadratic form over a number field $k$.
Then $Q$ is soluble over $k$ if and only if $Q$ is soluble over
$k_{p}$ for every place $p$.\end{thm}
\begin{proof}
Omitted\end{proof}
\begin{rem*}
{}
\begin{enumerate}
\item Lemma \ref{lem:7.1} implies if $n\geq3$, then local solubility is
automatic for all but finitely many primes.
\item When $n=2$ and $k=\qq$ this is very easy: $a\in\qq_{p}^{*2}$ ,
if and only if $v_{p}(a)$ is even. $a\in\rr^{*2}$, if and only if
$a>0$. Both of these implies $a\in\qq^{*2}$.
\item Using Rimenan-Roch one can show that any smooth and projective curve
of genus $0$ is over a number field $k$ is $k$-birationally equivalent
to a conic over $k$. Thus Theorem \ref{thm:7.2} implies that the
``Hasse principle'' holds for smooth and projective curves of genus
$0$.
\end{enumerate}
\end{rem*}

\subsection{Cubic forms}

Natural question: Is there an analogue of Lemma \ref{lem:7.1} for
a cubic forms?
\begin{thm}[Demyanov ($p\neq3$), Lewis, 1950's]
\label{thm:7.3} Suppose $[K:\qq_{p}]<\infty$. Assume $F=\sum_{i\leq j\leq k}x_{i}x_{j}x_{k}\in K[x_{1},\dots,x_{n}]$.
Then $F$ is soluble if $n\geq10$. \end{thm}
\begin{proof}
Treat case $k=\qq_{p}$. Let $\Delta=\Delta(F)$ be the discriminant
of $F$ (this is the resultant of $\frac{\partial F}{\partial x_{1}},\dots\frac{\partial F}{\partial x_{n}})$.
Then $\Delta$ is a non-zero form of degree $n2^{n-1}$ in the coefficients
of $F$. Moreover if $M\in\GL_{n}(\qq_{p})$ such that $x=My$, ($F(x)=F(My)=F^{*}(y)$)
then $F^{*}(y)=aF(y)$, then $\Delta(F^{*})=a^{n2^{n-1}}(\det M)^{3\cdot2^{n-1}}\Delta$. 

Since $\Delta$ is a non-zero form it can not vanish on any neighbourhood
of a point. Hence if $\Delta(F)=0$ then $\forall N\in\nn$ there
exists $c_{ijk}^{(N)}\in\qq_{p}$ such that $\Delta(F^{(N)})\neq0$
and $\left|c_{ijk}-c_{ijk}^{(N)}\right|_{p}<1/N$. Suppose $a^{(N)}$
is zero at $F^{(N)}$ in $\zz_{p}^{n}$. By compactness, these points
have an accumulation point in $\zz_{p}$ and since $F$ is continuous,
this point is a zero of $F$. Hence without loss of generality $\Delta(F)\neq0$.

Note if $F$ and $F^{*}$ are equivalent over $\qq_{p}$, then $\Delta(F)=0\iff\Delta(F^{*})=0$.
$F$ is equivalent over $\qq_{p}$ to a form with coefficients in
$\zz_{p}$. Then $\delta(F)=v_{p}(\Delta(F))\geq0$. We say that $F$
is \emph{reduced} if it has coefficients in $\zz_{p}$ and $\Delta(F)\neq0$
and $\delta(F)\leq\delta(F^{*})$ for all $F^{*}$ over $\zz_{p}$
equivalent to $F$ over $\zz_{p}$. 

It suffices to work with reduced $F$. Let $r\in\nn$ minimal such
that $F(x)\equiv F_{1}(L_{!},\dots,L_{r})\mod p$ where $F_{1}\in\zz_{p}[y_{1},\dots,y_{r}]$
and the $L_{i}$ are linear forms with coefficients in $\zz_{p}$,
and are linearly independent. Clearly $r\leq n$ and make unimodular
transformation $y_{i}=L_{i}$ for $1\leq i\leq r$, to obtain an equivalent
form $F^{*}$, where $F^{*}(y_{1},\dots,y_{n})\equiv F_{1}(y_{1},\dots,y_{r})$.
If $F$ is reduced then so is $F^{*}$. Let $F'(z_{1},\dots,z_{n})=p^{-1}F^{*}(pz_{1},\dots,pz_{r},z_{r+1},\dots,z_{n})$.
Then $F'$ has coefficients in $\zz_{p}$ and $\delta(F')=\delta(F^{*})+2^{n-1}(3r-n)$.
Hence $r\geq n/3$ since $F$ is reduced. Now $n\geq10$ implies $r\geq4$,
hence there exists $(b_{1},\dots,b_{r})\in\ff_{p}^{4}\setminus\{0\}$
such that $F_{1}(b)=0$ (by Chevaley-Warning: ``Over $\ff_{p}$ any
form of $n$ variables of degree $d$ is soluble if $n>d$''). Assume
without loss of generality $b_{1}=1$. Then $F^{*}(z_{1},b_{2}z_{1}+z_{2},\dots,b_{r}z_{1}+z_{r},z_{r+1},\dots,z_{n})\equiv z_{1}^{2}L+z_{1}Q+C\mod p$
where $L,Q,C$ are forms in $z_{2},\dots,z_{n}$. Since $r$ is minimal,
$L$ and $Q$ are not both identically zero modulo $p$. 
\begin{casenv}
\item $L$ not identically zero modulo $p$: Then $(1,0,\dots,0)$ is a
solution of $F^{*}\equiv0\mod p$ and some partial derivative of $F^{*}$
does not vanish modulo $p$ at $(1,0,\dots,0)$
\item $L$ is identically zero modulo $p$: There exists $d=(d_{2},\dots,d_{n})\in\zz^{n-1}$
such that $p\nmid(d_{2},\dots,d_{n})$ and such that $Q(d_{2},\dots,d_{n})\not\equiv0\mod p$.
Then $(-C(d),d_{2}Q(d),\dots,d_{n}Q(d))$ is a solution of $F^{*}\equiv0\mod p$
with $\frac{\partial F^{*}}{\partial x_{1}}\neq0\mod p$.
\end{casenv}
In either case Hensel's lemma yields the result\end{proof}
\begin{rem*}
~
\begin{enumerate}
\item $n\geq10$ is best possible in Theorem \ref{thm:7.3}. See problem
C.3
\item Artin's conjecture: $\qq_{p}$ is a $C_{2}$ field, i.e, any form
over $\qq_{p}$ in $n$ variables and degree $d$ is soluble over
$\qq_{p}$ if $n>d^{2}$. This is FALSE. 
\item What about an analogue of Theorem \ref{thm:7.2}? Let $k$ be a number
field and $F$ a cubic form over $k$. Then $F$ is soluble over $k$
if 


\begin{tabular}{|c|c|c|}
\hline 
$n$ & Conditions & Notes\tabularnewline
\hline 
\hline 
$n\geq16$ & None & Pleasants (1975)\tabularnewline
\hline 
$n\geq10$ & $F$ non-singular & Brawning and Vishe (2013)\tabularnewline
\hline 
\end{tabular}


However the Hasse principle can fail for cubic forms in fewer variables. 


For $n=4$, the first example was produced by Swinnerton-Dyer in 1962:
Let $K=\qq(\theta)$ where $\theta^{3}-7\theta^{2}+14\theta-7=0$.
Abelian cubic field of discriminant $49$ and $\mathcal{O}_{K}=\zz[\theta]$.
Here $(7)=P^{3}$ and $v_{P}(\theta)=1$. Consider 
\[
F(x_{1},\dots,x_{4})=N_{K/\qq}(x_{1}+\theta x_{2}+\theta^{3}x_{3})+x_{4}(x_{4}+x_{1})(2x_{4}+x_{1})
\]
Check: non-singular, soluble over $\qq_{p}$ for all $p$. But it
is \uline{not} soluble over $\qq$!
\begin{proof}
Note that if $N(\,)=0$ then $x_{1}=x_{2}=x_{3}=0$, hence $x_{4}=0$.
Contradiction as we want a non-zero solution. May assume that $x_{1},x_{4}$
are coprime integers and $x_{2},x_{3}\in\qq$. Now $7|N(\,)$ implies
$P$ divides $N(\,)$, hence $7|x_{1}$ and $7\nmid x_{4}$. Hence
$7\nmid x_{4}(x_{1}+x_{4})(2x_{4}+x_{1})$ which is a contradiction.
Hence $7\nmid N(\,)$.

Since $x_{4},x_{4}+x_{1}$ and $2x_{4}+x_{1}$ are all coprime, and
their product is a norm in $K$, each of them must separately be a
norm of an ideal. Now $p\neq7$ splits in $K$ if and only $p=\pm1\mod7$.
Hence each of the factors above is congruent to $\pm1$ modulo $7$.
This contradicts $x_{4}+(x_{4}+x_{1})=2x_{4}+x_{1}$. 
\end{proof}

c.f. Elsenhans-Jahnel. (Recent paper on the arXiv, they show there
is a Zariski dense set of counter examples).\end{enumerate}
\end{rem*}

\end{document}
