%&latex

% January 12, 2000

\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}
\newcommand{\rem}{\noindent{\textbf{Remark:}}}
\newtheorem{exa}[thm]{Example}



\begin{document}

\title{Lecture 2}

\author{Soad Ahmad}

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

\date{\today}

\email{sahmad02@bellsouth.net}

\maketitle

\section{Lecture 2}

\begin{pro} The sequence,
\[0\longrightarrow\Z_p\stackrel{p^n}\longrightarrow\Z_p\stackrel{\varepsilon_n}\longrightarrow\Z/{p^n}\longrightarrow0\]is an exact sequence.
\end{pro}

\begin{proof}

 Multiplication by $p$ (hence by $p^n$)is injective in $\Z_p$;
indeed, if $x=(x_n)$  is a $p$-adic integer such that $px=0$, we
have $px_{n+1}=0$ for all $n$, and $x_{n+1}$ is of the form
$p^ny_{n+1}$ with $y_{n+1}\in\Z/p^{n+1}$; since $x_n=x_{n+1}$ mod
$p^n$, we see that $x_n$ is divisible by $p^n$, hence, is zero.

It is clear that $p^n\Z_p \subseteq ker(\varepsilon_n)$;
conversely, let $x\in p^n\Z_p$ be a nonzero element, then there
exist a maximum $n$ such that $\varepsilon_n(x)=0$; so, $x=p^nu$
where $u\in\_p$ and $p\nmid u$. If $x\in ker(\varepsilon_n)$ then
$x=p^n u$ and therefore, $ker(\varepsilon_n)\subseteq p^n\Z_p$.

\end{proof}


\rem

\begin{enumerate}
\item $\Z$ is dense in $\Z_p$.
 Note that the map $\zeta:\Z\longrightarrow\Z_p$
given by: $\zeta(z)=(\cdots ,z,z,z)$ is an embedding of $\Z$ into
$\Z_p$. Let $x\in \Z_p$, then $x=(\cdots ,x_n,x_{n-1},\cdots
,x_2,x_1)$ and let $\{y\}^\infty_{n=1}\subseteq\Z$ such that
$x_n\equiv y_n$ mod $p^n$ for all $n$. So, $p^n|(y_n-x_n)$ for all
$n$ and because $x_{n+1}\equiv x_n$ mod $p^n$ we have that
$p^n\mid(x_{n+1}-y_n)$. If the distance between two p-adic
integers $x,y$ is defined by, $d(x,y)=e^{-\upsilon_p(x-y)}$. One
can check easily that this forms a metric and hence induces the
topology  on $\Z_p$. Now,


\[\begin{array}{ll}
 d((\cdots ,y_n,y_n,y_n),x)& = e^{-\upsilon_p(\cdots ,x_m-y_n,\cdots ,x_1-y_n)} \\
  & = e^{-(n+k)} < e^{-n} \longrightarrow 0, \mbox{ as }n
  \rightarrow \infty.
  \end{array}
\]

\


\item The ideal $p^n\Z_p$ forms a fundemental system of
            neighborhoods of {0}. these are open and closed  and
           \[ p^n\Z_p=\bigcap^n_{i=1}{\varepsilon_i}^{-1}(0)\]
 \item $\Z$ is an integral domain. To show that all we need to
 show is that $\Z_p$ has no zero divisors. Now let $xy$=0 in
 $\Z_p$. this means that $x_ny_n\equiv0\in \Z/p^n$ $\forall
 n\geq1$; since $\Z/p$ is a field, we have $x_1 $ or $ y_1$ must be zero.
\end{enumerate}


\begin{pro} \begin{enumerate} \item A non zero element $x\in\Z_p$ is invertible if and
only if $x$ is not in $p\Z/p$.

\item Every $x\in\Z_p$ can be written uniquely as $x=p^nu$ where
$u$ is invertible.
\end{enumerate}
\end{pro}


\begin{proof} \begin{enumerate}\item If $x\in\Z_p$ is invertible , then
$\varepsilon_n(x)$ is also invertible in $\Z/p^n$. So $p\nmid x_n$
for all $n$ and therefore; $x$ in not in $p\Z_p$.

\item  Let $n\geq0$ be the largest integer such that
$\varepsilon_n(x)=0$ (i,e $x\in p^n\Z_p$ but not in $p^{n+1}\Z_p$
). We know that $x=p^nu$ for a unique $u$ and $u$ is not in
$p\Z_p$. By 1. $u$ is invertible.
\end{enumerate}
\end{proof}


\end{document}