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\begin{document}

\title{Lecture 3}

\author{Michael W. Broome}

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

\date{\today}

\email{broome@math.lsu.edu} \

\maketitle

\section{Lecture 3}

\lem
\label{L:I.1.3}
$\Z_p$ is an integral domain.
\begin{proof}
We need to show that $xy=0$ and $x\ne0$, then $y=0$.  Let $n$ be the largest nonnegative integer such that $\varepsilon_n(x)=0$. Such an $n$ exists because $x\ne0$.  Thus, by Propostion 1.2, $x\in p^n\Z_p$, but $x\notin p^{n+1}\Z_p$.  Therefore, $x=p^nz$ for a unique $z\in\Z_p$.  So, we now have the following:
\[
0=xy=p^nzy.
\] 
>From the previous lecture, the function $p^n:\Z_p\longrightarrow\Z_p$ is injective, which implies that $zy=0$.  Note that $z\notin p\Z_p$.  Otherwise, $x\in p^{n+1}\Z_p$.  Thus, $z=(z_\nu)$ for each $z_\nu \notin p\Z /p^\nu$.  For $\nu=1$, $z_1\in\Z/p=\mathbb{F}_p.$  This implies that $z_1\neq0$ is in the field $\Z/p$.  Therefore, we have the following:
\[
z_1y_1=0 \Longrightarrow y_1=0.
\]  
Now for $z_\nu y_\nu=0$ in $\Z/p^{\nu}$, $z_\nu \notin p\Z/p^\nu \Z$ implies that $z_\nu$ is invertible in $\Z/p^\nu$.  So, $z_\nu^{-1}(z_\nu y_\nu)=z_\nu^{-1}(0)$ implies that $y_\nu=0$ for all $\nu$.  Therefore, $y=0$, and $\Z_p$ is an integral domain.
\end{proof}

\pro
\label{P:I.1.2} 
\item [(a)]An element $x\in\Z_p$ is invertible $\iff x\notin p\Z_p.$
\item [(b)]Every non-zero element $x\in \Z_p$ can be uniquely written as $x=p^nu$ for and integer $n\ge0$ and $u\in \Z_p^\times$.
\begin{proof}
Part (a) was proven in the last lecture.\\  
For part (b), let n be the largest nonnegative integer such that $\varepsilon_n(x)=0$.  That is, $x\in p^n\Z_p$, but $x\notin p^{n+1}\Z_p$.  Then, we know $x=p^nu$, for a unique $u\in\Z_p$.  Note that $u\notin p\Z_p$.  Therefore, $u$ is invertible by part (a).  That is, $u\in\Z_p^\times$.  To prove uniqueness,we use the fact that $\Z_p$ is an integral domain.  Note that $n$ is unique by the above argument.  If $p^nu=p^nu'$, then  we can use cancellation since $\Z_p$ is an integral domain.  Thus, we get that $u=u'$.  
\end{proof}

\Def
\label{D:I.1.3}
The function $v:\Z_p-\{0\}\longrightarrow\Z$ such that $v(x)$ is the largest $n$ so that $\varepsilon_n(x)=0$ (i.e. $x\in p^n\Z_p$ and $x\notin p^{(n+1)}\Z_p$) is well-defined by uniqueness.  This function is called the $p$-adic valuation.  Also, note that we will sometimes define $v(0)=+\infty$.\\
\\

\lem
\label{L:I.1.4}
The following are properties of the $p$-adic valuation function:\
\item [(1)]$v(xy)=v(x)+v(y)$
\item [(2)]$v(x+y)\ge inf\{v(x),v(y)\}$.
\begin {proof}
Let $x=p^\nu u$ and $y=p^\mu u'$ where $u,u'\in\Z_p^\times$ as in \ref{P:I.1.2}.  Assume $\nu\leq\mu$.  Then,
\[
xy=p^{\nu+\mu}uu',
\]
which shows (1).
Also,
\[
x+y=p^\nu u + p^\mu u'=p^\nu(u+p^{\mu-\nu}u'),
\]
which shows $v(x+y)\geq\nu$.  Similarly, if $\mu\leq\nu$, $v(x+y)\geq\mu$.  Therefore (2) holds.
\end{proof}

\pro
\label{P:I.1.3}
The topology of $\Z_p$ is defined by the distance function $|x-y|=e^{-v(x-y)}.$  Moreover, $\Z_p$ is a complete metric space.
\begin{proof}
Note that $p^n\Z_p$ form a fundamental system of neighborhoods of 0.  We just consider these neighborhoods.  Each neighborhood of 0 can be written as the following set:
\[
\{x:v(x)\geq n\},
\]
which if exponentiated, we get the following:
\[
\{x:d(x,0)=\vert x\vert\leq e^{-n}\}.
\] 
Note that $e$ works fine, but any number larger that one will work also.
>From this, we obtain the distance funtion
\[
\vert x-y\vert=e^{-v(x-y)}.
\]
$\Z_p$ is complete because it is compact.
\end{proof}

\Def
\label{D:I.1.4}
The fraction field $\Q_p of \Z_p$ is called the field of $p$-adic numbers.\\

Note:
Any integral domain has a fraction field.

\pro
\label{P:I.1.4}
The field $\Q_p$ with the topology given by $|x-y|=e^{-v(x-y)}$ is locally compact.  $\Z_p$ is an open compact subring.  Also, $\Q\subset\Q_p$ is a dense subset .
\begin{proof}
Since $\Z_p$ is compact and is system of neighborhoods of 0, then $\Q_p$ is locally compact.
Clearly, $\Z\subset\Q$.  Therefore, $\Z_p\subset\Q_p$.  So $\Z_p$ is an open compact subring. 
Since $\Z\subset\Z_p$, we can approximate $x,y\in\Z_p$ by $\zeta,\eta\in\Z$ in the $p$-adic topology.  Then, $\frac{x}{y}\in\Q_p$ can be approximated by $\frac{\zeta}{\eta}\in\Q$ more or less obviously.  Therefore, $\Q$ is a dense subet of $\Q_p$.\\\\\\ \\ 
\end{proof}

\rem
\item[(1)]The only ideals in $\Z_p$ are $p^n\Z_p$, where $n\ge 0.$
\begin {proof}
Let $I$ be an ideal such that $I\subset\Z_p$.  Let $n=inf\{v(x):x\in I\}$.  We claim that $I=p^n\Z_p.$  For any non-zero $y\in I$, $y=p^\nu u$, where $u$ is a unit in $Z_p$ and $n\leq\nu$.  Therefore, $y=p^n(p^{\nu-n}u)$.  This show that $I\subseteq(p^n)$.
Conversely, there is an $x\in I$ such that $x=p^n u$ where $u$ is again a unit in $Z_p$.  This implies that $p^n=u^{-1}x\in I$.  This shows that $(p^n)\subseteq I$.  So, $(p^n)=I$.
\end{proof}
\item[(2)]A sequence $(x_i)$ converges $\iff |x_{i+1}-x_i|\longrightarrow 0$ as $i\rightarrow \infty$  Note that this is certainly not true for $\R$, but this is true for $p$-adic numbers.
\begin{proof}
For necessity, the statement is true in any metric space.  For sufficiency, we just need to show that any sequence $(x_i)$ is a Cauchy sequence by completeness.  That is, $\vert x_i-x_j\vert<\varepsilon$ for $i,j\gg0$.  Also of note, $\Z_p$ has the ultrametric inequality, which is the following: $\vert x+y\vert\leq sup(\vert x\vert,\vert y\vert)$.  So, now we set up the following:
\[
\begin{array}{lll}
\vert x_{i+2}-x_i\vert&=&\vert(x_{i+2}-x_{i+1})+(x_{i+1}-x_i)\vert\\  &\leq& sup(\vert x_{i+2}-x_{i+1}\vert,\vert x_{i+1}-x_i\vert)    \\ 
&<&\varepsilon.                                    
\end{array}
\]
The rest of the proof can be done by induction.  Therefore, this is a Cauchy sequence.
\end{proof}

 
 
 



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\bibitem{sL}
S. Lang, \emph{Algebraic Number Theory}, Second Edition, Graduate  
Texts in Mathematics, {\bf 110}, Springer - Verlag, 1994.

\end{thebibliography}

\end{document}