%&latex

% January 29, 2000 Lecture Notes no1 prepared for Dr. Hoffman's Number Theory Class.

\documentclass[11pt]{amsart}

\usepackage{amsmath,amssymb,amscd}

\usepackage[mathscr]{eucal}

\newcommand{\Aut}{{\mathrm{Aut} }}
\newcommand{\End}{{\mathrm{End} }}
\newcommand{\GL}{{\mathbf{GL} }}
\newcommand{\Gal}{{\mathrm{Gal}\, }}
\newcommand{\G}{{\mathbf G}_{m}}
\newcommand{\Hom}{{\mathrm{Hom} }}



\newcommand{\Tr}{{\mathrm{Tr}\, }}

\newcommand{\Q}{\mathbf Q}
\newcommand{\Z}{\mathbf Z}
\newcommand{\R}{\mathbf R}
\newcommand{\C}{\mathbf C}
\newcommand{\F}[1]{\mathbf{F}_{#1}}
\newcommand{\Fbar}[1]{\overline{\mathbf{F}}_{#1}}
\newcommand{\fr}[1]{\,{\rm Frob}_{\, #1}}
\newcommand{\q}[1]{\mathbf{Q}_{#1}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\msr}[1]{\mathscr{#1}}
\newcommand{\mbf}[1]{\mathbf{#1}}
\newcommand{\mfr}[1]{\mathfrak{#1}}
\newcommand{\mr}[1]{\mathrm{#1}}

\newcommand{\ok}[1]{O_{#1}}

\newcommand{\re}[1]{{\rm Re} (#1)}
\newcommand{\im}[1]{{\rm Im} (#1)}

\theoremstyle{plain}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lem}[thm]{Lemma}
\newtheorem{pro}[thm]{Proposition}
\newtheorem{cor}[thm]{Corollary}
\theoremstyle{definition}
\newtheorem{Def}[thm]{Definition}
%\theoremstyle{remark}
\newtheorem{rem}[thm]{Remark}
\newtheorem{exa}[thm]{Example}
\newtheorem{note}[thm]{Note}
\newtheorem{prob}[thm]{Problem}

\begin{document}

\title{Lecture 4}

\author{Cem Guneri}

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

\date{\today}

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

\begin{rem}
\label{R I.1.3}
A series $\sum _{\nu=0}^{\infty} x_\nu$, $x_\nu\in\Q$ converges 
if and only if the sequence $(x_n)\longrightarrow 0$ in $p$-adic metric.\end{rem}

\proof Consider the difference $|S_n - S_{n-1}|_p = |x_n|_p$, where 
$S_n=\sum_{\nu=0}^{n} x_\nu$ is the sequence of partial sums. $(S_n)$ converges 
iff $|S_n - S_{n-1}|_p\longrightarrow 0$ (due to ultrametric property of $|.|_p$ ) 
iff $|x_n|_p\longrightarrow 0$.\endproof

\begin{note}A special case of the above remark is the fact that any series of 
the form $\sum_{i=0}^{\infty} x_ip^i$ with $x_i\in\Z_p$ converges to an element 
of $\Z_p$.\end{note}

\proof \[ |x_ip^i|_p=|x_i|_p|p^i|_p=|x_i|_p|p|_p^i=|x_i|_pe^{-i}\]
Note that $x_i$ is in $\Z_p$ and the norm of the elements in $Z_p$ are bounded 
by 1. Hence $|x_i|_p \leq 1$ and therefore as $i\longrightarrow\infty$ the 
product $|x_i|_pe^{-i}\longrightarrow0$. But this is enough to conclude that 
the series $\sum_{i=0}^{\infty} x_ip^i$ converges by the remark above. 
This limit is guaranteed to be in $\Z_p$ since the sequence of partial sums 
$S_n=\sum_{i=0}^{n} x_ip^i$ is a sequence in $\Z_p$ and $\Z_p$ is complete in 
the $p$-adic metric.\endproof

\begin{pro}
\label{P I.1.5}
Let $S\subset\Z_p$ be any set of coset representatives for 
$\Z_p/p\Z_p\cong\Z/p$. Then any $x\in\Q_p$ can be uniquely written as
\[x=p^n(s_0+s_1\cdot p+s_2\cdot p^2+...)\] where $n=v_p(x)$ and $s_i\in S$.\end{pro}

\proof First of all any expression of this form is a unique element in $\Q_p$.  
Because $p^n$ is clearly in $\Q_p$ and the sum inside the paranthesis is actually an 
element of $\Z_p$ due to the note above. Clearly the sum is a unique element in $\Z_p$
and multiplying this by $p^n$ gives a unique element in $\Q_p$.

Now let $0\not=x\in\Q_p$. We know we can write $x=p^nu$ where $n\in\Z$ and $u\in
\Z_p^*$. That means $u$ is not in $p\Z_p$ and therefore if we consider the map
$\varepsilon_1 : \Z_p \longrightarrow \Z/p$, this map will take $u$ to some element
in $\Z/p$ which is nonzero. In other words:
\[u \equiv s_0 \hspace{3pt}(mod p)\] 
for some $s_0 \in \Z_p$. Alternatively we can write $u=s_0+pu_1$ for some 
$u_1\in \Z_p$ which leads to
\[x=p^nu=p^n(s_0+pu_1)\]
Now if we do the same thing to $u_1$ and find $s_1\in\Z_p$ such that $u_1 \equiv s_1 (mod p)$, then $u_1=s_1+pu_2, u_2\in\Z_p$. Hence
\[x=p^nu=p^n(s_0+pu_1)=p^n(s_0+p(s_1+pu_2))=p^n(s_0+s_1\cdotp+p^2u_2)\]
Keep doing this to $u_i$'s to obtain the expansion we were trying to show.    Observe that the choice of $s_i$'s are so that they are actually representatives of elements of $\Z_p$. (Remember that $\varepsilon_1(u) \equiv
s_0$ in $\Z_p$) This finishes the proof.\endproof

\begin{exa} Consider $\Q\hookrightarrow\Q_3$, a dense subset, and take 
$\frac{1}{4}\in\Q\subset\Q_3$. Using the idea of geometric series
\[\frac{1}{1+3}=1-3+3^2-3^3+...\]
Observe that in this example $S\subset\Z_3$ is chosen as \{-1,0,1\}. If we take $S=\{0,1,2\}$, for example, then the expansion would be
\[\frac{1}{1+3}=1+2\cdot3+1\cdot3^2+2\cdot3^3+...\] \end{exa}

\begin{exa} This example shows the arithmetic of these power series in $\Z_{11}$. 
\[(1+6\cdot11+3\cdot11^2+8\cdot11^3+...)+(4+7\cdot11+2\cdot11^2+4\cdot11^3+...) = \]
\[5+2\cdot11+6\cdot11^2+1\cdot11^3+...\]
Observe that $S=\{0,1,2...,10\}$, the representatives of the elements of $\Z/11$. 

If you compute the inverse of the element $4+7\cdot11+2\cdot11^2+4\cdot11^3+...
$ , then the answer should be
\[\frac{1}{4+7\cdot11+2\cdot11^2+4\cdot11^3+...}=3+0\cdot11+9\cdot11^2+5\cdot11^3+...\] \end{exa}
\vskip 1cm

\section{P-adic Equations}
\begin{prob} (Diophantos)

Given a system of polynomial equations
\[f_\alpha(x_1,x_2,...,x_n)=0, \hspace{3pt}\alpha\in I.\]
with integer coefficients. We want to find a solution (or all solutions) 
$(x_1,x_2,...,x_n)\in\Z^n$. Another possible problem could be just to decide whether this system has any solution in $\Z^n$ at all.\end{prob}

One way to study this problem is to reduce it mod $N$. However, due to the Chinese Remainder Theorem $\Z/N \cong \prod _{i} \Z_{p_i}^{k_i}$ if $N=\prod _{i} p_i^{k_i}$. Hence we can reduce the problem even further to mod $p^k$ for a fixed prime $p$.

\begin{exa} Consider the equation $x^2+1=0$. This clearly has no solution in $\Z$. Let's check if it can have solutions in p-adic numbers. 

$\Q_3$: Suppose there exists $x\in\Q_3$ that solves this equation. Then we have the following:
\[x^2=-1 \hspace{3pt}\Rightarrow v_3(x^2)=v_3(-1) \hspace{3pt}\Rightarrow 2v_3(x)=0 \hspace{3pt}\Rightarrow v_3(x)=0\]
Observe that in the implication two above we are using the fact that $-1\in\Z_3$ is a unit and hence $v_3(-1)=0$. So what we have now is $v_3(x)=0$ which implies that $x\in\Z_3^* \subset \Z_3$. So if a solution exists in $\Q_3$, then it must be in $\Z_3$. 

Now if you recall the map $\varepsilon_1:\Z_3 \longrightarrow \Z/3$, a solution $x$ in $\Z_3$ would have to be taken to a solution in $\Z/3$ but $x^2+1=0$ has no solution in $\Z/3$. Hence $x^2+1=0$ can not be solved in $\Z_3$, which also means it can not be solved in $\Q_3$. (A solution in$\Q_3$ must be in $\Z_3$.)


$\Q_5$: The same equation over $\Q_5$. Due to the same reason as above if $x\in\Q_5$ solves $x^2+1=0$, then $x\in\Z_5$. This time $x^2+1=0$ has two solutions in $\Z/5$ which are $2$ and $3$. Let's pick $2$. That means a solution $x\in\Z_5$ of the equation $x^2+1=0$ is congruent to $2$ mod $5$. ($x\equiv2, mod 5$) Hence $x=2+5y$ for some $y$ in $\Z_5$. 

Next question is can we choose $y$ so that $x=2+5y$ is also a solution mod $5^2=25$? 

\[x^2+1\equiv 0 \hspace{3pt}(mod 25)\]
\[(2+5y)^2+1\equiv 0 \hspace{3pt}(mod 25)\]
\[4+20y+25y^2+1\equiv 0 \hspace{3pt}(mod 25)\]
\[5+20y\equiv 0 \hspace{3pt}(mod 25)\]
\[1+4y\equiv 0 \hspace{3pt}(mod 5)\]
\[y\equiv 1 \hspace{3pt}(mod 5)\]

This means $y\in\Z_5$ should be chosen as $y=1+5z$, where $z\in\Z_p$. Therefore:
\[x=2+5y=2+5(1+5z)=2+1\cdot5+5^2z\]
Continuing this way will produce a $5$-adic power series of the solution of $x^2+1=0$
in $\Z_5\subset \Q_5$. 

Doing the same thing to $3$ produces another solution in $\Z_5\subset \Q_5$, which actually 
means that the equation $x^2+1=0$ has two solutions in $\Q_5$.\end{exa}

\end{document}