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%% samplethesis.tex - template for an honors thesis in mathematics %%
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\documentclass[11pt,twoside]{report}
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% correctly, change the above line to
% \documentclass[11pt]{report}
% (without the %, of course)
% Import standard math macro packages
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{amsfonts}
% Declare theorem types. Add, subtract, and modify
% as you wish. I can't imagine you'll want the uconj
% type, for example.
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem*{uconj}{Uniform Boundedness Conjecture}
\theoremstyle{definition}
\newtheorem{definition} [theorem] {Definition}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{example} [theorem] {Example}
\newtheorem{remark} [theorem] {Remark}
% Set margins and spacing. Probably you shouldn't mess with these,
% unless your printer isn't aligned quite the same as mine and the
% margins are coming out wrong. (At least 1.5 inches on binding side,
% at least 1 inch on the other three sides.)
\setlength{\evensidemargin}{0in}
\setlength{\oddsidemargin}{0.5in}
\setlength{\textwidth}{6in}
\setlength{\topmargin}{0.2in}
\setlength{\textheight}{8.6in}
\setlength{\footnotesep}{14pt}
% User-defined macros. Add, subtract, modify as you wish.
\newcommand{\C}{{\mathbb{C}}}
\newcommand{\F}{{\mathbb{F}}}
\newcommand{\N}{{\mathbb{N}}}
\newcommand{\Q}{{\mathbb{Q}}}
\newcommand{\PP}{{\mathbb{P}}}
\newcommand{\R}{{\mathbb{R}}}
\newcommand{\Z}{{\mathbb{Z}}}
\newcommand{\cK}{{\mathcal{K}}}
\newcommand{\fm}{{\mathfrak{m}}}
\newcommand{\fo}{{\mathfrak{o}}}
\newcommand{\Cp}{\C_p}
\newcommand{\Cv}{\C_v}
\newcommand{\Qp}{\Q_p}
\newcommand{\Qpbar}{\bar{\Q}_p}
\newcommand{\Dbar}{\overline{D}}
% User-defined math operators. Add, subtract, modify as you wish.
\DeclareMathOperator{\rad}{rad}
\DeclareMathOperator{\diam}{diam}
\DeclareMathOperator{\divop}{div}
% Declare a counter variable, to be use later for a list environment
% You can probably delete this, unless you use a list environment
% or some other fancy LaTeX wizardry that requires a counter variable.
\newcounter{bean}
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% Let's get started %
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\begin{document}
% Set up double-spacing (must be done *after* \begin{document})
\setlength{\baselineskip}{21pt}
% Roman numeral page numbers for abstract, etc.
\pagenumbering{roman}
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% Title page %
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\begin{titlepage}
\mbox{}
\vspace{1in}
\begin{center}
\LARGE An Example of Thesis Formatting \\
with \LaTeX \\[2in]
\normalsize Charles Quincy Student \\[\medskipamount]
April 13, 2009 \\[1in]
Submitted to the \\
Department of Mathematics \\
of Amherst College \\
in partial fulfillment of the requirements \\
for the degree of \\
Bachelor of Arts with honors
\end{center}
\vfill
\begin{center}
Faculty Advisor: Professor Emily Dickinson
\end{center}
\vfill
\begin{center}
Copyright \copyright\ 2009 Charles Quincy Student
\end{center}
\end{titlepage}
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% Abstract %
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\chapter*{Abstract}
This short example thesis meets all of the formatting
requirements in Mathematics and shows how to use the
blackboard bold font $\Z, \R, \Q, \C, \N$.
Other fonts like calligraphic $\cK$ and fraktur $\fm$
also appear, as do user-defined math operators
like $\diam X$ and $\divop \vec{F}$.
We also model a writing style appropriate to
a math thesis, although a real thesis should
provide more exposition and justification than
appears in, say, Chapter~\ref{ch:dyn}, which
was taken largely from a published math research paper.
Throughout, the reader should look at both the
finished document and the raw LaTeX file to learn
how to use LaTeX effectively.
We recommend that the aspiring math thesis
writer also consult the document {\tt latextips.tex}
for more examples of typeset
mathematical equations, sectioning commands,
and related techniques.
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% Acknowledgements %
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\chapter*{Acknowledgements}
Thanks to Mom and Dad for everything.
Thanks to Lyle McGeoch, who wrote the original 1990
sample {\tt thesis.tex} file for CS theses on which
this sample math thesis was based.
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% Table of contents %
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\tableofcontents
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% If you also have a lot of figures
% or tables, un-comment one or both
% of the following lines to generate
% a list of them.
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%\listoffigures
%\listoftables
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% List of notation %
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\chapter*{List of Notation}
\noindent
\hspace{0.8in}
$\PP^1(\Q)$\dotfill p.\ \pageref{not:P1Q}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\F_p$ \dotfill p.\ \pageref{not:Fp}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$|\cdot|_v$ \dotfill p.\ \pageref{def:absval}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$|\cdot|_p$ \dotfill p.\ \pageref{not:vp}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$K_v$ \dotfill p.\ \pageref{not:Kv}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\bar{K}_v$ \dotfill p.\ \pageref{not:Kvbar}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\Cv$ \dotfill p.\ \pageref{def:Cv}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\Qp$ \dotfill p.\ \pageref{not:Qp}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\Qpbar$ \dotfill p.\ \pageref{not:Qpbar}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\Cp$ \dotfill p.\ \pageref{not:Cp}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$D(a,r)$ \dotfill p.\ \pageref{not:D}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\Dbar(a,r)$ \dotfill p.\ \pageref{not:D}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\diam X$ \dotfill p.\ \pageref{not:diam}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\rad X$ \dotfill p.\ \pageref{def:rad}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\cK_v$ \dotfill p.\ \pageref{not:cKv}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\cK_p$ \dotfill p.\ \pageref{not:cKp}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$\log_d x$ \dotfill p.\ \pageref{not:log}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$E(N,d)$ \dotfill p.\ \pageref{not:E}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$e(j,d)$ \dotfill p.\ \pageref{not:e}
\hspace{1.5in}
\noindent
\hspace{0.8in}
$L(x)$ \dotfill p.\ \pageref{not:L}
\hspace{1.5in}
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% Some magic code to leave an extra blank page
% if we are double-siding AND the front matter
% ends on an odd-numbered page.
% (Otherwise the even- and odd-side margins
% will be backwards for the rest of the thesis.)
\makeatletter
\if@twoside \ifodd\value{page}
\clearpage\mbox{}\thispagestyle{empty} \fi \fi
\makeatother
% Switch back to arabic page numbers AFTER the end of the current page.
\clearpage
\pagenumbering{arabic}
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% MAIN BODY OF THESIS %
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\chapter{Introduction}
\label{ch:intro}
Let $\phi\in \Q(z)$ be
a rational function with rational coefficients.
Let $\phi^n$ denote the $n^{\text{th}}$
iterate of $\phi$ under composition; that is,
$\phi^0$ is the identity function, and for $n\geq 1$,
$\phi^n = \phi\circ \phi^{n-1}$.
We will study the dynamics $\phi$ on the projective line
$\PP^1(\Q)=\Q\cup\{\infty\}$.\label{not:P1Q}
In particular, we say a point $x$ is
\emph{preperiodic} under $\phi$ if there are integers $n>m\geq 0$
such that $\phi^m(x)=\phi^n(x)$. The point
$y=\phi^m(x)$ satisfies $\phi^{n-m}(y)=y$
and is said to be \emph{periodic}
(of period $n-m$).
Note that $x\in\PP^1(\Q)$ is preperiodic if and only
if its orbit $\{\phi^n(x):n\geq 0\}$ is finite.
Although $\phi$ can have infinitely
many preperiodic points in $\C$, the same
is not true over $\Q$, as the following
result \cite{Nor} states.
\begin{theorem}[Northcott, 1950]
\label{thm:nor}
Let $\phi\in\Q(z)$ be a rational function
of degree $d\geq 2$. Then $\phi$ has
only finitely many preperiodic points in $\PP^1(\Q)$.
\end{theorem}
For example, let $\phi(z)=z^2 - 29/16$. Then the set
$\{5/4, -1/4, -7/4\}$ forms a periodic cycle
(of period $3$), and $-5/4$, $1/4$, $7/4$, and $\pm 3/4$ each
land on this cycle after one or two iterations. In addition,
the point $\infty$ is of course fixed. These nine rational
points are all preperiodic under $\phi$.
In fact,
$\phi$ has only these
nine rational preperiodic points \cite[Theorem~3]{Poon}.
Morton and Silverman proposed the following Conjecture \cite{MS1}.
\begin{uconj}
{\rm (Morton and Silverman, 1994)} \\
Given an integer $d\geq 2$, there is a constant
$\kappa=\kappa(d)$ such that no
rational function $\phi\in\Q(z)$ of
degree at least $d$ has more than
$\kappa$ preperiodic points in $\PP^1(\Q)$.
\end{uconj}
In this thesis we will study the Uniform Boundedness Conjecture
by analyzing filled Julia sets $\cK_v$, to be
introduced in Definition~\ref{def:jul}.
In Chapter~\ref{ch:fund} we will
set terminology and discuss the geometry
of subsets of certain fields.
In Chapter~\ref{ch:dyn} we will discuss
some topics related to dynamics
and prove our main results.
\chapter{Fundamentals}
\label{ch:fund}
\section{Absolute Values}
\begin{definition}
\label{def:absval}
Let $K$ be a field. An \emph{absolute value} $v$ on $K$ is
a real-valued
function $|\cdot|_v : K\rightarrow [0,\infty)$ such that
for any $x,y\in K$, $|x|=0$ if and only if $x=0$,
$|xy|_v=|x|_v |y|_v$, and $|x+y|_v \leq |x|_v + |y|_v$.
If $|\cdot|_v$ satisfies the stronger
triangle inequality $|x+y|_v \leq \max\{|x|_v,|y|_v\}$,
then we say $v$ is \emph{non-archimedean}; otherwise we say
$v$ is \emph{archimedean}.
\end{definition}
\begin{proposition}
\label{prop:easy}
Let $K$ be a field with absolute value $v$,
and let $x,y\in K$. Then:
\begin{list}{\rm \alph{bean}.}{\usecounter{bean}}
\item $|1|_v = 1$,
\item $\left|-x\right|_v = |x|_v$,
\item $|x-y|_v \leq |x|_v + |y|_v$,
\item If $v$ is \emph{non-archimedean} and $|x|_v < |y|_v$,
then $|x+y|_v = |x-y|_v = |y|_v$.
\end{list}
\end{proposition}
\begin{proof}
To prove part~(a), note that
$|1|_v = |1\cdot 1|_v = |1|_v |1|_v = |1|_v^2$,
and hence either $|1|_v=0$ or $|1|_v=1$.
However, since $1\neq 0$ in the field $K$, we also
have $|1|_v\neq 0$, according to Definition~\ref{def:absval}.
Thus, $|1|_v=1$, as desired.
For part~(b), we begin by noting that
$\left|-1\right|_v^2=|(-1)^2|_v=|1|_v=1$,
and therefore either
$\left|-1\right|_v =-1$ or $\left|-1\right|_v=1$.
However, $\left|-1\right|_v\in [0,\infty)$,
and hence $\left|-1\right|_v=1$.
Thus, $\left|-x\right|_v = \left|-1\right|_v |x|_v = |x|_v$.
Part~(c) is now immediate, since
$|x-y|_v \leq |x|_v + \left|-y\right|_v = |x|_v + |y|_v$.
Finally, to prove part~(d), we have
$$|y|_v = |(x+y) - x|_v \leq \max\{|x+y|_v, \left|-x\right|_v\}
= \max\{|x+y|_v, |x|_v\}
\leq \max\{|x|_v, |y|_v\} = |y|_v,$$
and therefore $\max\{|x+y|_v,|x|_v\} = |y|_v$.
However, $|x|_v <|y|_v$, implying that $|x+y|_v=|y|_v$.
The proof that $|x-y|_v=|y|_v$ is similar.
\end{proof}
The usual absolute value on $\Q$ is of course archimedean;
we will denote it by $v=\infty$ or $|\cdot|_{\infty}$.
Meanwhile, for any prime number $p$, the $p$-adic absolute
value $|\cdot|_p$\label{not:vp}
on $\Q$ is given by
$|0|_p=0$ and $|p^e\cdot(r/s)|_p=p^{-e}$
for any integers $e,r,s$ for which $p$ does not divide $r$ or $s$;
it is a non-archimedean absolute value.
See \cite{Gou,Rob} for more on
$p$-adic absolute values.
Given a field $K$ with an absolute value $v$, one can form
the completion $K_v$ \label{not:Kv} of $K$ with respect to $v$ as the set
of Cauchy sequences up to equivalence; $K_v$ is a field, and
$v$ extends to it in a natural way \cite[Proposition~I.3.2]{Rob}.
In addition, $v$
extends uniquely to an algebraic closure $\bar{K}_v$ \label{not:Kvbar}
of $K_v$ \cite[Proposition~II.3.3]{Rob}. However,
$\bar{K}_v$ need not be complete.
\begin{definition}
\label{def:Cv}
Let $K$ be a field with an absolute value $v$,
and let $\bar{K}_v$ be an algebraic closure of $K_v$.
We define $\Cv$ to be the completion of $\bar{K}_v$
with respect to $v$.
\end{definition}
Fortunately, $\Cv$ is both complete and algebraically closed
\cite[Theorem~III.3.3]{Rob}.
If $K=\Q$
and $v=\infty$ is the usual absolute value, then $K_v=\R$
and $\bar{K}_v=\C$ is already complete,
and hence $\Cv=\C$.
On the other hand,
if $K=\Q$ and $v=p$ is the $p$-adic
absolute value, then $K_v=\Qp$ \label{not:Qp} is the field
of $p$-adic rational numbers,
and the algebraic closure $\Qpbar$ \label{not:Qpbar}
of $\Qp$ is \emph{not} complete \cite[Corollary~III.1.4]{Rob}.
Thus, its completion $\Cv=\Cp$ \label{not:Cp}
is a strictly larger field.
\section{Radii and Diameters in $\Cv$}
\label{sec:disks}
In this section, we fix a field $\Cv$
as in Definition~\ref{def:Cv}.
Given a point $a\in\Cv$ and a real number $r>0$,
we define the open and closed disks of radius $r$
centered at $a$ to be
$$\label{not:D}
D(a,r) := \{x\in \Cv : |x-a|_v < r\}
\quad\text{and}\quad
\Dbar(a,r) := \{x\in \Cv : |x-a|_v \leq r\},
$$
respectively.
\begin{remark}
If $v$ is non-archimedean,
the closed unit disk $\Dbar(0,1)$ is often
denoted $\fo_v$. It is a commutative ring
with unity, since it is a subring of $K$;
that is, it contains $1$, is closed under
multiplication, and, thanks to the non-archimedean
property, is closed under addition and subtraction.
Meanwhile, the open unit disk $D(0,1)$ is
often denoted $\fm_v$, and it is a maximal
ideal of $\fo_v$. To see this,
first note that $\fm_v$ is
nonempty and closed under addition and subtraction,
again by the non-archimedean property.
Second,
given any $a\in\fo_v$ and $b\in\fm_v$,
we clearly have $|ab|_v=|a|_v |b|_v<1$,
and therefore $ba=ab\in \fm_v$, proving that
$\fm_v$ is an ideal. Finally, if $I\supsetneq \fm_v$
is an ideal of $\fo_v$ properly containing $\fm_v$,
then there
is some $c\in I$ with $|c|_v=1$. It follows
that $|c^{-1}|_v=1$, and hence $c^{-1}\in\fo_v$.
Therefore, for any $a\in\fo_v$, because $ac^{-1}\in\fo_v$,
we have $a=(ac^{-1})c \in I$. Thus, $I=\fo_v$,
proving that $\fm_v$ is in fact a maximal ideal.
\end{remark}
Just as in $\R$ or $\C$, the \emph{diameter}
of a set $X\subseteq\Cv$ is defined to be
$$\diam X := \sup\{ |x-y|_v : x,y\in X\}\label{not:diam}$$
if $X\neq\varnothing$, or $0$ if $X=\varnothing$.
If $X$ is bounded, i.e., if there is some $M>0$
such that $|x|_v\leq M$ for all $x\in X$, then
clearly $X$ has finite diameter, since
$|x-y|_v \leq |x|_v + |y|_v \leq 2M$.
However, we will actually be interested in
a different way to assign a size to $X$, as follows.
\begin{definition}
\label{def:rad}
Let $X\subseteq\Cv$. The \emph{radius} of $X$,
denoted $\rad X$, is the
the infimum of the radii of all closed disks containing $X$.
That is,
$$\rad X := \inf \{ r>0 : X\subseteq \Dbar(a,r)\text{ for some } a\in\Cv \}.$$
\end{definition}
Note that if $X$ is not a bounded set, then $\rad X = \inf\varnothing=\infty$.
\begin{proposition}
\label{prop:diamrad}
Let $X\subseteq\Cv$. Then
$$\rad X \leq \diam X \leq 2 \rad X.$$
\end{proposition}
\begin{proof}
Let $d=\diam X$ and $r=\rad X$.
If $X=\varnothing$,
then $r = 0= d$.
Similarly, if $X$ is unbounded, then
$r = \infty = d$.
Thus, it suffices to consider the
case that there is some $a\in X$,
and that $X\subseteq \Dbar(0,M)$
for some $M\geq 0$.
In particular, $r$ and $d$ are both finite.
To prove the first inequality, we simply observe
that $X\subseteq\Dbar(a,d)$, and therefore $r\leq d$.
For the second inequality, given any $\varepsilon>0$, there
is some point $b\in X$ and some real number $s\in (r,r+\varepsilon/2)$
such that $X\subseteq \Dbar(b,s)$. Therefore, for any $x,y\in X$,
\begin{equation}
\label{eq:xybound}
|x-y|_v = |(x-b)-(y-b)|_v \leq |x-b|_v + |y-b|_v \leq 2s
< 2r + \varepsilon.
\end{equation}
Since \eqref{eq:xybound} holds for all $\varepsilon>0$
and all $x,y\in X$, we have $d\leq 2r$.
\end{proof}
\chapter{Dynamics}
\label{ch:dyn}
\section{Filled Julia Sets}
\label{sec:jul}
The following definition originally appeared in \cite[p.98]{MS1}.
\begin{definition}
\label{def:reduc}
Let $\phi(z) \in \Q(z)$ be a rational function
with homogenous presentation
$$\phi\left( [x,y]\right) = [f(x,y),g(x,y)],$$
where $f,g\in\Z[x,y]$ are relatively prime homogeneous polynomials
of degree $d=\deg\phi$.
We say that $\phi$ has \emph{good reduction} at a prime $p$
if the reductions
$\bar{f}$ and $\bar{g}$ modulo $p$ have no common zeros in
$\F_p \times \F_p$ besides $(x,y)=(0,0)$.
\end{definition}
Here, $\F_p=\Z/p\Z$ \label{not:Fp}
denotes the field with $p$ elements.
Naturally, given a
homogeneous polynomial $f(x,y)=\sum_{i=0}^d a_i x^i y^{d-i}$,
the reduction $\bar{f}(x,y)$ in Definition~\ref{def:reduc}
means $\sum_{i=0}^d \bar{a}_i x^i y^{d-i}$.
Good reduction turns out to be closely related to
the notion of filled Julia sets.
The motivating idea for such sets is that for a polynomial $\phi$,
all of the interesting dynamics involves points that do not escape
to $\infty$ under iteration.
\begin{definition}
\label{def:jul}
Let $\Cv$ \label{not:Cv} be a complete, algebraically closed
field with absolute value $|\cdot|_v$, and let
$\phi(z) \in \Cv[z]$ be a polynomial of degree $d\geq 2$.
The \emph{filled Julia set} of $\phi$ at $v$ is
$$\cK_v = \big\{x\in\Cv: \{|\phi^n(x)|_v\}_{n\geq 1}
\text{ is bounded} \big\}.$$
\label{not:cKv}
\end{definition}
\begin{example}
\label{ex:cantor}
Fix a prime $p$, an integer $d\geq 2$
with $p\nmid (d-1)$,
and
$c\in\Cp$ with $|c|_p > 1$.
Set $r=|c|_p$ and $\phi(z)=z^d - c^{d-1}z$.
Note that for any $x\in\Cp$ with
$|x|_p>r$, we have $|\phi(x)|_p=|x|_p^d$,
so that $\phi^n(x)\to\infty$.
That is, $\cK_p\subseteq \Dbar(0,r)$;\label{not:cKp}
in particular, $\rad\cK_p\leq r$.
\end{example}
\begin{lemma}
\label{lem:minrad}
Let $p$ be a prime number, and let
$\phi(z) = a_d z^d + a_{d-1} z^{d-1} + \cdots +a_0 \in \Q[z]$
be a polynomial of degree $d\geq 2$.
Denote by $\cK_p$ the filled Julia set of $\phi$ in $\Cp$,
and set $r=|a_d|_p^{1/(d-1)} \rad\cK_p$.
If $r>1$ and $\cK_p\cap \Q\neq \varnothing$, then
$$
r\geq
\begin{cases}
p & \text{ if } d=2,
\\
p^{1/[(d-1)(d-2)]}
& \text{ if } d\geq 3.
\end{cases}
$$
\end{lemma}
\begin{proof}
Given $b\in \cK_p \cap \Q$, we may replace $\phi$ by
$\phi(z+b)-b\in \Q[z]$, which is a polynomial of the
same degree and lead coefficient as $\phi$, but with
filled Julia set translated by $-b$. In particular,
the radius $r$ is preserved; thus, we may assume without
loss that $0\in\cK_p$.
Choose $\alpha\in\Cp$ such that $\alpha^{d-1}=a_d$,
and let $j$ be the largest
index between $0$ and $d-1$ that maximizes
$\lambda_j:=|\alpha^{1-j} a_j|_p^{1/(d-j)}$.
\begin{claim}
\label{cl:lambda}
$\lambda_j >1$.
\end{claim}
\begin{proof}[Proof of Claim~\ref{cl:lambda}]
If $\lambda_j\leq 1$, then
$|a_i\alpha^{-i}|_p\leq |\alpha|_p^{-1} = |a_d \alpha^{-d}|_p$
for every
$i=0,\ldots, d-1$. Thus, for any $x\in \Cp$ with
$|x|_p > |\alpha|_p^{-1}$, we have $|\phi(x)|_p=|a_d x^d|_p$.
It follows by induction that
$$|\phi^n(x)|_p = |a_d|_p^{(d^n-1)/(d-1)} |x|_p^{d^n}
= |\alpha|_p^{-1} |\alpha x|_p^{d^n} \to \infty$$
as $n\to\infty$,
and hence $x\not\in \cK_p$. Thus,
as in Example~\ref{ex:cantor}, $\cK_p\subseteq \Dbar(0,|\alpha|_p^{-1})$,
contradicting the hypothesis that $r>1$.
\end{proof}
By Theorem~6.5.7 of \cite{Gou}, which considers
the so-called Newton polygon of $\phi$,
there is some $\beta\in\Cp$ with
$\phi(\beta)=0$ and $|\alpha\beta|_p = \lambda_j$.
We have $0,\alpha\beta\in\cK_p$;
hence, $r\geq \lambda_j$.
If $j=0$, then a simple induction shows that
$|\alpha \phi^n(0)|_p = |\alpha a_0|_p^{d^{n-1}}$ for $n\geq 1$.
However, that contradicts the hypothesis that $0\in\cK_p$,
since $|\alpha a_0|_p>1$.
Thus, $1\leq j\leq d-1$. Writing
$|a_d|_p= p^{e_1}$ and
$|a_j|_p = p^{e_2}$, where $e_1,e_2\in\Z$ and $e_2\geq 1$, we have
$$
r\geq \lambda_j
= |\alpha^{1-j} a_j|_p^{1/(d-j)} = p^f > 1,
\qquad \text{where}
f= \frac{1}{d-j} \bigg( \frac{(1-j)}{(d-1)} e_1 + e_2 \bigg) > 0.
$$
If $j=1$, then $f=e_2/(d-1) \geq 1/(d-1)$, which proves the
Lemma for the case $d=2$ and part of the case $d\geq 3$.
Finally, if $2\leq j\leq d-1$, then because $f>0$,
we must have $e_1 > (1-d)e_2$. However, $e_1$ and $e_2$
are integers, and therefore $e_1\geq 1 + (d-1) e_2$.
Thus,
$f \geq 1/[(d-1)(d-j)] \geq 1/[(d-1)(d-2)]$.
\end{proof}
\begin{remark}
The bounds of Lemma~\ref{lem:minrad} are sharp.
Indeed, one can check that they are attained by
$\phi(z)=z^2 - z/p$ for $d=2$
and by
$\phi(z)=p^d z^d - p z^2$ for $d\geq 3$.
\end{remark}
\section{Elementary Computations}
\label{sec:elem}
We will write $\log_d x$ \label{not:log}
to denote the logarithm of $x$
to base~$d$.
\begin{definition}
\label{def:E}
Let $N\geq 0$ and $d\geq 2$ be integers. We define $E(N,d)$
\label{not:E}
to be twice the sum of all base-$d$ coefficients of all integers
from $0$ to $N-1$. That is,
$$E(N,d) = 2\sum_{j=0}^{N-1} e(j,d),
\label{not:e}
\qquad\text{where}\qquad
e\left(\sum_{i=0}^M c_i d^i , d\right) = \sum_{i=0}^M c_i,$$
for $c_i\in\{0,1,\ldots, d-1\}$.
\end{definition}
%We declare $E(N,d)=0$ for $N\leq 1$.
We will need the following Lemma.
\begin{lemma}
\label{lem:Ejunk}
Let $d\geq 2$ and $N\geq 1$ be integers,
and write
$N=c+dk$ with $0\leq c\leq d-1$ and $k\geq 0$.
Then:
\begin{list}{\rm \alph{bean}.}{\usecounter{bean}}
\item
$E(N,d) = (d-c)E(k,d) + cE(k+1,d) + (d-1)N - c(d-c)$.
\item If $N\leq d$, then
$E(N,d) = N(N-1)$.
\item
$\displaystyle (d-c) \log_d\Big(\frac{dk}{N}\Big)
+ c \log_d\Big(\frac{dk+d}{N}\Big) \leq 0$.
\item
If $N\geq d$, then
$\displaystyle (d-1)\log_d\left(\frac{dk+d}{N}\right) - (d-c)\leq 0$.
\end{list}
\end{lemma}
\begin{proof}
Writing an arbitrary integer $j\geq 0$ as $j=i+d\ell$ for
$0\leq i\leq d-1$, we compute
\begin{align*}
E(N,d) & = 2 \sum_{j=0}^{N-1}e(j,d)
= 2\sum_{i=0}^{c-1}\sum_{\ell=0}^k e(i+d\ell,d)
+ 2\sum_{i=c}^{d-1}\sum_{\ell=0}^{k-1} e(i+d\ell,d)
\\
& = 2 \sum_{i=0}^{c-1}\sum_{\ell=0}^k (i + e(\ell,d))
+ 2 \sum_{i=c}^{d-1}\sum_{\ell=0}^{k-1} (i + e(\ell,d))
\\
& = \sum_{i=0}^{c-1} \left[2(k+1)i + E(k+1,d)\right]
+ \sum_{i=c}^{d-1} \left[2ki + E(k,d)\right]
\\
& =c E(k+1,d) + (d-c) E(k,d) + (k+1)c(c-1) + kd(d-1) - kc(c-1).
\end{align*}
Part (a) now follows by rewriting the last
three terms as
$$ c(c-1) + dk(d-1) = c(c-d) + (c+dk)(d-1) = (d-1)N -c(d-c).$$
For part (b), we simply observe that if $1\leq N\leq d$,
then
$$E(N,d)=2\big(1+\cdots +(N-1)\big) = N(N-1).$$
To prove part~(c), note that
the function $\log_d(x)$ is of course concave
down. Letting $x_1=dk/N$ and $x_2=(dk+d)/N$, then, we have
$x_1\leq 1< x_2$, and therefore $\log_d(1)\geq L(1)$, where
$$L(x) = \frac{1}{x_2 - x_1}
\left[(x_2 - x) \log_d(x_1) + (x-x_1)\log_d(x_2) \right]
\label{not:L}
$$
is the line through $(x_1,\log_d(x_1))$ and $(x_2,\log_d(x_2))$.
That is,
$$
0 \geq \frac{1}{d}\left[(d-c)\log_d\left(\frac{dk}{N}\right)
+ c \log_d\left(\frac{dk+d}{N}\right)\right].
$$
For part (d), we have
$$(d-1)\log_d\left(\frac{dk + d}{N}\right) =
\frac{(d-1)}{\log d} \cdot \log\left( 1 + \frac{d-c}{N} \right)
\leq \frac{(d-1)}{\log d} \cdot \frac{(d-c)}{N}.$$
However, $\log d = -\log [1 - (d-1)/d ] \geq (d-1)/d$,
and since $N\geq d$,
\[
(d-1)\log_d\left(\frac{dk+d}{N}\right)
\leq (d-1) \cdot \frac{d}{d-1}\cdot \frac{d-c}{N}
=\frac{d}{N}(d-c) \leq (d-c).
\qedhere
\]
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Note: the command \qedhere is needed to ensure that the
% end-of-proof box appears in the proper place (on the
% same line) if a proof ends with a displayed equation.
%
% In addition, the $ ... $ and $$ ... $$ commands for
% math mode are really just shorthand; the more proper
% commands are \( ... \) for in-text math, and
% \[ ... \] for displayed math. The \qedhere symbol
% doesn't work very well with $$ ... $$ (it doesn't put
% the qed box all the way to the far right), but it works
% properly with \[ ... \]. So I had to use it above.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem}
\label{thm:EFbd}
Let $d\geq 2$ and $N\geq 1$ be integers. Then
$E(N,d)\leq (d-1) N \log_d N$, with equality if $N$ is
a power of $d$.
\end{theorem}
\begin{proof}
The result is immediate for $N=1$ and $N=d$
by Lemma~\ref{lem:Ejunk}.b.
If $1 < N < d$, then because $(\log x)/(x-1)$
is a decreasing function, we have
$(\log d)/(d-1) \leq (\log N)/(N-1)$,
from which the desired inequality follows.
For $N\geq d+1$, we proceed by induction
on $N$, assuming the result holds for all positive integers up to $N-1$.
Write $N=c+ dk$, where $0\leq c\leq d-1$, so that
$1\leq k \leq N-2$.
By Lemma~\ref{lem:Ejunk}.a,
we have
\begin{align*}
E(N,d)
& =(d-c) E(k,d) + c E(k+1,d) + (d-1)N - c(d-c)
\\
& \leq (d-c)(d-1) k \log_d k + c(d-1)(k+1)\log_d(k+1)
+ (d-1)N - c(d-c)
\\
& = (d-c)(d-1)k\log_d(dk) + c(d-1)(k+1)\log_d(dk+d) - c(d-c),
\end{align*}
where the final equality is because
$N=(d-c)k + c(k+1)$, and
the inequality (which is equality if $N$ is a power of $d$)
is by the inductive hypothesis, since
$k,k+1\leq N-1$.
Adding and subtracting $(d-1)N\log_d N$, then,
\begin{align*}
E(N,d) & \leq (d-1)N\log_d N+
(d-c)(d-1)k\log_d\left(\frac{dk}{N}\right)
\\
& \phantom{{} \leq(d-1)N\log_d N}
+ c(d-1)(k+1)\log_d\left(\frac{dk+d}{N}\right) - c(d-c)
\\
& = (d-1)N\log_d N
+ c\left[ (d-1)\log_d\left(\frac{dk+d}{N}\right) - (d-c) \right].
\\
& \phantom{{} \leq(d-1)N\log_d N}
+ (d-1)k \left[ (d-c) \log_d\left(\frac{dk}{N}\right)
+ c \log_d\left(\frac{dk+d}{N}\right)\right]
\end{align*}
The quantities in square brackets are nonpositive
by Lemma~\ref{lem:Ejunk}.c--d, and the Theorem follows.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Bibliography
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Force LaTeX to list the bibliography in the table of contents.
\clearpage
\addcontentsline{toc}{chapter}{\protect\numberline{}{Bibliography}}
% Now the actual bibliography:
\begin{thebibliography}{9}
\bibitem{Gou}
F.~Gouv\^{e}a,
\emph{$p$-adic Numbers. An Introduction,} 2nd ed.,
Springer-Verlag, Berlin, 1997.
\bibitem{MS1}
P.~Morton and J.~Silverman,
Rational periodic points of rational functions,
\emph{Inter. Math. Res. Notices} \textbf{2} (1994), 97--110.
\bibitem{Nor}
D.~Northcott,
Periodic points of an algebraic variety,
\emph{ Ann. Math.\/} \textbf{51} (1950), 167--177.
\bibitem{Poon}
B.~Poonen,
The classification of rational preperiodic points of
quadratic polynomials over $\Q$: a refined conjecture,
\emph{Math. Z.} \textbf{228} (1998), 11--29.
\bibitem{Rob}
A.~Robert,
\emph{A Course in $p$-adic Analysis},
Springer-Verlag, New York, 2000.
\end{thebibliography}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Corrections
%
% This section should NOT be part of the
% original thesis. It is only part of the
% corrected version, to be handed in by the
% second-to-last day of classes.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter*{Corrections}
When originally submitted, this honors thesis contained some errors
which have been corrected in the current version. Here is a list of
the errors that were corrected.
\begin{description}
\item[Various places in the thesis.]
Approximately 20 spelling errors were corrected,
10 missing periods or commas were added in mathematical formulae,
and approximately 30 spacing and sizing changes
were made to mathematical formulae.
\item[Other changes:]
\item[p.~1, l.~7.]
The length of the period of $y$ was changed
from $n$ to $n-m$ in two places.
\item[p.~1, l.~--6.]
The reference to \cite[Theorem~3]{Poon} was added.
\item[p.~3.]
In the first paragraph of the Proof of Proposition~\ref{prop:easy},
``and hence $|1|_v=1$'' was changed to
``and hence either $|1|_v=0$ or$\ldots$ as desired.''
\item[p.~4, l.~9.]
The phrase ``is not complete'' was changed to ``need not be complete.''
\item[p.~5, l.~4.]
The formula ``$ab\in\fm_v$'' was changed to ``$ba=ab\in\fm_v$''.
\item[p.~5, l.~6.]
Two appearances of ``thus'' were changed to ``hence'' and ``Therefore''.
\item[p.~5, l.~10.]
The clause ``if $X\neq\varnothing$, or $0$ if $X=\varnothing$''
was added.
\item[p.~5.]
The sentence ``Note that if $X\ldots$'' was added after
Definition~\ref{def:rad}.
\item[p.~5.]
The four sentences, ``If $X=\varnothing$, then$\ldots$ are both finite''
were added to
the first paragraph of the Proof of Proposition~\ref{prop:diamrad}.
\item[p.~7, l.~1--3.]
The subscript $v$ was changed to to $p$ in four places.
\item[p.~7.]
The clauses ``let $r'=\rad\cK_p$, and set $r=|a_d|_p^{1/(d-1)} r'$''
were changed to ``set $r=|a_d|_p^{1/(d-1)} \rad\cK_p$.''
\item[p.~7, l.~13.]
The exponent $1/d-j$ was changed to $1/(d-j)$.
\item[p.~8, l.~8.]
The sentence, ``We will write $\log_d x\ldots$'' was added.
\item[p.~9, l.~2--4.]
On each of these three lines,
$\displaystyle \sum_{i=c-1}^{d-1}$ was changed to
to $\displaystyle \sum_{i=c}^{d-1}$.
\item[p.~9, l.~--4.]
In the first inequality of this line,
$\leq$ was changed to to $\geq$.
\item[p.~10, l.~--2.]
The word ``negative'' was changed to to ``nonpositive''.
\end{description}
\end{document}