%! TEX root = DA.tex
% vim: tw=50
% 11/10/2024 12PM

\newpage

\textbf{Fact:} If $n \neq m \in \Zbb$, then $|n -
m| \ge 1$.

Although this fact sounds very obvious, in this
course it will be one of our most used tools.

\newpage
\section{Diophantine Approximation}

\begin{fcthm}[Dirichlet]
  Assuming:
  - $\alpha$ is an irrational real number
  Then:
  there exist infinitely many $\frac{p}{q} \in
  \Qbb$ such that
  \[ \left| \alpha - \frac{p}{q} \right| \le
  \frac{1}{q^2} .\]
\end{fcthm}

\begin{proof}
  Consider the numbers $0, \alpha, 2\alpha,
  \ldots, N\alpha$ for some fixed $N \in
  \Zbb_{>0}$. Consider them in $\Rbb / \Zbb \equiv
  [0, 1]$. Note 
  \[
    \left[0, \frac{1}{N} \right) \sqcup \left[
    \frac{1}{N}, \frac{2}{N} \right) \sqcup \cdots
    \sqcup \left[ \frac{N - 1}{N}, 1 \right)
  .\]
  By the box principle (pigeonhole principle),
  there exists $N \ge n_2 > n_1 \ge 0$ such that
  $n_2 \alpha$ and $n_1 \alpha$ belong to the same
  interval. Then:
  \[ |n_2 \alpha - n_1 \alpha - p| \le \frac{1}{N} \]
  for some $p \in \Zbb$. Take $q = n_2 - n_1$.
  Then 
  \[
    \left| \alpha - \frac{p}{q} \right| \le
    \frac{1}{Nq} \le \frac{1}{q^2}
  .\]
  Take $N \to \infty$, then you get an infinite
  sequence of rationals. If $\alpha$ is not
  raitonal, then this sequence cannot stabilise,
  so we get infinitely many $\frac{p}{q}$ as
  desired.
\end{proof}

Can we do better?

In particular for $\alpha \in \ol{\Qbb}$.

\begin{theorem*}[Liouville]
  \label{liouville}
  Assuming:
  \begin{itemize}
    \item $\alpha$ is algebraic of degree $d$
  \end{itemize}
  Then there exists $c > 0$ such that for all
  $\frac{p}{q} \in \Qbb$ with $\alpha \neq
  \frac{p}{q}$, we have
  \[
    \left| \alpha - \frac{p}{q} \right| \ge
    \frac{c}{q^d}
  .\]
\end{theorem*}

\begin{proof}
  Let $P \in \Zbb[x]$ be the minimal polynomial of
 $\alpha$, so $P(\alpha) = 0$. Now note that $P
 \left( \frac{p}{q} \right) \neq 0$ (by
 irreducibility when $d \ge 2$, and for $d = 1$
 using the hypothesis that $\alpha \neq
 \frac{p}{q}$). Then
 \[
   \left| P \left( \frac{p}{q} \right)  \right|
   \ge \frac{1}{q^d}
 .\]
 Note that $P \left( \frac{p}{q} \right) $ is
 rational with denominator $q^d$. On the other
 hand,
 \[
   \left| P \left( \frac{p}{q} \right)  \right|
   \le \left( \max_{x \in [\alpha - 1, \alpha +
   1]} |P'(x)| \right) \cdot \left| \alpha -
   \frac{p}{q} \right|
 .\]
 provided $\left| \alpha - \frac{p}{q} \right|
 \le 1$, which we may assume. Hence
 \[ \left| \alpha - \frac{p}{q} \right|  >
 \frac{c}{q^d} .\]
\end{proof}

Improvements of the exponent $d$ in Liouville:
\begin{itemize}
  \item Thue: $\frac{d}{2} + 1 + \eps$
  \item Siegel: little better than $2\sqrt{d} +
    \eps$
  \item Dyson: $\sqrt{2d} + \eps$
\end{itemize}

\begin{fcthm}[Roth]
  Assuming:
  - $\alpha$ is an irrational real algebraic
  number
  Then:
  there exists $c = c(\alpha, \eps) > 0$ such that
  for all $\frac{p}{q} \in \Qbb$ we have
  \[
    \left| \alpha - \frac{p}{q} \right|  \ge
    \frac{c}{q^{2 + \eps}}
  .\]
\end{fcthm}

\begin{fcthm}[Thue]
  Assuming:
  - $P(X, Y) \in \Zbb[X, Y]$ homogeneous of degree $d \ge 3$
  - without repeated factors
  - $m \in \Zbb$
  Then:
  the equation
  \[
    P(X, Y) = m
  \]
  has only finitely many solutions in $\Zbb^2$
  with $\gcd(X, Y) = 1$.
\end{fcthm}

Liouville's theorem $\leftrightarrow$ $|P(p, q)|
\ge 1$.

\begin{fclemma}[]
  Assuming:
  - $P \in \Rbb[X, Y]$ be homogeneous of degree
  $d$
  - without repeated factors
  Then:
  for all $p, q \in \Zbb$, there exists $\alpha$
  root of $P(X, 1)$ such that
  \[
    cq^{-d} P(p, q) \le \left| \alpha -
    \frac{p}{q} \right|  \le C q^{-d} P(p, q)
  .\]
  Here $c$, $C$ depend on $P$, and a fixed compact
  set that contains $\frac{p}{q}$.
\end{fclemma}

\begin{proof}
  Let
  \[
    P(X, 1) = a(X - \alpha_1) \cdots (X - \alpha_d)
  ,\]
  with $\alpha_1, \ldots, \alpha_d$ distinct
  (since we assumed no repeated factors, and
  characteristic $0$ fields are always separable).
  Without loss of generality assume that
  $\alpha_1$ is the closest to $\frac{p}{q}$.

  Then $c_0 < \left| \frac{p}{q} - \alpha_j
  \right| < C_0$ for some constants depending on
  $P$ and the compact set for $j \neq 1$. So we
  get lower and upper bounds on $P \left(
  \frac{p}{q}, 1 \right)  = P(p, q) \cdot
  \frac{1}{q^d}$.
\end{proof}

\begin{proof}[Proof of Thue]
  Suppose $P(p, q) = m$. The lemma tells us that
  there exists $\alpha$ a root of $P$ such that
  \[
    \left| \frac{p}{q} - \alpha \right| < C \cdot
    q^{-d} |\ub{P(p, q)}_{m}| = C \cdot m \cdot
    q^{-d}
  .\]
  If the degree of $\alpha \ge 2$, then Roth or
  already Thue implies that $q$ must be bounded,
  hence only finitely many solutions.

  For $\alpha \in \Qbb$, we use Liouville.
\end{proof}

