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We will spend today's lecture discussing an
outline of the proof, discussing why certain parts
are necessary and also some intuition as to why
one would expect this method to work.

\begin{enumerate}[(1)]
  \item Suppose to the contrary that there are
    infinitely many $\frac{p_1}{q_1},
    \frac{p_2}{q_2}, \ldots$ such that $\left|
    \alpha - \frac{p_j}{q_j} \right| >
    \frac{1}{q^{\sqrt{2d} + \eps}}$.
  \item Choose two among these appropriately,
    which I will denote $\frac{p_1}{q_1},
    \frac{p_2}{q_2}$.
  \item Construct a polynomial $P \in \Zbb[X_1,
    X_2]$ that vanishes at $(\alpha, \alpha)$ to
    high order.
  \item Give a lower bound on $P \left(
    \frac{p_1}{q_1}, \frac{p_2}{q_2} \right)$.
  \item Give an upper bound on $P \left(
    \frac{p_1}{q_1}, \frac{p_2}{q_2} \right)$.
  \item Realise that they give a contradiction.
\end{enumerate}

1 variable is not enough: let $P(X)$ be of degree
$n$. Then $P$ may vanihs at $\alpha$ to order $n /
d$. Then we have a lower bound of
\[
  \left| P \left( \frac{p}{q} \right) \right| \ge
  \frac{1}{q^n}
,\]
and we might hope for an upper bound like
\[
  \left|P \left( \frac{p}{q} \right)\right| \lesssim \left| \alpha - \frac{p}{q}
  \right|^{n / d}
.\]

To get a contradiction, we need $\left| \alpha -
\frac{p}{q} \right|^{n / d} < \frac{1}{q^n}$, i.e.
$\left| \alpha - \frac{p}{q} \right| <
\frac{1}{q^d}$.

Lower bound
\[
  \left| P \left( \frac{p_1}{q_1}, \frac{p_2}{q_2}
  \right) \right| \ge \frac{1}{q_1^{n_1} q_2^{n_2}}
\]
where $n_1$ is the degree in $X_1$ and $n_2$ is
the degree in $n_2$.

Upper bound:
\[
  P \left( \frac{p_1}{q_1}, \frac{p_2}{q_2}
  \right) = \sum_{j_1, j_2} P_{j_1, j_2} (\alpha,
  \alpha) \left( \alpha - \frac{p_1}{q_1}
  \right)^{j_1} \left( \alpha - \frac{p_2}{q_2}
  \right)^{j_2}
\]
where $P_{j_1, j_2}(X_1, X_2) = \frac{1}{j_1!
j_2!} \frac{\partial^{j_1 + j_2}}{\partial
X_1^{j_1} \partial X_2^{j_2}} P(X_1, X_2)$.
Note
\[
  \left( \alpha - \frac{p_1}{q_1} \right)^{j_1}
  \left( \alpha - \frac{p_2}{q_2} \right)^{j_2}
  \le \frac{1}{q_1^{j_1(\sqrt{2d} + \eps)}} \cdot
  \frac{1}{q_2^{j_2(\sqrt{2d} + \eps)}} =
  \exp(-(\sqrt{2d} + \eps)(j_1 \log q_1 + j_2 \log
  q_j2))
.\]
Index of $P$ at $(\beta_1, \beta_2)$ with respect
to the weights $w_1, w_2$.
\[
  I_P(\beta_1, \beta_2; w_1, w_2) = \min(j_1 w_1 +
  j_2 w_2, P_{j_1, j_2}(\beta_1, \beta_2) \neq 0)
.\]
Use $w_1 = \log q_1$, $w_2 = \log q_2$. With this,
we get the upper bound
\[
  \left| P \left( \frac{p_1}{q_1}, \frac{p_2}{q_2}
  \right) \right| \lesssim \exp(-(\sqrt{2d} +
  \eps) \cdot I_P(\alpha, \alpha))
.\]
How big can $I_P(\alpha, \alpha)$ be made? We look
for $P$ in the form
\[
  P(X_1, X_2) = \sum_{i_1 = 0}^{n_1} \sum_{i_2 =
  0}^{n_2} a_{i_1, i_2} X_1^{i_1} X_2^{i_2}
.\]
The condition that $P_{j_1, j_2}(\alpha, \alpha) =
0$ is a linear equation for $a_{i_1, i_2}$ over
$\Qbb[\alpha]$.

By picking a basis of $\Qbb(\alpha)$ over $\Qbb$,
this becomes a system of $d$ linear equations. To find
$P$ such that $I_P(\alpha, \alpha) \ge I$ we need
to solve:
\[
  d \cdot |\{(j_1, j_2) : j_1 \log q_1 + j_2 \log
  q_2 \le i\}| \sim \frac{I^2}{2 \log q_1 \cdot
  \log q_2}
\]
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/b7b3f818dd274779.png}
\end{center}
I can choose $n_, n_2, I$, and I want to do the
following:
\[
  \frac{dI^2}{2\log q_1 \log q_2} \lesssim n_1 n_2
.\]
\[
  \exp(-(\sqrt{2d} + \eps) I) \lesssim
  \frac{1}{q_1^{n_1} q_2^{n_2}}
\]
\[
  (\sqrt{2d} + \eps) I \gtrsim n_1 \log q_1 + n_2
  \log q_2
\]
Take $n_k \sim \frac{\sqrt{2d} + \eps}{2} \cdot
\frac{I}{\log q_k}$ for some large $I$.

Subtleties that still need to be considered:
\begin{itemize}
  \item Siegel's Lemma will be needed to make sure that
    the $P_{j_1, j_2}$ are not too large.
  \item $P \left( \frac{p_1}{q_1}, \frac{p_2}{q_2}
    \right) \neq 0$.
\end{itemize}
$P_{j_1, j_2} \to$ coefficient of $x_1^{i_2}
x_2^{i_2}$ is $a_{i_1 + j_1, i_2 + j_2} \cdot {i_1
+ j_1 \choose i_1} {i_2 + j_2 \choose j_2}$, where
$a_{i_1 + i_2, j_1 + j_2}$ is the coefficient of
$X_1^{i_1 + j_1} X_2^{i_2 + j_2}$ in $P$.

$H(P_{j_1, j_2}) \le 2^{n_1 + n_2} H(P)$.
