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In the $K = \Qbb$ case, a key step is that for $L
\in \Zbb[X_1, \ldots, X_N]$ and $H(L) \le
\mathcal{H}$, the points $L(y_1, \ldots, y_N)$ are
integers confined in an interval of length $N
\mathcal{H} Y$ (where $y_1 ,\ldots, y_N = 0,
\ldots, N$).

In the general case, consider the map:
\begin{align*}
  \Phi : K
  &\to \Rbb^n \cdot \Cbb^s \cong \Rbb^D \\
  \alpha
  &\mapsto (v(\alpha))_{v \in M_{K, \infty}}
\end{align*}
The $v$-component of $\Phi(L(y_1, \ldots, y_N))$
is confined in an interval (or box) of size $NY
\cdot |L|_v$.

Let $\alpha = L(y_1, \ldots, y_N) - (z_1, \ldots,
z_N) \neq 0$. By the product formula,
\[
  \prod_{v \in M_{K, \infty}} |\alpha|_v^{d_v}
  = \prod_{_{v \in M_{K, f}}} |\alpha|_v^{-d_v}
  \ge \prod_{v \in M_{K, f}} |L|_v^{d_v}
.\]
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/7e34f8927d584022.png}
\end{center}
Make sure $\prod_v l_v \le$ RHS of above.

\textbf{Non-vanishing:}

\begin{proposition*}
  For every $\eps > 0$, there exists
  $C = C(\eps)$ such that the following holds. Let
  $n_1, n_2 \in \Zbb_{> 0}$, and let
  $\frac{p_1}{q_1}, \frac{p_2}{q_2} \in \Qbb$.
  Suppose that
  \[
    \exp(n_1 + n_2) < q_j^{n_j / C}
  \]
  for $j = 1, 2$, and that $\log q_2 > C \log
  q_1$.

  Let $P \neq 0 \in \Zbb[X_1, X_2]$ of degree in
  $n_j$ in $X_j$ for $j = 1, 2$ such that
  \[
    H(P) < q_j^{n_j / C}
  \]
  for $j = 1, 2$. Then
  \[
    I_P \left( \frac{p_1}{q_1}, \frac{p_2}{q_2},
    \log q_1, \log q_2 \right) \le \eps(n_1 \log
    q_1 + n_2 \log q_2)
  .\]
\end{proposition*}

Note: from now on, whenever we say $\frac{p}{q}
\in \Qbb$, we also mean $\gcd(p, q) = 1$.

When we apply this we will have $n_1 \log q_1 \sim
n_2 \log q_2$.

Without the asymmetry assumption ($\log q_2 > C
\log q_1$), we have the counterexample: $P = (X_1
- X_2)^n$, with $\frac{p_1}{q_1} =
\frac{p_2}{q_2}$.

Alternatively: $P = (R(X_1) - X_2 Q(X_1))^n$ (for
$R, Q$ some small degree polynomials) for
any $\frac{p_1}{q_1}, \frac{p_2}{q_2}$ such that
\[
  \frac{p_2}{q_2} = \frac{R \left( \frac{p_1}{q_1}
  \right)}{Q \left( \frac{p_2}{q_2} \right)}
\]

\begin{lemma*}
  Let $F, F^{(1)}, F^{(2)} \in \Zbb[X_1, X_2]$,
  and let $i_1, i_2 \in \Zbb_{\ge 0}$. Let
  $\alpha_1, \alpha_2 \in \Rbb$ and $w_1, w_2 \in
  \Rbb_{> 0}$. Then the following holds:
  \begin{align*}
    I_{F_{i_1, i_2}}(\alpha_1, \alpha_2)
    &\ge I_F(\alpha_1, \alpha_2) - i_1 w_1 - i_2
    w_2 \\
    I_{F^{(1)} + F^{(2)}}(\alpha_1, \alpha_2)
    &\ge \min_{j = 1, 2} I_{F^{(j)}}(\alpha_1,
    \alpha_2) \\
    I_{F^{(1)} F^{(2)}}
    &= I_{F^{(1)}}(\alpha_1,
    \alpha_2) + I_{F^{(2)}}(\alpha_1, \alpha_2)
  \end{align*}
\end{lemma*}

Baby case: $P(X_1, X_2) = F(X_1) G(X_2)$ for some
$F, G$ polynomials.

In this case if $I_P \ge \eps(n_1 \log q_1 + n_2
\log q_2)$ then either $I_F \ge \eps n_1 \log q_1$
or $I_G \ge \eps n_2 \log q_2$.

If $F$ vanishes at $\frac{p_1}{q_1}$ to order $m$
for some $m$, then
\[
  (q_1 X_1 - p_1)^m \mid F
.\]
The leading coefficient of $F$is divisible by
$q_1^m$. In particular, $H(F) > q_1^m$. Then $H(F)
> q_1^{\eps n_1}$ or $H(G) > q_2^{\eps n_2}$.

Hence $H(P) > \min(q_1^{\eps n_1}, q_2^{\eps
n_2})$, which contradicts the assumptions.

In general, we can always write
\[
  P(X_1, X_2) = F^{(1)}(X_1) G^{(1)}(X_2) + \cdots
  + F^{(h)}(X_1) G^{(h)}(X_2)
\]
with $h \le n_2$.

Consider $h = 2$.
\begin{align*}
  P(X_1, X_2)
  &= F^{(1)}(X_1) \cdot G^{(1)}(X_2) +
  F^{(2)}(X_1) G^{(2)}(X_2) \\
  \frac{\partial}{\partial X_2} P
  &= F^{(1)} \cdot \frac{\partial}{\partial x_2}
  G^{(1)} + F^{(2)} \cdot \frac{\partial}{\partial
  X_2} G_2
\end{align*}