%! TEX root = DA.tex
% vim: tw=50
% 18/11/2024 12PM

\begin{theorem*}[Nesterenko]
  Let $T_0, T_1, N, M \in \Zbb_{>0}$. Let
  $\Sigma_1, \Sigma_2 \subset \Cbb^2$ such that
  $|\Sigma_1| = N$, $|\Sigma_2| = M$, and the
  exponentials of the second coordinates of
  $\Sigma_1$ and the first coordinates of
  $\Sigma_2$ are distinct. Let $P \in \Cbb[X, Y]$
  of degree $\le T_0$ in $X$, and $\le T_1$ in
  $Y$. Suppose that $P(X, \exp(y))$ vanishes on
  $\Sigma_1 + \Sigma_2$. Then
  \[
    N \le T_1 \qquad \text{or} \qquad M \le T_0
    (T_1 + 1)
  .\]
\end{theorem*}

\begin{proof}
  If $P(X, Y) = \tilde{P}(X, Y) \cdot Y$, then
  $P(X, \exp(y))$ vanishes at exactly the same
  places as $\tilde{P}(X, \exp(y))$. So we may
  assume $Y \nmid P(X, Y)$. Suppose that $N >
  T_1$, and write $\Sigma_1 = \{(\xi_1, \eta_1),
  \ldots, (\xi_N, \eta_N)\}$. Then $P(\xi_j + X,
  \exp(\eta_j + y))$ vanishes on $\Sigma_2$ for
  all $j = 1, \ldots, N$. We write $P(X, Y) =
  R_1(X) Y^{k_1} + \cdots + R_K Y^{k_K}$ with $0 =
  k_1 < k_2 < \cdots < k_K \le T_1$.

  Then
  \[
    P(\xi_j + X, \exp(\eta_j + y))
    = R_1(\xi_j + X) \cdot \exp(\eta_j)^{k_1}
    \cdot \exp(y)^{k_1} + \cdots
  .\]
  Write
  \[
    Q_{i, j}(X) = R_i(\xi_j + X) \exp(\eta_j)^{k_i}
  .\]
  Then
  \[
    P(\xi_j + X, \exp(\eta_j + y))
    = \sum_{i = 1}^{n} Q_{i,j}(X) (\exp(y))^{k_i}
  .\]
  I look for polynomials $A_1, \ldots, A_k \in
  \Cbb[X]$ such that
  \[
    \sum_{j = 1}^{K} A_j(X) P(\xi_j + X,
    \exp(\eta_j + y)) = B(X) \in \Cbb[X]
    \tag{$*$}
    \label{lec17_eq}
  \]
  such that $\deg B \le T_0(T_1 + 1)$, and then
  since $B$ vanishes at the first coordinates of
  $\Sigma_2$, $M \le T_0(T_1 + 1)$ will follow.
\end{proof}

\begin{lemma*}
  Let $Q_{ij} \in \Cbb[X]$ for $i, j = 1, \ldots,
  K$ for some $K \in \Zbb_{> 0}$. Then there
  exists $A_1, \ldots, A_k \in \Cbb[X]$ such that
  \[
    \sum_i A_i Q_{ij} = \begin{cases}
      \det[Q_{ij}] & \text{if $j = 0$} \\
      0 & \text{otherwise}
    \end{cases}
  \]
\end{lemma*}

\begin{proof}
  Let $[\tilde{Q}_{ij}]$ be the adjugate of
  $[Q_{ij}]$. Then
  \[
    [\tilde{Q}_{ij}] \cdot [Q_{jk}]
    = \det[Q_{jk}] \cdot \id
  .\]
  Let $A_1, \ldots, A_k$ be the first row of
  $[\tilde{Q}_{ij}]$.
\end{proof}

\[
  \begin{pmatrix}
    Q_{11}(X)
    & \cdots
    & Q_{1k}(X) \\
    \vdots
    & \ddots
    & \vdots \\
    Q_{k1}(X)
    & \cdots
    & Q_{kk}(X)
  \end{pmatrix}
  \begin{pmatrix}
    \exp(y)^{e_1} \\
    \vdots \\
    \exp(y)^{e_k}
  \end{pmatrix}
  =
  \begin{pmatrix}
    P(\xi_1 + X, \exp(\eta_i + y)) \\
    \vdots \\
    P(\xi_k + X, \exp(\eta_k + y))
  \end{pmatrix}
\]
Premultiply this by the row vector $(A_1(X),
\ldots, A_k(X))$. We get \eqref{lec17_eq} with $B
= \det[Q_{ij}$.

$\deg B \le T_0 K \le T_0(T_1 + 1)$.

We need to make sure that $B \neq 0$

The leading term of $Q_{ij}$ is $a_i \cdot
\exp(\eta_j)^{k_i} \cdot X^{\deg R_i}$, where
$a_i$ is the leading coefficient of $R_i$.

To show $B \neq 0$, we will consider the leading
term of $B$:
\[
  \det[a_i \exp(\eta_j)^{k_i} X^{\deg R_i}]_{ij}
  = \det[\exp(\eta_j)^{k_i}]_{ij}
  X^{\sum \deg R_i} \prod a_i
.\]

\begin{lemma*}
  Let $K \in \Zbb_{\ge 1}$, wne let $0 = k_1 <
  \cdots < k_K \in \Zbb$. Let $A \subset \Cbb$
  such that $|\{\exp(\eta) : \eta \in A\}| > k_K$.
  Then there exists a choice of $\eta_1, \ldots, \eta_K
  \in A$ such that
  \[
    \det [\exp(\eta_i)^{k_j}] \neq 0
  .\]
\end{lemma*}

\begin{proof}
  By induction on $K$. $K = 1$ is true.

  Suppose $K > 1$, and the claim holds for $K -
  1$. Consider the determinant:
  \[
    \begin{vmatrix}
      \exp(\eta_1)^{k_1}
      & \cdots
      & \exp(\eta_K)^{k_K} \\
      \vdots
      & \ddots
      & \vdots \\
      \exp(\eta_{K - 1})^{k_1}
      & \cdots
      & \exp(\eta_{K - 1}^{k_K} \\
      z^{k_1}
      & \cdots
      & z^{k_K}
    \end{vmatrix}
    = D(z)
  \]
  which has the property that the upper left $(K -
  1) \times (K - 1)$ minor is $\neq 0$.

  Now $D$ is a polynomial which is $\neq 0$ of
  degree $k_K$, so it has at most $k_K$ many $0$s.
  Choose $\eta_K$ such that $\exp(\eta_K)$ is not
  one of them.
\end{proof}
