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% 11/11/2024 12PM

  So we find $P \in \Zbb[X_1, X_2]$ such that
  $I_P(\alpha, \alpha; \log q_1, \log q_2) \ge I$
  and $H)(P) \le (4H(\alpha))^{\delta^{-1} (n_1 +
  n_2)}$. We need:
  \[
    H(P), \exp(n_1 + n_2) \le q_j^{n_j / C} \sim
    q_1^{n_1 / C}
  \]
  for $j = 1, 2$ and $\log q_2 > C \log q_1$. This
  will be fine if $(4H(\alpha))^{\delta^{-1}} <
  q_1^C$. This is fine if $\eps_0$ is sufficiently
  small with respect to $\alpha$ and $\delta$.

  Then $I_P \left( \frac{p_1}{q_1},
  \frac{p_2}{q_2} \right) \le \eps_0(n_1 \log q_1
  + n_2 \log q_2)$. Then there exists $\tilde{P}$
  a partial derivative of $P$ such that
  \[
    H(\tilde{P}) \le (8H(\alpha))^{\delta^{-1}(n_1
    + n_2)}
  ,\]
  \[
    I_{\tilde{D}}(\alpha, \alpha)
    \ge I - \eps_0(n_1 \log q_1 + n_2 \log q_2)
    \ge \frac{n_1 \log q_1 + n_1 \log
    q_2}{\sqrt{2d} + \frac{\eps}{5}}
  ,\]
  if $\eps_0$ is sufficiently small.

  $\tilde{P} \left( \frac{p_1}{q_1},
  \frac{p_2}{q_2} \right) \neq 0$. Then
  \[
    \left| \tilde{P} \left( \frac{p_1}{q_1},
    \frac{p_2}{q_2} \right) \right| >
    \frac{1}{q_1^{n_1} q_2^{n_2}}
  .\]
  Taylor's formula:
  \[
    \tilde{P} \left( \frac{p_1}{q_1},
    \frac{p_2}{q_2} \right) = \sum_{i_1, i_2}
    \tilde{P}_{i_1, i_2}(\alpha, \alpha) \left(
    \alpha - \frac{p}{q} \right)^{i_1} \left(
    \alpha - \frac{p_2}{q_2} \right)^{i_2}
  \]
  If $i_1, i_2$ are such that $P_{i_1,
  i_2}(\alpha, \alpha) \neq 0$, then
  \[
    i_1 \log q_1 + i_2 \log q_2 > \frac{n_1 \log
    q_1 + n_2 \log q_2}{\sqrt{2d} + \frac{\eps}{5}}
  \]
  hence
  \begin{align*}
    \left| \alpha - \frac{p_1}{q_1} \right|^{i_1}
    \left| \alpha - \frac{p_2}{q_2} \right|^{i_2}
    &< \exp \left( -(\sqrt{2d} + \eps) \cdot
    \frac{n_1 \log q_1 + n_2 \log q_2}{\sqrt{2d} +
    \frac{\eps}{5}} \right) \\
    &< \left( q_1^{n_1} q_2^{n_2}
    \right)^{-\frac{\sqrt{2d} + \eps}{\sqrt{2d} +
    \frac{\eps}{5}}}
  \end{align*}
  The exponent is smaller than $-1$!

  Now estimate the coefficients:
  \begin{align*}
    \tilde{P}_{i_1, i_2}(\alpha, \alpha)
    &\le (n_1
    + 1)(n_2 + 1)(8H(\alpha))^{\delta^{-1}(n_1 +
    n_2)} \cdot \max(1, |\alpha|)^{n_1 + n_2} \\
    &< C_1(\alpha, \eps)^{n_1 + n_2}
  \end{align*}
  and
  \begin{align*}
    \tilde{P} \left( \frac{p_1}{q_1},
    \frac{p_2}{q_2} \right)
    &\le (n_1 + 1)(n_2 + 1) C_1(\alpha, \eps)^{n_1
    + n_2} \cdot (q_1^{n_1}
    q_2^{n_2})^{-\frac{\sqrt{2d} + \eps}{\sqrt{2d}
    + \frac{\eps}{5}}} \\
    &\le (2C_1(\alpha, \eps))^{n_1
    + n_2} \cdot (q_1^{n_1}
    q_2^{n_2})^{-\frac{\sqrt{2d} + \eps}{\sqrt{2d}
    + \frac{\eps}{5}}} \\
    &< (q_1^{n_1} q_2^{n_2})^{-1}
  \end{align*}
  Contradiction.
\end{proof}

\begin{theorem*}[Gelfond-Schneider]
  \label{gs}
  Let $\lambda_1, \lambda_2$ be logarithms of
  non-zero algebraic numbers. Then $\lambda_1,
  \lambda_2$ are linearly independent over
  $\ol{\Qbb}$ if and only if they are linearly
  independent over $\Qbb$.
\end{theorem*}

We will prove this by assuming
$\frac{\lambda_1}{\lambda_2} \in \ol{\Qbb}
\setminus \Qbb$, and then showing that a
particular determinant is both equal to zero and
not equal to zero, hence getting a contradiction.

Before doing this, we will discuss how the
previous proof could have been instead been
phrased using determinants.

We considered some functions $\varphi_1, \ldots,
\varphi_L$ which were some enumeration of
$X_1^{j_1} X_2^{j_2}$. Then we used
\nameref{siegel} to find $a_1, \ldots, a_L$ such
that $D = a_1 \varphi_1 + \cdots + a_L \varphi_L$
vanishes at $u_1 = (\alpha, \alpha)$ to some
order. (Note that $P$ also vanishes at all
Galois-conjugates of $(\alpha, \alpha)$:
$u_2, \ldots, u_d$). Then we find an argument to
show that $P$ also vanishes at $u_{d + 1} = \left(
\frac{p_1}{q_1}, \frac{p_2}{q_2} \right)$ to some
order.

This means that for $i = 1, \ldots, L$ there
exists $k(i) \in \{1, \ldots, d + 1\}$ and some
partial differentiation operator $\partial_i$ such
that $\partial_i P(u_{k(i)}) = 0$. We also showed
that $P$ with so much vanishing cannot exist.

Let:
\[
  M = \begin{pmatrix}
    \partial_1 \varphi_1(u_{k(1)})
    & \cdots
    & \partial_L \varphi_1(u_{k(l)}) \\
    \vdots
    & \ddots
    & \vdots \\
    \partial_1 \varphi_L(u_{k(1)})
    & \cdots
    & \partial_L \varphi_L(u_{k(L)})
  \end{pmatrix}
\]
Then $P$ having all that vanishing is equivalent
to
\[
  (a_1, \ldots, a_L) M = (0, \ldots, 0)
.\]
Now the existence of $P$ is equivalent to $\det M
= 0$.
