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\begin{theorem*}
  $\alpha$ algebraic of degree $d \le 3$. Then
  there exists $C = C(\alpha) > 0$, $\eps =
  \eps(\alpha) > 0$ such that for all $\frac{p}{q}
  \in \Qbb$:
  \[
    \left| \alpha - \frac{p}{q} \right| > q^{-(d -
    \eps)}
  .\]
\end{theorem*}

\begin{proof}
  Fix some $\alpha$ and $\eps > 0$ small enough.
  Suppose that
  \[
    \left| \alpha - \frac{p}{q} \right| < q^{-(d -
    \eps)}
  \]
  for some $\frac{p}{q} \in \Qbb$. We aim to show
  that $q < C = C(\alpha)$. We assume as we may
  that $\alpha$ is an algebraic integer.

  Let
  \[
    P(X) = (X - \alpha_1) \cdots (X - \alpha_d)
  \]
  be the minimal polynomial of $\alpha =
  \alpha_1$. Then:
  \[
    (p - \alpha_1 q) \cdots (p - \alpha_d q) = Q <
    Cq^\eps
  .\]
  With $Q \in \Zbb$. Then
  \[
    N_{\Qbb(\alpha_j) / \Qbb} (p - \alpha_j q)
    \mid Q^d
  .\]
  In particular:
  \[
    N(p - \alpha_j q) < Cq^{d\eps}
  .\]
  Therefore: $\exists \tilde{\alpha}_j$, $u_1,
  \ldots, u_r, b_1, \ldots, b_r \in \Zbb$ such
  that $p - \alpha_j q = \tilde{\alpha}_j
  u_1^{b_1} \cdots u_r^{b_r}$. Then
  \begin{align*}
    H(\tilde{\alpha}_j)
    &< C \cdot q^\eps \\
    |b_j|
    &< C \cdot \log q
  \end{align*}
  Use $|p - \alpha_1 q| < q^{-(d - 1 - \eps)}$.

  Then $p - \alpha_j q$ is very close to
  $(\alpha_1 - \alpha_j) q$. Consider: $(\alpha_1
  - \alpha_2) (\alpha_1 - \alpha_3) q$, which is
  similar to both of $(p - \alpha_2 q)(\alpha_1 -
  \alpha_3)$ and $(\alpha_1 - \alpha_2)(p -
  \alpha_3 q)$.

  Now more formally:
  \begin{align*}
    \left| 1 - \frac{(p - \alpha_2 q) (\alpha_1 -
    \alpha_3)}{(\alpha_1 - \alpha_2) (p - \alpha_3
    q)} \right|
    &= \left| 1 - \frac{((p - \alpha_1 q) +
    (\alpha_1 - \alpha_2) q) (\alpha_1 -
    \alpha_3)}{(\alpha_1 - \alpha_2) ((p -
    \alpha_1 q) + (\alpha_1 - \alpha_3) q)}
    \right| \\
    &< Cq^{-(d - \eps)}
  \end{align*}
  \[
    \left| \frac{A - \kappa_1}{B - \kappa_2} -
    \frac{A}{B} \right|
    < \frac{\max(\kappa_1, \kappa_2)}{q}
  .\]
  $A \sim B \sim q$.
  Now use the proposition:
  \begin{align*}
    p - \alpha_2 q
    &= \tilde{\alpha}_2 u_1^{b_1} \cdots u_r^{b_r}
    \\
    p - \alpha_3 q
    &= \tilde{\alpha}_3 w_1^{e_1} \cdots w_r^{e_r}
    \\
    H(\tilde{\alpha}_2), H(\tilde{\alpha}_3)
    &\le C \cdot q^\eps \\
    |b_1|, \ldots, |b_r|, |e_1|, \ldots, |e_r|
    &< C \cdot \log q
  \end{align*}
  Writing $\alpha^* = \frac{\tilde{\alpha}_2
  (\alpha_1 - \alpha_3)}{\tilde{\alpha}_3
  (\alpha_1 - \alpha_2)}$ we have:
  \[
    |1 - \alpha^* u_1^{b_1} \cdots u_r^{b_r}
    w_1^{-e_1} \cdots w_r^{-e_r}| < Cq^{-(d - \eps)}
  .\]
  $H(\alpha^*) < C \cdot q^{2\eps}$. Take $\log$
  to be the principal branch, that is $|\Im
  \log(\blank)| \le \pi$. Warning: $\log(xy) \neq
  \log(x) + \log(y)$ in general. This is Lipschitz
  around $1$, so we get
  \begin{align*}
    |\log(\alpha^*) + b_1 \log(u_1) + \cdots + b_r
    \log(u_r) - e_1 \log(w_1) - \cdots - e_r
    \log(w_r) + 2 k \cdot \ub{\log(-1)}_{=\pi i}|
    &< Cq^{-(d - \eps)}
  \end{align*}
  for a suitable $k \in \Zbb$, and $|k| < C \log
  q$.

  Reminder:
  \begin{theorem*}
    Let $n \in \Zbb_{\ge 1}$. Let $\alpha_1, \ldots,
    \alpha_n \in \ol{\Qbb}_{\neq 0}$, and let $\log
    \alpha_j$ be any choice of the $\log$ of
    $\alpha_j$. Let $b_1, \ldots, b_n \in \Zbb$ and
    let $\Lambda = b_1 \log \alpha_1 + \cdots + b_n
    \log \alpha_n$. Let
    \begin{align*}
      A_j
      &= \max(H(\alpha_j), \exp(|\log \alpha_j|),
      10) \\
      B^*
      &= \max \left( \frac{|b_1|}{\log A_n}, \ldots,
      \frac{|b_{n - 1}|}{\log A_n}, |b_n|, 10 \right)
    \end{align*}
    Then there exists an effective constant $C$ (a
    function of $n$ and the degree of
    $\Qbb(\alpha_1, \ldots, \alpha_n)$) such that
    $\Lambda \neq 0$ implies
    \[
      |\Lambda| > \exp(-C\log(A_1) \cdots \log(A_n)
      \log(B^*))
    .\]
  \end{theorem*}

  So the lower bound gives:

  We apply the theorem with $\alpha_n = \alpha^*$.
  $A_1, \ldots, A_{n - 1} < C$, $A_n < C \cdot
  q^{2\eps}$. $B^* \le \frac{C \log q}{\log A_n}
  \le \frac{C}{\eps}$. $|k| < C \log q$. So
  \[
    |\Lambda| > \exp(-C \cdot \eps \log q \cdot
    \log \eps^{-1})
    > q^{-C\eps\log \eps^{-1}}
  .\]
  We still need to consider $\Lambda = 0$. This is
  equivalent to:
  \[
    1 = \frac{(p - \alpha_2 q) (\alpha_1 -
    \alpha_3)}{(\alpha_1 - \alpha_2) (p - \alpha_3
    q)}
  .\]
  Solving this equation gives $\alpha_2 =
  \alpha_3$ or $p = \alpha_1 q$. Neither is the
  case.

  If we use the weaker bound for $|\Lambda|$, then
  we would prove:
  \[
    \left| \alpha - \frac{p}{q} \right| > C \cdot
    q^{-(d - \frac{\eps}{\log \log q})}
    .
    \qedhere
  \]
\end{proof}