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\subsection{Transcendence}

Liouville proved $\alpha = \sum_{n = 0}^{\infty}
\frac{1}{10^{n!}}$ is transcendental.

What about $e$, $\pi$, $2^{\sqrt{2}}$?

Hermite: $e$ is transcendental.

Lindemann: If $\alpha \neq 0$, then at least one
of $\alpha$ or $e^\alpha$ is transcendental.

\begin{fcthm}[Lindemann-Weierstrass]
  Assuming:
  - $\alpha_1, \ldots, \alpha_n \in \Cbb$
  distinct
  Then:
  $e^{\alpha_1}, \ldots, e^{\alpha_n}$ are
  linearly independent over $\ol{\Qbb}$ (algebraic
  closure of $\Qbb$).
\end{fcthm}

Hilbert's 7th problem: Let $\alpha \neq 0, 1$,
algebraic, $\beta$ irrational algebraic. Then
$\alpha^\beta$ is transcendental.

Note (for this problem): $\alpha^\beta = \exp(\beta \cdot
\log\alpha)$ where $\log\alpha$ is any complex
number with $e^{\log \alpha} =\alpha$. So in the
above problem we can think of ``$\alpha^\beta$ is
transcendental'' as meaning ``any choice for
$\alpha^\beta$ is transcendental''.

Convention: If $\alpha \in \Rbb_{> 0}$, then
$\log\alpha \in \Rbb$.

\begin{theorem*}
  Let $\alpha_1, \alpha_2$ be non-zero algebraic
  numbers. Then $\log \alpha_1, \log \alpha_2$ are
  linearly independent over $\ol{\Qbb}$ if and
  only if they are linearly independent over
  $\Qbb$.
\end{theorem*}

\begin{proof}[Proof of Hilbert's 7th $\iff$
  above Theorem is true]
  \phantom{}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] Suppose $\log \alpha_1, \log \alpha_2$ are
      dependent over $\Qbb$. Then $\exists \beta \in
      \ol{\Qbb}$ such that $\beta \log \alpha_1 = \log
      \alpha_2$. Then $\alpha_1^\beta = \alpha_2$
      either $\beta \in \Qbb$ or $\alpha_1 = 1$.
    \item[$\Leftarrow$] Suppose there exists
      $\alpha_1$, $\alpha_2$ non-zero algebraic
      such that $\alpha_1^\beta = \alpha_2$ for
      some $\beta \in \ol{\Qbb}$. Then $\beta \log
      \alpha_1 = \log \alpha_2$ for some choice of
      the logarithms. If the logarithms are $0$,
      then we deduce $\alpha_1 = 1$, a
      contradiction. Otherwise, we
      deduce that $\beta \in \Qbb$ (by the above
      theorem), which is also a contradiction.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{theorem*}[Baber]
  Let $\log \alpha_1, \ldots, \log \alpha_n$ be
  $\Qbb$-linearly independent logarithms of
  algebraic numbers.

  Then $1, \log \alpha_1, \ldots, \log \alpha_n$
  are linearly independent over $\ol{\Qbb}$.
\end{theorem*}

\begin{conjecture}[Schanuel]
  Let $\alpha_1, \ldots, \alpha_n \in \Cbb$ be
  linearly independent over $\Qbb$. Then the
  transcendence degree of $\Qbb(\alpha_1, \ldots,
  \alpha_n, e^{\alpha_1}, \ldots, e^{\alpha_n})$
  is at least $n$.
\end{conjecture}

Let $\alpha_1, \ldots, \alpha_n \in \Qbb_{> 0}$,
and $b_1, \ldots, b_n \in \Zbb$. Let $A_j$ be the
max of the numerator and the denominator of $a_j$.

Let $B = \max(|b_1|, \ldots, |b_n|)$. Then
\[ b_1 \log a_1 + \cdots b_n \log a_n \text{ close
to $0$} \iff a_1^{b_1} \cdots a_n^{b_n} \text{
close to $1$} .\]
\[ |a_1^{b_1} \cdots a_n^{b_n} - 1| \ge A_1^{-b}
\cdots A_n^{-B} = \exp(-(\log A_1 + \log A_n)B) .\]
\[ |b_1 \log a_1 + \cdots + b_n \log a_n| \ge
\half \exp(-(\log A_1 + \cdots + \log A_n)B) .\]

\begin{notation*}
  \glssymboldefn{minp}%
  \glssymboldefn{height}%
  Let $\alpha \in \ol{\Qbb}$, denote its minimal
  polynomial in $\Zbb[X]$ by $f_\alpha$.

  If $f \in \Cbb[X]$, then $H(f)$ (the height of
  $f$) is the maximal absolute value of its
  coefficients.
\end{notation*}

\begin{theorem*}
  Let $\alpha_1, \ldots, \alpha_n \in
  \ol{\Qbb}_{\neq 0}$, $\beta_0, \ldots, \beta_n
  \in \ol{\Qbb}$. Fix some choices of $\log
  \alpha_j$. Let $A_j = \max(\height(\minp
  {\alpha_j}) \exp(|\log \alpha_j|), 10)$.

  Let $\Lambda = \beta_0 + \beta_1 \log \alpha_1 +
  \cdots + \beta_n \log \alpha_n$. Then there
  exists an \emph{effective} constant $C$ depending on $n$
  and the degree of
  $\Qbb(\alpha_1, \ldots, \alpha_n, \beta_0,
  \beta_1, \ldots, \beta_n)$ such that either
  $\Lambda = 0$ or
  \[
    |\Lambda| > \exp(-C (\log A_1)\cdots (\log
    A_n)(\log B))
  .\]
\end{theorem*}

Conjecturially: this should be
\[
  |\Lambda| > \exp(-C \max(\log A_1, \ldots, \log
  A_n, \log B))
.\]
