%! TEX root = Galois.tex
% vim: tw=50
% 10/10/2023 12AM

\begin{flashcard}[transcendental-defn]
\begin{definition}[Transcendental number]
  \glsdefn{transc_el}{transcendental}{N/A}
  \cloze{
  Let $L \extendover K$ be a \gls{field_ex}. $\alpha \in
  L$ is \emph{transcendental} over $K$ if it is
  not \gls{alg_el}. In this case $\ringadj{K}{\alpha} \cong
  \ringadj{K}{X}$ and $\fieldadj{K}{\alpha} \cong
  \fieldadj{K}{X}$.
  }
\end{definition}
\end{flashcard}

\begin{remark*}
  Since $\ringadj{K}{X}$ is not a \gls{field} and $1, X, X^2,
  \ldots$ are linearly independent over $K$, we
  have $\fieldadj{K}{\alpha} \neq
  \ringadj{K}{\alpha}$ and $\degover{\fieldadj{K}{\alpha}
  }{K} = \infty$.
\end{remark*}

\begin{flashcard}[algebraic-defn]
\begin{definition*}[Algebraic Field]
  \glsdefn{alg_ex}{algebraic}{N/A}
  \cloze{
  We say a \gls{field_ex} $L \extendover K$ is
  \emph{algebraic} if every $\alpha \in L$ is
  \gls{alg_el} over $K$.
  }
\end{definition*}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item We have $\degover{\fieldadj{K}{\alpha}}{K} < \infty$ if
      and only if $\alpha$ is \gls{alg_el} over $K$.
    \item If $\degover{L}{K} < \infty$ then for any
      $\alpha \in L$ we certainly have
      $\degover{\fieldadj{K}{\alpha}}{K} < \infty$.
      So any \gls{finite_ex} \gls{ex} is
      \gls{alg_ex}, but the inverse is not true.
      \begin{example}
        \vspace{0.5em}
        $K = \QQ$, $L = \bigcup_{n = 1}^\infty
        \ringadj{\QQ}{\sqrt[2^n]{2}} \subset \RR$. This is a
        union of a nested sequence of
        \glspl{field}
        \[ \fieldadj{\QQ}{\sqrt{2}} \subset
        \fieldadj{\QQ}{\sqrt[4]{2}}
        \subset \fieldadj{\QQ}{\sqrt[8]{2}} \subset \cdots \]
        and so this is a \gls{field}.
        $\degover{L}{K} = \infty
        = \infty$ since
        $\degover{\fieldadj{\QQ}{\sqrt[2^n]{n}}}{\QQ}
        = 2^n$ is unbounded. But every $\alpha \in
        L$ belongs to a \gls{finite_ex} \gls{ex} of $\QQ$
        and so is \gls{alg_el} over $\QQ$. Therefore
        $L \extendover \QQ$ is \gls{alg_ex}.
      \end{example}
  \end{enumerate}
\end{remark*}

\begin{remark}
  \glsdefn{transc_Q}{transcendental}{N/A}
  \glsdefn{alg_Q}{algebraic}{N/A}
  Classically $\alpha \in \CC$ is called algebraic
  / transcendental if it is \gls{alg_el} /
  \gls{transc_el} over $\QQ$.

  A countability argument (see IA Numbers and
  Sets) shows that \gls{transc_Q} numbers exist.
  Liouville showed that $\sum_{n = 1}^\infty
  \frac{1}{10^{n!}}$ is \gls{transc_Q}.

  It was proved in the 19th century that $e$ and
  $\pi$ are \gls{transc_Q} (Hermite \& Lindeman).
\end{remark}

\begin{example}
  \label{lec3_last_example}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Let $f(X) = X^d - n$ ($d, n \in \ZZ$, $d
      \ge 2$, $n \neq 0$). Suppose there exists a
      prime $p$ such that when we write $n = p^e
      m$ with $p \nmid m$, then $(d, e) = 1$. We
      claim that $f$ is irreducible in $\ringadj{\QQ}{X}$.
      Equivalently we show that
      $\degover{\fieldadj{\QQ}{\alpha}}{
      \QQ} = d$ where $\alpha = \sqrt[d]{n}$. By
      Euclid's algorithm there exist $r, s \in
      \ZZ$ such that $rd + se = 1$ (we may arrange
      $s > 0$). Then $p^{dr} n^s = p^{dr} (p^e
      m)^s = pm^s$. Let $\beta = p^r \alpha^s$ so
      that $\beta^d = p m^s$. Then $\beta$ is a
      root of $g(X) = X^d - p m^s$, which is
      irreducible in $\ringadj{\ZZ}{X}$ by Eisenstein's
      criterion, so irreducible in
      $\ringadj{\QQ}{X}$ by
      Gauss' Lemma, therefore
      $\degover{\fieldadj{\QQ}{\beta}}{\QQ}
      = d$. But $\fieldadj{\QQ}{\beta} \subset
      \fieldadj{\QQ}{\alpha}$
      and $\degover{\fieldadj{\QQ}{\alpha}}{\QQ} \le \deg f = d$.
      This gives $\degover{\fieldadj{\QQ}{\alpha}}{\QQ} = d$ as
      required.
    \item Let $p$ be an odd prime, $\zeta_p =
      e^{2\pi i / p}$ and
      \[ \alpha = 2\cos \left( \frac{2\pi}{p}
      \right) = \zeta_p + \zeta_p^{-1} .\]
      Let's compute $\degover{\fieldadj{\QQ}{\alpha}}{\QQ}$.
      $\zeta_p$ is a root of
      \[ f(X) = \frac{X^p - 1}{X - 1} = X^{p - 1}
      + \cdots + X^2 + X + 1 .\]
      which is irreducible by Eisenstein's
      criterion applied to $f(X + 1)$. Therefore
      $\degover{\fieldadj{\zeta_p}}{\QQ} = \deg f = p - 1$.
      Note that $\fieldadj{\QQ}{\zeta_p}
      \extendover \fieldadj{\QQ}{\alpha}
      \extendover \QQ$, so we will try to use
      \nameref{tower_law}
      to find the \gls{ex_deg} of the \gls{ex}
      $\fieldadj{\QQ}{\alpha} \extendover \QQ$. $\zeta_p$ is a root
      of
      \[ g(X) = (X - \zeta_p)(X - \zeta_p^{-1}) =
      X^2 - \alpha X + 1 \in
      \ringadj{\fieldadj{\QQ}{\alpha}}{X} \]
      Therefore
      $\degover{\fieldadj{\QQ}{\zeta_p}}{\fieldadj{\QQ}{\alpha}} \le
      \deg g = 2$. But $\alpha \in \RR$ and
      $\zeta_p \not\in \RR$ so $\zeta_p \not\in
      \fieldadj{\QQ}{\alpha} \subset \RR$. So
      $\degover{\fieldadj{\QQ}{\zeta_p}}
      {\fieldadj{\QQ}{\alpha}} = 2$, so by
      \nameref{tower_law}
      $\degover{\fieldadj{\QQ}{\alpha}}{\QQ} =
      \frac{p - 1}{2}$.
    \item Suppose $m, n$ and $mn$ are not perfect
      squares. Let $\alpha = \sqrt{m} + \sqrt{n}$.
      Let's compute
      $\degover{\fieldadj{\QQ}{\alpha}}{\QQ}$. Clearly
      $\fieldadjoin\QQ(\alpha) \subset
      \fieldadjoin\QQ(\sqrt{m},
      \sqrt{n})$. Conversely we have
      \[ m = (\alpha - \sqrt{n})^2 = \alpha^2 -
      2\alpha \sqrt{n} + n \]
      \[ \implies \sqrt{n} = \frac{\alpha^2 - m +
      n}{2\alpha} \]
      so $\sqrt{n} \in \fieldadjoin\QQ(\alpha)$, and similarly
      $\sqrt{m} \in \fieldadjoin\QQ(\alpha)$. So therefore
      $\fieldadjoin\QQ(\alpha) = \fieldadjoin\QQ(\sqrt{m}, \sqrt{n})$.
      Note that $\fieldadjoin\QQ(\sqrt{m},
      \sqrt{n}) \extendover
      \fieldadjoin\QQ(\sqrt{n}) \extendover \QQ$, with
      $\exdeg[\fieldadjoin\QQ(\sqrt{n}) : \QQ] = 2$. We also know
      $\exdeg[\fieldadj{\fieldadjoin\QQ(\sqrt{n})}{\sqrt{m}}
      : \fieldadjoin\QQ(\sqrt{n})]
      \le 2$. Suppose $\sqrt{m} \in
      \fieldadjoin\QQ(\sqrt{n})$. Then $\sqrt{m} = r +
      s\sqrt{n}$ for some $r, s \in \QQ$.
      Therefore $m = r^2 + 2rs\sqrt{n} + s^2n$.
      Since $\sqrt{n} \not\in \QQ$ we must have
      $rs = 0$. If $r = 0$ then $mn$ will be a
      square, and if $s = 0$ then $m$ is a square.
      Either way we get a contradiction. So
      $\exdeg[\fieldadjoin\QQ(\sqrt{m}, \sqrt{n})
      : \fieldadjoin\QQ(\sqrt{n})] =
      2$. So by \nameref{tower_law},
      $\exdeg[\fieldadjoin\QQ(\alpha)
      : \QQ] = 4$.
  \end{enumerate}
\end{example}