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

\begin{flashcard}[algebraic-iff-finite-dim-lemma]
\begin{lemma}
  \label{algebraic_if_finite_dimension}
  Let $L \extendover K$ be a \gls{field_ex} and
  $\alpha_1, \ldots, \alpha_n \in L$. Then
  \[ \text{$\alpha_1, \ldots, \alpha_n$ are
  \gls{alg_el} over $K$} \iff \exdeg[\fieldadjoin K(\alpha_1, \ldots,
  \alpha_n) : K] < \infty .\]
\end{lemma}

\begin{proof}
  \fcscrap{The case $n = 1$ was a remark in the
  previous lecture.}\phantom{}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] \cloze{By induction on on $n$
      using \nameref{tower_law}.}
    \item[$\Leftarrow$] \cloze{Clear since
        $\fieldadjoin K(\alpha_1)
      \subset \fieldadjoin K(\alpha_1, \ldots, \alpha_n)$.}
      \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[algebraic-is-subfield-coro]
\begin{corollary}
  \label{algebraic_is_subfield_coro}
  Let $L \extendover K$ be any \gls{field_ex}. Then the
  set
  \[ \{\alpha \in L \mid \text{$\alpha$ is
  \gls{alg_ex} over $K$}\} \]
  is a subfield of $L$.
\end{corollary}

\begin{proof}
  \cloze{
  If $\alpha, \beta$ are \gls{alg_el} over $K$ then
  by \cref{algebraic_if_finite_dimension}
  $\exdeg[\fieldadjoin K(\alpha, \beta) : K] < \infty$. Let $\gamma =
  \alpha \pm \beta$, or $\alpha \beta$ or (if
  $\alpha \neq 0$) $\frac{1}{\alpha}$. Then
  $\gamma \in \fieldadjoin K(\alpha, \beta)$, so since
  $\exdeg[\fieldadjoin K(\alpha, \beta, \gamma) : K] < \infty$, by
  \cref{algebraic_if_finite_dimension}, we get
  that $\gamma$ is \gls{alg_el} over $K$.
  }
\end{proof}
\end{flashcard}

\begin{example*}
  Taking $K = \QQ$ and $L = \CC$ we see that
  \[ \algclos{\QQ} = \{\text{\gls{alg_Q} numbers}\} \]
  is a \gls{field}.

  Since $\algclos{\QQ} \subset \fieldadj \QQ(\sqrt[2]{2})$ and
  $\exdeg[\fieldadj \QQ(\sqrt[d]{2}) : \QQ] = d$ for all $d$, we
  see that $\exdeg[\algclos{\QQ} : \QQ] = \infty$.
\end{example*}

\begin{flashcard}[algebraic-tower-prop]
\begin{proposition}
  \label{alg_tower_prop}
  Let $M \extendover L \extendover K$ be a
  \gls{field_ex}. Then
  \[ \text{$M \extendover K$ is \gls{alg_ex}} \iff \text{$M
  \extendover L$ and $L \extendover K$ are both
  \gls{alg_ex}} \]
\end{proposition}

\begin{proof}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] \cloze{Every element of $M$ is
      \gls{alg_el} over $K$, hence \gls{alg_el} over
      $L$, so $M \extendover L$ is \gls{alg_ex}. Also, as $L
      \subset M$, $L$ is \gls{alg_ex} over $K$.}
    \item[$\Leftarrow$] \cloze{Let $\alpha \in M$. We
      must show that $\alpha$ is \gls{alg_el} over
      $K$. Since $M \extendover L$ is \gls{alg_ex}, $\alpha$
      is a root of some
      \[ f(X) = c_n X^n + \cdots + c_1 X + c_0 \in
      \ringadjoin L[X] \]
      Let $L_0 = \fieldadjoin K(c_0, c_1, \ldots, c_n)$. Since
      each $c_i$ is \gls{alg_el} over $K$,
      \cref{algebraic_if_finite_dimension} gives
      $\exdeg[L_0 : K] < \infty$. But $f$ has
      coefficients in $L_0$, so $\exdeg[L_0(\alpha) :
      L_0] \le \deg f < \infty$. By
      \nameref{tower_law} $\exdeg[\fieldadjoin L_0(\alpha) : K] <
      \infty$. Therefore $\alpha$ is \gls{alg_el}
      over $K$.} \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\newpage
\section{Ruler and Compass Constructions}
\label{sec2}

We use our results on \glspl{field_ex} to show
that certain classical problems cannot be solved.

\begin{flashcard}[ruler-compass-construction]
\begin{definition*}[Ruler and Compass Construction]
  \glsdefn{constructible_point}{constructible}{N/A}
  \glsalias{constructed_point}{constructed}{N/A}
  \cloze{
  Let $S \subset \RR^2$ be a finite set of points.
  We may:
  \begin{enumerate}[(i)]
    \item Draw a straight line through any 2
      distinct points in $S$.
    \item Draw a circle with centre any point in
      $S$ and radius the distance between 2 points
      in $S$.
    \item Enlarge $S$ by adjoining any point of
      intersection of 2 distinct lines or circles.
  \end{enumerate}
  A point $(x, y) \in \RR^2$ is
  \emph{constructible} from $S$ if we can enlarge
  $S$ to contain $(x, y)$ by a finite sequence of
  the above operations.

  \glsdefn{constructible_real}{constructible}{N/A}
  A number $x \in \RR$ is constructible if $(x,
  0)$ can be \gls{constructed_point} from $\{(0, 0), (1,
  0)\}$.
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

We will relate this to the following:

\begin{flashcard}[constructible-field-defn]
\begin{definition*}[Constructible Field]
  \glsdefn{constructible_field}{constructible}{N/A}
  \cloze{
  Let $K \subset \RR$ be a subfield. We say that
  $K$ is constructible if there exists integer $n
  \ge 0$ and a sequence of subfields of $\RR$
  \[ \QQ = F_0 \subset F_1 \subset F_2 \subset
  \cdots \subset F_n \]
  such that $\exdeg[F_i : F_{i - 1}] = 2$ for all $i$
  and $K \subset F_n$.
  }
\end{definition*}
\end{flashcard}

\begin{remark*}
  We see by \nameref{tower_law} that $K$
  \gls{constructible_field} $\implies$ $\exdeg[K : \QQ]$ is a power
  of 2.
\end{remark*}

\begin{flashcard}[compass-construction-is-constructible-field]
\begin{theorem}
  % Theorem 2.1
  \label{compass_construction_is_constructible_field}
  If $x \in \RR$ is \gls{constructible_real}, then
  $\fieldadjoin \QQ(x)$
  is a \gls{constructible_field} subfield of $\RR$.
\end{theorem}

\begin{proof}
  \cloze{
  Suppose $S \subset \RR^2$ is a finite set of
  points all of whose coordinates belong to a
  \gls{constructible_field} \gls{field} $K$.

  It suffices to show that if we adjoin $(x, y)
  \in \RR^2$ to $S$ using (iii), then $K(x, y)$ is
  also \gls{constructible_field}. Since $K$ is
  \gls{constructible_field},
  there exists a sequence $\QQ = F_0 \subset F_1
  \subset \cdots \subset F_n$ with $\exdeg[F_i : F_{i -
  1}] = 2$ for all $1 \le i \le n$ and $K \subset
  F_n$.

  The lines and circles in
  (i) and (ii) have equations of the form $ax + by
  = c$ and $(x - a)^2 + (y - b)^2 = c$ with $a, b,
  c \in K$. If $(x, y)$ is a point of intersection
  of 2 such lines or circles then $x = r +
  s\sqrt{v}$, $y = t + u\sqrt{v}$ for some $r, s,
  t, u, v \in K$. Therefore $x, y \in \fieldadjoin K(\sqrt{v})
  \subset \fieldadjoin F_n(\sqrt{v})$. Since
  $\exdeg[\fieldadjoin F_n(\sqrt{v}) :
  F_n]$ is 1 or 2, it follows that $\fieldadjoin K(x, y)$ is
  \gls{constructible_field}.
  }
\end{proof}
\end{flashcard}

\begin{remark*}
  It can be shown that $(x \pm y, 0)$, $(x / y,
  0)$ and $(\sqrt{x}, 0)$ are
  \gls{constructible_point} from
  the set $\{(0, 0), (1, 0), (x, 0), (y, 0)\}$.

  Using this, one can also prove that the converse
  of
  \cref{compass_construction_is_constructible_field}
  holds, i.e. $\fieldadjoin \QQ(x)$ is
  \gls{constructible_field} implies
  $x$ is \gls{constructible_real}.
\end{remark*}

\begin{corollary}
  % Corollary 2.2
  \label{constructible_requires_power_of_2}
  If $x \in \RR$ is \gls{constructible_real} then $x$ is
  \gls{alg_el} over $\QQ$ and $\exdeg[\fieldadjoin \QQ(x) : \QQ]$ is a
  power of 2.
\end{corollary}

\subsubsection*{Some Classical Problems}

\begin{enumerate}[(1)]
  \item \textbf{Squaring the circle:} One
    classical problem is to
    \glsref[constructible_point]{construct} a square
    whose area is the same as that as a circle
    with unit radius. This amounts to
    \glsref[constructible_real]{constructing}
    the real number $\sqrt{\pi}$. This is
    impossible by
    \nameref{constructible_requires_power_of_2}
    and the fact that $\pi$ is \gls{transc_Q}
    (Lindeman).
  \item \textbf{Duplicating the cube:}
    \glsref[constructible_point]{Construct} a
    cube whose volume is twice that of a given
    cube. This amounts to
    \glsref[constructible_real]{construction} of
    $\sqrt[3]{2}$. But $\sqrt[3]{2}$ has
    \gls{min_poly} $X^3 - 2$ so
    $\exdeg[\fieldadjoin \QQ(\sqrt[3]{2}) :
    \QQ] = 3$, which is not a power of 2.
  \item \textbf{Trisecting the angle:} One is to
    divide a given angle into 3 equal angles. Let
    us suppose the given angle is $120^\circ =
    \frac{2\pi}{3}$. Since this is itself
    \gls{constructible_real}, if the trisection problem can
    be solved, then the angle $\frac{2\pi}{9}$ is
    \gls{constructible_point}, in
    other words the real numbers $\cos \left(
    \frac{2\pi}{9} \right)$ and $\sin \left(
    \frac{2\pi}{9} \right)$ are
    \gls{constructible_real}.
    From the formula $\cos 3\theta = 4\cos^3
    \theta - 3\cos\theta$, we see that
    $2\cos\left( \frac{2\pi}{9} \right)$ is a root
    of $f(X) = X^3 - 3X + 1$. Noting that $f(\pm
    1) \neq 0$, and that $f$ is monic, we can use
    Gauss' lemma to observe that $f$ is
    irreducible in $\ringadjoin \QQ[X]$. Therefore
    $\exdeg[\fieldadj{\QQ}{\cos(2\pi/9)} : \QQ] = 3$, which is not
    a power of 2. So this construction is
    impossible.
\end{enumerate}
