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\begin{remark}
  % Remark 6.3
  \label{lec12_first_remark}
  We saw in the proof of (i) $\implies$ (ii) that
  if $L \extendover K$ is \gls{galois}, $G =
  \Gal(L \extendover K)$, and $\alpha
  \in L$ then the \gls{min_poly} of $\alpha$ over
  $K$ is
  \[ \prod_{i = 1}^m (X - \alpha_i) \]
  where $\alpha_1, \ldots, \alpha_m$ are the
  distinct elements of $\operatorname{ord}_G(\alpha) =
  \{\sigma(\alpha) : \sigma \in G\}$.
\end{remark}

\begin{flashcard}[ft-of-galois-theory]
\begin{theorem}[Fundamental Theorem of Galois
  Theory]
  % Theorem 6.4
  \label{ft_of_gal}
  \cloze{Let $L \extendover K$ be a \gls{finite_ex}
  \gls{galois} \gls{ex}, $G = \Gal(L \extendover
  K)$.}
  \begin{enumerate}[(a)]
    \item \cloze{Let $F$ be an intermediate
      \gls{field}, i.e.
      $K \subset F \subset L$. Then $L
      \extendover F$ is \gls{galois} and $\Gal(L
      \extendover F)$ is a subgroup of $G$.}
    \item \cloze{There is an inclusion reversing
      bijection
      \begin{align*}
        \{\text{intermediate \glspl{field} $K \subset F
        \subset L$}\}
        &\to \{\text{subgroups $H
        \subset G$}\} \\
        F
        &\mapsto \Gal(L \extendover F) \\
        \fixedf L^H
        &\mapsfrom H
      \end{align*}
      }
    \item \cloze{Let $F$ be an intermediate \gls{field},
      i.e. $K \subset F \subset L$. Then
      \begin{align*}
        \text{$F \extendover K$ \gls{galois}}
        &\iff \sigma F = F ~\forall \sigma \in G
        \\
        &\iff \text{$H = \Gal(L \extendover F)$ is
        a normal subgroup of $G$}
      \end{align*}
      In this case the restriction map
      \begin{align*}
        G
        &\to \Gal(F \extendover K) \\
        \sigma
        &\mapsto \sigma|_F
      \end{align*}
      is surjective with kernel $H$, and so
      \[ \Gal(F \extendover K) \cong G / H \]
      (a quotient of $G$).
      }
  \end{enumerate}
\end{theorem}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(a)]
    \item \cloze{By
      \cref{classification_of_finite_galois_extensions},
      $L$ is the \gls{splitting_field} over $K$ of
      some \gls{sep_poly} polynomial $f \in \radj
      K[X]$. Then $L$ is the \gls{splitting_field}
      of $f$ over $F$. So $L \extendover F$ is
      \gls{galois}. $\Gal(L \extendover F)$ is a
      subgroup of $\Gal(L \extendover K)$ since
      any \gls{automorph} of $L$ acting as the
      identity on $F$ also acts as the identity on
      $K$.}
    \item \cloze{To show we have a bijection, we need to
      check both compositions are the identity.
      \begin{enumerate}[(i)]
        \item $F = \fixedf L^{\Gal(L \extendover
          F)}$: This holds since $L \extendover F$
          is \gls{galois}.
        \item $\Gal(L \extendover \fixedf L^H) =
          H$: we certainly have $H \subset \Gal(L
          \extendover \fixedf L^H)$ so it suffices
          to show $\#\Gal(L \extendover \fixedf
          L^H) \le \#H$. Let $F = \fixedf L^H$. As
          $L \extendover F$ is \gls{finite_ex} and
          \gls{sep_ex}, the \nameref{prim_el_thm}
          tells us that $L = \fadj F(\alpha)$ for
          some $\alpha \in L$. Then $\alpha$ is a
          root of
          \[ f(X) = \prod_{\sigma \in H} (X -
          \sigma(\alpha)) \]
          which has coefficients in $\fixedf L^H =
          F$. Therefore $\#\Gal(L \extendover
          \fixedf L^H) = \exdeg[L : \fixedf L^H] =
          \exdeg[\fadj F(\alpha) : F] \le \deg f =
          \#H$.
      \end{enumerate}
      If $F_1 \subset F_2$ then $\Gal(L
      \extendover F_2) \subset \Gal(L \extendover
      F_1)$, so the bijection reverses
      inclusions.}
    \item \cloze{We first show
      \[ \text{$F \extendover K$ \gls{galois}} \iff
      \sigma F = F ~\forall \sigma \in G .\]
      \begin{enumerate}[$\Rightarrow$]
        \item[$\Rightarrow$] Let $\alpha \in F$
          have \gls{min_poly} $f$ over $K$. For
          any $\sigma \in G$, $\sigma(\alpha)$ is
          a root of $f$. Since $F \extendover K$
          is \gls{norm_ex} we have $\sigma(\alpha)
          \in F$, so $\sigma F \subset F$. As
          $\exdeg[\sigma F : K] = \exdeg[F : K]$,
          it follows that $\sigma F = F$.
        \item[$\Leftarrow$] Let $\alpha \in F$. By
          \cref{lec12_first_remark}, its
          \gls{min_poly} over $K$ is
          \[ f(X) = \prod_{i = 1}^m (X - \alpha_i) \]
          where $\alpha_1, \ldots, \alpha_m$ are
          the distinct elements of
          $\{\sigma(\alpha) : \sigma \in G\}$. The
          assumption $\sigma(F) = F ~\forall
          \sigma \in G$ tells us that $\alpha_1,
          \ldots, \alpha_m \in F$. This shows $F
          \extendover K$ is \gls{norm_ex}. But
          also, $L \extendover K$ \gls{galois}, so
          so $L \extendover K$ is \gls{sep_ex}, so
          $F \extendover K$ is \gls{sep_ex}. Hence
          $F \extendover K$ is \gls{norm_ex} and
          \gls{sep_ex}, so by
          \cref{classification_of_finite_galois_extensions},
          $F \extendover K$ is \gls{galois}.
      \end{enumerate}
      Suppose $H \subset G$ corresponds to $F =
      \fixedf L^H$. For $\sigma \in G$,
      \begin{align*}
        \fixedf L^{\sigma H\sigma^{-1}}
        &= \{x \in
        L \st \sigma \tau \sigma^{-1}(x) = x
        ~\forall \tau \in H\} \\
        &= \{x \in L \st \tau \sigma^{-1}(x) =
        \sigma^{-1}(x) ~\forall \tau \in H\} \\
        &= \{x \in L \st \sigma^{-1}(x) \in
        \fixedf L^H = F\} \\
        &= \sigma F
      \end{align*}
      So
      \begin{align*}
        \sigma F = F ~\forall \sigma \in G
        &\iff \fixedf L^{\sigma H\sigma^{-1}} =
        \fixedf L^H ~\forall \sigma \in G \\
        &\iff \sigma H \sigma^{-1} = H ~\forall
        \sigma \in G \\
        &\iff \text{$H \subset G$ is a normal
        subgroup}
      \end{align*}
      Consider the restriction map
      \begin{align*}
        G = \Gal(L \extendover K)
        &\mapsto \Gal(F \extendover K) \\
        \sigma
        &\mapsto \sigma|_F
      \end{align*}
      Then
      \[ \Ker(\operatorname{res}) = \{\sigma
      \in \Gal(L \extendover K) \st \sigma(x) = x
      ~\forall x \in F\} = \Gal(L \extendover F) = H \]
      Therefore $G / H \cong
      \Im(\operatorname{res}) \le \Gal(F
      \extendover K)$. But,
      \[ \#(G / H) = \frac{\#G}{\#H} =
      \frac{\exdeg [L : K]}{\exdeg [L : F]} =
      \exdeg [F : K] = \#\Gal(F \extendover K) .\]
      Therefore $\operatorname{res}$ is surjcetive
      and $\Gal(F \extendover K) \cong G / H$.
      } \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{example}
  Let $K = \fadj \QQ(\sqrt{2}, \sqrt{3})$. $K
  \extendover \QQ$ is the \gls{splitting_field} of
  the polynomial $(X^2 - 2)(X^2 - 3)$. Therefore
  $K \extendover \QQ$ is \gls{norm_ex}.
  \glsref[sep_ex]{Separability} is automatic in
  $\characteristic = 0$, hence $K \extendover \QQ$
  is \gls{galois}. If $\sigma \in \Gal(K
  \extendover \QQ)$, then it is uniquely
  determined by $\sigma(\sqrt{2})$ and
  $\sigma(\sqrt{3})$. Since $\sigma(\sqrt{2}) =
  \pm \sqrt{2}$ and $\sigma(\sqrt{3}) = \pm
  \sqrt{3}$, we have
  \[ \#\Gal(K \extendover \QQ) \le 4 .\]
  We saw in \cref{lec3_last_example} that $\exdeg
  [K : \QQ] = 4$. Hence $\#\Gal(K \extendover \QQ)
  = 4$. Let
  \[ \sigma : \sqrt{2} \mapsto \sqrt{2}; \sqrt{3}
  \mapsto -\sqrt{3} \]
  \[ \tau : \sqrt{2} \mapsto -\sqrt{2}; \sqrt{3}
  \mapsto \sqrt{3} \]
  Then $\sigma^2 = \tau^2 = \id$ and $\sigma \tau
  = \tau \sigma$, so $\Gal(K \extendover \QQ)
  \cong C_2 \times C_2$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/753fa742a0a149f6.png}
  \end{center}
\end{example}