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% 02/11/2023 14PM

\begin{example}
  % Example 6.6
  \label{example_6_6}
  Let $K = \fadj \QQ(\alpha)$ where $\alpha =
  \sqrt{2 + \sqrt{2}}$. Then $(\alpha^2 - 2)^2 =
  2$, so $\alpha$ is a root of $f(X) = X^4 - 4X^2
  + 2$. This is irreducible in $\radj \ZZ[X]$ by
  Eisenstein's criterion with $p = 2$. Therefore
  it is irreducible in $\radj \QQ[X]$ by Gauss'
  Lemma, so $\exdeg [K : \QQ] = 4$.

  Now $(2 + \sqrt{2}) (2 - \sqrt{2}) = 2$, so $f$
  has roots $\pm \alpha, \pm
  \frac{\sqrt{2}}{\alpha}$ (note $\sqrt{2} =
  \alpha^2 - 2$). Therefore $K$ is a
  \gls{splitting_field} for $f$ over $\QQ$, so $K
  \extendover \QQ$ is \gls{norm_ex} hence
  \gls{galois} (since
  \glsref[sep_ex]{separability} is automatic in
  $\characteristic 0$).

  If $\sigma \in \Gal(K \extendover \QQ)$ then it
  is uniquely determined by $\sigma(\alpha)$. But
  $\sigma(\alpha) \in \{\pm \alpha, \pm \sqrt{2} /
  \alpha\}$, and $\#\Gal(K \extendover \QQ) =
  \exdeg [K : \QQ] = 4$, so all possibilities must
  occur (could see this more directly using
  \cref{automorphisms_bound}). We fix $\sigma \in
  \Gal(K \extendover \QQ)$ with $\sigma(\alpha) =
  \frac{\sqrt{2}}{\alpha}$, hence
  $\sigma(\alpha^2) = \frac{2}{\alpha^2}$, so
  $\sigma(2 + \sqrt{2}) = 2 - \sqrt{2}$, so
  $\sigma(\sqrt{2}) = -\sqrt{2}$.

  Therefore
  \[ \sigma^2(\alpha) = \sigma \left(
  \frac{\sqrt{2}}{\alpha} \right) =
  -\frac{\sqrt{2}}{\left( \frac{\sqrt{2}}{\alpha}
  \right)} = -\alpha \]
  Therefore $\sigma^2 \neq \id$, but $\sigma^4 =
  \id$. So
  \[ \Gal(K \extendover \QQ) \cong C_4 \]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/0952bc9405d640b4.png}
  \end{center}
\end{example}

\begin{flashcard}[composite-subfield-defn]
\begin{definition*}[Composite subfield]
  \glsadjdefn{comp_field}{composite}{\gls{field}}
  \glssymboldefn{comp_field}{$L_1 L_2$}{$L_1 L_2$}
  \cloze{
  Let $L_1$, $L_2$ be sub\glspl{field} of a
  \gls{field} $M$. The \emph{composite} $L_1 L_2$
  is the smallest sub\gls{field} of $M$ to contain both
  $L_1$ and $L_2$. (This exists since the
  intersection of any collection of
  sub\glspl{field} is a
  sub\gls{field}).
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[composite-subfield-galois-thm]
\begin{theorem}
  % Theorem 6.7
  \label{comp_subf_gal_thm}
  Let $\exdeg [M : k] < \infty$ and let $L_1$,
  $L_2$ be intermediate \glspl{field} i.e. $K \subset L_i
  \subset M$ for $i = 1, 2$.
  \begin{enumerate}[(i)]
    \item \cloze{If $L_1 \extendover K$ is \gls{galois}
      then $\compfield{L_1}{L_2} \extendover L_2$ is
      \gls{galois} and there is an injective group
      homomorphism
      \[ \Gal(\compfield{L_1}{L_2} \extendover L_2)
      \hookrightarrow \Gal(L_1 \extendover K) \]
      This is surjective if and only if $L_1 \cap
      L_2 = K$.
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/fd6a9a0d2b084ef1.png}
      \end{center}
      }
    \item \cloze{If $L_1 \extendover K$ and $L_2
      \extendover K$ are both \gls{galois} then
      $\compfield{L_1}{L_2} \extendover K$ is
      \gls{galois} and there is an injective group
      homomorphism
      \[ \Gal(\compfield{L_1}{L_2} \extendover K)
      \hookrightarrow \Gal(L_1 \extendover K)
      \times \Gal(L_1 \extendover K) \]
      This is surjective if and only if $L_1 \cap
      L_2 = K$.}
  \end{enumerate}
\end{theorem}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item \cloze{$L_1 \extendover K$ \gls{galois}
      $\implies$ $L_1$ is the
      \gls{splitting_field} of some \gls{sep_poly}
      polynomial $f \in \radj K[X]$. Then
      $\compfield{L_1}{L_2}$ is a
      \gls{splitting_field} for $f$ over $L_2$.
      Therefore $\compfield{L_1}{L_2} \extendover
      L_2$ is \gls{galois}. If $\sigma \in
      \Gal(\compfield{L_1}{L_2} \extendover L_2)$,
      then $\sigma|_K = \id$ and since $L_1
      \extendover K$ is \gls{norm_ex} we have
      $\sigma(L_1) \subset L_1$. We consider the
      group homomorphism
      \begin{align*}
        \Gal(\compfield{L_1}{L_2} \extendover L_2)
        &\stackrel{\operatorname{res}}{\longrightarrow}
        \Gal(L_1 \extendover K) \\
        \sigma
        &\longmapsto \sigma|_{L_1}
      \end{align*}
      It is injective since if $\sigma|_{L_1} =
      \id$ then $\sigma$ acts trivially on both
      $L_1$ and $L_2$ and hence on
      $\compfield{L_1}{L_2}$. Now suppose $L_1
      \cap L_2 = K$. Then $L_1 \extendover K$ is
      \gls{finite_ex} and \gls{sep_ex}. So by
      \nameref{prim_el_thm}, $L_1 = \fadj
      K(\alpha)$ for some $\alpha \in L_1$. Let $f
      \in \radj K[X]$ be the \gls{min_poly} of
      $\alpha$ over $K$. Suppose $f = gh$ for some
      $g, h \in \radj L_2[X]$, $\deg g, \deg h >
      0$. Then $f$ splits into linear factors over
      $L_1$, so $g$ and $h$ have coefficients in
      $L_1 \cap L_2 = K$, which contradicts the
      fact that $f$ is irreducible over $K$.
      Therefore $f$ is irreducible in $\radj
      L_2[X]$. Therefore
      \[ \exdeg [L_1 : K] = \deg f = \exdeg [L_1 L_2 :
      L_2] \]
      The map $\operatorname{res}$ is therefore an
      isomorphism. Conversely, if
      $\Im(\operatorname{res}) \subset \Gal(L_1
      \extendover L_1 \cap L_2) \subset \Gal(L_1
      \extendover K)$. So if $\operatorname{res}$
      is surjective then $L_1 \cap L_2 = K$.}
    \item \cloze{$L_i \extendover K$ \gls{galois}
      $\implies$ $L_i$ is a \gls{splitting_field}
      of some \gls{sep_poly} polynomial $f_i \in
      \radj K[X]$. Then $\compfield{L_1}{L_2}$ is
      the \gls{splitting_field} of the
      \gls{sep_poly} $\lcm(f_1, f_2)$. Therefore
      $\compfield{L_1}{L_2} \extendover K$ is
      \gls{galois}.
      \begin{align*}
        \text{It is surjective}
        &\iff \exdeg[\compfield{L_1}{L_2} : K]
        = \exdeg [L_1 : k] \exdeg[L_2 : K] \\
        &\iff [\compfield{L_1}{L_2} : L_2]
        \cancel{\exdeg[L_2 : K]}
        = \exdeg [L_1 : K] \cancel{[L_2 : K]} \\
        &\stackrel{\text{(i)}}{\iff} L_1 \cap L_2
        = K \qedhere
      \end{align*}
      }
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[galois-closure-existence-thm]
\begin{theorem}
  \glsnoundefn{gal_clos}{Galois closure}{Galois
  closures}
  Let $L \extendover K$ be \gls{finite_ex} and
  \gls{sep_ex}. Then there exists a
  \gls{finite_ex} \gls{ex} $M \extendover L$ such
  that
  \begin{enumerate}[(i)]
    \item \cloze{$M \extendover K$ is
      \gls{galois}.}
    \item \cloze{If $L \subset M' \subset M$ and $M'
      \extendover K$ is \gls{galois} then $M' =
      M$.}
  \end{enumerate}
  We say $M \extendover K$ is a \emph{Galois
  closure} of $L \extendover K$.
\end{theorem}

\begin{proof}
  \cloze{
  By \nameref{prim_el_thm}, $L = \fadj K(\alpha)$
  for some $\alpha \in L$. Let $f$ be the
  \gls{min_poly} of $\alpha$ over $K$. Then $L
  \extendover K$ \gls{sep_ex} implies $f$ is
  \gls{sep_poly}. Let $M$ be a
  \gls{splitting_field} for $f$ over $L$. Since $L
  = \fadj K(\alpha)$ where $\alpha$ is a root of
  $f$, it follows that $M$ is a
  \gls{splitting_field} of $f$ over $K$. Now
  \cref{classification_of_finite_galois_extensions}
  implies that $M \extendover K$ is \gls{galois}.
  Let $M'$ as (ii). As $\alpha \in M'$ and $M'
  \extendover K$ is \gls{norm_ex}, $f$ splits into
  linear factors over $M'$. Hence $M' = M$.
  }
\end{proof}
\end{flashcard}