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\section{Galois Extensions}

\begin{flashcard}[automorphism-defn]
\begin{definition*}[automorphism]
  \glsnoundefn{automorph}{automorphism}{automorphisms}
  \glssymboldefn{automorph}{Aut}{Aut}
  \cloze{
  An \emph{automorphism} of a \gls{field} $L$ is a
  bijective homomorphism $\sigma : L \to L$. We
  write $\Aut(L)$ for the group of automorphisms
  of $L$ under composition, i.e.
  \[ (\sigma \tau)(x) = \sigma(\tau(x)) .\]
  }
\end{definition*}
\end{flashcard}

\textbf{Exercise:} Check inverses (i.e. check that
$\sigma^{-1}$ is a homormorphism).

\begin{flashcard}[Kautomorph-notation]
\begin{definition*}
  \glsnoundefn{Kautomorph}{$K$-automorphism}{$K$-automorphisms}
  \glssymboldefn{Kautomorph}{$Aut(L / K)$}{$Aut(L
  / K)$}
  Let $L \extendover K$ be a \gls{field_ex}. A
  $K$-automorphism of $L$ is an \gls{automorph}
  $\sigma \in \Aut(L)$ whose restriction to $K$ is
  the identity map. The $K$-automorphisms of $L$
  form a subgroup $\cloze{\KAut(L \extendover K)} \subset
  \Aut(L)$.
\end{definition*}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\Aut(\QQ)$ and $\Aut(\FF_p)$ are both
      trivial. Therefore $\Aut(L) = \KAut(L
      \extendover K)$
      where $K$ is the \gls{prime_subfield} of $L$.
    \item If $\exdeg[L : K] < \infty$ then any
      \Kembedding{K} $L \to L$ is surjective (by
      rank-nullity), i.e.
      \[ \Hom_k(L, L) = \KAut(L \extendover K) \]
  \end{enumerate}
\end{remark*}

\begin{flashcard}[finite-automorphisms-bound]
\begin{lemma}
  % Lemma 6.1
  \label{finite_automorph_bound}
  Let $L \extendover K$ be a \gls{finite_ex}
  \gls{ex}. Then
  \[ \#\Aut(L \extendover K) \le \cloze{\exdeg[L :
  K]} \]
\end{lemma}

\begin{proof}
  Take $M = L$ in \cref{hom_bound_general_thm}.
\end{proof}
\end{flashcard}

\begin{flashcard}[aut-fixed-points-defn]
\begin{definition*}
  \glsnoundefn{fixed_field}{fixed field}{fixed
  fields}
  \glssymboldefn{fixed_field}{$L^S$}{$L^S$}
  If $S \subset \Aut(L)$ is any subset we define
  the \emph{fixed field} of $S$ to be
  \[ \cloze{L^S = \{x \in L \st \sigma(x) = x
  ~\forall \sigma \in S\}} \]
  This is a subfield of $L$.
\end{definition*}
\end{flashcard}

\begin{flashcard}[galois-extension-defn]
\begin{definition*}
  \glsadjdefn{galois}{Galois}{\gls{ex}}
  A \gls{field_ex} $L \extendover K$ is
  \emph{Galois} if \cloze{it is \gls{alg_ex} and
  \[ K = \fixedf L^{\KAut(L \extendover K)} \]}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\KAut(\CC \extendover \RR) = \{1,
      \tau\}$ where $\tau$ is complex conjugation.
      If $z \in \CC$, then
      \[ z \in \RR \iff \tau(z) = z \]
      therefore $\CC \extendover \RR$ is
      \gls{galois}.
    \item Let $L = \fadj \QQ(\sqrt{2})$. $f(X) =
      X^2 - 2$.
      \[ \KAut(L \extendover \QQ) \leftrightarrow
      \{\text{roots of $f$ in $L$}\} \]
      Therefore $\KAut(L \extendover \QQ) = \{1,
      \tau\}$ where $\tau : \fadj \QQ(\sqrt{2})
      \to \fadj \QQ(\sqrt{2})$, $a + b\sqrt{2}
      \mapsto a - b\sqrt{2}$, $a, b \in \QQ$.
      \begin{align*}
        \fixedf L^\tau
        &= \{a + b\sqrt{2} \st a + b\sqrt{2} = a -
        b\sqrt{2}\} \\\
        &= \{a + b\sqrt{2} \st b = 0\} \\
        &= \QQ
      \end{align*}
      Therefore $L \extendover \QQ$ is
      \gls{galois}.
    \item Let $L = \fadj \QQ(\sqrt[3]{2})$, $f(X)
      = X^3 - 2$. Then
      \[ \KAut(L \extendover \QQ) \leftrightarrow
      \{\text{roots of $f$ in $L$}\} \]
      Since $L \subset \RR$ we see that $\#\KAut(L
      \extendover \QQ) = 1$. Therefore $L
      \extendover \QQ$ is not \gls{galois}.
    \item Let $K \extendover \FF_p$ be a
      \gls{finite_ex} \gls{ex} ($\implies |K| <
      \infty$). Let $\phi : K \to K$, $x \mapsto
      x^p$. By \cref{frob_hom_prop}, $\phi \in
      \KAut(K \extendover \FF_p)$. Then
      \begin{align*}
        \fixedf K^\phi
        &= \{x \in K \st \phi(x) = x\} \\
        &= \{\text{roots of $X^p - X$ in $K$}\} \\
        &\supset \FF_p
      \end{align*}
      (with equality in the $\supset$ by
      \cref{roots_bound_lemma}). Therefore $K
      \extendover \FF_p$ is \gls{galois}.
  \end{enumerate}
\end{example*}

\begin{flashcard}[classification-of-finite-galois-extensions-thm]
\begin{theorem}[Classification of finite Galois
  extensions]
  % Theorem 6.2
  \label{classification_of_finite_galois_extensions}
  Let $\exdeg[L : K] < \infty$ and $G = \KAut(L
  \extendover K)$. Then the following are
  equivalent:
  \begin{enumerate}[(i)]
    \item \cloze{$L \extendover K$ is \gls{galois}, i.e.
      $K = \fixedf L^G$.}
    \item \cloze{$L \extendover K$ is
      \gls{norm_ex} and \gls{sep_ex}.}
    \item \cloze{$L$ is the \gls{splitting_field}
      of a \gls{sep_poly} over $K$.}
    \item \cloze{$\#G = \exdeg[L : K]$ (i.e.
      equality holds in \cref{finite_automorph_bound}).}
  \end{enumerate}
\end{theorem}

\begin{proof}
  \begin{enumerate}[(i) $\implies$ (ii)]
    \item[(i) $\implies$ (ii)] \cloze{Let $\alpha \in L$
      and $\{\sigma(\alpha) : \sigma \in G\} =
      \{\alpha_1, \ldots, \alpha_m\}$ with
      $\alpha_1, \ldots, \alpha_m$ distinct. Let
      $f(X) = \prod_{i = 1}^m (X - \alpha_i)$. We
      let $\sigma \in G$ act on $\radj L[X]$ via
      \[ \sigma \left( \sum c_i X^i \right) = \sum
      \sigma(c_i) X^i \]
      Since $G$ permutes the $\alpha_i$ we have
      $\sigma f = f ~\forall \sigma \in G$. $L
      \extendover K$ \gls{galois} implies $f \in
      \radj K[X]$. Let $g$ be the \gls{min_poly}
      of $\alpha$ over $K$. Since $f(\alpha) = 0$
      we have $g \mid f$. Since $g(\sigma(\alpha))
      = \sigma(g(\alpha)) = 0 ~\forall \sigma \in
      G$, every root of $f$ is a root of $g$. By
      construction $f$ is \gls{sep_poly} and
      monic, so $f = g$. Therefore the
      \gls{min_poly} of $\alpha$ over $K$ splits
      into distinct linear factors over $L$. Since
      $\alpha \in L$ is arbitrary, this shows
      that $L \extendover K$ is \gls{norm_ex} and
      \gls{sep_ex}.}
    \item[(ii) $\implies$ (iii)] \cloze{By
      \cref{hom_bound_general_thm}, $L$ is the
      \gls{splitting_field} of some $f \in \radj
      K[X]$. Write $f = \prod_{i = 1}^m f_i^{e_i}$
      where ther $f_i \in \radj K[X]$ are distinct
      and irreducible, and $e_i \ge 1$. Since $L
      \extendover K$ is \gls{sep_ex}, each $f_i$
      is \gls{sep_poly}. Moreover, $\gcd(f_i, f_J)
      = 1$ in $\radj K[X]$ so by
      \cref{gcd_same_in_extension_lemma},
      $\gcd(f_i, f_j) = 1$ in $\radj L[X]$.
      Therefore $g = \prod_{i = 1}^m f_i$ is
      \gls{sep_poly}, and $L$ is a
      \gls{splitting_field} for $g$ over $K$.}
    \item[(iii) $\implies$ (iv)] \cloze{Let $L$ be the
      \gls{splitting_field} of a \gls{sep_poly} $f
      \in \radj K[X]$. Then $L = \fadj K(\alpha_1,
      \ldots, \alpha_n)$ where $\alpha_1, \ldots,
      \alpha_n$ are the roots of $f$. The
      \gls{min_poly} $f_i$ of each $\alpha_i$
      divides $f$ and so splits into distinct
      linear factors over $L$. Taking $M = L$ in
      \cref{hom_bound_general_thm} gives
      $\#\KAut(L \extendover K) = \exdeg[L : K]$.}
    \item[(iv) $\implies$ (i)] \cloze{$G \subset \KAut(L
      \extendover \fixedf L^G) \subset \KAut(L
      \extendover K) = G$. Therefore $G = \KAut(L
      \extendover \fixedf L^G)$.
      \[ \implies \exdeg[L : K] \stackrel{\text{by
      (iv)}}{=} \#G = \#\KAut(L
      \extendover \fixedf L^G)
      \stackrel{\text{\cref{finite_automorph_bound}}}{\le} \exdeg[L :
      \fixedf L^G] \]
      So by \nameref{tower_law},
      \[ \cancel{\exdeg[L : \fixedf L^G]}
      \exdeg[\fixedf L^G : K] \le \cancel{\exdeg[L
      : L^G]} \]
      so $\fixedf L^G = K$.
      } \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[galois-group-defn]
\begin{definition*}[Galois group]
  \glsnoundefn{galois_group}{Galois group}{Galois
  groups}
  \glssymboldefn{galois_group}{Gal}{Gal}
  \cloze{
  If $L \extendover K$ is a \gls{galois} \gls{ex}
  then we write $\Gal(L \extendover K)$ for
  $\KAut(L \extendover K)$ (we call this the
  \emph{Galois group of $L$ over $K$}).
  }
\end{definition*}
\end{flashcard}