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% 09/11/2023 12PM

\newpage
\section{The Galois Group of a Polynomial}

\begin{flashcard}[galois-group-defn]
\begin{definition*}[Galois group of a polynomial]
  \glssymboldefn{galois_group_poly}{Gal$(f / K)$}{Gal$(f /
  K)$}
  \cloze{
  Let $f \in \radj K[X]$ be a \gls{sep_poly} of
  degree $n$. Let $L$ be a \gls{splitting_field}
  for $f$ over $K$. The action of $G = \Gal(L
  \extendover K)$ on the roots $\alpha_1, \ldots,
  \alpha_n$ of $f$ determines an injective group
  homomorphism $\iota : G \to S_n$. It's image is
  the \emph{Galois group of $f$ over $K$}, written
  $\Galpoly(f)$ or $\Galpoly(f \galover K)$.
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[f-irred-iff-gal-transitive]
\begin{lemma}
  % Lemma 9.1
  \label{f_irred_iff_gal_transitive}
  Let $f \in \radj K[X]$ be a \gls{sep_poly}. Then
  \[ \cloze{\text{$f$ irreducible}} \iff
  \cloze{\text{$\Galpoly(f \galover K)$ is
  transitive}} \]
  \cloze{(Recall  $H \subset S_n$ is transitive if
  $\forall i, j \in \{1, \ldots, n\}$, there
  exists $\sigma \in H$ such that $\sigma(i) =
  j$).}
\end{lemma}

\begin{proof}
  \phantom{}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] \cloze{If $f = gh$, $g, h \in
      \radj K[X]$, $\deg g > 0$, $\deg h > 0$ then
      $\Galpoly(f \galover K)$ sends roots of $g$
      to roots of $g$ (and not to roots of $h$),
      and so cannot act transitively on the roots
      of $f$.}
    \item[$\Leftarrow$] \cloze{Without loss of
      generality, $f$ is monic. Let $\alpha \in L$
      be a root of $f$. Then $f$ is the
      \gls{min_poly} of $\alpha$ over $K$. Then by
      \cref{lec12_first_remark},
      \[ \{\sigma(\alpha) : \sigma \in \Gal(L
      \extendover K)\} = \{\text{roots of $f$ in
      $L$}\} \]
      Therefore $\Gal(L \extendover K)$ acts
      transitively on $\alpha_1, \ldots,
      \alpha_n$. Therefore $\Galpoly(f
      \galover K) \subset S_n$ is a transitive
      subgroup.}
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{remark*}[Alternative proof of $\Rightarrow$]
  By \cref{automorphisms_bound}, there exists
  \Kisom[K] $\fadj K(\alpha_i) \cong \fadj
  K(\alpha_j)$, $\alpha_i \mapsto \alpha_j$. This
  extends to an automorphis mof $L$ by
  \nameref{uniqueness_of_splitting_fields}.
\end{remark*}
\vspace{-1em}

Let $f
\in \radj K[X]$ be a monic \gls{sep_poly}
polynomial with roots $\alpha_1, \ldots,
\alpha_n$ in a \gls{splitting_field} $L$. Recall
from \cref{sec_4},
\[ \Disc(f) = \prod_{i < j} (\alpha_i -
\alpha_j)^2 \]

\begin{flashcard}[fixed-field-disc-lemma]
\begin{lemma}
  % Lemma 9.2
  \label{fixed_field_disc_lemma}
  \cloze{Assume $\characteristic K \neq 2$. Let
  $\Delta = \Disc(f)$. }The \gls{fixed_field} of
  $\Galpoly(f \galover K) \cap A_n$ is \cloze{$\fadj
  K(\sqrt{\Delta})$. In particular}
  \[ \cloze{\Galpoly(f \galover K) \subset A_n \iff
  \text{$\Delta$ is a square in $K$}} \]
\end{lemma}

\begin{proof}
  \cloze{The sign of a permutation $\pi \in S_n$ is
  defined so that (as an identiy in $\radj
  \ZZ[X_1, \ldots, X_n]$) we have
  \[ \prod_{i < j} (X_{\pi(i)} - X_{\pi(j)}) =
  \sign(\pi) \prod_{i < j} (X_i - X_j) \]
  We put $\delta = \prod_{i < j} (\alpha_i -
  \alpha_j)$ so that $\delta^2 = \Delta$. So if
  $\sigma \in G = \Galpoly(f \galover K) =
  \Gal(L \extendover K)$ then
  \[ \sigma(\delta) = \sign(\sigma) \delta .\]
  As $f$ is \gls{sep_poly} and $\characteristic K
  \neq 2$, $\delta \neq -\delta$. Therefore
  \begin{align*}
    G \cap A_n
    &= \{\sigma \in G \st \sigma(\sigma) = 1\} \\
    &= \{\sigma \in G \st \sigma(\delta) =
    \delta\} \\
    &= \Gal(L \extendover \fadj K(\delta))
  \end{align*}
  Therefore $\fixedf L^{G \cap A_n} = \fadj
  K(\delta)$. In particular,
  \begin{align*}
    G \subset A_N
    &\iff G \cap A_n = G \\
    &\iff \fadj K(\sqrt{\Delta}) = K \\
    &\iff \text{$\Delta$ is a square in $K$}
    \qedhere
  \end{align*}}
\end{proof}
\end{flashcard}

\begin{remark*}
  $G = \Galpoly(g \galover K) \subset S_n$ is
  really only defined up to conjugacy, since if we
  reorder $\alpha_1, \ldots, \alpha_n$ using
  $\sigma \in S_n$ then $G$ changes to $\sigma
  G\sigma^{-1}$. But we \emph{can} distinguish
  between
  \[ \langle (12), (34) \rangle \subset S_4 \qquad
  \text{and} \qquad \langle (12)(34), (13)(24)
  \rangle \subset S_4 \]
  even though both are isomorphic to $C_2 \times
  C_2$.
\end{remark*}
\vspace{-1em}

What is $G \hookrightarrow S_n$ up to conjugacy?
\begin{itemize}
  \item $n = 2$: The only transitive subgroup of
    $S_2$ is itself.
  \item $n = 3$: The transitive subgroups of $S_3$
    are $S_3$ and $A_3 \cong C_3$. So if $f \in
    \radj K[X]$ is irreducible then $\Galpoly(f
    \galover K)$ is $A_3$ or $S_3$. By
    \cref{fixed_field_disc_lemma}, $\Galpoly(f
    \galover K) = A_3$ if and only if $\Disc(f)$
    is a square in $K$. Taking $n = 3$ in
    \es[3]{2} gives
    \begin{flashcard}[cubic-disc-formula]
    \[ \Disc(X^3 + aX + b) = \cloze{-4a^3 - 27b^2} \]
    \end{flashcard}
    (LEARN THIS FORMULA).
    \begin{example*}
      \vspace{0.8em}
      $f(X) = X^3 - 3X + 1$ (see \cref{sec2} and
      \es{1}).
      \[ \Disc(f) = -4(-3)^3 - 27 = 81 = 9^2 \]
      Therefore $\Galpoly(f \galover \QQ)$ is $1$ or
      $A_3$. We checked in \cref{sec2} that $f$ is
      irreducible over $\QQ$, so therefore $\Galpoly(f
      \galover \QQ) = A_3$.
    \end{example*}
  \item $n = 4$: The transitive subgroups of $S_4$
    are
    \[ S_4, A_4, D_8, C_4, V \cong C_2 \times C_2 \]
    $S_4$, $A_4$, $V$ are normal subgroups where
    \[ V = \{\id, (12)(34), (13)(24), (14)(23)\} \]
    There are 3 conjugate copies of each of $C_4$
    and $D_8$.

    Let $S_4$ act on $V \setminus \{\id\}$ by
    conjugation since $g(12)(34)g^{-1} =
    (g(1)g(2))(g(3)g(4))$ it would be equivalent
    to let $S_4$ act on the set of ways of
    partitioning the set $\{1, 2, 3, 4\}$ into 2
    subsets of size $2$. The corresponding
    permutation representation is a group
    homomorphism $\pi : S_4 \to S_3$. If $H =
    \{\sigma \in S_4 \st \sigma(1) = 1\} = \langle
    (234), (23) \rangle \subset S_4$ then $\pi|_H
    : H \to S_3$ is an isomorphism. So $\pi$ is
    surjective and $\#\ker\pi = 4$. $V$ abelian
    $\implies$ $V \subset \ker\pi$. Hence $V =
    \ker \pi$.

    If $G \subset S_4$ then applying the
    isomorphism theorem to $\pi|_G$ gives $G /
    G \cap V \cong \pi(G) \subset S_4$.
    \begin{center}
      \begin{tabular}{c|c}
        transitive subgroup $G \subset S_4$ &
        $\pi(G) \subset S_3$ \\
        \hline
        $S_4$ & $S_3$ \\
        $A_4$ & $A_3$ \\
        \text{$C_4$ or $D_8$} & $C_2$ \\
        $V$ & $1$
      \end{tabular}
    \end{center}
\end{itemize}