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% 11/11/2023 12PM

\vspace{-2em}
\begin{itemize}
  \item[] Let $f(X) = \prod_{i = 1}^4 (X -
    \alpha_i)$ be a monic quartic polynomial.
    Define
    \begin{align*}
      \beta_1
      &= (\alpha_1 + \alpha_2)(\alpha_3 +
      \alpha_4) \\
      \beta_2
      &= (\alpha_1 + \alpha_3)(\alpha_2 +
      \alpha_4) \\
      \beta_3
      &= (\alpha_1 + \alpha_4)(\alpha_2 + \alpha_3)
    \end{align*}
    \begin{definition*}
      \vspace{0.8em}
      \glsnoundefn{resolvent}{resolvent cubic}{N/A}
      The \emph{resolvent cubic} is
      \[ g(X) = \prod_{i = 1}^3 (X - \beta_i) .\]
    \end{definition*}
    \begin{theorem}
      % Thereom 9.3
      \label{resolvent_facts}
      \vspace{0.8em}
      Let $f, g$ as above.
      \begin{enumerate}[(i)]
        \item If $f \in \radj K[X]$ then $g \in
          \radj K[X]$.
        \item If $f$ is \gls{sep_poly} then $g$ is
          \gls{sep_poly}.
        \item If (i) and (ii) hold then
          \[ \pi(\Galpoly(f \galover K)) =
          \Galpoly(g \galover K) \]
          In particular if $f \in \radj K[X]$ is
          irreducible then $\Galpoly(g \galover
          K)$ determines $\Galpoly(f \galover K)$
          up to ambiguity between $C_4$ and $D_8$
          when $\#\Galpoly(g \galover K) = 2$.
      \end{enumerate}
    \end{theorem}
    \begin{proof}
      \phantom{}
      \begin{enumerate}[(i)]
        \item More generally each coefficient of
          $g$ is a symmetric polynomial in $\radj
          \ZZ[\beta_1, \beta_2, \beta_3]$, hence a
          symmetric polynomial in $\radj
          \ZZ[\alpha_1, \alpha_2, \alpha_3,
          \alpha_4]$ and so by the
          \nameref{symm_func_thm}, is a
          $\ZZ$-coefficient polynomial in the
          coefficients of $f$.
        \item \begin{align*}
            \beta_1 - \beta_2
            &= \alpha_1 \alpha_3
            + \cancel{\alpha_1 \alpha_4}
            + \cancel{\alpha_2 \alpha_3}
            + \alpha_2 \alpha_4
            - \alpha_1 \alpha_2
            - \cancel{\alpha_1 \alpha_4}
            - \cancel{\alpha_2 \alpha_3}
            - \alpha_3 \alpha_4 \\
            &= \alpha_1 \alpha_3
            + \alpha_2 \alpha_4
            - \alpha_1 \alpha_2
            - \alpha_3 \alpha_4 \\
            &= (\alpha_1 - \alpha_4)(\alpha_3 -
            \alpha_2)
          \end{align*}
          If $f$ \gls{sep_poly} then $\alpha_1,
          \ldots, \alpha_4$ are distinct, hence
          $\beta_1 \neq \beta_2$. Similar
          calculation shows $\beta_1, \beta_2,
          \beta_3$ are all distinct. Hence $g$ is
          \gls{sep_poly}.
        \item Let $M$ be a \gls{splitting_field}
          of $f$ over $K$. Let $\alpha_1, \ldots,
          \alpha_4 \in M$ be the roots of $f$.
          Then $L \defeq \fadj K(\beta_1, \beta_2,
          \beta_3) \subset M$ is a
          \gls{splitting_field} for $g$ over $K$.
          If an element of $\Gal(M \extendover K)$
          permutes $\alpha_1, \ldots, \alpha_4$
          according to $\sigma \in S_4$, then it
          restricts to an element of $\Gal(L
          \extendover K)$ permuting $\beta_1,
          \beta_2, \beta_3$ according to
          $\pi(\sigma)
          \in S_3$. In other words, there is a
          commutative diagram
          \begin{center}
            \begin{tikzcd}
              \Gal(M \extendover K) \ar[r, "\res"]
              \ar[d, "\iota_4"] &
              \Gal(L \extendover K) \ar[d, "\iota_3"] \\
              S_4 \ar[r, "\pi"] & S_3
            \end{tikzcd}
          \end{center}
          By \cref{ft_of_gal}(c), the map $\res :
          \Gal(M \extendover K) \to \Gal(L
          \extendover K)$ is surjective. Therefore
          $\pi(\Im \iota_4) = \Im \iota_3$.
          Therefore $\pi(\Galpoly(f \galover K)) =
          \Galpoly(g \galover K)$. \qedhere
      \end{enumerate}
    \end{proof}
    \begin{proposition}
      % Proposition 9.4
      \vspace{0.8em}
      Let $f$ be a monic quartic polynomial with
      \gls{resolvent} $g$. Then
      \begin{enumerate}[(i)]
        \item $\Disc(f) = \Disc(g)$.
        \item If
          \[ f(X) = X^4 + pX^2 + qX + r \]
          then
          \[ g(X) = X^3 - 2pX^2 + (p^2 - 4r)X +
          q^2 \]
      \end{enumerate}
    \end{proposition}
    \begin{proof}
      \phantom{}
      \begin{enumerate}[(i)]
        \item Exercise (see proof of \cref{resolvent_facts}(ii)).
        \item We must show
          \begin{align*}
            \beta_1 + \beta_2 + \beta_3
            &= 2p \tag{1} \\
            \beta_1 \beta_2 + \beta_2 \beta_3
            &= p^2 - 4r \\
            \beta_1 \beta_2 \beta_3
            &= -q^2
          \end{align*}
          We have $\beta_1 + \beta_2 + \beta_3 =
          2 \sum_{i < j} \alpha_i \alpha_j = 2p$,
          which proves (1). Since $\alpha_1 +
          \alpha_2 + \alpha_3 + \alpha_4 = 0$, we
          have
          \[ \label{lec17_l144_eq} \begin{cases}
            \beta_1
            = -(\alpha_1 + \alpha_2)^2 \\
            \beta_2
            = -(\alpha_1 + \alpha_3)^2 \\
            \beta_3
            = -(\alpha_1 + \alpha_4)^2
          \end{cases} \tag{$*$} \]
          \[ (\alpha_1 + \alpha_2)(\alpha_1 +
          \alpha_3)(\alpha_1 + \alpha_4) =
          \alpha_1^2 \cancel{(\alpha_1 + \alpha_2 +
          \alpha_3 + \alpha_4)} + \ob{\sum_{i < j < k}
          \alpha_i \alpha_j \alpha_k}^{=-q} \]
          Therefore $\beta_1 \beta_2 \beta_3 =
          -q^2$, which proves (3). (2) is left as
          an exercise.
      \end{enumerate}
    \end{proof}
    \begin{example*}
      \vspace{0.8em}
      $f(X) = X^4 - 4X^2 + 2$. Irreducible in
      $\QQ[X]$ by Eisenstein ($p = 2$) and
      Gauss' Lemma. $g(X) = X(X^2 + 8X + 8)$.
      \[ \Disc(f) = \Disc(g) = 8^2 \Disc(X^2 + 8X
      + 8) = 2^{11} \]
      \[ \Galpoly(g \galover \QQ) = C_2 \implies
      \Galpoly(f \galover \QQ) = C_4 \text{ or }
      D_8 .\]
      But $f(X) = (X^2 - 2 + \sqrt{2})(X^2 - 2 -
      \sqrt{2})$. Therefore $\Galpoly(f \galover
      \QQ) \cap A_4 = \Galpoly(f \galover
      \fadj\QQ(\sqrt{2}))$ is not a transitive
      subgroup of $S_4$. Therefore $\Galpoly(f
      \galover \QQ) \cong C_4$ (compare with
      \cref{example_6_6}).
    \end{example*}
    \vspace{-1em}

    Now we find a formula for the roots of a quartic
    polynomial.
    \begin{enumerate}[(i)]
      \item Replace $f(X)$ by $f(X + c)$ such that
        $f$ has no $X^3$ term ($\implies$
        $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4
        = 0$).
      \item Find the roots of $\beta_1, \beta_2,
        \beta_3$ of the \gls{resolvent} using the
        method of \cref{sec_4}.
    \end{enumerate}
    By \eqref{lec17_l144_eq} we have
    \[ \alpha_1 = \half (\sqrt{-\beta_1} +
    \sqrt{-\beta_2} + \sqrt{-\beta_3}) \]
    where we choose square roots such that
    $\sqrt{-\beta_1} \sqrt{-\beta_2}
    \sqrt{-\beta_3} = -q$.
\end{itemize}

Recall $\sigma \in S_n$ has cycle type $(n_1,
\ldots, n_r)$ if when written as a product of
disjoint cycles, these cycles have lengths $n_1,
n_2, \ldots, n_r$.

\begin{flashcard}[poly-irred-factors-galois-group-structure-lemma]
\begin{lemma}
  % Lemma 9.5
  \label{poly_ired_factors_gal_structure}
  Let $f \in \radj \FF_p[X]$ be a \gls{sep_poly}
  polynomial with irreducible factors of degrees
  $n_1, \ldots, n_r$ ($n = \deg f = \sum n_i$).
  Then \cloze{$\Galpoly(f \galover \FF_p) \subset S_n$ is
  generated by a single element of cycle type
  $(n_1, \ldots, n_r)$. In particualr, $\Galpoly(f
  \galover \FF_p)$ is cyclic of order $\lcm(n_1,
  \ldots, n_r)$.}
\end{lemma}

\begin{proof}
  \cloze{Let $L$ be a \gls{splitting_field} of $f$ over
  $\FF_p$. Let $\alpha_1, \ldots, \alpha_n$ be the
  roots of $f$ in $L$. Then
  \cref{Fpn_over_Fp_is_gal} implies that $G =
  \Gal(L \extendover \FF_p)$ is cyclic generated
  by Frobenius $\phi : x \mapsto x^p$. Write $f =
  \prod_i f_i$, where $f_i \in \radj \FF_p[X]$ is
  irreducible of degree $n_i$. Since $G$ permutes
  the roots of each $f_i$ transitively, the action
  of $\phi$ on the roots of $f_i$ is given by a
  single $n_i$ cycle.}
\end{proof}
\end{flashcard}