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% 21/11/2023 12PM

\begin{flashcard}[Kummer-thm]
\begin{theorem}[Kummer]
  % Theorem 10.8
  \cloze{Assume $\characteristic K \nmid n$ and
  $\runities_n \subset K$. Then every \gls{ex_deg}
  $n$ \gls{galois} \gls{ex} $L \extendover K$ with
  cyclic \gls{galois_group} is of the form $L =
  \fadj K(\sqrt[n]{a})$ for some $a \in
  \mult{K}$.}
\end{theorem}

\begin{proof}
  \cloze{Write $\Gal(L \extendover K) = \{\sigma^i : 0
  \le i < n\}$. By \cref{lindep_of_fembeddings},
  there exists $x \in L$ such that
  \[ \ub{\sum_{j = 0}^{n - 1} \runity_n^j
  \sigma^{j(x)}}_{= \alpha}
  \neq 0 \]
  (Lagrange resolvent). Then
  \[ \sigma(\alpha) = \sum_{j = 0}^{n - 1} \runity_n^j
  \sigma^{j + 1}(x) = \sum_{j = 0}^{n - 1}
  \runity_n^{j - 1} \sigma^j(x) = \runity_n^{-1}
  \alpha \]
  The Galois conjugates $\sigma^j(\alpha) =
  \runity_n^{-j} \alpha$. So $\exdeg[\fadj
  K(\alpha) : K] = n$ and $L = \fadj K(\alpha)$.
  Finally $\sigma(\alpha^n) = (\runity_n^{-1}
  \alpha)^n = \alpha^n$, so $\alpha^n \in K$.}
\end{proof}
\end{flashcard}

Now let $K$ be a \gls{field} with $\characteristic
K = 0$. Let $f \in \radj K[X]$ be a polynomial.

\begin{flashcard}[soluble-by-radicals]
\begin{definition*}[Soluble by radicals]
  \glsadjdefn{sol_rad}{soluble by
  radicals}{polynomial}
  \cloze{$f$ is \emph{soluble by radicals} over $K$ if
  there exist fields
  \[ K = K_0 \subset K_1 \subset K_2 \subset \cdots
  \subset K_m \]
  such that $f$ has a root in $K_m$ and for each
  $1 \le i \le m$, $K_i = \fadj K_{i -
  1}(\alpha_i)$ with $\alpha_i^{d_i} \in K_{i -
  1}$ for some $d_i \ge 1$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[soluble-group]
\begin{definition*}[Soluble group]
  \glsadjdefn{sol_gp}{soluble}{group}
  \cloze{A finite group $G$ is \emph{soluble} if there
  exist subgroups
  \[ \{1\} = H_0 \subset H_1 \subset H_2 \subset
  \cdots \subset H_m = G \]
  such that for each $1 \le i \le m$, $H_{i - 1}
  \normalsub H_i$ and $H_i / H_{i - 1}$ is
  abelian.}
\end{definition*}
\end{flashcard}

\begin{remark*}
  The definition is unchanged if we replace
  ``abelian'' by ``cyclic'' or ``cyclic of prime
  order''.
\end{remark*}

\begin{example*}
  $S_4$ is soluble:
  \[ \{1\} \subset V \subset A_4 \subset S_4 \]
  with $V \cong C_2 \times C_2$, $A_4 / V \cong
  C_3$, $S_4 / A_4 \cong C_2$.
\end{example*}

\begin{lemma}
  % Lemma 10.9
  \label{subgp_and_quot_of_sol_is_sol}
  If $G$ is soluble then so is every subgroup and
  quotient of $G$.
\end{lemma}

\begin{proof}
  Exercise (\es[7]{4}).
\end{proof}

\begin{flashcard}[f-soluble-if-gal-soluble]
\begin{theorem}
  % Theorem 10.10
  \label{f_soluble_iff_gal_soluble}
  Let $f \in \radj K[X]$ irreducible. Then
  \[ \text{$f$ is \gls{sol_rad}} \iff
  \cloze{\text{$\Galpoly(f \galover K)$ is
  \gls{sol_gp}}} \]
\end{theorem}
\end{flashcard}

\begin{flashcard}[f-adj-roots-of-gal-conjs-is-gal]
\begin{lemma}
  % Lemma 10.11
  \label{f_adj_roots_of_gal_conjs_is_gal}
  Let $L \extendover K$ be a \gls{finite_ex}
  \gls{galois} \gls{ex}. \cloze{Let
  \[ \Gal(L \extendover K) = \{\sigma_1, \ldots,
  \sigma_m\} \]
  (say $\sigma_1 = \id$).} Let $a \in \mult{L}$ and
  $n \ge 1$. Then
  \[ M = \cloze{\fadj L (\runities_n,
  \sqrt[n]{\sigma_1(a)}, \ldots,
  \sqrt[n]{\sigma_m(a)})} \]
  is a \gls{galois} \gls{ex} of $K$.
\end{lemma}

\begin{proof}
  \cloze{Let
  \[ f(X) = \prod_{i = 1}^m (X^n - \sigma_i(a))
  \in \radj K[X] \]
  Then $m$ is the \gls{comp_field} of $L$ and the
  \gls{splitting_field} of $f$ over $K$. Therefore
  $M \extendover K$ is \gls{galois} by
  \cref{comp_subf_gal_thm}(ii).}
\end{proof}
\end{flashcard}

\begin{flashcard}[proof-of-f-sol-iff-gal-sol]
\prompt{Proof that $f$ (irreducible) soluble if and only if
$\Galpoly(f \galover K)$ soluble?}
\begin{proof}[Proof of
  \cref{f_soluble_iff_gal_soluble}]
  \phantom{}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] \cloze{By definition there exist
      \glspl{field} $K = K_0 \subset K_1 \subset
      \cdots K_m$ such that $f$ has a root in
      $K_m$ and for each $1 \le i \le m$, $K_i =
      K_{i - 1}(\alpha_i)$ with $\alpha_i^{d_i}
      \in K_{i - 1}$ for some $d_i \ge 1$.
      Repeatedly applying
      \cref{f_adj_roots_of_gal_conjs_is_gal}, we
      may assume $K_m \extendover K$ is
      \gls{galois}. By adjoining suitable
      \glsref[prim_n_root]{roots of unity} first,
      we may further assume each $K_i \extendover
      K_{i - 1}$ is either \gls{cycl_ex} or
      \glsref[kumm_ex]{Kummer}. By
      \cref{cyclotomic_ex_hom_thm} and
      \cref{kummer_gal_is_cyclic} each $\Gal(K_i
      \extendover K_{i - 1})$ is abelian. So by
      the \nameref{ft_of_gal}, $\Gal(K_m \extendover
      K)$ is soluble. Since $f$ has a root in
      $K_m$ and $K_m \extendover K$ is
      \gls{norm_ex}, we know that $f$ slpits in
      $K_m$. Therefore $\Galpoly(f \galover K)$
      is a quotient of $\Gal(K_m \extendover K)$,
      and hence $\Galpoly(f \galover K)$ is
      \gls{sol_gp} by
      \cref{subgp_and_quot_of_sol_is_sol}.}
    \item[$\Leftarrow$] \cloze{By the
      \hyperref{ft_of_gal}{Galois
      correspondence} there exists
      $K = K_0 \subset K_1 \subset \cdots
      \subset K_m$
      such that $f$ has a root in $K_m$ and each
      $K_i \extendover K_{i - 1}$ is \gls{galois}
      with cyclic \gls{galois_group}. Let $n =
      \lcm_{1 \le i \le m} \exdeg [K_i : K_{i -
      1}]$. Then
      \[ K = K_0 \subset \fadj K_0(\runity_n)
      \subset \fadj K_1(\runity_n) \subset \cdots
      \subset \fadj K_m(\runity_n) \]
      By \cref{comp_subf_gal_thm}(i), $\fadj
      K_i(\runity_n) \extendover \fadj K_{i -
      1}(\runity_n)$ is \gls{galois} and
      \[ \Gal(\fadj K_i(\runity_n) \extendover
      \fadj K_{i - 1}(\runity_n)) \hookrightarrow
      \Gal(K_i \extendover K_{i - 1}) \]
      Therefore $\Gal(\fadj K_i(\runity_n)
      \extendover \fadj K_{i - 1}(\runity_n))$ is
      cyclic of order dividing $n$.\qedhere}
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[poly-gal-gp-Sn-or-An-not-sol]
\begin{corollary}
  % Corollary 10.12
  If $f \in \radj K[X]$ is a polynomial of degree
  $n \ge 5$ with \gls{galois_group} $A_n$ or
  $S_n$, then $f$ is \emph{not} \gls{sol_rad} over
  $K$.
\end{corollary}

\begin{proof}
  \cloze{$A_5$ is non abelian and simple, hence not
  \gls{sol_gp}. By
  \cref{subgp_and_quot_of_sol_is_sol}, $A_n$ and
  $S_n$ are not soluble for all $n \ge
  5$\fcscrap{ (in fact
  $A_n$ is simple for all $n \ge 5$)}.}
\end{proof}
\end{flashcard}

\begin{example*}
  $K = \QQ$, $f(X) = X^5 - X + a$, with $a \in
  \ZZ$, $\gcd(a, 10) = 1$. Then
  \[ f \equiv X^5 + X + 1 = (X^2 + X + 1)(X^3 +
  X^2 + 1) \pmod{2} .\]
  Therefore $\Galpoly(f \galover \QQ)$ contains an
  element $\sigma$ with cycle type $(2, 3)$. Then
  $\sigma^3$ is a transposition.

  Using the trick in \es[5]{4}, we find
  $\modred{f} \in \FF_5[X]$ is irreducible, hence
  $\Galpoly(f \galover \QQ)$ contains a $5$-cycle.

  Now \es[7(i)]{3}: Let $p$ be a prime. If $G
  \subset S_p$ is a subgroup containing both a
  $p$-cycle and a transposition, then $G = S_p$.
  Therefore $\Galpoly(f \galover \QQ) = S_5$ and
  $f$ is not \gls{sol_rad}.
\end{example*}