%! TEX root = Galois.tex
% vim: tw=50
% 14/11/2023 12PM

\begin{flashcard}[reduction-mod-p-thm]
\begin{theorem}[``Reduction modulo $p$'']
  % Theorem 9.6
  \label{red_mod_p_thm}
  \glssymboldefn{modred}{$\ol{f}$}{$\ol{f}$}
  \cloze{Let $f \in \radj \ZZ[X]$ be a monic
  \gls{sep_poly} polynomial of degree $n \ge 1$.
  Let $p$ be a prime and suppose the reduction of
  $f$ modulo $p$, say $\modred{f} \in \radj \FF_p[X]$
  is also \gls{sep_poly}. Then $\Galpoly(\modred{f}
  \galover \FF_p) \subset \Galpoly(f \galover
  \QQ)$ as subgroups of $S_N$ (up to conjugacy).}
\end{theorem}
\end{flashcard}

\begin{proof}[Proof (non-examinable)]
  See below.
\end{proof}

\begin{flashcard}[reduction-mod-p-factors-lemma]
\begin{corollary}
  % Corollary 9.7
  With the same assumptions, suppose $\modred{f} = g_1
  g_2 \cdots g_r$ \cloze{where $g_i \in \radj \FF_p[X]$
  is irreducible of degree $n_i$. Then $\Galpoly(f
  \galover \QQ) \subset S_n$ contains an element
  with cycle type $(n_1, n_2, \ldots, n_r)$.}
\end{corollary}

\begin{proof}
  \cloze{Combine \cref{poly_ired_factors_gal_structure}
  and \cref{red_mod_p_thm}.}
\end{proof}
\end{flashcard}

\begin{example*}
  $f(X) = X^4 - 3X + 1$. Modulo $2$, $\modred{f} = X^4
  + X + 1 \in \radj \FF_2[X]$ is irreducible.
  Modulo $5$, $\modred{f} = (X + 1)(X^3 - X^2 + X
  + 1)$ (noting that the second factor is
  irreducible in $\radj \FF_5[X]$). Therefore
  $\Galpoly(f \galover \QQ)$ contains a $3$-cycle
  and a $4$-cycle. Hence $\Galpoly(f \galover \QQ)
  = S_4$.
\end{example*}
\vspace{-1em}

Let $f \in \radj K[X]$ be a monic \gls{sep_poly}
polynomial of degree $n$ with
\gls{splitting_field} $L$ and roots $\alpha_1,
\ldots, \alpha_n \in L$. Let
\begin{align*}
  F(\tau_1, \ldots, \tau_n, X)
  &= \prod_{\sigma \in S_n} (X - (\alpha_1
  T_{\sigma(1)} + \cdots + \alpha_n
  T_{\sigma(n)})) \\
  &\in \radj K[T_1, \ldots, T_n, X]
\end{align*}
Indeed, the coefficients of this polynomial are in
$L$, and are fixed by $\Gal(L \extendover K)$
hence are in $K$.

We define an action $*$ of $S_n$ on $\radj K[T_1,
\ldots, T_n, X]$ by permuting the $T_i$, i.e.
\[ (\sigma * h) (T_1, \ldots, T_n, X) =
h(T_{\sigma(1)}, \ldots, T_{\sigma(n)}, X) \]
We note that $\sigma * F = F$ for all $\sigma \in
S_n$.

\begin{flashcard}[irred-factor-of-symm-prod-lemma]
\begin{lemma}
  % Lemma 9.8
  \label{irred_factor_of_symm_prod_lemma}
  Let $F_1 \in \radj K[T_1, \ldots, T_n, X]$ be an
  irreducible factor of $F$. Then \cloze{$\Galpoly(f
  \galover K) \subset S_n$ is conjugate to
  $\Stab(F_1) = \{\tau \in S_n \st \tau * F_1 =
  F_1\}$.}
\end{lemma}

\begin{proof}
  \cloze{Without loss of generality $F_1$ is monic in
  $X$. Replacing $F_1$ by $\tau * F_1$ for
  suitable $\tau \in S_n$ we may suppose it has a
  factor
  \[ X - (\alpha_1 T_1 + \cdots + \alpha_n T_n) .\]
  Then for each $\sigma \in G = \Galpoly(f
  \galover K)$, it had a factor
  \[ X - (\alpha_{\sigma(1)} T_1 + \cdots
  \alpha_{\sigma(n)} T_n) \]
  Now
  \[ \prod_{\sigma \in G} (X - (\alpha_{\sigma(1)}
  T_1 + \cdots + \alpha_{\sigma(n)} T_n)) \]
  has coefficients in $K$, and divides $F_1$,
  hence is equal to $F_1$ (sine $F_1$ irreducible
  and monic in $X$). For $\tau \in S_n$ we have
  \begin{align*}
    \tau * F_1
    &= \prod_{\sigma \in G} (X -
    (\alpha_{\sigma(1)} T_{\tau(1)} + \cdots +
    \alpha_{\sigma(n)} T_{\tau(n)})) \\
    &= \prod_{\sigma \in G} (X - (\alpha_{\sigma
    \tau^{-1}(1)} T_1 + \cdots + \alpha_{\sigma
    \tau^{-1}(n)} T_n)) \\
    &= \prod_{\sigma \in G\tau^{-1}} (X -
    (\alpha_{\sigma(1)} T_1 + \cdots +
    \alpha_{\sigma(n)} T_n))
  \end{align*}
  So $\tau * F_1 = F_1$ if and only if $G =
  G\tau^{-1}$, which happens if and only if $\tau
  \in G$.}
\end{proof}
\end{flashcard}

\begin{proof}[Proof of \nameref{red_mod_p_thm}
  (non-examinable)]
  By the \nameref{symm_func_thm} the coefficients
  of $F$ are $\ZZ$-coefficient polynomials in the
  coefficients of $f$. So if $f \in \radj \ZZ[X]$
  then $F \in \radj \ZZ[T_1, \ldots, T_n, X]$. Let
  $\modred{f} \in \radj \FF_p[X]$ and $\modred{F}
  \in \radj \FF_p[T_1, \ldots, T_n, X]$ be the
  polynomials obtained by reducing all
  coefficients modulo $p$. We may equally
  construct $\modred{F}$ from $\modred{f}$ in the
  same way we constructed $F$ from $f$. Write $F =
  F_1 F_2 \cdots F_s$, $F_i \in \radj \ZZ[T_1,
  \ldots, T_n, X]$ distrinct irreducibles (also
  irreducible in $\radj \QQ[T_1, \ldots, T_n,
  X]$). Let $\modred{F} = \phi_1 \phi_2 \cdots
  \phi_t$, $\phi_i \in \radj \FF_p[T_1, \ldots,
  T_n, X]$ distinct irreducibles. Without loss of
  generality $\phi_1 \mid \modred{F_1}$ (hence
  $\phi_1 \nmid \modred{F}_j$ for all $j > 1$).
  Then
  \[ \{\tau \in S_n \st \tau * \phi_1 = \phi_1\}
  \subset \{\tau \in S_n \st \tau * F_1 = F_1\} .\]
  \cref{irred_factor_of_symm_prod_lemma} shows
  that up to conjugacy,
  \[ \Galpoly(\modred{f} \galover \FF_p) \subset
  \Galpoly(f \galover \QQ). \qedhere \]
\end{proof}

\newpage
\section{Cyclotomic and Kummer extensions}

\glssymboldefn{mu_n}{$\mu_n$}{$\mu_n$}
Let $K$ be a \gls{field}, and $n \ge 1$ an
integer. We suppose $\characteristic K \nmid n$ (i.e. either
$\characteristic K = 0$, or $\characteristic K = p
> 0$ and $p \nmid n$). Let $L \extendover K$ be a
\gls{splitting_field} of $f(X) = X^n - 1$. Since
$f\fd(X) = nX^{n - 1}$ and $n \cdot 1_K \neq 0$ we
have $\gcd(f, f\fd) = 1$ and so $f$ is
\gls{sep_poly}. By
\cref{classification_of_finite_galois_extensions},
$L \extendover K$ is \gls{galois}. Let $\mu_n =
\{x \in L \st x^n = 1\}$ be the group of $n$-th
roots of unity. This is a subgroup of $\mult{L}$
of order $n$ (since $f$ splits into distinct
linear factors over $L$) and is cyclic by
\cref{cyclic_mult_subgroup}.

\begin{definition*}
  \glsnoundefn{prim_n_root}{primitive $n$-th root of
  unity}{primitive $n$-roots of unity}
  \glssymboldefn{zeta_n}{$\zeta_n$}{$\zeta_n$}
  $\zeta_n \in \mu_n$ is a \emph{primitive $n$-th
  root of unity} if it has order exactly $n$ in
  $\mult{L}$.
\end{definition*}

\begin{example*}
  If $K \subset \CC$ then
  we can take $\zeta_n = e^{2\pi i / n}$.
\end{example*}
\vspace{-1em}

Then
\[ \mu_n = \langle \zeta_n \rangle = \{1, \zeta_n,
\zeta_n^2, \ldots, \zeta_n^{n - 1}\} \]
and
\[ L = \fadj K(1, \zeta_n, \zeta_n^2, \ldots,
\zeta_n^{n - 1}) = \fadj K(\zeta_n) .\]

\begin{definition*}
  \glsadjdefn{cycl_ex}{cyclotomic}{\gls{field_ex}}
  $\fadj K(\zeta_n) \extendover K$ is called a
  \emph{cyclotomic extension}.
\end{definition*}
\vspace{-1em}

Next time: we show
\[ \Gal(\fadj K(\zeta_n) \extendover K) \subset
\mult{(\ZZ / n\ZZ)} .\]
(with equality when $K = \QQ$).
