%! TEX root = Galois.tex
% vim: tw=50
% 07/11/2023 12PM

We generalise
\nameref{galois_ex_trace_norm_formulae} to $L
\extendover K$ \gls{sep_ex}. Let $\algclos{K}$ be an
\gls{alg_closure} of $K$. Then
\nameref{finite_sep_ex_tfae} implies that
$\#\Hom_K(L, \algclos{K}) = \exdeg[L : K]$.

\begin{flashcard}[finite-sep-ex-trace-norm-thm]
\begin{theorem}
  % Theorem 7.6
  Let $L \extendover K$ be a \gls{finite_ex}
  \gls{sep_ex} \gls{ex} of \gls{ex_deg} $d$. Let
  $\sigma_1, \ldots, \sigma_d$ be the
  \Kembeddingpl{K} $L \hookrightarrow
  \algclos{K}$. Let $\alpha \in L$. Then
  \[ \aTrace_{L \extendover K} (\alpha) = \cloze{\sum_{i
    = 1}^d \sigma_i(\alpha),} \qquad \aNorm_{L
  \extendover K}(\alpha) = \cloze{\prod_{i = 1}^d
  \sigma_i(\alpha)} .\]
\end{theorem}

\begin{proof}
  \cloze{
  Let $f$ be the \gls{min_poly} of $\alpha$ over
  $K$. Let $\alpha_1, \ldots, \alpha_n$ be the
  roots of $f$ in $\algclos{K}$. By
  \nameref{automorphisms_bound},
  \begin{align*}
    \Hom_K(\fadj K(\alpha), \algclos{K})
    &\leftrightarrow \{\alpha_1, \ldots,
    \alpha_n\} \\
    \sigma
    &\mapsto \sigma(\alpha)
  \end{align*}
  Since $L \extendover \fadj K(\alpha)$ is
  \gls{sep_ex}, each \Kembedding{K} $\fadj
  K(\alpha) \hookrightarrow \algclos{K}$ extends
  to an \gls{embedding} $L \hookrightarrow \algclos{K}$
  in exactly $m = \exdeg[L : \fadj K(\alpha)]$
  ways. Therefore
  \begin{align*}
    \aTrace_{L \extendover K}(\alpha)
    &= m \sum_{j = 1}^n \alpha_j
    = \sum_{i = 1}^d \sigma_i(\alpha) \\
    \aNorm_{L \extendover K}(\alpha)
    &= \left( \prod_{j = 1}^n \alpha_j \right)^m
    = \prod_{i = 1}^d \sigma_i(\alpha)
  \end{align*}
  (the first equality in each line holds by
  \nameref{min_poly_norm_trace_lemma}, and the
  second equality in each line holds since $\#\{1
  \le i \le d \st \sigma_i(\alpha) = \alpha_j\} =
  m$).
  }
\end{proof}
\end{flashcard}

\newpage
\section{Finite Fields}

Fix $p$ a prime number. Recall $\FF_p = \ZZ /
p\ZZ$. We describe all finite \glspl{field} of
\gls{char} $p$ (these are necessarily
\gls{finite_ex} \glspl{ex} of $\FF_p$) and their
Galois theory. We will use
\cref{finite_field_sizes_prop},
\cref{cyclic_mult_subgroup} and
\cref{frob_hom_prop},
so it is worth revisiting these.

\begin{note*}
  $\phi$ in \cref{frob_hom_prop} is an
  automorphism of $K$ (injective since a
  homomorphism of \glspl{field}, hence surjective since
  $|K| < \infty$).
\end{note*}

\begin{flashcard}[finite-field-existence-characterisation-thm]
\begin{theorem}
  % Theorem 8.1
  \label{finite_field_unique_existence_thm}
  Let $q = p^n$ for some $n \ge 1$. Then:
  \begin{enumerate}[(i)]
    \item \cloze{There exists a \gls{field} with $q$
      elements.}
    \item \cloze{Any \gls{field} with $q$ elemenst is a
      \gls{splitting_field} of $X^q - X$ over
      $\FF_p$.}
  \end{enumerate}
  \cloze{
  In particular, any two finite \glspl{field} with the
  same order are isomorphic (by \nameref{uniqueness_of_splitting_fields}).}
\end{theorem}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item \cloze{Let $L$ be a \gls{splitting_field} of
      $f(x) = X^q - X$ over $\FF_p$. Let $K
      \subset L$ be the \gls{fixed_field} of
      $\phi^n : L \to L$ (note $\phi^n(x) =
      x^{q}$). Then
      \[ K = \{\alpha \in L \st \phi^n(\alpha) =
      \alpha\} = \{\alpha \in L \st f(\alpha) = 0\} \]
      Therefore $\#K \le \deg f = q$. But $f\fd(X)
      = -1$ so $\gcd(f, f\fd) = 1$ so $f$ is
      \gls{sep_poly} (by
      \cref{simple_root_if_fd_lemma}). Therefore
      $\#K = q$.}
    \item \cloze{Suppose $K$ is a \gls{field} with $\#K =
      q$. Then Lagrange's theorem (group theory)
      implies that $\alpha^{q - 1} = 1 ~\forall
      \alpha \in \mult{K}$, hence $\alpha^q =
      \alpha ~\forall \alpha \in K$.
      So
      \[ f(X) = X^q - X
      = \prod_{\alpha \in K} (X - \alpha) \]
      splits into linear factors over $K$, but
      clearly not over any proper subfield (since
      $f$ \gls{sep_poly} as mentioned in (i)). So $K$
      is a \gls{splitting_field} for $f$ over
      $\FF_p$.}
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{notation*}
  \glssymboldefn{finite_field}{$F_q$}{$F_q$}
  We write $\FF_q$ for any field with $q$
  elements. By
  \cref{finite_field_unique_existence_thm}, any
  two such are isomorphic, although there is no
  canonical choice of isomorphism.
\end{notation*}

\begin{flashcard}[Fpn-over-Fp-is-Galois-thm]
\begin{theorem}
  % Theorem 8.2
  \label{Fpn_over_Fp_is_gal}
  $\qFF_{p^n} \extendover \FF_p$ is \cloze{\gls{galois}
  with $\Gal(\qFF_{p^n} \extendover \FF_p)$ cyclic
  of order $n$, generated by the
  \hyperref[frob_hom_prop]{Frobenius} $\phi : x
  \mapsto x^p$.}
\end{theorem}

\begin{proof}
  \cloze{Let $L = \qFF_{p^n}$. Let $G \subset \Aut(L
  \extendover \FF_p)$ be the subgroup generated by
  \hyperref[frob_hom_prop]{Frobenius}. Then
  \begin{align*}
    \# \fixedf L^G
    &= \# \fixedf L^\phi \\
    &= \#\{\alpha \in L \st \alpha^p - \alpha =
    0\} \\
    &\le p
    &&\text{(\cref{roots_bound_lemma})}
  \end{align*}
  But $\FF_p \subset \fixedf L^G$, so $\fixedf L^G
  = \FF_p$. So
  \[ \FF_p \subset \fixedf L^{\Aut(L \extendover
  \FF_p)} \subset \fixedf L^G = \FF_p \]
  so we get $\FF_p = \fixedf L^{\Aut(L \extendover
  \FF_p)}$, i.e. $L \extendover \FF_p$ is
  \gls{galois}. Also, we get $\fixedf L^{\Aut(L
  \extendover \FF_p)} = \fixedf L^G$, which
  implies $\Aut(L \extendover \FF_p) = G$.
  Therefore
  \[ \Gal(L \extendover \FF_p) = G = \langle \phi
  \rangle \]
  and it has order $\exdeg[L : \FF_p] = n$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[any-finite-field-extension-galois-coro]
\begin{corollary}
  % Corollary 8.3
  Let $L \extendover K$ be any \gls{ex} of finite
  \glspl{field} with $\#K = q$. Then \cloze{$L
  \extendover K$ is \gls{galois} with $\Gal(L
  \extendover K)$ cyclic, generated by the
  $q$-power Frobenius $x \mapsto x^q$.}
\end{corollary}

\begin{proof}
  \cloze{
  Let $L = \qFF_{p^n}$. We have $\FF_p \subset K
  \subset L$. By \cref{Fpn_over_Fp_is_gal}, $L
  \extendover \FF_p$ is \gls{galois} with
  \[ G = \Gal(L \extendover \FF_p) = \langle \phi
  \rangle \cong C_n \]
  where $\phi : x \mapsto x^p$.
  \nameref{ft_of_gal} gives that $L \extendover K$
  is \gls{galois} and $H = \Gal(L \extendover K)
  \subset G$. Since $G = \langle \phi \rangle
  \cong C_n$ we have $H = \langle \phi^m \rangle$
  for some $m \mid n$. Then
  \[ \label{lec15_l200_eq} \exdeg[K : \FF_p] = \frac{\exdeg[L :
  \FF_p]}{\exdeg[L : K]} = \frac{\#G}{\#H} = (G :
  H) = m \tag{$*$} \]
  Therefore $q = \#K = p^m$ and $\phi^m : x
  \mapsto x^q$.
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[Fpn-unique-subfield-order-pm-coro]
\begin{corollary}
  % Corollary 8.4
  $\qFF_{p^n}$ has a unique subfield of order $p^m$
  for each $m \mid n$ and no others.
\end{corollary}

\begin{proof}
  \cloze{
  We apply the \nameref{ft_of_gal}. The subgroups
  of $G = \Gal(\qFF_{p^n} \extendover \FF_p) =
  \langle \phi \rangle \cong C_n$ are the
  subgroups $H = \langle \phi^m \rangle$ for $m
  \mid n$ (and no others). If $K = \fixedf
  {\qFF_{p^n}}^H$ then $H = \Gal(\qFF_{p^n}
  \extendover K)$ and $\exdeg[K : \FF_p] = (G : H)
  = m$ (see \eqref{lec15_l200_eq}). Therefore $\#K
  = p^m$.
  }
\end{proof}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/6268c66bd6d94d95.png}
  \end{center}
  All \glspl{ex} are \gls{galois} with cyclic
  \gls{galois} group of orders indicated.
\end{example*}