%! TEX root = Galois.tex
% vim: tw=50
% 26/10/2023 12PM

\begin{remark*}
  \cref{irred_implies_sep} and
  \cref{prim_el_thm} show that if $\exdeg[K : \QQ]
  < \infty$ then $K = \QQ(\alpha)$ for some
  $\alpha \in K$.
\end{remark*}
\vspace{-1em}

\textbf{One aim for today:} Show that if $L
\extendover K$ is a \gls{field_ex} and $\alpha_1,
\ldots, \alpha_n \in L$, then
\[ \text{$\alpha_1, \ldots, \alpha_n$ are
\gls{sep_el} over $K$} \implies \text{$\fadj
K(\alpha_1, \ldots, \alpha_n) \extendover K$ is
\gls{sep_ex}} \]

\begin{flashcard}[HomK-notation]
\begin{notation*}
  \glssymboldefn{homk}{HomK}{HomK}
  Let $L \extendover K$ and $M \extendover K$ be
  \glspl{field_ex}. We write
  \[ \fHom_k(L, M) = \cloze{\{\text{\Kembeddingpl{K} $L
  \hookrightarrow M$}\}} \]
\end{notation*}
\end{flashcard}

\begin{flashcard}[HomK-bound-lemma]
\begin{lemma}
  % Lemma 5.7
  \label{homk_bound_lemma}
  Let $\exdeg[L : K] < \infty$. Suppose $L = \fadj
  K(\alpha)$ and $f$ is the \gls{min_poly} of
  $\alpha$ over $K$. Let $M \extendover K$ be any
  \gls{field_ex}. Then
  \[ \#\fHom_K(L, M) \le \cloze{\exdeg[L : K]} ,\]
  with equality if and only if \cloze{$f$ splits into
  linear factors over $M$.}
\end{lemma}

\begin{proof}
  \cloze{
  By \cref{automorphisms_bound},
  \begin{align*}
    \#\fHom_K(L, M)
    &= \#\{\text{roots of $f$ in $M$}\} \\
    &\le \deg f = \exdeg[L : K]
  \end{align*}
  with equality if and only if $f$ splits into
  discinct linear factors over $M$.
  }
\end{proof}
\end{flashcard}

\begin{theorem}
  % Theorem 5.8
  \label{hom_bound_general_thm}
  Let $\exdeg[L : K] < \infty$.
  Write $L = \fadj K(\alpha_1, \ldots, \alpha_n)$
  and let $f_i$ be the \gls{min_poly} of
  $\alpha_i$ over $K$. Let $M \extendover K$ be
  any \gls{field_ex}. Then
  \[ \#\fHom_K(L, M) \le \exdeg[L : K] \]
  with equality if and only if each $f_i$ splits
  into distinct linear factors over $M$.
\end{theorem}
\vspace{-1em}

Note that \cref{homk_bound_lemma} is the case $n =
1$.

\textbf{Obvious variant of
\cref{hom_bound_general_thm}:} Let $\sigma : K
\hookrightarrow M$ be an embedding. Then
\[ \#\{\text{\glspl{sigma_hom} $L \to M$}\} \le
\exdeg[L : K] \]
with equality if and only if each $\sigma(f_i)$
splits into distinct linear factors over $M$.

We'll use this variant in the induction argument.

\begin{proof}[Proof of \cref{hom_bound_general_thm}]
  By induction on $n$. The case $n = 1$ is proved
  in \cref{homk_bound_lemma}. So suppose $n > 1$.
  Let $K_1 = \fadj K(\alpha_1)$. Then
  \cref{homk_bound_lemma} implies
  \[ \label{lec10_l88_eq} \#\fHom_K(K, M) \le \exdeg[K_1 : K] \tag{1} \]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/7257fcb751b04ff5.png}
  \end{center}
  Let $\sigma \in \fHom_K(K_1, M)$. By the
  induction hypothesis,
  \[ \label{lec10_l94_eq} \#\{\text{\glspl{sigma_hom} $L = \fadj
  K_1(\alpha_2, \ldots, \alpha_n) \to M$}\} \le
  \exdeg[L : K_1] \tag{2} \]
  Then \nameref{tower_law} with
  \eqref{lec10_l88_eq} and
  \eqref{lec10_l94_eq} gives
  \[ \#\fHom_K(L, M) \le \exdeg[L : K_1]
  \exdeg[K_1 : K] = \exdeg[L : K] \]
  If equality holds then equality holds in both
  \eqref{lec10_l88_eq} and \eqref{lec10_l94_eq}.

  Equality in \eqref{lec10_l88_eq} implies $f_1$
  splits into distinct linear factors over $M$.
  Reordering the $\alpha_i$ gives the same
  conclusion for all the $f_i$. Conversely, if
  each $f_i$ splits into distinct linear factors
  over $M$ then \cref{homk_bound_lemma} gives
  equality in \eqref{lec10_l88_eq}.

  For $2 \le i \le n$, the \gls{min_poly} of
  $\alpha_i$ over $K_1$ divides $f_i$ and so
  splits into distinct linear factors over $M$.
  Then the induction hypothesis implies that
  equality holds in \eqref{lec10_l94_eq}. Since we
  now have equality in both \eqref{lec10_l88_eq}
  and \eqref{lec10_l94_eq}, it follows that
  $\#\fHom_K(L, M) = \exdeg[L : K]$.
\end{proof}

\begin{corollary}
  % Cororollary 5.9
  \label{finite_sep_ex_tfae}
  Let $\exdeg [L : K] < \infty$. Write $L = \fadj
  K(\alpha_1, \ldots, \alpha_n)$ and let $f_i$ be
  the \gls{min_poly} of $\alpha_i$ over $K$. Let
  $M \extendover K$ be any \gls{field_ex} in which
  $\prod f_i$ splits as a product of linear
  factors (for example $M = \algclos{K}$). Then
  the following are equivalent:
  \begin{enumerate}[(i)]
    \item $L \extendover K$ is \gls{sep_ex}.
    \item Each $\alpha_i$ is \gls{sep_el} over
      $K$.
    \item Each $f_i$ splits into distinct linear
      factors over $M$.
    \item $\#\fHom_K(L, M) = \exdeg[L : K]$.
  \end{enumerate}
\end{corollary}

\begin{proof}
  \phantom{}
  (i) $\implies$ (ii) $\implies$ (iii) by
  definition.

  (iii) $\implies$ (iv) see
  \cref{hom_bound_general_thm}.

  (iv) $\implies$ (i) Let $\beta \in L$. Applying
  \cref{homk_bound_lemma} to $L = \fadj
  K(\alpha_1, \ldots, \alpha_n, \beta)$ shows that
  $\beta$ is separable over $K$.
\end{proof}

\begin{remark*}
  (ii) $\implies$ (i) is the result promised at
  the start of the lecture.

  (i) $\iff$ (iv) is a useful characterisation of
  \gls{sep_ex} \glspl{ex}.
\end{remark*}

\begin{example}
  % Example 5.10
  Let $K$ be any \gls{field}. The polynomial $T^n
  - Y \in \radj K[Y, T]$ is irreducible (it
  suffices to consider factorisations of the form
  $f(T)(g(T) + Yh(T))$ where $f, g, h \in \radj
  K[T]$).

  Since $\radj K[Y]$ is a UFD with field of
  fractions $\fadj K(Y)$, it follows by Gauss's
  Lemma that
  \[ \label{lec10_l173_eq} T^n - Y \in \radj \fadj K(Y)[T] \tag{$*$} \]
  is irreducible. The \gls{field_ex} $\fadj K(X) /
  \fadj K(X^n)$ is generated  by $X$ which is a
  root of $T^n - X^n \in \radj \fadj K(X^n)[T]$.
  Putting $Y = X^n$ in \eqref{lec10_l173_eq} shows
  this is irreducible. Therefore $\exdeg[\fadj
  K(X) : \fadj K(X^n)] = n$. Now take $K = \FF_p$
  and $n = p$ ($p$ a prime). We claim that $\fadj
  \FF_p(X) / \fadj \FF_p(X^p)$ is an
  \glsref[sep_ex]{inseparable} \gls{ex} of
  \gls{ex_deg} $p$. Indeed, the \gls{min_poly} of
  $X$ over $\fadj \FF_p(X^p)$ is $f(T) = T^p -
  X^p \in \radj \fadj \FF_p(X^p)[T]$, which is
  \glsref[sep_poly]{inseparable} since $f(T) = (T
  - X)^p$ (compare to \cref{frob_hom_prop}).
\end{example}