%! TEX root = Galois.tex
% vim: tw=50
% 14/10/2023 12AM

\begin{remark*}
  The last example shows that a regular $9$-gon cannot be
  \glsref[constructible_point]{constructed} with
  \glsref[constructible_point]{ruler and compass}.

  Later we will prove the result of Gauss which
  says that a regular $n$-gon is
  \gls{constructible_point} if
  and only if $\phi(n)$ is a power of 2.
\end{remark*}

\newpage
\section{Splitting Fields}

\textbf{Question:} Let $f \in \ringadjoin K[X]$ be a
nonconstant polynomial. Is there a \gls{field_ex}
$L \extendover K$ in which $f$ has a root? (or even
better: an \gls{ex} in which $f$ splits into
linear factors)?

If $K \subset \CC$ then the fundamental theorem of
algebra shows we can factor $f$ in $\ringadjoin \CC[X]$ as a
product of linear polynomials. But what about $K =
\FF_p$?

\begin{flashcard}[K-hom-defn]
\begin{definition}[$K$-homomorphism]
  % Definition 3.1
  \glsdefn{K_hom}{$K$-homomorphism}{$K$-homomorphisms}
  \glsalias{K_embedding}{$K$-embedding}{$K$-embeddings}
  \cloze{Let $L \extendover K$ and $M \extendover K$ be
  \glspl{field_ex}. A \emph{$K$-homomorphism} (or
  \emph{$K$-embedding}) of $L$ into $M$ is a ring
  homomorphism $L \to M$ which is the identity on
  $K$.}
\end{definition}
\end{flashcard}

\begin{flashcard}[simple-extension-automorphisms-bound]
\begin{theorem}
  % Theorem 3.2
  \label{automorphisms_bound}
  Let $L = \fieldadjoin K(\alpha)$ where $\alpha$
  is \gls{alg_el}
  over $K$ with \gls{min_poly} $f$. Let $M
  \extendover K$ be any \gls{field_ex}. Then there is a
  bijection
  \cloze{
  \begin{align*}
    \{\text{\Khom{K} $L \to M$}\}
    &\leftrightarrow \{\text{roots of $f$ in
    $M$}\} \\
    \tau
    &\mapsto \tau(\alpha)
  \end{align*}
  }
  In particular (by \cref{roots_bound_lemma}),
  \[ \cloze{\#\{\text{\Khompl{K} $L \to M$}\} \le
  \deg f} \]
\end{theorem}

\begin{proof}
  \cloze{
  Write $f = \sum_{i = 0}^d c_i X^i$, $c_i \in K$.
  Let $\tau : L \to M$ be a \Khom{K}. Then
  \begin{align*}
      f(\tau(\alpha))
      &= \sum_i c_i \tau(\alpha)^i \\
      &= \tau \left( \sum_i c_i \alpha^i \right)
      \\
      &= \tau(f(\alpha)) \\
      &= 0
  \end{align*}
  Therefore $\tau(\alpha) \in M$ is a root of $f$.

  \textbf{Injectivity:} Since $L = \fieldadjoin K(\alpha)$, and
  \Khom{K} $\tau : L \to M$ is uniquely
  determined by $\tau(\alpha)$.

  \textbf{Surjectivity:} We saw earlier that
  evaluation at $\alpha$ gives an isomorphism
  $\ringadjoin K[X] / (f) \to \fieldadjoin K(\alpha) = L$ where $X + (f)
  \mapsto \alpha$. Now let $\beta \in M$ be a root
  of $f$. Since $f$ is irreducible it is the
  \gls{min_poly} for $\beta$ over $K$.
  Evaluation at $\beta$ gives a ring homomorphism
  \begin{align*}
    \ringadjoin K[X]
    &\to M \\
    g
    &\mapsto g(\beta)
  \end{align*}
  with kernel $(f)$. By the isomorphism theorem we
  get
  \begin{align*}
    \frac{\ringadjoin K[X]}{(f)}
    &\stackrel{\psi}{\hookrightarrow} M \\
    X + (f)
    &\mapsto \beta
  \end{align*}
  Since $\phi, \psi$ are \Khompl{K} and
  $\phi$ is an isomorphism it follows that
  \[ \tau = \psi \cdot \phi^{-1} : L \to M \]
  is a \Khom{K} with $\tau(\alpha) =
  \beta$.
  }
\end{proof}
\end{flashcard}

\begin{example*}
  There are exactly 2 \Khom{\QQ}
  $\fieldadjoin \QQ(\sqrt{2}) \to \RR$ given by $a + b\sqrt{2}
  \mapsto b\sqrt{2}$ and $a + b\sqrt{2} \mapsto a
  - b\sqrt{2}$, $a, b \in \QQ$. We actually need a
  slight variant of \cref{automorphisms_bound}.
  The proof is exactly the same, but the extra
  generality is useful for inductive proofs.
\end{example*}

\begin{flashcard}[sigma-embedding]
\begin{definition}[$\sigma$-embedding]
  \glsdefn{sigma_embedding}{$\sigma$-embedding}{$\sigma$-embeddings}
  \glsnoundefn[sigma_embedding]{sigma_hom}{$\sigma$-homomorphism}{$\sigma$-homomorphisms}
  \cloze{
  Let $L \extendover K$ and $M \extendover K'$ be
  \glspl{field_ex}. Let $\sigma : K \to K'$ be a
  \gls{field} \gls{embedding}. A $\sigma$-embedding $\tau : L \to M$
  is an \gls{embedding} which extends $\sigma$, i.e.
  $\tau(x) = \sigma(x) ~\forall x \in K$.
  Equivalently, $\sigma = \tau|_K$ is the
  restriction of $\tau$ to $K$.
  }
\end{definition}
\end{flashcard}

\begin{center}
  \includegraphics[width=0.6\linewidth]
  {images/c74081546a8611ee.png}
\end{center}

\begin{note*}
  Taking $\sigma = \id : K \to K$ we recover the
  definition of a \glsref[K_hom]{$K$-embedding}.
\end{note*}

\begin{theorem}
  % Theorem 3.4
  \label{simple_extension_sigma_hom_bijection}
  Let $L = \fieldadjoin K(\alpha)$ where $\alpha$ is
  \gls{alg_el}
  over $K$ with \gls{min_poly} $f$. Let
  $\sigma : K \to K'$ be a \gls{field}
  \gls{embedding}, and $M
  \extendover K'$ any \gls{field_ex}. Then there is a
  bijection
  \begin{align*}
    \{\text{\glspl{sigma_embedding} $L \to M$}\}
    &\to \{\text{roots of $\sigma f$ in $M$}\} \\
    \tau
    &\mapsto \tau(\alpha)
  \end{align*}
  (where $\sigma f$ is the polynomial in
  $\ringadjoin K'[X]$
  obtained by applying $\sigma$ to each
  coefficient of $f$). In particular,
  \[ \#\{\text{\glspl{sigma_embedding} $L \to M$}\}
  \le \deg f \]
\end{theorem}

\begin{example*}
  Let $K = \fieldadjoin \QQ(\sqrt{2})$ and let $L =
  \fieldadjoin \QQ(\alpha)$ where $\alpha = \sqrt{1 +
  \sqrt{2}}$ (exercise: check that $1 + \sqrt{2}$
  is not a square in $K$, so that we get $\exdeg[L : K]
  = 2$). There are exactly 2 \glspl{K_embedding} $L \to
  \RR$ given by $\alpha \mapsto \pm \sqrt{1 +
  \sqrt{2}}$.

  However, if $\sigma : K \to K$, $a + b \sqrt{2}
  \mapsto a - b\sqrt{2}$ then there are no
  \glspl{sigma_embedding} $L \to \RR$ (since $1 -
  \sqrt{2} < 0$).
\end{example*}

\begin{flashcard}[splitting-field-defn]
\begin{definition}[splitting field]
  \glsdefn{splitting_field}{splitting
  field}{splitting fields}
  \cloze{
  Let $K$ be a \gls{field}. Let $0 \neq f \in
  \ringadjoin K[X]$. An
  extension $L \extendover K$ is a \emph{splitting
  field} of $f$
  over $K$ if
  \begin{enumerate}[(i)]
    \item $f$ splits into linear factors over $L$.
    \item $L = \fieldadjoin K(\alpha_1, \ldots, \alpha_n)$
      where $\alpha_i$ are the roots of $f$ in
      $L$.
  \end{enumerate}
  }
\end{definition}
\end{flashcard}

\begin{remark*}
  (ii) is equivalent to saying $f$ does not split
  into linear factors over any (proper) subfield of $L$
  containing $K$.
\end{remark*}