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\begin{note*}
  (ii) implies that $\exdeg[L : K] < \infty$.
\end{note*}

\begin{flashcard}[existence-of-splitting-fields]
\begin{theorem}[Existence of splitting fields]
  \label{existence_of_splitting_fields}
  Let $0 \neq f \in \ringadjoin K[X]$. Then there exists a
  \gls{splitting_field} for $f$ over $K$.
\end{theorem}

\begin{proof}
  \cloze{
  (We adjoin roots of $f$ one at a time).

  The proof is by induction on the degree of $f$.
  If $\deg f \le 1$ then $L = K$ is the
  \gls{splitting_field}.

  Now assume every polynomial of degree $< \deg f$
  has a \gls{splitting_field}. Let $g$ be an irreducible
  factor of $f$. Let $K_1 = \ringadjoin K[X] / (g)$ and
  $\alpha_1 = X + (g) \in \ringadjoin K[X]$, with $K_1 =
  \fieldadjoin K(\alpha_1)$. Then $f(\alpha_1)
  = 0$. So $f(X) = (X - \alpha_1) f_1(X)$ for some
  $f_1 \in \ringadjoin K_1[X]$ with $\deg f_1 < \deg f$.

  By
  induction hypothesis there exists a
  \gls{splitting_field} $L$ for $f_1$ over $K_1$, so $L =
  \fieldadjoin K_1(\alpha_2, \ldots, \alpha_n)$ where
  $\alpha_2, \ldots, \alpha_n$ are the roots of
  $f_1$ in $L$. We claim that $L$ is a
  \gls{splitting_field} for $f$ over $K$. Since $f_1$ splits in
  $L$, so does $f(X) = (X - \alpha) f_1(X)$.
  Moreover $L = \fieldadjoin K_1(\alpha_2, \ldots, \alpha_n) =
  \fieldadjoin K(\alpha_1, \ldots, \alpha_n)$ and $\alpha_1,
  \ldots, \alpha_n$ are roots of $f$. Therefore
  $L$ satisfies (i) and (ii).
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[uniqueness-of-splitting-fields]
\begin{theorem}[Uniqueness of splitting fields]
  \label{uniqueness_of_splitting_fields}
  \cloze{
  Let $0 \neq f \in \ringadjoin K[X]$. Let $L$ be a
  \gls{splitting_field} of $f$ over $K$. Let $\sigma : K
  \hookrightarrow M$ be any \gls{field_embedding} such
  that $\sigma f \in \ringadjoin M[X]$ splits. Then
  \begin{enumerate}[(i)]
    \item There exists a \gls{sigma_embedding} $\tau
      : L \hookrightarrow M$.
    \item If $M$ is a \gls{splitting_field} for $\sigma
      f$ over $\sigma K$ then any $\tau$ as in (i)
      is an isomorphism.
  \end{enumerate}
  In particular any two \glspl{splitting_field} for $f$
  over $K$ are \glsref[K_hom]{$K$-isomorphic}.
  }
\end{theorem}

\fcscrap{
\begin{center}
  \includegraphics[width=0.6\linewidth]
  {images/cfe26d7a6ce011ee.png}
\end{center}
}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item \cloze{By induction on $n = \exdeg[L : K]$. If
      $n = 1$ then $L = K$ and there is nothing to
      prove. So suppose $n > 1$ and let $g \in
      \ringadjoin K[X]$
      be an irreducible factor of $f$, of degree $>
      1$. Let $\alpha \in L$ be a root of $g$. Let
      $\beta \in M$ be a root of $\sigma g$. By 
      \cref{simple_extension_sigma_hom_bijection},
      $\sigma$ extends to an \gls{embedding} $\sigma_1 :
      \fieldadjoin K(\alpha) \to M$, $\alpha \mapsto \beta$.
      Then $\exdeg [L : \fieldadjoin K(\alpha)] < \exdeg[L : K]$. $L$ is a
      \gls{splitting_field} of $f$ over $\fieldadjoin K(\alpha)$ and
      $\sigma_1 f = \sigma f$ splits in $M$. So by
      the induction hypothesis $\sigma_1$ extends
      to an \gls{embedding} $\tau : L \to M$.}
    \item \cloze{Pick any $\tau : L
      \hookrightarrow M$ as in (i). Let $\alpha_1,
      \ldots, \alpha_n$ be the roots of $f$ in
      $L$. The roots of $\sigma f$ in $M$ are
      $\tau \alpha_1$, \ldots, $\tau \alpha_n$
      (consider splitting $\sigma f$ as a product
      of linear factors in two ways, and then use
      the fact that $\ringadjoin M[X]$ is a UFD). So
      if $M$ is a \gls{splitting_field} for $\sigma f$
      over $\sigma K$ then
      \[ M = \sigma \fieldadjoin K(\tau\alpha_1, \ldots,
      \tau\alpha_n) = \tau(\fieldadjoin K(\alpha_1, \ldots,
      \alpha_n)) = \tau(L) \]
      So $\tau$ is surjective, so it's an
      isomorphism (recall \glspl{field_embedding} are
      always injective).}
  \end{enumerate}
  \cloze{For the final statement, suppose $L \extendover
  K$ and $M \extendover K$ are
  \glspl{splitting_field} for $f$
  over $K$, and let $\sigma : K \hookrightarrow M$
  be the inclusion map. Then (i) and (ii) give a
  \glsref[K_hom]{$K$-isomorphism} $L \to M$.}
\end{proof}
\end{flashcard}

\begin{warning*}
  The previous theorem means that we can say ``the
  \gls{splitting_field} of $f$ over $K$'' since all such
  \glspl{field} are isomorphic.

  However, the isomorphism between such
  \glspl{field} is
  not necessarily unique, an in fact in some cases
  we can use a non-identity automorphism.
\end{warning*}

\begin{example*}
  If $K \subset \CC$ then by the fundamental
  theorem of algebra, one \gls{splitting_field} for $f$
  over $K$ is the subfield $\fieldadjoin K(\alpha_1, \ldots,
  \alpha_n) \subset \CC$ where $\alpha_1, \ldots,
  \alpha_n \in \CC$ are the roots of $f$. 
  \begin{enumerate}[(i)]
    \item Consider
      \[ f(X) = X^3 - 2 = (X - \sqrt[3]{2})(X -
      \omega \sqrt[3]{2})(X - \omega^2
      \sqrt[3]{2}) \in \ringadjoin \QQ[X] \]
      Then $\fieldadjoin \QQ(\omega, \sqrt[3]{2})$ is a
      \gls{splitting_field} for $f$ over $\QQ$.
      \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/cf0594c06ce211ee.png}
      \end{center}
      By the \nameref{tower_law}
      $\exdeg[\fieldadjoin\QQ(\omega,
      \sqrt[3]{2}) : \QQ]$ is $\le 6$ and is
      divisible by both $2$ and $3$. Since
      $\gcd(2, 3) = 1$, we get
      $\exdeg[\fieldadjoin \QQ(\omega,
      \sqrt[3]{2}) : \QQ] = 6$.
    \item Let $p$ be an odd prime and
      \begin{align*}
        f(X)
        &= \frac{X^p - 1}{X - 1} = X^{p - 1}
        + \cdots + X^2 + X + 1 \in \ringadjoin \QQ[X] \\
        &= \prod_{r = 1}^{p - 1} (X - \zeta_p^r)
      \end{align*}
      where $\zeta_p = e^{2\pi i / p}$. Then $f$
      has \gls{splitting_field}
      \[ \fieldadjoin \QQ(\zeta_p, \zeta_p^2, \ldots,
      \zeta_p^{p - 1}) = \fieldadjoin \QQ(\zeta_p) \]
      So in this case the \gls{splitting_field} is
      obtained by adjoining just one root.
    \item Let $f(X) = X^3 - 2 \in \ringadjoin \FF_7[X]$. Then
      $f$ is irreducible (since $2$ isn't a cube
      modulo $7$). Let $L = \ringadjoin \FF_7[X] / (f)$, so $L
      = \fieldadjoin \FF_7(\alpha)$ with $\alpha^3 = 2$. Noting
      that $2^3 \equiv 1 \pmod{7}$, we get that
      $f(X) = (X - \alpha)(X - 2\alpha)(X -
      4\alpha)$. So $L = \fieldadjoin \FF_7(\alpha)$ is a
      \gls{splitting_field} for $f$ over $\FF_7$.
  \end{enumerate}
\end{example*}