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% 25/11/2023 12PM

Now we want to prove uniqueness of
\glspl{alg_closure}.

\begin{flashcard}[kembedding-into-alg-closure]
\begin{proposition}
  % Proposition 11.2
  \label{kembedding_into_alg_closure}
  Let $L \extendover K$ be an \gls{alg_ex}
  \gls{ex} and $M \extendover K$ be a
  \gls{field_ex} with $M$ \gls{alg_closed}. Then
  there exists a \Kembedding{K} $L \hookrightarrow
  M$.
\end{proposition}

\begin{proof}
  \cloze{
  Let
  \[ S = \{(F, \sigma) : (\sigma : F
  \hookrightarrow M) \text{ a \Kembedding{K}}, K
  \subset F \subset L\} ,\]
  with \gls{porder} $(F_1, \sigma_1) \le (F_2,
  \sigma_2)$ if $F_1 \subset F_2$ and
  $\sigma_2|_{F_1} = \sigma_1$. Then $(S, \le)$ is
  a \gls{poset}. It is non-empty as $(K, \id) \in
  S$. Suppose $T = \{(F_i, \sigma_i) : i \in I\}$
  is a \gls{chain} (where $I$ is some indexing
  set). Let $F = \bigcup_{i \in I} F_i$ (a field
  since $T$ is \gls{tot_orded}). Define
  \begin{align*}
    \sigma : F
    &\to M \\
    x
    &\mapsto \sigma_i(x)
    \quad\text{if $x \in F_i$}
  \end{align*}
  This is well-defined since $\sigma_i$ and
  $\sigma_j$ agree on $F_i \cap F_j$ (again since
  $T$ is \gls{tot_orded}).

  Then $(F, \sigma) \in S$ is an \gls{upp_bd} for
  $T$. Hence by \nameref{zorn}, $S$ has a
  \gls{max_el} element $(F, \sigma)$.

  Let $\alpha \in L$. Then $\alpha$ is \gls{alg_el}
  over $K$, hence \gls{alg_el} over $F$. By
  \cref{simple_extension_sigma_hom_bijection},
  we may extend $\sigma : F \hookrightarrow M$ to
  $\tau : \fadj F(\alpha) \hookrightarrow M$
  (using here that $M$ is \gls{alg_closed}). Then
  $(F, \sigma) \le (\fadj F(\alpha), \tau)$. Since
  $(F, \sigma)$ is \gls{max_el}, we must have
  $\fadj F(\alpha) = F$, so $\alpha \in F$.
  Therefore $F = L$ and $\sigma : L
  \hookrightarrow M$ is a \Kembedding{K} as
  required.
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[uniqueness-of-alg-closure]
\begin{corollary}[Uniqueness of algebraic closure]
  % Corollary 11.3
  \cloze{
  Let $K$ be a \gls{field}. Let $L_1$ and $L_2$ be
  \gls{alg_closure} of $K$. Then there exists a
  \Kisom{K} $\phi : L_1 \simeqto L_2$.}
\end{corollary}

\fcscrap{
\begin{note*}
  $\phi$ is not necessarily unique.
\end{note*}
}

\begin{proof}
  \cloze{Since $L_1 \extendover K$ is \gls{alg_ex} and
  $L_2 \extendover K$ is a \gls{field_ex} with
  $L_2$ \gls{alg_closed},
  \cref{kembedding_into_alg_closure} gives a
  \Kembedding{K} $\phi : L_1 \hookrightarrow L_2$.
  Any $\alpha \in L_2$ is \gls{alg_el} over $K$,
  hence \gls{alg_el} over $\phi(L_1)$. But
  $\phi(L_1) \cong L_1$ is \gls{alg_closed}, and
  therefore $\alpha \in \phi(L_1)$. This shows
  that $\phi$ is surjective.}
\end{proof}
\end{flashcard}

\newpage
\section{Artin's Theorem}

\begin{flashcard}[artins-thm-on-invts]
\begin{theorem}[Artin's Theorem on Invariants]
  % Theorem 12.2
  \label{artins_thm}
  \cloze{
  Let $L$ be a \gls{field} and $G \subset \Aut(L)$
  a finite subgroup. Then $L \extendover \fixedf
  L^G$ is a \gls{finite_ex} \gls{galois} \gls{ex}
  with \gls{galois_group} $G$. In particular,
  \[ \exdeg[L : \fixedf L^G] = \#G .\]
  }
\end{theorem}

\fcscrap{
\begin{remark*}
  Let $K = \fixedf L^G$. Then $G \subset \Aut(L
  \extendover K)$ and
  \[ K \subset \fixedf L^{\Aut(L \extendover K)}
  \subset \fixedf L^G = K .\]
  Therefore $K = \fixedf L^{\Aut(L / K)}$.
  \textit{If} we knew $L \extendover K$ is
  \gls{alg_ex}, then it would follow (by
  definition) that $L \extendover K$ is
  \gls{galois}. \textit{If} moreover we knew $L
  \extendover K$ is \gls{finite_ex} then
  \[ \fixedf L^G = \fixedf L^{\Gal(L \extendover
  K)} \stackrel{\text{\cref{ft_of_gal}}}{\implies}
  G = \Gal(L \extendover K) .\]
\end{remark*}
}

\begin{proof}
  \cloze{
  Let $K = \fixedf L^G$. Pick any $\alpha \in L$.
  Let
  \[ f(X) = \prod_{i = 1}^m (X - \alpha_i) \]
  where $\alpha_1, \ldots, \alpha_m$ are the
  distinct elements of $\orb_G(\alpha) =
  \{\sigma(\alpha) : \sigma \in G\}$. Then $\sigma
  f = f$ for all $\sigma \in G$, hence $f \in
  \radj K[X]$. Therefore $\alpha$ is \gls{alg_el}
  and \gls{sep_el} over $K$. Hence $L \extendover
  K$ is \gls{alg_ex} and \gls{sep_ex} and
  \[ \exdeg[\fadj K(\alpha) : K] \le \#G \qquad
  \forall \alpha \in L .\]
  Now pick $\alpha \in L$ with $\exdeg[\fadj
  K(\alpha) : K]$ maximal.

  \textbf{Claim:} $L = \fadj K(\alpha)$.

  \textit{Proof of Claim:} Let $\beta \in L$. Then
  $\fadj K(\alpha, \beta) \extendover K$ is
  \gls{finite_ex}  and \gls{sep_ex} so by
  \nameref{prim_el_thm}, $\fadj K(\alpha, \beta) =
  \fadj K(\theta)$ for some $\theta \in L$. By our
  choice of $\alpha$,
  \[ \exdeg[\fadj K(\theta) : K] \le \exdeg[\fadj
  K(\alpha) : K] .\]
  Since $\fadj K(\alpha) \subset \fadj K(\theta)$,
  this gives $\fadj K(\alpha) = \fadj K(\theta)$
  and hence $\beta \in \fadj K(\alpha)$. This
  proves the claim.

  Now
  \[ \#\Aut(L \extendover K) \le \exdeg[L : K]
  = \exdeg[\fadj K(\alpha) : K] \le \# G .\]
  Since $G \subset \Aut(L \extendover K)$, it
  follows that
  \[ \#\Aut(L \extendover K) = \exdeg[L : K] ,\]
  so by
  \cref{classification_of_finite_galois_extensions},
  $L \extendover K$ is \gls{galois}
  and $G = \Aut(L \extendover K)$.
  }
\end{proof}
\end{flashcard}

\begin{example*}
  Let $L = \fadj \CC(X_1, X_2)$. Define $\sigma,
  \tau \in \Aut(L)$ by
  \begin{align*}
    (\sigma f)(X_1, X_2)
    &= f(i X_1, -iX_2) \\
    (\tau f)(X_1, X_2)
    &= f(X_2, X_1)
  \end{align*}
  Let $G = \langle \sigma, \tau \rangle \cong
  D_8$. Aim: compute $\fixedf L^G$. We spot that
  $X_1 X_2, X_1^4 + X_2^4 \in \fixedf L^G$. So we
  have:
  \begin{center}
    \begin{tikzcd}
      L \ar[no head]{d}{8}
      \ar[swap,no head,bend right=60]{dd}{\le 8} \\
      \fixedf L^G \ar[no head]{d} \\
      \fadj \CC(X_1 X_2, X_1^4 + X_2^4)
    \end{tikzcd}
  \end{center}
  Let
  \[ f(T) = (T^4 - X_1^4)(T^4 - X_2^4) = T^8 -
  (X_1^4 + X_2^4)T^4 + (X_1 X_2)^4 \in \radj \fadj
  \CC(X_1 X_2, X_1^4 + X_2^4)[T] \]
  hence
  \[ \exdeg[L : \fadj \CC(X_1 X_2, X_1^4 + X_2^4)]
  \le 8 \]
  Then \nameref{artins_thm} implies $\exdeg[L :
  \fixedf L^G] = \#G = 8$. Then by the
  \nameref{tower_law},
  \[ \fixedf L^G = \fadj \CC(X_1 X_2, X_1^4 +
  X_2^4) .\]
\end{example*}
