%! TEX root = MT.tex
% vim: tw=50
% 15/02/2025 11AM

\newpage
\section{Algebraically closed fields}

\begin{fcdefnstar}[Algebraically closed]
  Suppose $(K, +, \cdot)$ is a field (in
  $\mathcal{L}_{\text{rings}}$). It is
  \emph{algebraically closed} if every
  non-constant polynomial over $K$ has a root in
  $K$.
\end{fcdefnstar}

\begin{fcdefn}[ACF]
  \glssymboldefn{acf}%
  \label{defn:9.1}
  $\ACF$ is the $\mathcal{L}_{\text{rings}}$ theory
  axiomatising algebraically closed fields. It
  consists of:
  \begin{itemize}
    \item field axioms
    \item for every $d \ge 1$, we add an axiom:
    \[
      \forall v_0, \ldots, v_d, \exists x (v_d
      \neq 0 \implies v_0 + v_1 x + \cdots + v_d
      x^d = 0)
    .\]
  \end{itemize}
\end{fcdefn}

\begin{note*}
  This is an infinite axiomatisation.
\end{note*}

\begin{fcdefn}[ACF with characteristic]
  \label{defn:9.2}
  For $n \ge 1$, let $\chi_n$ be the sentence
  \[
    \ub{1 + \cdots + 1}_{\text{$n$ times}}
    = 0
  .\]
  Set
  \begin{align*}
    \ACF_0
    &= \ACF \cup \{\neg \chi_n : n \in
    \Nbb\} \\
    \ACF_p
    &= \ACF \cup \{\chi_p\}
    &&\text{(for $p$ a prime)}
  \end{align*}
\end{fcdefn}

\begin{fcthm}[]
  \label{thm:9.3}
  $\ACF_0$ and $\ACF_p$ are
  \cloze{$\kappa$-\gls{kcat} for $\kappa >
  \aleph_0$.}
\end{fcthm}

\begin{proof}
  The Transcendence degree of $k \models \acf$
  (and algebraically closed field) is the
  cardinality of the largest algebraically
  independent subset of $k$.

  For example,
  \begin{align*}
    \trdeg_{\Qbb}(\ol{\Qbb})
    &= 0 \\
    \trdeg_{\Qbb}(\ol{\Qbb(\pi)})
    &= 1 \\
    \trdeg_{\Qbb}(\Cbb)
    &= 2^{\aleph_0} \\
    \trdeg_{\Qbb}(\ol{\Qbb(x_i)_{i < \kappa}})
    &= \kappa
  \end{align*}
  From algebra we know:
  \begin{enumerate}[(1)]
    \item If $k, k' \models \acf$ then $k \cong
      k'$ if and only if
      \begin{itemize}
        \item $\trdeg(k) = \trdeg(k')$
        \item $\charr(k) = \charr(k')$
        \item $|k| = |k'|$
      \end{itemize}
    \item If $k \models \acf$, $\lambda =
      \trdeg_{\Qbb}(k)$, then $|\kappa| = \aleph_0
      + \lambda$. Thus if $k, k' \models \acf_0$
      or $\acf_p$ are uncountable and $|k| = |k'|$
      then $\trdeg_{\Qbb}(k) = \trdeg_{\Qbb}(k')$
      so $k \cong k'$. \qedhere
  \end{enumerate}
\end{proof}

\begin{corollary}[]
  \label{coor:9.4}
  $\acf_0$ and $\acf_p$ are complete.
\end{corollary}

\begin{proof}
  \nameref{thm:1.8}.
\end{proof}

\begin{remark*}
  $\acf_0$ and $\acf_p$ are not
  $\aleph_0$-\gls{kcat}. (Consider fields with
  different finite transcendence degrees).
\end{remark*}

\begin{fcdefn}[Polynomial map]
  \glsnoundefn{polymap}{polynomial map}{polynomial
  maps}%
  \label{defn:9.5}
  Let $k$ be a field. We say a function $\phi :
  K^m \to K^m$ is a \emph{polynomial map} if
  \[
    \Phi(\ol{x})
    = (p_1(x_1, \ldots, x_m), \ldots, p_n(x_1,
    \ldots, x_m))
  \]
  where each $p_1, \ldots, p_n$ is a polynomial.
\end{fcdefn}

\begin{fcthm}[Ax-Grothendieck]
  \label{thm:9.6}
  Assuming:
  - $k$ an algebraically closed field
  - $\Phi : K^n \to K^n$ an injective
  \gls{polymap}
  Then:
  $\Phi$ is surjective.
\end{fcthm}

\begin{proof}
  First suppose $K = \ol{\Fbb_p}$ for some prime
  $p$. $\ol{\Fbb_p} = \bigcup_k \Fbb_{p^k}$. Fix
  an $m$ such that the coefficients of $\Phi$ all
  lie in $\Fbb_{p^m}$. Note: $\ol{\Fbb_p} =
  \bigcup_k \Fbb_{p^{mk}}$.

  For any $k \ge 1$, $\Phi$ induces an injective
  \gls{polymap} $\Fbb_{p^{km}} \to \Fbb_{p^{km}}$
  which has to be surjective (as finite field).
  Hence
  \begin{align*}
    \Phi(\ol{\Fbb_p}^n)
    &= \Phi\left(\bigcup_k \Fbb_{p^{mk}}^n\right)
    \\
    &= \bigcup_k \Phi(\Fbb_{p^{mk}}^n) \\
    &= \bigcup_k \Fbb_{p^{mk}}^n \\
    &= \ol{\Fbb_p}^n
  \end{align*}
  So $\Phi$ is surjective.

  Given $n, d \ge 1$, let $\psi_{n, d}$ be the
  $\mathcal{L}$-sentence expressing ``every
  injective \gls{polymap} with $n$-coordinates,
  each of which is a polynomial in $n$ variables
  with degree at most $d$, is surjective''.

  Exercise: Show that this is a first order
  $\mathcal{L}_{\text{rings}}$ sentence.

  Now $\ol{\Fbb_p} \models \psi_{n, d}$ for all $n, d
  \ge 1$ and $p$ prime.

  $\acf_p$ is \gls{compt}, so $\acf_p \models
  \psi_{n, d}$. Now consider $\acf_0$. Suppose for
  contradiction that $\acf_0 \not\models \psi_{n,
  d}$ for some $n, d$, so $\acf_0 \models
  \neg\psi_{n, d}$.

  By compactness, there exists $\Sigma
  \subseteq \acf_0$ finite such that $\Sigma
  \models \neg \psi_{n, d}$.

  In particular, $\Sigma \subseteq \acf \cup \{\neg
  \chi_0, \ldots, \neg \chi_m\}$ for some $m$.
  Choose a prime $p$ such that $p > m$ and $\acf_p
  \vdash \Sigma$.

  So must have $\acf_p \models \neg \psi_{n, d}$,
  contradiction.
\end{proof}

\begin{fcthm}[Lipschiptz principal]
  \label{thm:9.7}
  Assuming:
  - $\phi$ an $\mathcal{L}_{\text{rings}}$
  sentence
  Thens:[(1)]tfae
  - $\acf_0 \models \phi$, i.e. $\phi$ in every $k
  \models \acf_0$
  - $\acf_0 \cup \{\phi\}$ is consistent
  - there exists some $n > 0$ such that $\acf_p
  \models \phi$ for any $p > n$
  - for all $n > 0$, there exists some $p >
  n$ such that $\acf_p \cup \{\phi\}$ is
  consistent
\end{fcthm}

\newpage
\section{Diagrams}

Let $\mathcal{N}, \mathcal{M}$ be
$\mathcal{L}$-structures.

\begin{remark}
  \label{remark:10.1}
  If $h : \mathcal{M} \to \mathcal{N}$ is an
  (elementary) $\mathcal{L}$-embedding then after
  identifying $a \in M$ with $h(a) \in N$ we can
  view $M$ as an (elementary) \gls{substruc} of
  $N$.
  \begin{center}
    \includegraphics[width=0.4\linewidth]{images/f32ef80cefb14b88.png}
  \end{center}
  Given $A \subseteq M$, let $\mathcal{L}_A =
  \mathcal{L} \cup \{\ul{a} : a \in A\}$ where
  $\ul{a}$ is a new constant symbol. Then $M$ is
  an $\mathcal{L}_A$-structure. Interpret $\ul{a}$
  as $a$.
\end{remark}