%! TEX root = MT.tex
% vim: tw=50
% 01/03/2025 11AM

\begin{proof}
  By assumption $p \cup \Th_A(\mathcal{M})$ is
  consistent.

  Need to show $p \cup \Th_M(\mathcal{M})$ is
  consistent.

  Fix $\Sigma \subseteq p \cup \Th_M(\mathcal{M})$
  finite. $\Sigma \subseteq p \cup \{\varphi_1,
  \ldots, \varphi_t\}$, $\varphi_i$ an
  $\mathcal{L}_M$-sentence with $M \models
  \varphi_i$.

  Let $\varphi^*$ be $\bigwedge_{i = 1}^t
  \varphi_i$, then $\varphi^*$ can be written
  $\varphi^*(b_1, \ldots, b_m)$ where $b_1,
  \ldots, b_m \in M \setminus A$ and $\varphi^*(x_1,
  \ldots, x_m)$ an $\mathcal{L}_A$-formula.

  Since $\mathcal{M} \models \varphi^*(b_1,
  \ldots, b_m)$ we get $\mathcal{M} \models
  \exists \ol{v}, \varphi^*(v_1, \ldots, v_m)$, so
  $\exists \ol{v} \varphi^*(\ol{v}) \in
  \Th_A(\mathcal{M})$.

  So as $\Th_A(\mathcal{M}) \cup p$ is consistent,
  we have $N \models \Th_A(\mathcal{M}) \cup p$
  with
  \begin{itemize}
    \item $\ol{c} \in N^m$ with $\mathcal{N}
      \models \varphi^*(\ol{c})$.
    \item $\ol{a} \in N^m$ with $\mathcal{N}
      \models p(\ol{a})$.
  \end{itemize}
  Expand $\mathcal{N}$ to an
  $\mathcal{L}_M$-structure, i.e. let
  \begin{itemize}
    \item $b_i^{\mathcal{N}} = c_i$ for $i \in
      \{1, \ldots, m\}$.
    \item $\ol{b}^{\mathcal{N}}$ arbitary for $b
      \in M \setminus \langle (A \cup \{b_1,
      \ldots, b_m\})\rangle$.
  \end{itemize}
  Then $\mathcal{N} \models \varphi(b_1, \ldots,
  b_m)$. So $\mathcal{N} \models \varphi^*$, so
  $\mathcal{N} \models \Sigma$.
\end{proof}

\begin{remark}
  \label{remark:13.5}
  If $\mathcal{M} \esub \mathcal{N}$ and $A
  \subseteq M$ then $\Snm_n^{\mathcal{M}}(A) =
  \Snm_n^{\mathcal{N}}(A)$ since $\Th_A(\mathcal{M}) =
  \Th_A(\mathcal{N})$.
\end{remark}

\begin{remark}
  \label{remark:13.6}
  $p$ is an \gls{ntype} over $A$ with respect to
  $M$ if and only if $\forall q \subseteq p$
  finite, $\exists \ol{a} \in M^n$ such that
  $\ol{a} \not\models q$.

  \begin{iffproof}
    \rightimpl
    Clear.

    \leftimpl
    Choose $N \esup M$ realising $p$. Fix $q
    \subseteq p$ finite, $\varphi(\ol{x})$ the
    conjunction of all $\mathcal{L}_A$-formulas in
    $q$. Then $\mathcal{N} \models \exists \ol{x},
    \varphi(\ol{x})$. So $\mathcal{M} \models
    \exists \ol{x}, \varphi(\ol{x})$, i.e. $q$ is
    realised in $\mathcal{M}$.
  \end{iffproof}
\end{remark}

\begin{example}
  \label{eg:13.7}
  Suppose $K \models \acf$, $A \subseteq K$. We
  want to describe $\Snm_n^K(A)$. Fix $p \in
  \Snm_n^K(A)$. By \gls{qe} we only need to
  consider quantifier free formulas.

  Moreover,
  \begin{align*}
    \varphi \wedge \psi \in p
    &\iff \varphi, \psi \in p \\
    \neg \psi \in p
    &\iff \varphi \notin p
  \end{align*}
  So we can concentrate on atomic formulas
  $\varphi$, polynomials in variables $x_1,
  \ldots, x_n$ over the field generated by $A$,
  say $F$ (i.e. $F[\ol{x}]$).

  Let $I_p = \{f(\ol{x}) \in F[\ol{x}] : f(\ol{x})
  = 0 \in p\}$. Then $I_p$ is a prime ideal and $p
  \mapsto I_p$ is a bijection $\Snm_n^K(A) \mapsto
  \Spec F[\ol{x}]$ ($\Spec F[\ol{x}]$ is the set
  of prime ideals of $F[\ol{x}]$).
  So $\Snm_1^K(A)$ consists of
  \[
    \{p_a : a \in A\} \cup \{q\}
  ,\]
  where $p_a$ contains (and thus is determined by)
  $x = a$ and $q = \{x \neq a : a \in F\}$.

  $|\Snm_1^K(K)| = |K|$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/002a4b2779094708.png}
  \end{center}
\end{example}
