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\begin{notation*}
  \glssymboldefn{mequiv}%
  Given $\mathcal{M}$, $\ol{a}, \ol{b} \in M^n$,
  write $\ol{a} \equiv^M \ol{b}$ if $\tp^M(\ol{a})
  \equiv^M \ol{b}$.

  So $\mathcal{M}$ is $\aleph_0$-\gls{homog} if and only if
  whenever $\ol{a} \equiv^M \ol{b}$ and $c \in M$,
  there exists $d \in M$ with $\ol{a} c \mequiv
  \ol{b} d$.
\end{notation*}

\begin{fclemma}[]
  \label{lemma:15.6}
  Assuming:
  - $T$ a complete $\mathcal{L}$-theory with infinite models
  - $\mathcal{M} \models T$
  Then:
  there is an $\mathcal{N} \esup \mathcal{M}$ with
  $|N| \le |M| + |L|$ and $\mathcal{N}$ is
  $\aleph_0$-\gls{homog}.
\end{fclemma}

\begin{proof}
  \textbf{First claim:} For any $M \models T$,
  there is $\mathcal{N} \esup \mathcal{M}$ with
  $|N| \le |M| + |L|$ and for any $\ol{a}, \ol{b},
  c$ from $M$ such that $\ol{a} \mequiv \ol{b}$
  there is some $d \in N$ with $\ol{a}c \mequiv[N]
  \ol{b} d$.

  Proof of claim: Enumerate all $(\ol{a}, \ol{b},
  c)$ as $(\ol{a}_\alpha, \ol{b}_\alpha,
  c_\alpha)_{\alpha \le |M|}$. Now let
  $\mathcal{M}_0 = \mathcal{M}$, and use
  transfinite induction to form a chain
  $(\mathcal{M}_\alpha)_{\alpha < |M|}$.
  \begin{itemize}
    \item $\alpha$ is a limit ordinal: set
      $\mathcal{M}_\alpha = \bigcup_{i < \alpha}
      \mathcal{M}_i$ (then $|M_\alpha| \le
      |\alpha|(|M| + |L|) = |M| + |L|$ as $\alpha
      \le |M|$).
    \item Given $\alpha$ (not a limit ordinal),
      look at $(a_\alpha, b_\alpha, c_\alpha)$.
      Assume $\ol{a}_\alpha \mequiv[M_\alpha]
      \ol{b}_\alpha$ so $f_\alpha : \ol{a}_\alpha
      \to \ol{b}_\alpha$ \gls{partelem}. Then we
      apply \cref{prop:13.4} (Note the elementary
      super structure construnted here is of size
      $\le |M| + |L|$) to find
      $\mathcal{M}_{\alpha + 1} \esup
      \mathcal{M}_\alpha$ with
      $|\mathcal{M}_{\alpha + 1}| \le
      |\mathcal{M}_\alpha| + |L| \le |M| + |L|$
      with a $d \in \mathcal{M}_{\alpha + 1}$
      realising $f_\alpha(\tp(c_\alpha /
      a_\alpha))$. Then by construction,
      $(\ol{a}_\alpha c)
      \mequiv[\mathcal{M}_{\alpha + 1}]
      (\ol{b}_\alpha d)$.
  \end{itemize}
  Now let $\mathcal{N} = \bigcup_{\alpha <
  |M|} \mathcal{M}_\alpha$. Note that we might
  have introduced new elements. Note $|N| \le
  |M|(|M| + |L|) = |M| + |L|$.

  We build a new chain $\mathcal{M} =
  \mathcal{N}_0 \esub \mathcal{N}_1 \esub
  \mathcal{N}_2 \esub \cdots$ of countably
  many steps with $|N_1| \le |M| + |L|$ and
  such that for any $\ol{a}, \ol{b}, \ol{c}
  \in N$, if $\ol{a} \mequiv \ol{b}$ then
  there is $d \in N_{i + 1}$ such that $\ol{a}
  c \mequiv[N_{i + 1}] \ol{b} d$. We do this
  by iterating the claim.

  FInally let $\mathcal{N} = \bigcup_{i <
  \aleph_0} \mathcal{N}_i$. Then:
  \begin{itemize}
    \item $|N| \le |M| + |L|$.
    \item $\mathcal{N}$ is
      $\aleph_0$-\gls{homog} as any $\ol{a},
      \ol{b}, c$ from $\mathcal{N}$ lie in
      $\mathcal{N}_i$ for some $i$.
  \end{itemize}
\end{proof}

\begin{fcdefn}[Saturated]
  \glsadjdefn{satur}{saturated}{model}%
  \label{defn:15.7}
  We say $\mathcal{M}$ is \emph{saturated} if it
  is $|M|$-saturated.
\end{fcdefn}

\begin{fcthm}[]
  \label{thm:15.8}
  Assuming:
  - $T$ a complete $\mathcal{L}$-theory with infinite models
  - $\mathcal{L}$ is countable
  Then:
  \begin{iffc}
    \lhs
    $T$ has a countable \gls{satur} model
    \rhs
    $S_n(T)$ is countable for every $n \ge 1$.
  \end{iffc}
\end{fcthm}

\begin{iffproof}
  \rightimpl
  $\mathcal{M} \models T$ countable, \gls{satur}.
  \begin{itemize}
    \item $M^n$ is countable for all $n \in \Nbb$.
    \item We have a map $p \to \ol{a} \models p$
      ($p \in S_n(T)$, $\ol{a}$ some realisation).
      This is map since $\mathcal{M}$ \gls{satur},
      and injective (because complete types).
  \end{itemize}
  So $S_n(T)$ is countable.

  \leftimpl
  Enumerate $\bigcup_{n \ge 1} S_n(T) =
  \{p_1, p_2, p_3, \ldots\}$. Fix a countable
  $M_0 \models T$, and build a chain
  $\mathcal{M}_0 \esub \mathcal{M}_1 \esub
  \mathcal{M}_2 \esub \cdots$ such that
  $\mathcal{M}_i$ realises $p_i$ and is
  countable, using \cref{prop:13.4}.

  Get $\mathcal{N} = \bigcup_{i \in \Nbb}
  \mathcal{M}_i$. Then $\mathcal{N} \models T$
  and is countable. $\mathcal{N}$ realises all
  types over $\emptyset$.

  Apply \cref{lemma:15.6} to get $M \esup
  \mathcal{N}$ countable and
  $\aleph_0$-\gls{homog} structure.

  So by \cref{prop:15.5} is is
  $\aleph_0$-\gls{satur}.
\end{iffproof}

\begin{example}
  \label{eg:15.9}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Let $T = \acf_p$ and let
      \[
        F = \begin{cases}
          \Qbb
          & \text{if $p = 0$} \\
          \Fbb_p
          & \text{if $p > \infty$}
        \end{cases}
      .\]
      Then
      \[
        S_n(T)
        = \Spec(F[x_1, \ldots, x_n])
      .\]
      Thus $S_n(T)$ is countable, since every
      ideal in $\Spec(F[x_1, \ldots, x_n])$ is
      finitely generated (Hilbert's basis
      theorem). So by \cref{thm:15.8}, $\acf_p$
      has a countable \gls{satur} model which is
      the model of transcendence degree
      $\aleph_0$.

      Note: if $F \models \acf_p$ has
      transcendence degree $n$, the type
      determined by ``$x_1, \ldots, x_{n + 1}$''
      is an algebraically independent set.
    \item Let $T = \mathrm{TFDAG}$ (torsion free,
      divisible abelian groups). This has a
      countable \gls{satur} model, which is the
      $\Qbb$-vector space of dimension $\aleph_0$.
    \item Let $T = \Th(\Zbb, +, 0)$. For $n \ge
      1$, let $\delta_n(x)$ be the
      $\mathcal{L}$-formula $\exists y (x = ny)$,
      and let $\Pbb$ be the set of primes. Given
      $X \subseteq \Pbb$ finite,
      \[
        q_x = \{\delta_n(x) : n \in X\} \cup
        \{\neg \delta_n(x) : n \in \Pbb \setminus
        X\}
      .\]
      $q_X$ is satisfiable in $\Zbb$, thus
      $\exists p_X \in S_1(T)$ with $q_x \le p_x$.

      If $X \neq Y$, then $p_X \neq p_Y$, so
      $|S_1(T)| \ge 2^{\aleph_0}$. By
      \cref{thm:15.8}, $T$ doesn't have a
      countable \gls{satur} model.
  \end{enumerate}
\end{example}

\begin{example}
  \label{eg:15.10}
  Let $M \models \RG$. We describe $\Snm_1^M(M)$.
  For $a \in M$, let $p_a \in \Snm_1^M(M)$ be the
  type containing ``$x = a$'' (exercise: why is
  this unique).

  FOr $V \le M$ set
  \[
    p_V = \{x \neq a : a \in M\} \cup \{F(x, q) :
    a \in V\} \cup \{\neg E(x, a) : a \in M
    \setminus V\}
  .\]
  $p_V$ is a $1$-type with respect to $M$, not
  realised in $M$, determines a complete $1$-type,
  as we have determined all atomic formula, by $a
  \in p_V$ determines a complete type.

  So $\Snm_1^M(M) = \{p_a : a \in M\} \cup \{p_V :
  V \subseteq M\}$.

  $|\Snm_1^M(M)| = 2^{|M|}$.

  Note: in general $|\Snm_1^M(A)| \le 2^{(|A| +
  |L| + \aleph_0)}$.

  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/094828c7338347fa.png}
  \end{center}
\end{example}
