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    \item[(iii) $\implies$ (ii)]
      Let $\mathcal{M}, \mathcal{N} \models T$, $A
      \subseteq \mathcal{M}$, $A \subseteq
      \mathcal{N}$. Let $\varphi(\ol{x}, y)$ be a
      quantifier free formula and let $\ol{a} \in
      A$ such that $M \models \exists y
      \varphi(\ol{a}, y)$.

      As $A \subseteq \mathcal{M}$, $A \subseteq
      \mathcal{N}$, we have $\mathcal{M},
      \mathcal{N} \models T \cup \Diag(A)$, so by
      (iii) we have $\mathcal{M}
      \ee_{\mathcal{L}_A} \mathcal{N}$ so
      $\mathcal{N} \models \exists \varphi(\ol{a},
      y)$.
    \item[(ii) $\implies$ (i)] We want to show
      \gls{qe}.

      By \cref{lemma:11.3}, it is sufficient to
      show for $\varphi(\ol{x}, y)$ quantifier
      free, we can find $\psi(\ol{x})$ quantifier
      free such that
      \[
        T \models \forall \ol{x}(\exists y,
        \varphi(\ol{x}, y) \leftrightarrow
        \psi(\ol{x}))
      .\]
      Let $\mathcal{L}^* = \mathcal{L} \cup \{c_1,
      \ldots, c_n\}$ where each $c_i$ is a new
      constant (with $n = |\ol{x}|$). Let
      \[
        \Gamma
        = \{\psi(\ol{x}) : \psi(\ol{x}) \text{
        quantifier free formula such that } T
        \models \forall \ol{x}, (\exists y,
        \varphi(\ol{x}, y) \to \psi(\ol{x}))\}
      .\]
      In definable sets:
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/3981abbde52f4ca3.png}
      \end{center}
      Claim: $T \cup \Gamma \models \exists y,
      \varphi(\ol{x}, y)$.

      Proof: Suppose not. Then there is an
      $\mathcal{L}^*$-structure
      \[
        \mathcal{N} \models T \cup \Gamma \cup
        \{\neg \exists y, \varphi(\eps, y)\}
      .\]
      Let $a_i = c_i^N$, and let $A \subseteq
      \mathcal{N}$ be the \gls{substruc} generated
      by $a_1, \ldots, a_n$ in $N$.

      Then $\mathcal{N} \models T$, $A \subseteq
      \mathcal{N}$, $\mathcal{N} \models \neg
      \exists y, \varphi(\ol{c}, y)$. Any $b \in
      A$ is of the form $t^N(\ol{a})$ for an
      $\mathcal{L}$-term $t$ (exercise).

      So $\Diag(A)$ can be viewed as an
      $\mathcal{L}^*$-structure by replacing
      $\ul{b}$ ($b = t^N(\ol{a})$) with
      $t(\ol{c})$ in $\mathcal{L}^*$. Let
      \[
        \Sigma = T \cup \Diag(A) \cup \{\exists y,
        \varphi(\ol{x}, y)\}
      .\]
      IDEA: build $M \models \Sigma$, $M \models T
      \cup \Diag(A)$ and $M \models \exists y,
      \varphi(\ol{a}, Y)$, contradicting (ii).

      Suffices to prove $\Sigma$ is consistent.

      Assume $\Sigma$ is inconsistent. Then by
      compactness we have $\psi_1(\ol{x}), \ldots,
      \psi_n(\ol{x})$ quantifier free
      $\mathcal{L}$-formulas with
      \begin{itemize}
        \item $\psi_1(\ol{c}), \ldots,
          \psi_m(\ol{c}) \in \Diag(A)$.
        \item $T \cup \left\{ \bigwedge_{i = 1}^m
          \psi_i(\ol{c})\right\} \cup \{\exists y,
          \varphi(\ol{c}, y)\}$ is unsatisfiable.
      \end{itemize}
      In definable sets:
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/5e79c9c01721419d.png}
      \end{center}
      Let $\psi(\ol{x})$ be $\neg \bigwedge_{i =
      1}^m \psi_i(\ol{x})$, then $T \models
      \exists y, \varphi(\ol{c}, y) \to
      \psi(\ol{c})$.

      So $T \models \forall \ol{x} (\exists y,
      \varphi(\ol{x}, y) \to \psi(\ol{x}))$ so
      $\psi(\ol{c}) \in \Gamma$. So $\mathcal{N}
      \models \psi(\ol{c})$ but also $\mathcal{N}
      \models \Diag(A)$ ($A \subseteq
      \mathcal{N}$). Then
      \[
        N \models \ub{\bigwedge_{i = 1}^m
        \psi_i(\ol{c})}_{\in \Diag(A)} = \neg\psi(\ol{c})
      \]
      contradiction. So by compactness, $\Sigma$
      is consistent.

      This proves the claim.

      Reminder: the claim was that $T \cup \Gamma
      \models \exists y, \varphi(\ol{x}, y)$. So
      by compactness, there are $\psi_1(\ol{x}),
      \ldots, \psi_m(\ol{x})$ quantifier free such
      that
      \[
        T \cup \ub{\{\psi(\ol{c}), \ldots,
        \psi_m(\ol{c})\}}_{\subseteq \Gamma}
        \models \exists y, \varphi(\ol{x}, y)
      .\]
      Recall
      \[
        T
        \models \forall \ol{x} (\exists y,
        \varphi(\ol{x}, y) \to \bigwedge_{i = 1}^m
        \psi_i(\ol{x}))
      \]
      by choice of $n$. Let
      \[
        \psi(\ol{x}) = \bigwedge_{i = 1}^m
        \psi_i(\ol{x})
      .\]
      Then
      \[
        T \models \psi(\ol{c}) \to \exists y,
        \varphi(\ol{c}, y)
      \]
      hence
      \[
        T \models \forall \ol{x} (\psi(\ol{x}) \to
        \exists y, \varphi(\ol{x}, y))
      .\]
      Thus $T \models \forall \ol{x},
      (\psi(\ol{x}) \leftrightarrow \exists y,
      \varphi(\ol{x}, y))$.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item In (iii) we may assume $A \subseteq M
      \models T$, as otherwise $T \cup \Diag(A)$
      is inconsistent and hence trivially
      \gls{compt}.
    \item In (ii) and (iii) we may assume
      $\mathcal{A}$ is finitely generated.
  \end{enumerate}
\end{remark*}

\newpage
\section{Examples}

\begin{fcthm}[]
  \label{thm:12.1}
  $\acf$ has \cloze{\gls{qe}}.
\end{fcthm}

\begin{proof}
  We will show that condition \cref{thm:11.4}(iii)
  holds.

  So we need to show that for any
  finitely generated $A$, $T \cup \Diag(A)$
  is \gls{compt}.

  Fix a finitely generated $\mathcal{L}$-structure
  $A$ and show $\acf \cup \Diag(A)$ is
  \gls{compt}. Use \nameref{losvaught}. Fix $K_1,
  K_2 \models \acf \cup \Diag(A)$ uncountable,
  wiht $|K_1| = |K_2|$. $A$ is a finitely
  generated integral domain contained in $K_1,
  K_2$.

  So since $A$ contains $1$, it determines the
  characteristic. So $\charr(K_1) = \charr(K_2)$. So
  $K_1 \mcong K_2$ (in
  $\mathcal{L}_{\text{rings}}$).

  Need an $\mathcal{L}_A$-isomorphism, i.e. an
  isomorphism $\Phi : K_1 \to K_2$ preserving $A$.
  Consider $F_i$ the fraction field of $A$ in
  $K_i$. The field of fractions of an integral
  domain is unique up to isomorphism, i.e.
  $\exists \tau : F_1 \to F_2$ preserves $A$
  pointwise.

  $A$ is finitely generated (hence
  finite $\trdeg$), so
  $\trdeg(K_1 / F_1) = \trdeg(K_2 / F_2)$.
  Therefore $\tau$ extends to $\tau^* : K_1 \to
  K_2$ fixing $A$ pointwise. So $K_1
  \mcong_{\mathcal{L}_A} K_2$.
\end{proof}

\begin{fcdefn}[Constructible]
  \label{defn:12.2}
  Let $F$ be a field. We say that $X \subseteq
  F^n$ is \emph{constructible} if it is a boolean
  combination of subsets of $F^n$ defined by
  $p(x_1, \ldots, x_n) = 0$ for $p(\ol{x}) \in
  F[\ol{x}]$.
\end{fcdefn}

\begin{fccoro}[Chevalley's Theorem]
  Assuming:
  - $K \models \acf$
  - $X \subseteq K^n$ a constructible set
  Then:
  the projection
  \[
    Y = \{(a_1, \ldots, a_{n - 1}) \in K^{n - 1} :
    (\ol{a}, b) \in X \text{ for some $b \in K$}\}
  \]
  of $X$ is also constructible.
\end{fccoro}