[proofplan] The closure axioms follow directly from algebraicity: algebraic formulas use only finitely many parameters, monotonicity is immediate, and idempotence is obtained by replacing algebraic parameters by their finite defining sets. The only substantial axiom is exchange. Given $a \in \operatorname{acl}(A b) \setminus \operatorname{acl}(A)$, we build an $A$-definable relation $R(x,y)$ with uniformly finite $x$-fibres and $R(a,b)$. Strong minimality then forces the $y$-fibre $R(a,M)$ to be finite, because otherwise a saturation argument would produce one $y$ with too many $x$-preimages, contradicting the uniform fibre bound. [/proofplan]

[step:Verify the closure axioms before exchange] Let $A \subseteq M$. Here $\operatorname{acl}_M(A)$ denotes the set of elements of $M$ algebraic over $A$, meaning that $c \in \operatorname{acl}_M(A)$ iff there is a formula with finitely many realizations in $M$, using only parameters from $A$, that is satisfied by $c$. Since every element $a \in A$ is defined over $A$ by the formula $x = a$, we have $A \subseteq \operatorname{acl}_M(A)$. If $A \subseteq B \subseteq M$ and $c \in \operatorname{acl}_M(A)$, then some formula $\varphi(x,\alpha)$ with parameter tuple $\alpha \in A^r$ has finitely many realizations and satisfies $\varphi(c,\alpha)$. Since $\alpha$ is also a tuple from $B$, the same formula witnesses $c \in \operatorname{acl}_M(B)$. For finite character, suppose $c \in \operatorname{acl}_M(A)$. Then there are a formula $\varphi(x,\alpha)$, a finite parameter tuple $\alpha = (\alpha_1,\dots,\alpha_r) \in A^r$, and an integer $N \in \mathbb{N}$ such that $\varphi(c,\alpha)$ holds and $\varphi(x,\alpha)$ has at most $N$ realizations. Let $A_0 := \{\alpha_1,\dots,\alpha_r\}$. Then $A_0 \subseteq A$ is finite and $c \in \operatorname{acl}_M(A_0)$. It remains to prove idempotence. The inclusion $\operatorname{acl}_M(A) \subseteq \operatorname{acl}_M(\operatorname{acl}_M(A))$ follows from extensivity and monotonicity. Conversely, let $c \in \operatorname{acl}_M(\operatorname{acl}_M(A))$. Then there are a formula $\varphi(x,d)$, a tuple $d = (d_1,\dots,d_r) \in \operatorname{acl}_M(A)^r$, and an integer $N \in \mathbb{N}$ such that $\varphi(c,d)$ holds and $\varphi(x,d)$ has at most $N$ realizations. For each index $i \in \{1,\dots,r\}$, choose a formula $\psi_i(z_i,\alpha_i)$ with finite parameter tuple $\alpha_i$ from $A$ and an integer $N_i \in \mathbb{N}$ such that $\psi_i(d_i,\alpha_i)$ holds and $\psi_i(z_i,\alpha_i)$ has at most $N_i$ realizations. Let $\alpha$ be the concatenation of the tuples $\alpha_1,\dots,\alpha_r$. We use $\exists^{\leq N}w\,\varphi(w,z_1,\dots,z_r)$ to denote the first-order assertion that at most $N$ elements $w$ satisfy $\varphi(w,z_1,\dots,z_r)$. Define the formula \begin{align*} \theta(x,\alpha) :\Longleftrightarrow \exists z_1 \cdots \exists z_r \left(\bigwedge_{i=1}^{r} \psi_i(z_i,\alpha_i) \wedge \varphi(x,z_1,\dots,z_r) \wedge \exists^{\leq N}w\,\varphi(w,z_1,\dots,z_r)\right). \end{align*} The element $c$ satisfies $\theta(c,\alpha)$ by taking $z_i = d_i$ for each $i \in \{1,\dots,r\}$, because $\varphi(x,d)$ has at most $N$ realizations. The tuple $(z_1,\dots,z_r)$ has at most $\prod_{i=1}^{r} N_i$ possible values, since each formula $\psi_i(z_i,\alpha_i)$ has at most $N_i$ realizations. For each such tuple satisfying the final bounded-fibre condition, the set of possible $x$ has at most $N$ elements. Hence $\theta(x,\alpha)$ has at most $N\prod_{i=1}^{r}N_i$ realizations in $M$. Since $\alpha$ is a tuple from $A$ and $M \models \theta(c,\alpha)$, this witnesses $c \in \operatorname{acl}_M(A)$. Thus idempotence holds. [/step]

[step:Build a uniformly finite relation from algebraicity of $a$ over $A b$]Let $A \subseteq M$ and let $a,b \in M$ satisfy \begin{align*} a \in \operatorname{acl}_M(A \cup \{b\}) \setminus \operatorname{acl}_M(A). \end{align*} By algebraicity over $A \cup \{b\}$, there are a formula $\varphi(x,y,\alpha)$, a finite tuple $\alpha \in A^r$, and an integer $n \in \mathbb{N}$ such that $M \models \varphi(a,b,\alpha)$ and the set \begin{align*} \{x \in M : M \models \varphi(x,b,\alpha)\} \end{align*} has at most $n$ elements. We use $\exists^{\leq n}u\,\varphi(u,y,\alpha)$ to denote the first-order assertion that at most $n$ elements $u$ satisfy $\varphi(u,y,\alpha)$. Define an $A$-definable binary relation $R(x,y)$ by \begin{align*} R(x,y) :\Longleftrightarrow \varphi(x,y,\alpha) \wedge \exists^{\leq n} u\, \varphi(u,y,\alpha). \end{align*} Then $M \models R(a,b)$. Also, for every $y \in M$, the fibre \begin{align*} R(M,y) := \{x \in M : M \models R(x,y)\} \end{align*} has cardinality at most $n$.[/step]

[guided]We start from the assumption that $a$ is algebraic over $A \cup \{b\}$. This means that $a$ lies in a finite set definable using parameters from $A$ together with the parameter $b$. Concretely, there are a formula $\varphi(x,y,\alpha)$, where $\alpha = (\alpha_1,\dots,\alpha_r)$ is a finite tuple from $A$, and an integer $n \in \mathbb{N}$ such that $M \models \varphi(a,b,\alpha)$ and \begin{align*} \left|\{x \in M : M \models \varphi(x,b,\alpha)\}\right| \leq n. \end{align*} The difficulty is that this bound is initially known only for the single parameter value $y=b$. To use strong minimality uniformly, we replace $\varphi$ by a relation whose $x$-fibres are bounded by $n$ for every value of $y$. Define \begin{align*} R(x,y) :\Longleftrightarrow \varphi(x,y,\alpha) \wedge \exists^{\leq n} u\, \varphi(u,y,\alpha). \end{align*} The notation $\exists^{\leq n}u\,\varphi(u,y,\alpha)$ denotes the first-order assertion that at most $n$ elements $u$ satisfy $\varphi(u,y,\alpha)$. Because the original fibre over $b$ has at most $n$ elements and $M \models \varphi(a,b,\alpha)$, we have $M \models R(a,b)$. Moreover, for every $y \in M$, if $M \models R(x,y)$, then $\varphi(x,y,\alpha)$ holds and the formula $\varphi(u,y,\alpha)$ has at most $n$ realizations. Hence \begin{align*} |R(M,y)| \leq n \end{align*} for every $y \in M$. This is the uniform finite-fibre condition that will contradict strong minimality if too many distinct $x$ have large $R$-fibres in the $y$-direction.[/guided]

[step:Show the fibre $R(a,M)$ is finite] We claim that \begin{align*} R(a,M) := \{y \in M : M \models R(a,y)\} \end{align*} is finite. Suppose instead that $R(a,M)$ is infinite. Let $S \succeq M$ be an elementary extension that is countably saturated over the parameter set $A \cup \{a\}$. This amount of saturation is enough to realize the countable partial type in the variables $x_0,\dots,x_n$ constructed below. We interpret $R$ in $S$ using the same parameter tuple $\alpha \in A^r$. For each $m \in \mathbb{N}$, define the $A$-definable set \begin{align*} X_m := \{x \in S : S \models \exists y_1 \cdots \exists y_m \left(\bigwedge_{1 \leq i < j \leq m} y_i \neq y_j \wedge \bigwedge_{i=1}^{m} R(x,y_i)\right)\}. \end{align*} Since $R(a,M)$ is infinite, for every $m \in \mathbb{N}$ the model $M$ satisfies $a \in X_m$, and therefore $S$ also satisfies $a \in X_m$ by elementarity. Because $a \notin \operatorname{acl}(A)$, no finite $A$-definable subset of $S$ contains $a$. Hence each $X_m$ is infinite. Since $T$ is strongly minimal, every $A$-definable subset of $S$ in one variable is finite or cofinite; therefore each $X_m$ is cofinite in $S$. Consider the partial type in variables $x_0,\dots,x_n$ \begin{align*} \Sigma(x_0,\dots,x_n) := \{x_i \neq x_j : 0 \leq i < j \leq n\} \cup \{X_m(x_i) : m \in \mathbb{N},\ 0 \leq i \leq n\}. \end{align*} This type is finitely satisfiable in $S$: for a finite subcollection, let $m_0 \in \mathbb{N}$ be the largest index $m$ that occurs among the formulas $X_m(x_i)$. Since $X_{m_0}$ is cofinite in the infinite set $S$, it contains at least $n+1$ distinct elements, and these elements satisfy all formulas in the finite subcollection. By saturation, choose pairwise distinct elements $x_0,\dots,x_n \in S$ realizing $\Sigma$. For each $i \in \{0,\dots,n\}$, membership of $x_i$ in every $X_m$ means that \begin{align*} R(x_i,S) := \{y \in S : S \models R(x_i,y)\} \end{align*} is infinite. By strong minimality, each set $R(x_i,S)$ is cofinite in $S$. The intersection \begin{align*} \bigcap_{i=0}^{n} R(x_i,S) \end{align*} is a finite intersection of cofinite subsets of the infinite set $S$, hence is nonempty. Choose $y \in S$ in this intersection. Then $S \models R(x_i,y)$ for every $i \in \{0,\dots,n\}$, so the fibre $R(S,y)$ contains the $n+1$ distinct elements $x_0,\dots,x_n$. This contradicts the uniform bound $|R(S,y)| \leq n$, which holds by the definition of $R$ and elementarity. Therefore $R(a,M)$ is finite. [/step]

[step:Use the finite fibre to exchange $a$ and $b$] Since $M \models R(a,b)$, we have $b \in R(a,M)$. The set $R(a,M)$ is definable over $A \cup \{a\}$ and is finite by the previous step. Therefore $b$ is algebraic over $A \cup \{a\}$, so \begin{align*} b \in \operatorname{acl}_M(A \cup \{a\}). \end{align*} This proves exchange. Together with the closure, monotonicity, idempotence, and finite character verified above, $\operatorname{acl}_M$ is a pregeometry on $M$. [/step]