[proofplan] We use the pregeometry on models of a strongly minimal theory induced by [algebraic closure](/page/Algebraic%20Closure). In a countable language, the algebraic closure of every finite parameter set is countable, so the algebraic closure of a basis $B$ has size $\max(|B|,\aleph_0)$, where $\aleph_0$ denotes the cardinality of $\mathbb{N}$. Since every model is generated by any of its bases under algebraic closure, an uncountable model of cardinality $\kappa$ must have basis size $\kappa$. Finally, the classification of strongly minimal models by basis cardinality implies that two models of the same uncountable cardinality are isomorphic. [/proofplan] [step:Count the algebraic closure of every finite parameter set] Let $M \models T$, and let $A \subset M$ be finite. We write $\operatorname{acl}_M(A)$ for the algebraic closure of $A$ inside $M$, namely the set of all elements $b \in M$ such that $b$ belongs to a finite $A$-definable subset of $M$. Since $L$ is countable and $A$ is finite, the expanded language $L(A)$ obtained by adding constant symbols for the elements of $A$ is countable. Hence there are only countably many $L(A)$-formulas $\varphi(x)$ with one free variable. For each such formula whose realization set \begin{align*} \varphi(M) := \{m \in M : M \models \varphi(m)\} \end{align*} is finite, the set $\varphi(M)$ contributes only finitely many elements to $\operatorname{acl}_M(A)$. Therefore \begin{align*} \operatorname{acl}_M(A) \subset \bigcup_{\substack{\varphi(x) \in L(A)\\ \varphi(M)\ \text{finite}}} \varphi(M). \end{align*} The right-hand side is a countable union of finite sets, hence is countable. Thus $\operatorname{acl}_M(A)$ is countable for every finite $A \subset M$. [guided] The point of this step is that countability of the language forces algebraicity over finitely many parameters to be a countable phenomenon. Fix a model $M \models T$ and a finite set $A \subset M$. By definition, an element $b \in M$ lies in $\operatorname{acl}_M(A)$ exactly when there is an $L(A)$-formula $\varphi(x)$ such that $M \models \varphi(b)$ and the definable set \begin{align*} \varphi(M) := \{m \in M : M \models \varphi(m)\} \end{align*} is finite. Because $L$ is countable and $A$ is finite, the language $L(A)$ is still countable. The collection of first-order formulas in one free variable over a countable language is countable, since formulas are finite strings from a countable alphabet. For every formula $\varphi(x) \in L(A)$ with finite realization set, $\varphi(M)$ contains only finitely many elements. Hence all algebraic elements over $A$ are contained in the countable union \begin{align*} \bigcup_{\substack{\varphi(x) \in L(A)\\ \varphi(M)\ \text{finite}}} \varphi(M). \end{align*} A countable union of finite sets is countable. Therefore $\operatorname{acl}_M(A)$ is countable. [/guided] [/step] [step:Compute the size of the algebraic closure of an arbitrary basis] Let $\aleph_0 := |\mathbb{N}|$ denote the least infinite cardinal. Let $B \subset M$ be a basis for the algebraic-closure pregeometry on $M$, meaning that $B$ is a maximal algebraically independent subset of $M$. By maximality in the pregeometry, $B$ spans $M$ under algebraic closure, so \begin{align*} M = \operatorname{acl}_M(B). \end{align*} Since algebraicity is witnessed by a formula using only finitely many parameters, every element of $\operatorname{acl}_M(B)$ lies in $\operatorname{acl}_M(A)$ for some finite subset $A \subset B$. Hence \begin{align*} M = \operatorname{acl}_M(B) = \bigcup_{A \in [B]^{<\omega}} \operatorname{acl}_M(A), \end{align*} where $[B]^{<\omega}$ denotes the set of finite subsets of $B$. By the previous step, each set $\operatorname{acl}_M(A)$ is countable. Therefore \begin{align*} |M| \le |[B]^{<\omega}| \cdot \aleph_0 = \max(|B|,\aleph_0). \end{align*} Since $B \subset M$, we also have $|B| \le |M|$. Thus \begin{align*} |M| = \max(|B|,\aleph_0). \end{align*} [guided] We now pass from finite parameter sets to an arbitrary basis. Define $\aleph_0 := |\mathbb{N}|$, the least infinite cardinal. Let $B \subset M$ be a basis for the algebraic-closure pregeometry on $M$, meaning that $B$ is a maximal algebraically independent subset of $M$. In a pregeometry, maximal independent sets span; in the present notation this means \begin{align*} M = \operatorname{acl}_M(B). \end{align*} This is the precise sense in which a basis generates the model. The key point is that algebraicity over a possibly large set still uses only finitely many parameters. If $b \in \operatorname{acl}_M(B)$, then there is a formula witnessing that $b$ belongs to a finite definable set over parameters from $B$. A formula contains only finitely many parameter symbols, so there is a finite subset $A \subset B$ such that $b \in \operatorname{acl}_M(A)$. Therefore \begin{align*} M = \operatorname{acl}_M(B) = \bigcup_{A \in [B]^{<\omega}} \operatorname{acl}_M(A), \end{align*} where $[B]^{<\omega}$ denotes the set of finite subsets of $B$. The preceding step proves that $\operatorname{acl}_M(A)$ is countable for every finite $A \subset M$, and hence for every finite $A \subset B$. The number of finite subsets of $B$ is $|[B]^{<\omega}|$, which is $\max(|B|,\aleph_0)$ when $B$ is infinite and finite when $B$ is finite; in both cases the product estimate gives \begin{align*} |M| \le |[B]^{<\omega}| \cdot \aleph_0 = \max(|B|,\aleph_0). \end{align*} The reverse comparison needed for equality comes from inclusion: since $B \subset M$, we have $|B| \le |M|$, and since $M$ is infinite as a model of a strongly minimal theory in this context, $\aleph_0 \le \max(|M|,\aleph_0)$. Combining the displayed upper bound with $|B| \le |M|$ yields \begin{align*} |M| = \max(|B|,\aleph_0). \end{align*} [/guided] [/step] [step:Show that an uncountable model has basis size equal to its cardinality] Assume now that $M \models T$ is uncountable and $|M| = \kappa$. Let $B \subset M$ be a basis for the algebraic-closure pregeometry on $M$. From the preceding step, \begin{align*} \kappa = |M| = \max(|B|,\aleph_0). \end{align*} Since $\kappa$ is uncountable, $\kappa > \aleph_0$. Therefore the maximum above can equal $\kappa$ only if \begin{align*} |B| = \kappa. \end{align*} So every model of $T$ of uncountable cardinality $\kappa$ has basis size $\kappa$. [guided] Let $M \models T$ be an uncountable model and write $|M| = \kappa$. Choose a basis $B \subset M$ for the algebraic-closure pregeometry on $M$. The previous step gives the exact cardinal computation \begin{align*} \kappa = |M| = \max(|B|,\aleph_0). \end{align*} Because $M$ is uncountable, $\kappa > \aleph_0$. If $|B| < \kappa$, then the maximum $\max(|B|,\aleph_0)$ would also be strictly below $\kappa$ when $\kappa$ is the displayed maximum. Hence the equality above forces \begin{align*} |B| = \kappa. \end{align*} Thus, in uncountable cardinalities, the size of the model and the dimension of its algebraic-closure basis coincide. [/guided] [/step] [step:Apply classification by basis cardinality to obtain the isomorphism] Let $M \models T$ and $N \models T$ satisfy $|M| = |N| = \kappa$, where $\kappa$ is uncountable. Choose bases $B_M \subset M$ and $B_N \subset N$ for the algebraic-closure pregeometries on $M$ and $N$, respectively. By the previous step, \begin{align*} |B_M| = \kappa = |B_N|. \end{align*} Hence there exists a bijection \begin{align*} f_0: B_M &\to B_N. \end{align*} We use the classification theorem for models of a strongly minimal complete theory: if $T$ is strongly minimal and complete, $M,N \models T$, $B_M$ and $B_N$ are bases for the algebraic-closure pregeometries on $M$ and $N$, and $f_0: B_M \to B_N$ is a bijection, then $f_0$ extends to an elementary $L$-isomorphism \begin{align*} f: M &\to N. \end{align*} The hypotheses apply here because the theorem statement assumes that $T$ is strongly minimal and complete, because $M,N \models T$ by choice, because $B_M$ and $B_N$ were chosen as bases, and because the displayed map $f_0$ is a bijection. Therefore $M \cong N$. Therefore any two models of $T$ of cardinality $\kappa$ are isomorphic. Since $\kappa$ was an arbitrary uncountable cardinal, $T$ is categorical in every uncountable cardinal. [guided] Let $M \models T$ and $N \models T$ be models with the same uncountable cardinality $\kappa$. Choose bases $B_M \subset M$ and $B_N \subset N$ for their algebraic-closure pregeometries. The previous step shows that every uncountable model has basis size equal to its underlying cardinality, so \begin{align*} |B_M| = \kappa = |B_N|. \end{align*} Since two sets with the same cardinality admit a bijection, choose a map \begin{align*} f_0: B_M &\to B_N \end{align*} that is bijective. Now we invoke the classification theorem for models of a strongly minimal complete theory. The theorem says: if $T$ is strongly minimal and complete, $M,N \models T$, $B_M$ and $B_N$ are bases for the algebraic-closure pregeometries on $M$ and $N$, and $f_0: B_M \to B_N$ is a bijection, then $f_0$ extends to an elementary $L$-isomorphism \begin{align*} f: M &\to N. \end{align*} We verify each hypothesis. The theory $T$ is strongly minimal and complete by the theorem statement. The structures $M$ and $N$ are models of $T$ by the first sentence of this step. The sets $B_M$ and $B_N$ were chosen to be bases for the relevant algebraic-closure pregeometries. Finally, $f_0$ is a bijection by construction from the equality $|B_M| = |B_N|$. Thus the classification theorem applies and gives an $L$-isomorphism $f: M \to N$. Hence any two models of $T$ of cardinality $\kappa$ are isomorphic. Since $\kappa$ was an arbitrary uncountable cardinal, $T$ is categorical in every uncountable cardinal. [/guided] [/step]