[proofplan] We compare the given uncountable model $M$ with the prime model $P$ over the basis $A \cup I$. Primeness gives an elementary embedding of $P$ into $M$ over $A \cup I$, so it remains to show that no element of $M$ lies outside the image. The uncountable categoricity hypothesis is used through the stated Morley and unidimensional analysis assumptions: in particular, the dimension-transfer conclusion is internal to the model $M$. If an element outside $P$ existed, that transfer would produce a new element of $D(M)$ independent from the basis $I$, contradicting the defining spanning property of $I$. Finally, prime models over bases of the same cardinality are compared by a bijection of bases, and the usual countable-language bound for prime models over a parameter set gives the cardinal computation. [/proofplan]

[step:Embed the prime model over the basis into $M$]Let $I \subset D(M)$ be a $\operatorname{cl}_A$-basis of $D(M)$. By the assumed existence of [prime models](/page/Prime%20Model) over sets of the form $A \cup I$, let $P \models T(A)$ be prime over $A \cup I$. Since $A \cup I \subseteq M$ and $M \models T(A)$, the defining property of primeness gives an elementary embedding \begin{align*} e:P \to M \end{align*} such that $e(a)=a$ for every $a \in A$ and $e(i)=i$ for every $i \in I$. Replacing $P$ by its elementary image $e(P) \preccurlyeq M$, we may assume \begin{align*} A \cup I \subseteq P \preccurlyeq M. \end{align*}[/step]

[guided]The first move is to place the prime model and the given model inside the same ambient structure. The set over which we build the prime model is $A \cup I$, where $A$ is already named in the language of $T(A)$ and $I$ is the chosen basis of $D(M)$. By hypothesis, a prime model over $A \cup I$ exists; denote it by $P$. The phrase “$P$ is prime over $A \cup I$” means that whenever $N \models T(A)$ contains $A \cup I$, there is an elementary embedding of $P$ into $N$ fixing $A \cup I$ pointwise. Applying this with $N=M$ gives an elementary embedding \begin{align*} e:P \to M \end{align*} with $e|_{A \cup I}=\operatorname{id}_{A \cup I}$, where $e|_{A \cup I}:A \cup I \to A \cup I$ denotes the restriction of $e$ to $A \cup I$ and $\operatorname{id}_{A \cup I}:A \cup I \to A \cup I$ denotes the identity map on $A \cup I$. The image $e(P)$ is an elementary submodel of $M$ because $e$ is elementary. Since $e$ fixes $A \cup I$, the image still contains the base $A \cup I$. We may therefore identify $P$ with this image and work as though \begin{align*} A \cup I \subseteq P \preccurlyeq M. \end{align*} This identification is only a notational simplification: it replaces $P$ by an isomorphic copy inside $M$ over the prescribed base.[/guided]

[step:Assume an element outside the prime submodel and apply model-internal dimension transfer]Suppose, toward a contradiction, that there exists an element \begin{align*} b \in M \setminus P. \end{align*} Since $b$ is realized in the uncountable model $M$ and $P$ is prime over $A \cup I$, the assumed model-internal dimension-transfer conclusion of the Morley analysis applies to the pair $P \preccurlyeq M$ and the element $b$. It gives an element \begin{align*} d \in D(M) \end{align*} such that \begin{align*} d \notin \operatorname{cl}_A(I). \end{align*}[/step]

[guided]We argue by contradiction. Assume that the elementary submodel $P$ does not exhaust $M$, and choose \begin{align*} b \in M \setminus P. \end{align*} The hypothesis is deliberately stated in the model-internal form needed here. It says that whenever $P$ is prime over $A \cup I$ inside an uncountable model $M$ and an element $b \in M$ is not in $P$, the Morley analysis transfers this failure of primeness to the strongly minimal set inside the same model. Applying that conclusion to the present $P \preccurlyeq M$ and $b \in M \setminus P$ gives an element \begin{align*} d \in D(M) \end{align*} whose $D$-dimension is not accounted for by the basis $I$, equivalently \begin{align*} d \notin \operatorname{cl}_A(I). \end{align*} The membership $d \in D(M)$ is part of the stated transfer conclusion and is essential: the next step uses that $I$ spans the interpretation $D(M)$, so the transferred element must lie in $M$, not merely in a larger ambient structure.[/guided]

[step:Contradict the spanning property of the basis]Since $I$ is a basis of $D(M)$ for $\operatorname{cl}_A$, it spans $D(M)$: \begin{align*} D(M) \subseteq \operatorname{cl}_A(I). \end{align*} The element $d$ produced above lies in $D(M)$, hence \begin{align*} d \in \operatorname{cl}_A(I). \end{align*} This contradicts \begin{align*} d \notin \operatorname{cl}_A(I). \end{align*} Therefore no element $b \in M \setminus P$ exists, and hence \begin{align*} M=P. \end{align*} Thus $M$ is prime over $A \cup I$.[/step]

[guided]The contradiction uses the exact meaning of “basis” for the closure operator \begin{align*} \operatorname{cl}_A:\mathcal{P}(D) \to \mathcal{P}(D), \qquad \operatorname{cl}_A(X)=D \cap \operatorname{acl}(A \cup X). \end{align*} A basis is independent and spanning. The spanning part says that every element of $D(M)$ belongs to the closure of the basis $I$ over $A$: \begin{align*} D(M) \subseteq \operatorname{cl}_A(I). \end{align*} The previous step produced an element \begin{align*} d \in D(M) \end{align*} with \begin{align*} d \notin \operatorname{cl}_A(I). \end{align*} Because $d \in D(M)$ and $I$ spans $D(M)$, the same element must also satisfy \begin{align*} d \in \operatorname{cl}_A(I). \end{align*} This is impossible. Hence the assumption that there exists $b \in M \setminus P$ is false. Therefore $M \setminus P=\varnothing$, so \begin{align*} M=P. \end{align*} Since $P$ was chosen prime over $A \cup I$, this proves that the original uncountable model $M$ is prime over $A \cup I$.[/guided]

[step:Compute cardinality and compare prime models over bases of the same size]The language of $T(A)$ is countable because the original language is countable and $A$ is finite. Let $\kappa=|I|$. Since $A$ is finite, the parameter set satisfies \begin{align*} |A \cup I|=\max\{\kappa,|A|\}=\kappa \end{align*} when $I$ is infinite, and in all cases there are at most $\max\{|I|,\aleph_0\}$ formulas with parameters from finite tuples of $A \cup I$. We use the standard [prime model](/page/Prime%20Model) counting theorem for countable languages: if $L$ is countable, $B$ is a parameter set, and $P_B$ is prime over $B$, then \begin{align*} |P_B| \leq \max\{|B|,\aleph_0\}. \end{align*} Indeed, in the atomic construction of a prime model over $B$, each element realizes an isolated type over a finite tuple from $B$; there are at most $\max\{|B|,\aleph_0\}$ formulas with parameters from finite tuples of $B$, and isolated types are determined by isolating formulas. Applying this with $B=A \cup I$ gives \begin{align*} |P| \leq \max\{|I|,\aleph_0\}. \end{align*} Conversely, $I \subseteq P$, so \begin{align*} |P| \geq |I|. \end{align*} Also every infinite model in a countable language has cardinality at least $\aleph_0$. Hence \begin{align*} |P|=\max\{|I|,\aleph_0\}. \end{align*} Since $M=P$, we obtain \begin{align*} |M|=\max\{|I|,\aleph_0\}. \end{align*} Because $M$ is uncountable, this maximum cannot be $\aleph_0$, so $|I|$ is uncountable and \begin{align*} |M|=|I|. \end{align*} It remains to record the promised dependence on dimension at the level of isomorphism type. We use the homogeneity component of the stated strongly minimal Morley analysis: for bases of the same cardinality in the [strongly minimal set](/page/Strongly%20Minimal%20Set) $D$, any bijection of bases extends over $A$ to a partial elementary map, and prime models over corresponding parameter sets are unique over that map. Let $M_0$ and $M_1$ be uncountable models of $T(A)$ with $\operatorname{cl}_A$-bases $I_0 \subset D(M_0)$ and $I_1 \subset D(M_1)$ satisfying $|I_0|=|I_1|$. Choose a bijection \begin{align*} f:I_0 \to I_1. \end{align*} Extend it by the identity on $A$ to a map \begin{align*} f_A:A \cup I_0 \to A \cup I_1. \end{align*} The homogeneity hypothesis applies because $I_0$ and $I_1$ are independent spanning sets for the same $A$-definable strongly minimal geometry and $f_A$ fixes $A$ pointwise. Hence $f_A$ is partial elementary. By the first part of the proof, $M_0$ is prime over $A \cup I_0$ and $M_1$ is prime over $A \cup I_1$. The uniqueness of prime models over corresponding parameter sets then extends $f_A$ to an isomorphism \begin{align*} M_0 \cong M_1 \end{align*} over $A$. Thus the uncountable model is determined, up to the fixed countable language and finite named parameter set $A$, by the dimension $|I|$ of $D(M)$; in particular its cardinal is $|I|$.[/step]

[guided]We now compute the size of $M$ from the basis. The language of $T(A)$ is countable: the original language is countable, and naming the finite parameter set $A$ adds only finitely many constant symbols. Let \begin{align*} \kappa=|I|. \end{align*} The parameter set $A \cup I$ has cardinal $\max\{\kappa,|A|\}$, and since $A$ is finite, the number of formulas with parameters from finite tuples of $A \cup I$ is bounded by \begin{align*} \max\{|I|,\aleph_0\}. \end{align*} This is the correct counting object: the parameter set itself is not enlarged by the countable language, but the collection of formulas over that parameter set is counted using both the countably many formula schemes and the available parameters. We apply the standard [prime model](/page/Prime%20Model) counting theorem for countable languages. It says that if $L$ is countable, $B$ is a parameter set, and $P_B$ is prime over $B$, then \begin{align*} |P_B| \leq \max\{|B|,\aleph_0\}. \end{align*} The reason is that a prime model over $B$ is built atomically over $B$: every element realizes an isolated type over some finite tuple from $B$. There are only $\max\{|B|,\aleph_0\}$ formulas with parameters from finite tuples of $B$, and an isolated type is determined by an isolating formula. With $B=A \cup I$, this gives \begin{align*} |P| \leq \max\{|I|,\aleph_0\}. \end{align*} The reverse lower bound comes from the inclusion $I \subseteq P$: \begin{align*} |P| \geq |I|. \end{align*} Also, because $P \models T(A)$ is an infinite model in a countable language, it has cardinality at least $\aleph_0$. Combining the upper and lower bounds yields \begin{align*} |P|=\max\{|I|,\aleph_0\}. \end{align*} The previous step proved $M=P$, so \begin{align*} |M|=\max\{|I|,\aleph_0\}. \end{align*} Since $M$ is assumed uncountable, the maximum is not $\aleph_0$; therefore $|I|$ is uncountable and \begin{align*} |M|=|I|. \end{align*} Finally we compare models with bases of the same size. Let $M_0$ and $M_1$ be uncountable models of $T(A)$ with $\operatorname{cl}_A$-bases $I_0 \subset D(M_0)$ and $I_1 \subset D(M_1)$, and assume \begin{align*} |I_0|=|I_1|. \end{align*} Choose a bijection \begin{align*} f:I_0 \to I_1. \end{align*} Define the extension over the named parameters by \begin{align*} f_A:A \cup I_0 &\to A \cup I_1,\\ f_A(a)&=a \quad \text{for } a \in A,\\ f_A(i)&=f(i) \quad \text{for } i \in I_0. \end{align*} The relevant homogeneity fact from the strongly minimal Morley analysis is not merely exchange. It says that independent tuples of the same length in the $A$-definable strongly minimal geometry have the same type over $A$, and hence a bijection between bases extends to a partial elementary map over $A$. The hypotheses are satisfied here because $I_0$ and $I_1$ are independent bases for the same closure operator $\operatorname{cl}_A$ on the same $A$-definable set $D$, and $f_A$ fixes every element of $A$. Thus $f_A$ is partial elementary. By the first part of the proof, $M_0$ is prime over $A \cup I_0$ and $M_1$ is prime over $A \cup I_1$. The uniqueness theorem for prime models over corresponding parameter sets applies to the partial elementary map $f_A$: it extends $f_A$ to an isomorphism between the prime models over those parameter sets. Therefore there is an isomorphism \begin{align*} M_0 \cong M_1 \end{align*} over $A$. This proves that, after fixing the countable language and the finite named parameter set $A$, the uncountable model is determined by the dimension $|I|$ of $D(M)$, and in particular its cardinality is $|I|$.[/guided]