= Solution
Let $D=\operatorname{dom}(f)$ and take a formula $\varphi(x,\bar d)$ over $D$. Since $T$ is a <strongly minimal theory>, the set $\varphi(\mathcal U,\bar d)$ is finite or cofinite. If $\varphi(b,\bar d)$ holds, it cannot be finite because $b\notin\operatorname{acl}(D)$, so it is cofinite. Elementarity of $f$ makes $\varphi(\mathcal U,f\bar d)$ cofinite. Its finite complement is algebraic over $f(D)$, so $c\notin\operatorname{acl}(\operatorname{ran}f)$ lies in the set. Applying the same argument to $\neg\varphi$ gives the reverse implication. Hence \b[$f\cup\{(b,c)\}$ is elementary]: both elements realize the <generic type in a strongly minimal theory> over the corresponding domains.
If models $M,N$ have bases $B,C$ of the same <dimension of a pregeometry>, choose a bijection $B\to C$. Repeated use of the one-point result makes it elementary on every finite subset and hence on $B$. Every model is the algebraic closure of a basis. Extending the map back and forth across algebraic elements produces an isomorphism $M\to N$. Thus \b[models of a strongly minimal theory with the same dimension are isomorphic].
Back to article page
