Let and take a formula over . Since is a strongly minimal theory, the set is finite or cofinite. If holds, it cannot be finite because , so it is cofinite. Elementarity of makes cofinite. Its finite complement is algebraic over , so lies in the set. Applying the same argument to gives the reverse implication. Hence is elementary: both elements realize the generic type in a strongly minimal theory over the corresponding domains.
If models have bases of the same dimension of a pregeometry, choose a bijection . Repeated use of the one-point result makes it elementary on every finite subset and hence on . Every model is the algebraic closure of a basis. Extending the map back and forth across algebraic elements produces an isomorphism . Thus models of a strongly minimal theory with the same dimension are isomorphic.
Let be a finite partial elementary map in . In a countable language, the algebraic closures of finite sets are countable. An alternating back-and-forth construction extends to an isomorphism
At each step, an element algebraic over the current domain has a finite algebraic type, and elementarity provides a matching realization on the other side.
Choose a basis of a pregeometry over the first closed set and a basis of a pregeometry over the second. The dimension of a pregeometry is the same for the two extensions: the isomorphism preserves the finite ranks already contributed by and , while both sides have the same ambient dimension. Choose a bijection . Part (a) and uniqueness of the generic type in a strongly minimal theory make the enlarged map elementary. Since an elementary submodel is algebraically closed in the monster,
and the map extends over algebraic closure to an automorphism of . Thus every finite partial elementary map extends to an automorphism: every model is homogeneous, as stated in homogeneity of models of a countable strongly minimal theory.