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.
In a strongly minimal theory, model-theoretic algebraic closure is a pregeometry. If is algebraically closed, every element outside realizes the generic type in a strongly minimal theory: every one-variable definable set is finite or cofinite, and a point outside belongs to none of the finite ones.
Enumerate a finite tuple of distinct elements of the independent set . Independence says
Its image tuple under the bijection has the same property. Inductively, and realize the same generic type over the algebraic closures of the preceding tuples, so every finite restriction of is elementary. First-order formulas involve only finitely many parameters, hence the entire bijection is an elementary map. This is the independent set in a strongly minimal theory principle.
A complete theory is strongly minimal when, in every model of , every definable subset of the home sort in one variable, allowing parameters, is finite or cofinite.