The Ax-Grothendieck Theorem says that if is a variety over an algebraically closed field , and is a morphism of varieties such that is injective, then is bijective. Here, "variety" can be interpreted as finite-type scheme over .

The Ax-Grothendieck theorem has a relatively straightforward proof using model theory, and is often listed as an example of a theorem that is easy to prove using mathematical logic, and harder to prove directly using algebraic geometry.

## Proof sketch

Varieties can be seen as (special) definable sets, and morphisms of varieties yield definable maps. Therefore, it suffices to show that for every model of ACF, the following condition holds:

• (*) If is definable (with parameters) and is definable (with parameters), and injective, then is a bijection.

Conditition (*) is equivalent to a small conjunction of first-order statements (an easy exercise). In other words, the set of models of ACF satisfying (*) is an elementary class.

Suppose has the property that the definable closure of any finite subset is finite. Then (*) holds. Indeed, suppose is definable and injective, and . Let be a finite set over which are defined. Then induces an injective map from to itself. Since is finite, is a bijection, by the pigeonhole principle. So , and is surjective.

The algebraic closure of is a model of ACF in which every finite set has finite definable closure. (The perfect field generated by any finite set is finite.) So (*) holds in . Every characteristic model of ACF (every model of ) is elementarily equivalent to . Since (*) is a conjunction of first-order statements, (*) holds in all models of . Then by compactness, it also holds in at least one model of , hence in all models of . So (*) holds in all models of ACF. ::: ::: :::