## FANDOM

78 Pages

The Ax-Grothendieck Theorem says that if $V$ is a variety over an algebraically closed field $K$, and $f : V \to V$ is a morphism of varieties such that $V(K) \to V(K)$ is injective, then $V(K) \to V(K)$ is bijective. Here, "variety" can be interpreted as finite-type scheme over $K$.

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 Edit

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

• (*) If $D \subset K^n$ is definable (with parameters) and $f : D \to D$ is definable (with parameters), and injective, then $f$ 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 $K \models ACF$ has the property that the definable closure of any finite subset is finite. Then (*) holds. Indeed, suppose $f : D \hookrightarrow D$ is definable and injective, and $p \in D$. Let $S$ be a finite set over which $f, D, p$ are defined. Then $f$ induces an injective map from $D(S) := D \cap \operatorname{dcl}(S)$ to itself. Since $D(S)$ is finite, $f$ is a bijection, by the pigeonhole principle. So $p \in f(D)$, and $f$ is surjective.

The algebraic closure $\overline{\mathbb{F}_p}$ of $\mathbb{F}_p$ 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 $\overline{\mathbb{F}_p}$. Every characteristic $p$ model of ACF (every model of $ACF_p$) is elementarily equivalent to $\overline{\mathbb{F}_p}$. Since (*) is a conjunction of first-order statements, (*) holds in all models of $ACF_p$. Then by compactness, it also holds in at least one model of $ACF_0$, hence in all models of $ACF_0$. So (*) holds in all models of ACF.