Ax–Grothendieck theorem

Remarks

For fixed $n$ and a bound $d$ on the degrees of the $f_i$, the statement is a single first-order sentence in the language of rings. It holds for every finite field, because an injective map from a finite set to itself is surjective. It also holds in the algebraic closure $\overline{\F_p}$, the union of the finite fields $\F_{p^k}$. Given an injective $F$ over $\overline{\F_p}$ and a point $b$, the coefficients of the $f_i$ and the coordinates of $b$ lie in one finite field $\F_{p^k}$. The map $F$ sends $\F_{p^k}^n$ injectively into itself, so it is onto $\F_{p^k}^n$, and $b$ has a preimage. By the Lefschetz principle, it then holds in every algebraically closed field of characteristic $0$, such as $\C$.

Example: over $\C$, the map $F(x, y) = (x, y + x^2)$ is injective, and it is surjective, since $(a, b) = F(a, b - a^2)$.

The hypothesis that $K$ is algebraically closed is necessary. Over $\Q$, the polynomial map $x \mapsto x^3$ is injective but not surjective, since $2$ is not the cube of a rational number.

Going deeper

For a student

The theorem was proved by James Ax with model theory and by Alexander Grothendieck with algebraic geometry. The model-theoretic proof is the standard showcase of the Lefschetz principle: a statement checked over finite fields transfers to $\C$. The converse fails: a surjective polynomial map need not be injective, as $z \mapsto z^2$ on $\C$ shows. It is related to the open Jacobian conjecture, which asks whether a polynomial map $\C^n \to \C^n$ with constant nonzero Jacobian determinant has a polynomial inverse.