Ax-Grothendieck Theorem

Let be an Algebraically Closed Field. Suppose that the function is an injective Polynomial Map. Then is surjective.

Proof

First suppose

for some prime . Fix some such that all coefficients of come from . Then for any , induces an injective polynomial map which has to be surjective since is finite. Hence

so is surjective.

Now we consider . First, for any define to be:

Quot

Every injective polynomial map with coordinates, each of which is a polynomial in variables and degree at most , is surjective

Suppose that

for some . As is a Complete Theory we have

By Compactness Theorem, there is some finite such that

In particular, cannot encode infinite characteristic of its models, and thus there is a prime such that

which is a contradiction.