The Ax-Grothendieck theorem

The Ax-Grothendieck theorem says the following: Let \(f: \mathbb{C}^n \to \mathbb{C}^n\) be a polynomial function. If it’s injective, then it’s surjective as well.

Here’s how to prove it:

  1. The statement can be formulated as a first-order statement in the language of fields
  2. If a statement like that fails for \(\mathbb{C}\), there’s a disproof in the first-order theory of algebraically closed fields of characteristic zero.
  3. Such a proof is finite, so it only uses finitely many of the assumptions \(p \neq 0\) - hence the theorem also fails in algebraically closed fields of sufficiently high characteristic.
  4. Hence it fails in the algebraic closure of \(\mathbb{F}_p\). The specific counterexample is in some finite extension of \(\mathbb{F}_p\), which is a finite field. But clearly the theorem is true for finite fields, just by counting.

I think this is a pretty cool proof - it uses model theory in a really surprising way, and the step where you use the fact that proofs are finite is just totally bonkers.

First-order logic and completeness

Recall that first-order logic works with statements built up out of the logical connectives \(\wedge, \vee, \neg, \Rightarrow \bot, \top, \forall, \exists, =\). The language of fields augments these symbols with \(+, -, \cdot, 1, 0\) (it’s really just the language of commutative rings - being a field just means that additional axioms hold).

A statement in the first-order language of fields could be something like this:

\(\forall a,b: \neg (a = 0) \Rightarrow \exists x: a\cdot x + b = 0\)

This statement says that all nonconstant linear polynomials have a solution, which is true.

We could also write down a statement like

\(\forall a, b, c, d, e: \neg (a = 0) \Rightarrow \exists x: a x^4 + bx^3 + cx^2 + dx + e = 0\)

This says that all fourth-order polynomials have a root, which is not always true, but true sometimes, like in \(\mathbb{C}\).

Similarly, we can form a statement \(R_n\) for all \(n \in \mathbb{N}\), saying that all \(n\)th order polynomials have a root. Note that the conjunction of all these statements is not a first-order statement in itself - we can’t take conjuctions of infinitely many statements, and there’s no way to quantify over all polynomials in the theory of fields.

Similarly, we can form statements like \(1+1 \neq 0\). This is also true in \(\mathbb{C}\), but fails in fields of characteristic \(2\). We can form statements \(C_n\) for all \(n\), saying that \(n \neq 0\) - that the field does not have characteristic \(n\).

The theory of algebraically closed fields of characteristic zero is the theory consisting of the axioms of a field augmented with the statements \(C_n\) and \(R_n\) for all \(n\). A model of this theory is precisely an algebraically closed field of characteristic zero (hence the name).

Now here is a highly nontrivial fact about this theory: it is complete. This means that any statement expressible in the first-order language of fields is either provable or disprovable in it. This means that, from the perspective of first-order logic, there are really no differences between fields of this type - even though there are of course many different such fields, they have the same first-order properties.

Ax-Grothendieck in first order

We want to phrase the Ax-Grothendieck theorem as a first-order statement. It turns out this isn’t quite possible, because we can’t quantify over “all degree \(d\) polynomials” or “all \(n \in \mathbb{N}\)”. Therefore we instead formulate the statement \(P(n,d)\): “Any polynomial function \(f : \mathbb{C}^n \to \mathbb{C}^n\) of degree at most \(d\) is surjective if it’s injective”. Then the Ax-Grothendieck statement says that \(P(n,d)\) holds for all \(n \geq 1, d \geq 0\).

(By the way: by “polynomial \(\mathbb{C}^n \to \mathbb{C}^n\)”, I mean a function where each coordinate is an \(n\)-variable polynomial)

How does \(P(n,d)\) look as a first-order formula? It’s quite complicated - mostly because a degree \(d\) polynomial like \(f\) involves a large number of coefficients. The main point is that there is a definite number of coefficients, once \(n\) and \(d\) is fixed, and so we can write “for all choices of coefficients for \(f\)…” in first-order logic. For instance, if \(d = 2, n=1\), this looks like \(\forall a_{11},a_{12},a_{21},a_{22},b_1,b_2 \dots\) Here \(a_{12}\) is the coefficient for \(x_1\) in the second coordinate polynomial of \(f\), and so on.

Now of course we can also make sense of injectivity: it’s just the statement \(f(x_1,\dots x_n) = f(x'_1, \dots x'_n) \Rightarrow x_1=x'_1 \wedge \dots \wedge x_n = x'_n\) - where the equality on the left of the implication is shorthand for a complicated expression involving the coefficients of \(f\).

And we can make sense of surjectivity: it just means \(\forall y_1 \dots y_n, \exists x_1, \dots x_n: f(x_1, \dots x_n) = (y_1, \dots y_n)\)

Hence we have all the ingredients to make sense of \(P(n,d)\).

Finishing up the proof

We now have most of the ingredients to fill out the proof above. Since Ax-grothendieck is a (family of) first-order statements, if it fails, there is a disproof in the first-order theory of algebraically closed characteristic zero fields. To be specific, if Ax-Grothendieck is false, then \(P(n,d)\) is false for some specific \(n,d\), and hence there is a disproof of that statement, i.e a proof of \(\neg P(n,d)\).

Observe that such a proof is finite. Hence it can use at most finitely many of the statements \(C_n\). Hence for \(p\) large enough (larger than any \(n\) where \(C_n\) is used in the proof), Ax-Grothendieck fails for algebraically closed fields of characteristic \(p\) - like the algebraic closure of \(\mathbb{F}_p\). (How do we know it fails? We have a proof that it fails!) Hence there exists some polynomial function \(f: \overline{\mathbb{F}_p}^n \to \overline{\mathbb{F}_p}^n\) which is injective and not surjective. But now we can take the algebraic extension of \(F_p\) by all the coefficientsof \(f\) and the coordinates of a point \(y \in \overline{\mathbb{F}_p}^n\) which is not in the image. Call this extension \(L\). Then \(f: L^n \to L^n\) is a well-defined injective polynomial which is not surjective. And since \(L\) is a finite extension of a finite field, it’s finite. But then we clearly have a contradiction: injective fuctions between finite sets of the same cardinality are bijections, just by counting.

This concludes the proof.