From: AndrEs Eduardo Caicedo <acaicedo@math.berkeley.edu>
Subject: Re: Injective polynomial mapping
Date: Sat, 17 Feb 2001 01:01:36 -0800
Newsgroups: sci.math
Summary: Injective polynomials from C^n to C^n are also surjective

John Chamish wrote:

> Maybe the right question to ask is if such an f must be surjective
> mapping from C^n -> C^n.
>
> I suspect that the answer is yes.
>
> John

It is, indeed (as Peter van Rossum pointed out already). W. Rudin
presented a nice proof in the Monthly a few years ago. My favorite
argument is model theoretic and due to Ax.

The idea is that in a field F the statement p_d="every polynomial
mapping f:F^n->F^n of degree at most d, if injective is surjective" is
first order. The theory of algebraically closed fields of characteristic
zero is complete, so C satisfies p_d iff p_d is a consequence of this
theory. Since only finitely many axioms are used in any proof, this
holds iff p_d is true in any algebraically closed field of
characteristic p prime (except, possibly, for finitely many primes).
Again by completeness of the theory, it suffices to prove p_d in the
algebraic closure of the field with p elements.

Now, if K is such closure, and f:K^n->K^n is a polynomial, f is
definable over a finite subfield K_0. Given a tuple x in K^n, there is a
finite subfield K_1 extending K_0 and containing all the elements of x.
Restricting to K_1, we see f:K_1^n->K_1^n and since K_1 is finite and f
is injective, it must be onto. Thus, x is in the range of f, and since x
in K was arbitrary, f:K^n->K^n is onto.

AndrEs