## Injective polynomial map

Let $k$ be a field, $R = k[x_1,...,x_n]$ and $\Phi: R \rightarrow R$ be an endomorphism. It is known that if $\Phi$ is surjective then it is also injective (see http://mathoverflow.net/questions/71185/a-proof-for-a-statement-about-polynomial-automorphism).

For the inverse problem, it is known that if $k$ is a finite field or an algebraically closed field then $\Phi$ is injective implies $\Phi$ is surjective (see http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/ for the proof when $k$ is finite or $\mathbb{C}$).

The remaining question is an counter-example in which $k$ is not finite nor algebraically closed (let $k = \mathbb{R}$, for instance), $\Phi$ is injective but not surjective.

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You are misinterpreting the Grothendieck-Ax theorem. The theorem talks not about a map of polynomial rings, but about the induced map of affine spaces. Case in point: the map $k[x]\to k[x]$ given by $x\mapsto x^2$ is injective and not surjective, for any field $k$. – Jack Huizenga Jul 26 2011 at 18:32
You are right. Too bad. – mr.bigproblem Jul 27 2011 at 3:11