most recent 30 from http://mathoverflow.net 2013-05-19T15:12:07Z http://mathoverflow.net/feeds/question/71336 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71336/injective-polynomial-map mr.bigproblem 2011-07-26T18:06:50Z 2011-07-26T18:06:50Z

<p>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 <a href="http://mathoverflow.net/questions/71185/a-proof-for-a-statement-about-polynomial-automorphism" rel="nofollow">http://mathoverflow.net/questions/71185/a-proof-for-a-statement-about-polynomial-automorphism</a>). </p> <p>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 <a href="http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/" rel="nofollow">http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/</a> for the proof when $k$ is finite or $\mathbb{C}$). </p> <p>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.</p>