Timeline for A proof for a statement about polynomial automorphism
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jul 25, 2011 at 9:19 | vote | accept | mr.bigproblem | ||
| Jul 25, 2011 at 9:19 | |||||
| Jul 25, 2011 at 8:50 | vote | accept | mr.bigproblem | ||
| Jul 25, 2011 at 9:18 | |||||
| Jul 25, 2011 at 8:41 | comment | added | Francesco Polizzi | You are right. But really only the definition of dimension sufficies to prove this statement. For a short proof, see [Shafarevich, Algebraic Geometry 1, Chapter I, Section 6, Theorem 1]. By the way, also Jack Huizenga's nice proof uses some kind of "dimension argument": in fact $M/M^2$ can be seen as the cotangent space of $\mathbb{A}^n$ at $(0, \ldots, 0)$. | |
| Jul 25, 2011 at 8:27 | comment | added | mr.bigproblem | How do you prove the fact that the only closed subset of $\mathbb{A}^n_k$ isomorphic to $\mathbb{A}^n_k$ is itself? To prove the similar algebraic fact for polynomial rings, I had to use dimension. – mr.bigproblem 0 secs ago | |
| Jul 25, 2011 at 8:01 | history | answered | Francesco Polizzi | CC BY-SA 3.0 |