Timeline for A proof for a statement about polynomial automorphism
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
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Jul 25, 2011 at 9:19 | vote | accept | mr.bigproblem | ||
Jul 26, 2011 at 0:55 | |||||
Jul 25, 2011 at 9:18 | vote | accept | mr.bigproblem | ||
Jul 25, 2011 at 9:19 | |||||
Jul 25, 2011 at 7:41 | comment | added | mr.bigproblem | Yes, now it's a correct proof. Thanks! | |
Jul 25, 2011 at 7:39 | vote | accept | mr.bigproblem | ||
Jul 25, 2011 at 8:50 | |||||
Jul 25, 2011 at 7:26 | comment | added | Jack Huizenga | You are right, there were some issues with the original. I think it's been fixed now. | |
Jul 25, 2011 at 7:24 | history | edited | Jack Huizenga | CC BY-SA 3.0 |
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Jul 25, 2011 at 7:11 | history | edited | Jack Huizenga | CC BY-SA 3.0 |
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Jul 25, 2011 at 7:09 | comment | added | mr.bigproblem | Dear Jack, how do you imply that $\Phi_*: M/M^2 \rightarrow N/N^2$ is isomorphic? I guess, to verify this, one needs the condition that $Ker \Phi|_M = 0$, which is equivalent to $Ker \Phi = 0$. Moreover, why does it contradict when one has $\Phi_*(f) = 0$? Do you mean that this implies $f \in M^2$ and then using induction implies $f \in M^n$ and finally by Krull's intersection theorem, $f = 0$, a contradiction? | |
Jul 25, 2011 at 7:03 | vote | accept | mr.bigproblem | ||
Jul 25, 2011 at 7:03 | |||||
Jul 25, 2011 at 6:11 | history | edited | Jack Huizenga | CC BY-SA 3.0 |
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Jul 25, 2011 at 6:05 | history | edited | Jack Huizenga | CC BY-SA 3.0 |
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Jul 25, 2011 at 6:00 | history | answered | Jack Huizenga | CC BY-SA 3.0 |