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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
edited body
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
added 246 characters in body
Jul 25, 2011 at 6:05 history edited Jack Huizenga CC BY-SA 3.0
added 273 characters in body; edited body
Jul 25, 2011 at 6:00 history answered Jack Huizenga CC BY-SA 3.0