Timeline for When quotients of polynomial rings are isomorphic to polynomial rings?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 4, 2014 at 11:58 | comment | added | Lev Borisov | I am not an expert, but I have not heard about cancellation conjecture being related to the Jacobian conjecture. | |
| Sep 3, 2014 at 20:54 | comment | added | Abdelmalek Abdesselam | @Lev: interesting comment. Would your argument for failure in high dimension apply to the Jacobian conjecture too? | |
| S Sep 3, 2014 at 17:53 | history | suggested | user26857 | CC BY-SA 3.0 |
edited title to reflect better the question
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| Sep 3, 2014 at 17:34 | review | Suggested edits | |||
| S Sep 3, 2014 at 17:53 | |||||
| Jul 17, 2014 at 18:44 | answer | added | zcn | timeline score: 1 | |
| Nov 16, 2013 at 19:03 | comment | added | Lev Borisov | The statement of arxiv.org/abs/1208.0483 is in positive characteristic. But it certainly makes the conjecture unlikely to hold (for high enough dimension) in characteristic zero. | |
| Feb 9, 2013 at 8:35 | comment | added | Neil Epstein | Martin: I'm not sure the cancellation problem is so wide open any more. Check out arxiv.org/abs/1208.0483. | |
| Feb 9, 2013 at 1:07 | comment | added | Richard Stanley | If the ideal $I$ is generated by homogeneous polynomials, then the condition is that $I$ is generated by polynomials of degree one. | |
| Feb 9, 2013 at 1:05 | history | edited | Richard Stanley | CC BY-SA 3.0 |
missing ] inserted
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| Feb 8, 2013 at 17:07 | comment | added | Mohan | There are probably no such non-trivial criterion. One deep result is the Abhyankar-Moh theorem which says that $\mathbb{C}[x,y]/f$ is the polynomial ring in one variable if and only if there is a $\mathbb{C}$- algebra automorphism of $\mathbb{C}[x,y]$ which transforms $f$ into a variable. | |
| Feb 8, 2013 at 16:41 | comment | added | Martin Brandenburg | I doubt that there are is a (useful and nontrivial) criterion. One could use it to attack the wide open Zariski cancellation problem. | |
| Feb 8, 2013 at 12:16 | comment | added | Allen Knutson | And the quotient ring has to be regular; check out mathoverflow.net/questions/79956/… | |
| Feb 8, 2013 at 11:56 | comment | added | Emil Jeřábek | An obvious necessary condition is that the ideal has to be prime, which is enough to rule out $\langle Y^2\rangle$. | |
| Feb 8, 2013 at 11:43 | history | asked | Alex | CC BY-SA 3.0 |