Timeline for These rings are isomorphic?
Current License: CC BY-SA 3.0
8 events
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| Jun 9, 2016 at 6:38 | comment | added | Ilya Bogdanov | Well, it affects a bit: I've provided an automorphism of $\mathbb C[[x,z,u]]$ mapping the first ideal to the second. To get the isomorphism in the initial variables, you need to conjugate this automorphism by the change of variables. The result seems to be $x\mapsto x\root4\of{1+u}$, $u\mapsto u$, $y\mapsto y+x^2-x^2\sqrt{1+u}$. | |
| Jun 9, 2016 at 2:05 | comment | added | Otoniel Silva | Thank you for your answer, could you please write the isomorphism explicitly? The definition of $x^{'}$ result that $z$ also depend on $x^{'}$ , this may not affect the construction of the isomorphism? | |
| Jun 8, 2016 at 16:59 | vote | accept | Otoniel Silva | ||
| Jun 8, 2016 at 10:38 | comment | added | Ilya Bogdanov | Fedor: Thanks, surely. znt: Something between; I did not try to find a Groebner basis, but followed some similar way, trying just to simplify the relations. | |
| Jun 8, 2016 at 10:37 | history | edited | Ilya Bogdanov | CC BY-SA 3.0 |
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| Jun 8, 2016 at 10:34 | comment | added | znt | Does one just come up with this from playing around or is there some sort of Groebner Basis strategy for proving this? Well done, by the way. | |
| Jun 8, 2016 at 10:32 | comment | added | Fedor Petrov | You mean $z=y+x^2$? | |
| Jun 8, 2016 at 10:25 | history | answered | Ilya Bogdanov | CC BY-SA 3.0 |