Timeline for These rings are isomorphic?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 8, 2016 at 16:59 | vote | accept | Otoniel Silva | ||
| Jun 8, 2016 at 16:59 | vote | accept | Otoniel Silva | ||
| Jun 8, 2016 at 16:59 | |||||
| Jun 8, 2016 at 10:25 | answer | added | Ilya Bogdanov | timeline score: 9 | |
| Jun 8, 2016 at 10:23 | answer | added | Neil Strickland | timeline score: 2 | |
| Jun 8, 2016 at 9:57 | comment | added | znt | ...but $xy$ is in the square of the maximal ideal... | |
| Jun 8, 2016 at 9:57 | comment | added | Ilya Bogdanov | $x^3y=0$, so $xy$ is a nilpotent in $B$... | |
| Jun 8, 2016 at 9:55 | comment | added | znt | Even stronger -- does $B$ contain any non-zero nilpotent element which is in the maximal ideal but not its square? | |
| Jun 8, 2016 at 9:53 | comment | added | Ilya Bogdanov | Does $B$ contain any non-zero nilpotent element? | |
| Jun 8, 2016 at 9:39 | comment | added | Neil Strickland | @znt If I did this correctly (using Maple) then these dimensions are the same for $n\leq 12$. More precisely, if we use a pure lexicographic term order with $y<x<u$ then the Grobner bases for $A/\mathfrak{m}^n$ and $B/\mathfrak{m}^n$ have the same leading terms. | |
| Jun 8, 2016 at 9:25 | comment | added | znt | To try and prove they're not isomorphic I might grit my teeth (or pull out a computer algebra package) and compute the dimension of $A/m^n$ and $B/m^n$ for the first few values of $n$, where here $m$ is the maximal ideal $(x,y,u)$. | |
| Jun 8, 2016 at 9:23 | comment | added | znt | @Daniel Larsson -- it's not clear that the isomorphism is supposed to send $y$ to $y$. Otoniel Silva -- there are definitely morphisms in each direction because you can just send all the variables to zero. | |
| Jun 8, 2016 at 7:15 | comment | added | Daniel Larsson | In $A$ $y$ is a zero-divisor but not in $B$? | |
| Jun 8, 2016 at 4:11 | history | asked | Otoniel Silva | CC BY-SA 3.0 |