Timeline for Decomposition of a quotient module
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 27, 2013 at 12:15 | vote | accept | TmobiusX | ||
| Jun 27, 2013 at 12:15 | vote | accept | TmobiusX | ||
| Jun 27, 2013 at 12:15 | |||||
| Jun 27, 2013 at 12:09 | vote | accept | TmobiusX | ||
| Jun 27, 2013 at 12:15 | |||||
| Jun 26, 2013 at 18:19 | answer | added | Graham Leuschke | timeline score: 1 | |
| Jun 26, 2013 at 17:57 | comment | added | Steven Landsburg | Graham: Thanks to your email, I withdraw my claim! | |
| Jun 26, 2013 at 15:26 | comment | added | Graham Leuschke | As sure as one can ever be with computer algebra -- I checked it on Singular and got the same answer. | |
| Jun 26, 2013 at 13:16 | comment | added | Steven Landsburg | Graham: I haven't tried to compute the Betti numbers, but I did compute the cokernel, and it appears to have a direct summand of $k$. Are you sure of your calculation? | |
| Jun 26, 2013 at 11:18 | comment | added | Graham Leuschke | The point is that the Betti numbers (ranks of free modules in the resolution) of $\mathrm{coker} M$ would have to be greater than or equal to those of $k$, if $k$ were a direct summand. They aren't, so it isn't. | |
| Jun 26, 2013 at 7:06 | comment | added | TmobiusX | Thank you, Graham. Your idea is good. But I know little of Macaulay2. | |
| Jun 25, 2013 at 15:20 | answer | added | Steven Landsburg | timeline score: 2 | |
| Jun 25, 2013 at 14:39 | comment | added | Graham Leuschke | I don't think this is true. According to Macaulay2, the Betti numbers of the cokernel of your matrix are $2,2,4,11,32,95,...$, while those of the residue fields are $1,4,13,40,121,364,...$. | |
| S Jun 25, 2013 at 13:04 | history | suggested | agt | CC BY-SA 3.0 |
the matrix was not rendered correctly
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| Jun 25, 2013 at 12:50 | review | Suggested edits | |||
| S Jun 25, 2013 at 13:04 | |||||
| Jun 25, 2013 at 12:42 | history | asked | TmobiusX | CC BY-SA 3.0 |