Timeline for Structure theorem for finitely generated Z[G] modules
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 11, 2010 at 17:58 | vote | accept | B. Naskrecki | ||
| Nov 11, 2010 at 7:40 | comment | added | Kevin Buzzard | Martin says that $\mathbf{Z}[\mathbf{Z}/2]$-modules are hard to classify. What goes wrong with the following geometric approach: if $R=\mathbf{Z}[\mathbf{Z}/2]$ then $\mathrm{Spec}(R)$ is two copies of $\mathrm{Spec}(\mathbf{Z})$ glued at the point $(2)$, so a f.g. module for $\mathrm{Spec}(R)$ is something like two f.g. abelian groups $M_1$ and $M_2$ (which we understand) equipped with an isomorphism $M_1/2=M_2/2$? This seems to be some sort of a classification, unless I made a slip. | |
| Nov 11, 2010 at 3:50 | answer | added | Emerton | timeline score: 13 | |
| Nov 11, 2010 at 2:41 | comment | added | Alex B. | Exactly. My point was that the hardness of the problem has nothing to do with zero-divisors. I was merely objecting to the heuristic in your first comment. | |
| Nov 11, 2010 at 2:30 | comment | added | Harry Gindi | @Alex: Sure, but being over a field changes everything, especially an algebraically closed one of characteristic zero (Schur's lemma, Maschke's theorem, etc)! | |
| Nov 11, 2010 at 2:06 | comment | added | Alex B. | Harry, note that $\mathbb{C}[G]$ also has zero-divisors, but I think it is fair to say that $\mathbb{C}[G]$-modules are pretty well understood. | |
| Nov 11, 2010 at 1:17 | answer | added | Alex B. | timeline score: 5 | |
| Nov 11, 2010 at 1:12 | answer | added | user631 | timeline score: 7 | |
| Nov 11, 2010 at 0:31 | comment | added | Martin Brandenburg | A finitely generated $Z[Z/2]$-module is the same as a fintely generated abelian group together with an involution. I think these are already hard to classify. | |
| Nov 11, 2010 at 0:10 | comment | added | Harry Gindi | Also note that $Z[Z/nZ]\cong Z[x]/(x^n-1)$. | |
| Nov 11, 2010 at 0:08 | comment | added | Harry Gindi | Z[G=Z/nZ] has zero-divisors, so I wouldn't bet on it. | |
| Nov 11, 2010 at 0:00 | history | asked | B. Naskrecki | CC BY-SA 2.5 |