Timeline for A ring such that all projectives are stably free but not all projectives are free?
Current License: CC BY-SA 4.0
21 events
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| Feb 19 at 22:06 | comment | added | Mohan | @MatthewMorrow $K_0(k[t_1,\ldots, t_n])=\mathbb{Z}$ (not zero) is precisely the statement of Hilbert syzygy theorem. | |
| Feb 19 at 21:15 | answer | added | William Thomas | timeline score: 0 | |
| Jun 24, 2022 at 21:01 | history | edited | Glorfindel | CC BY-SA 4.0 |
broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jun 28, 2010 at 2:09 | vote | accept | Hailong Dao | ||
| Jun 9, 2010 at 19:34 | answer | added | Mohan | timeline score: 21 | |
| Jun 9, 2010 at 19:09 | answer | added | Seamus | timeline score: 4 | |
| Jun 8, 2010 at 23:46 | comment | added | Victor Protsak | Swan constructed examples of Noetherian rings of dimension $m$ such that all projective modules of rank $\ne m$ are free, for all $m\equiv 2 (\mod 4).$ This may not be quite what you wanted, because the construction is topological and the rings are difficult to describe explicitly, but the conclusion is very strong. See Swan, Richard G. Topological examples of projective modules. Trans. Amer. Math. Soc. 230 (1977), 201--234 MR0448350 | |
| Jun 8, 2010 at 19:21 | comment | added | Hailong Dao | @Matthew: you should have put it on MO! Such question is (I think) exactly what MO is meant for. | |
| Jun 8, 2010 at 18:56 | history | edited | Hailong Dao | CC BY-SA 2.5 |
added 1746 characters in body; edited tags; edited tags
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| Jun 8, 2010 at 9:12 | comment | added | Matthew Morrow | (typo: $0=\mathbb{Z}$). | |
| Jun 8, 2010 at 9:11 | comment | added | Matthew Morrow | Very nice question! I was asking people in the dept this a couple of weeks ago, and I we couldn't answer it. My interest was the following: by homotopy invariance of K-theory we know that $K_0(k[T_1,\dots,T_n])=0$, and I wondered if this implied Serre's conjecture (even though elementary, direct proofs of Serre's conjecture are now known). Seems not. | |
| Jun 8, 2010 at 8:05 | answer | added | Simon Wadsley | timeline score: 2 | |
| Jun 8, 2010 at 6:53 | answer | added | Torsten Ekedahl | timeline score: 12 | |
| Jun 8, 2010 at 6:38 | comment | added | Hailong Dao | Kevin: for Noetherian ring I don't think it matters. One can always resolves by finite free modules, and the high syzygy would be projective. For free, it matters. | |
| Jun 8, 2010 at 6:13 | comment | added | Kevin Ventullo | Ahh, FD didn't mean what I thought. So in general, does the projective dimension of a f.g. module always mean the shortest resolution by finite projectives, or just projectives? | |
| Jun 8, 2010 at 5:50 | history | edited | Hailong Dao | CC BY-SA 2.5 |
added 134 characters in body
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| Jun 8, 2010 at 5:40 | comment | added | Hailong Dao | Kevin: Let R=Z/6Z, M=Z/2Z. M is projective but does not have a finite free resolution. | |
| Jun 8, 2010 at 5:34 | comment | added | Kevin Ventullo | My answer to the question you linked to shows that (1) holds for all rings. It is not equivalent to (3) since being stably free means you can direct sum with a finite free module to get a free module. | |
| Jun 8, 2010 at 4:27 | answer | added | Tyler Lawson | timeline score: 9 | |
| Jun 8, 2010 at 4:08 | history | asked | Hailong Dao | CC BY-SA 2.5 |