Timeline for What is the right definition of the Picard group of a commutative ring?
Current License: CC BY-SA 2.5
16 events
| when toggle format | what | by | license | comment | |
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| Oct 3, 2021 at 21:20 | comment | added | Cam McLeman | @LSpice It took 30 years for all the ingredients of that joke to mature from inception to delivery, but I think we can all agree it was worth the wait. | |
| Sep 30, 2021 at 19:44 | comment | added | LSpice | @CamMcLeman, re, maybe he's born with it … maybe it's non-Abelian? | |
| Mar 26, 2018 at 18:36 | answer | added | Georges Elencwajg | timeline score: 8 | |
| Apr 8, 2016 at 0:32 | comment | added | David Handelman | How about K${}_0 (R)$ factored out by the subgroup generated by the free on one generator module? | |
| Dec 29, 2012 at 0:04 | answer | added | Akhil Mathew | timeline score: 31 | |
| Feb 2, 2010 at 11:25 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Feb 2, 2010 at 3:47 | vote | accept | Pete L. Clark | ||
| Feb 2, 2010 at 3:44 | answer | added | Clark Barwick | timeline score: 19 | |
| Feb 2, 2010 at 2:44 | answer | added | Martin Brandenburg | timeline score: 4 | |
| Feb 2, 2010 at 2:28 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Feb 2, 2010 at 2:24 | comment | added | Pete L. Clark | (...and you have to check that the tensor product of finite rank projectives is finite rank projective. Unless I have made some silly mistake, this seems to come out immediately from the characterization of such a guy as a direct summand of a finitely generated free module.) | |
| Feb 2, 2010 at 2:20 | comment | added | Pete L. Clark | @VA: I think so. The dual of a finite rank projective module is again finite rank projective. Is there something else to check? | |
| Feb 2, 2010 at 2:13 | comment | added | VA. | Now (2) is a group. But is (1) a group? | |
| Feb 2, 2010 at 1:58 | comment | added | Cam McLeman | I can't help but wonder what makes a person non-commutative. Were you born like that? | |
| Feb 2, 2010 at 1:45 | comment | added | Mariano Suárez-Álvarez | As a non-commutative person, let me add that one can also consider the invertible $R$-$R$-bimodules, and/or the group of self-equivalences of the category of, say, left $R$-modules. | |
| Feb 2, 2010 at 1:42 | history | asked | Pete L. Clark | CC BY-SA 2.5 |