Timeline for Is every locally free module of rank $1$ over a commutative ring concretely invertible?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Dec 27, 2019 at 1:00 | comment | added | Jesse Elliott | I don't think the solution is great, because the ring $B$ depends on $P$. It would be much better if there were a single ring that worked for all invertible $R$-modules $P$ (which is what I thought was being asked). | |
| Dec 26, 2019 at 22:41 | comment | added | Georges Elencwajg | What an amazingly crisp, elegant and brilliant solution: I can't begin to tell you how grateful I am for this definitive answer, dear darx! | |
| Dec 26, 2019 at 22:37 | vote | accept | Georges Elencwajg | ||
| Dec 26, 2019 at 20:45 | review | First posts | |||
| Dec 26, 2019 at 20:59 | |||||
| Dec 26, 2019 at 20:44 | history | answered | darx | CC BY-SA 4.0 |