Timeline for Are the tensor-invertible coherent sheaves on an algebraic space (Zariski) locally free of rank one?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 2, 2022 at 10:28 | vote | accept | Jason Starr | ||
| Sep 2, 2022 at 7:00 | answer | added | Laurent Moret-Bailly | timeline score: 8 | |
| Sep 1, 2022 at 23:53 | history | edited | Jason Starr | CC BY-SA 4.0 |
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| Sep 1, 2022 at 23:52 | comment | added | Jason Starr | @LSpice Yes, indeed. I will add a remark. | |
| Sep 1, 2022 at 21:06 | comment | added | LSpice | Presumably this is motivated by your answer to Quotients of schemes by connected groups? | |
| Sep 1, 2022 at 20:47 | history | edited | Jason Starr | CC BY-SA 4.0 |
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| Sep 1, 2022 at 20:43 | comment | added | Jason Starr | @Z.M An algebraic space has a Zariski cover by algebraic spaces of the form $\text{Spec}(R)$ if and only if the algebraic space is a scheme. | |
| Sep 1, 2022 at 20:12 | comment | added | Z. M | Sorry for my confusion. If you have something like $\mathcal M\otimes_{\mathcal O_X}\mathcal N\cong\mathcal O_X$ for quasicoherent sheaves $\mathcal M,\mathcal N$, you can always pullback along a map $\operatorname{Spec}(R)\to X$, and invoking the affine result? | |
| Sep 1, 2022 at 19:57 | history | edited | Jason Starr | CC BY-SA 4.0 |
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| Sep 1, 2022 at 19:45 | history | asked | Jason Starr | CC BY-SA 4.0 |