Timeline for Rigidity of the category of schemes
Current License: CC BY-SA 4.0
11 events
when toggle format | what | by | license | comment | |
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Dec 23, 2019 at 18:33 | comment | added | Martin Brandenburg | @user131755 Where did Laurent point out that "nilpotent thickenings" are preserved? | |
Dec 23, 2019 at 18:29 | comment | added | Martin Brandenburg | @R.vanDobbendeBruyn van d This sounds also very interesting. Can one find this article somewhere? | |
Dec 23, 2019 at 18:27 | vote | accept | Martin Brandenburg | ||
Dec 23, 2019 at 23:15 | |||||
Dec 23, 2019 at 18:27 | comment | added | Martin Brandenburg | @user131755 I didn't check the details yet, but this looks like a promising proof to me. Great! | |
Jan 7, 2019 at 12:38 | comment | added | user131755 | @TimCampion Serre's criterion also assumes quasicompactness of the scheme, though. | |
Dec 27, 2018 at 6:05 | comment | added | Tim Campion | Let $X$ be a scheme defined over a field of characteristic zero (I don't know if this is necessary). If $QCoh(X)$ has a compact projective generator, then $X$ is affine. For if $P \twoheadrightarrow \mathcal O_X$ is an epimorphism from a compact projective, then evaluation and coevaluation form maps $\mathcal O_X{}^\to_\leftarrow P \otimes P^\vee$ whose composite is $n \cdot id_{\mathcal O_X}$ where $n$ is the rank of $P$. Since we're in characteristic 0, this is invertible, so $\mathcal O_X$ is a retract of the projective $P$ and hence also projective. So $X$ is affine by Serre's criterion. | |
Nov 30, 2018 at 15:03 | comment | added | user131755 | @R.vanDobbendeBruyn I sent you an email | |
Nov 27, 2018 at 18:58 | comment | added | R. van Dobben de Bruyn | Dear user131755, I am finishing up an article proving more generally that $\operatorname{\underline{Isom}}(\operatorname{\underline{Sch}}_S,\operatorname{\underline{Sch}}_{S'}) = \operatorname{\underline{Isom}}(S',S)$, which is a non-Noetherian version of a result by Mochizuki (in particular answering the question of the OP). Your argument looks correct, and the observation that thickenings are automatically affine would give a substantial shortcut to my existing argument. Could you please get in touch with me so that we can discuss how to proceed? | |
Nov 25, 2018 at 17:05 | review | Late answers | |||
Nov 25, 2018 at 17:46 | |||||
Nov 25, 2018 at 16:50 | review | First posts | |||
Nov 25, 2018 at 16:50 | |||||
Nov 25, 2018 at 16:47 | history | answered | user131755 | CC BY-SA 4.0 |