Timeline for Is every "nice" abelian category with enough projectives an additive presheaf category?
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 4, 2019 at 19:13 | comment | added | Aurélien Djament | Neeman's example in his famous paper is very interesting, but if I remember correctly, it is not a Grothendieck category (in Roos's article correcting the one for which Neeman found this counter-example, I think that it is proven that less than Grothendieck category is enough to make sure that all works). It would be interesting to have strange phenomena like the previous one, but for Grothendieck categories (or theorems saying that it is impossible...). For example, can one find a Grothendieck category with enough projectives, locally finitely generated, but beeing not a presheaf category? | |
| Jan 31, 2018 at 18:39 | history | rollback | Tim Campion |
Rollback to Revision 8
|
|
| Jan 31, 2018 at 18:23 | history | edited | Tim Campion | CC BY-SA 3.0 |
added 6830 characters in body
|
| Jan 27, 2018 at 1:46 | history | undeleted | Tim Campion | ||
| Jan 27, 2018 at 1:46 | history | edited | Tim Campion | CC BY-SA 3.0 |
Again, the old argument hinged on misunderstanding what happens when the iterative process stabilizes.
|
| Jan 27, 2018 at 1:34 | history | deleted | Tim Campion | via Vote | |
| Jan 23, 2018 at 19:53 | history | edited | Tim Campion | CC BY-SA 3.0 |
added 2361 characters in body
|
| Jan 23, 2018 at 17:35 | comment | added | Tim Campion | To close the gap: Let $i: U \to X$ be the inclusion of an open affine. Then $i$ is a quasicompact, quasiseparated morphism of schemes. Hence the direct image $i_\ast$ takes quasicoherent sheaves to quasicoherent sheaves. It's easy to check that the inverse image $i^\ast$ (which is just restriction) also preserves quasicoherence. Moreover, because $U$ is open, $i_\ast$ is exact. So $i^\ast: \mathrm{QCoh}(X) \to \mathrm{QCoh}(Y)$ has an exact right adjoint, and hence preserves projectives. Every projective over the affine $U$ is locally free, and so every projective over $X$ is locally free. | |
| Jan 18, 2018 at 17:37 | comment | added | Tim Campion | @DenisNardin I only need the implication (compact) projective $\Rightarrow$ locally free. My thinking was that this follows from the fact that projective implies free over a local ring. But now I'm suddenly doubting the implicit assumption that localization preserves the property of being projective... | |
| Jan 18, 2018 at 8:45 | comment | added | Denis Nardin | How do you go from compact projective to locally free of finite rank? As far as I know the equivalence projective=locally free is not usually true for non-affine schemes | |
| Jan 18, 2018 at 2:21 | history | edited | Tim Campion | CC BY-SA 3.0 |
added 12 characters in body
|
| Jan 18, 2018 at 1:57 | history | undeleted | Tim Campion | ||
| Jan 18, 2018 at 1:56 | history | edited | Tim Campion | CC BY-SA 3.0 |
Reworked argument; old argument was fundamentally wrong.
|
| Jan 15, 2018 at 20:29 | history | deleted | Tim Campion | via Vote | |
| Jan 10, 2018 at 9:02 | history | edited | Tim Campion | CC BY-SA 3.0 |
deleted 2 characters in body
|
| Jan 10, 2018 at 8:19 | comment | added | Denis Nardin | Unfortunately, I don't know any. In general, there aren't that many schemes with enough projectives, so it might very well be that the only such schemes are affines. | |
| Jan 10, 2018 at 5:35 | history | edited | Tim Campion | CC BY-SA 3.0 |
added 994 characters in body
|
| Jan 10, 2018 at 0:10 | history | edited | Tim Campion | CC BY-SA 3.0 |
added 789 characters in body
|
| Jan 9, 2018 at 20:22 | comment | added | Tim Campion | @DenisNardin Shoot -- you're right. I don't know much about noncommutative geometry -- what's an example of a noncommutative ring $R$ (not Morita equivalent to a commutative one) such that $Mod(R) = QCoh(X)$ for a scheme $X$? | |
| Jan 9, 2018 at 20:18 | comment | added | Denis Nardin | I'm not sure I understand your proof. $\mathcal{C}$ having a compact projective generator is not saying anything about $X$ being affine. It is only saying that $\mathcal{C}$ is the module category for a (possibly noncommutative) ring. | |
| S Jan 9, 2018 at 20:13 | history | answered | Tim Campion | CC BY-SA 3.0 | |
| S Jan 9, 2018 at 20:13 | history | made wiki | Post Made Community Wiki by Tim Campion |