Timeline for Is quotient by maximal destabilizing sheaf, torsion-free?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 13, 2017 at 14:57 | comment | added | Thomas Poguntke | I don't think there is such a thing as geometrically (non-)torsion-free. | |
| Mar 10, 2017 at 8:59 | comment | added | user45397 | @Henri Thanks for the answer. I had a non-algebraically closed field in mind. I have specified it more clearly in the question now. Does the same argument still work for non-algebraically closed field? If I understand corrently, the Serre's conditions on normality is not preserved under base change, meaning isn't it possible that the quotient is torsion-free on $X$ but not geometrically torsion-free? | |
| Mar 10, 2017 at 8:57 | history | edited | user45397 | CC BY-SA 3.0 |
added 39 characters in body
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| Mar 10, 2017 at 8:57 | history | undeleted | user45397 | ||
| Mar 10, 2017 at 1:27 | history | deleted | user45397 | via Vote | |
| Mar 10, 2017 at 0:28 | comment | added | Henri | If $k=\mathbb C$, then you can argue the following: as $F'$ is saturated in $F$ (saturation increases slope), $F/F'$ is torsion free, so it is locally free on an open subset whose complement has codimension at least two. In particular, if the base is a smooth curve, $F/F'$ is locally free. For a reference, see for instance Kobayashi, Differential Geometry of Complex Vector Bundles, Corollary 5.15. | |
| Mar 9, 2017 at 23:52 | history | asked | user45397 | CC BY-SA 3.0 |