Timeline for Can the support of a coherent sheaf be a numerically trivial divisor?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2017 at 4:56 | comment | added | naf | Yes, the point is that on a projective variety a nonzero effective divisor is never numerically trivial. | |
| Oct 24, 2017 at 16:09 | comment | added | user52991 | @ulrich, I have been trying to understand what you said. Do we have that - if we have a numerically trivial divisor $D$, then it cannot be the support of any sheaf. Is it because the $D$ is not effective? | |
| S Oct 24, 2017 at 15:00 | history | suggested | Mike Pierce | CC BY-SA 3.0 |
Added the link to the MathSE question
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| Oct 24, 2017 at 14:41 | review | Suggested edits | |||
| S Oct 24, 2017 at 15:00 | |||||
| Oct 24, 2017 at 11:14 | comment | added | user52991 | @ulrich, thank you for the clarification. It would be helpful if you can elaborate a bit. | |
| Oct 24, 2017 at 11:13 | comment | added | naf | Your hypothesis that $F$ has rank $0$ and $c_1(F)$ is numerically trivial is enough to conclude that $F$ cannot be supported on a divisor. (The Picard number 1 hypothesis is not necessary for this.) | |
| Oct 24, 2017 at 11:07 | history | asked | user52991 | CC BY-SA 3.0 |