Timeline for Moving a divisor on a (reducible, non-reduced) curve
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 11, 2015 at 17:55 | vote | accept | Question Mark | ||
| Jan 11, 2015 at 7:21 | answer | added | Sándor Kovács | timeline score: 5 | |
| Jan 10, 2015 at 6:49 | comment | added | Sándor Kovács | I'll try to write an answer tomorrow to explain what I mean. | |
| Jan 10, 2015 at 0:12 | comment | added | Question Mark | Thanks, but that is precisely my question: how to find a right principal divisor to move away from the intersection points? The intersection points are not DVRs, so it is not clear to me what "right amount of zeros or poles" means. Could you elaborate on your comment? | |
| Jan 9, 2015 at 23:24 | comment | added | Sándor Kovács | As long as you are not requiring $D$ to be effective (which you can't require for a general $\mathscr L$) you can move the support by adding a principal divisor with the right amount of zeros or poles at the questionable points. Or you could just start with choosing representatives for your divisor that are non-zero and invertible in a neighbourhood of any intersection points. You can always do this for any finite set of points. | |
| Jan 9, 2015 at 23:14 | history | asked | Question Mark | CC BY-SA 3.0 |