Timeline for sequence of sheaves for studying intersection
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 27, 2011 at 9:35 | comment | added | Srks | @quim Thank you! this is just what I was missing. | |
| Sep 27, 2011 at 7:18 | vote | accept | Srks | ||
| Sep 26, 2011 at 13:35 | comment | added | quim | The point is that you know $H|_B=3P$, and then the isomorphism tells you that $3P\in |3P|$, being a single divisor, has a single divisor in the preimage. In other words, the isomorphism in $H^0$ induces an isomorphism in the projectivizations $|H|={\mathbb P}(H^0({\mathcal O}_{X}(d)))\cong {\mathbb P}(H^0({\mathcal O}_{B}(3P)))=|3P|$. | |
| Sep 26, 2011 at 12:04 | comment | added | Srks | thank you quim for your answer. For my problem I can assume $P$ to be a point of order $3d$ for all $d$ so I have no problems about flexes. Anyway I can see the existence form the isomorphism of $H^0$ but how can you show that is unique? | |
| Sep 26, 2011 at 10:55 | comment | added | naf | One can find such conics passing through any point of order $6$ on $B$ (where we choose a flex as the zero for the group law). | |
| Sep 26, 2011 at 9:42 | comment | added | quim | BTW: I don't know offhand if there exist other points P (not flexes) where there is an irreducible conic Q with Q.B=6P. My guess is yes. | |
| Sep 26, 2011 at 9:40 | history | answered | quim | CC BY-SA 3.0 |