Timeline for Is there any rational curve on an Abelian variety?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21, 2014 at 17:32 | comment | added | Matthieu Romagny | @Felipe Voloch: but then you end up using the fact that you wanted to prove (that a map from $P^1$ to an ab. var. is constant). | |
| Jan 10, 2010 at 17:34 | comment | added | Felipe Voloch | @Anweshi Ha! Here is a short proof that, for curves, the Jacobian is the Albanese, using that maps from $P^1$ to an abelian variety are constant. A map from a curve to an abelian variety extends to divisors by linearity and is constant on linear equivalence classes because it's constant on $P^1$'s, so the map gives a well-defined map from the jacobian to the abelian variety. If you want to develop things this way from first principles, you have to avoid my answer to the original question and use one of the other answers. | |
| Jan 10, 2010 at 16:27 | comment | added | Anweshi | Mumford avoids the Jacobian altogether. Lang's book is not very organized and uses the old language. But I suppose I will indeed find Albanese and Picard there. Thanks. | |
| Jan 10, 2010 at 16:11 | comment | added | Felipe Voloch | @Anweshi Reference: Lang, Abelian Varieties. Probably in Mumford too. | |
| Jan 10, 2010 at 15:18 | comment | added | Anweshi | Fabulous answer. Could you please provide some reference in which the Albanese and Picard functorialities of Jacobian are compared with rigor and full proof? | |
| Dec 16, 2009 at 6:25 | vote | accept | Fei YE | ||
| Dec 16, 2009 at 6:25 | vote | accept | Fei YE | ||
| Dec 16, 2009 at 6:25 | |||||
| Dec 16, 2009 at 3:49 | history | answered | Felipe Voloch | CC BY-SA 2.5 |