Timeline for dual isogeny for abelian varieties over a general field
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 26, 2011 at 22:36 | history | edited | user565739 | CC BY-SA 3.0 |
added 137 characters in body
|
| Jul 26, 2011 at 22:28 | history | edited | user565739 | CC BY-SA 3.0 |
added 2791 characters in body
|
| Jul 26, 2011 at 16:41 | vote | accept | user565739 | ||
| Jul 26, 2011 at 13:52 | history | edited | Qing Liu |
add tag
|
|
| Jul 26, 2011 at 13:50 | answer | added | Qing Liu | timeline score: 6 | |
| Jul 26, 2011 at 0:40 | comment | added | Joe Silverman | It's been a while since I looked (and I don't have a copy handy), but doesn't Mumford give a construction of the dual isogeny over arbitrary fields in his book on Abelian Varieties? Probably Lang does, too, in his *Abelian Varieties" book, but that's in the older Weil-style language, so probably harder to read. | |
| Jul 25, 2011 at 21:43 | history | edited | user565739 | CC BY-SA 3.0 |
added 189 characters in body
|
| Jul 25, 2011 at 21:30 | history | edited | user565739 | CC BY-SA 3.0 |
added 930 characters in body
|
| Jul 5, 2011 at 10:14 | comment | added | naf | As I have already mentioned, you can get a proof for abelian varieties by working with group schemes. One cannot use just Frobenius in this case; for example, by taking products of elliptic curves it is easy to find inseparable isogenies of degree $p$ between abelian varieties of dimension $g$ for any $g \geq 1$. | |
| Jul 5, 2011 at 9:52 | history | edited | user565739 | CC BY-SA 3.0 |
added 1661 characters in body
|
| Jul 4, 2011 at 10:37 | comment | added | user565739 | @François: You are right. Since I read this on a book on Elliptic Curves, so $E$ is isomorphic to it's dual ( I guess so ). @Jason: Thanks. For an elliptic curve over a perfect field $k$, the existence of $\hat{\phi}$ is for all isogeny. One needs to decompose $\phi$ as a separably isogeny followed by a Frobenius morphism. I have to think about why this doesn't work when $k$ is not perfect (Hence in this case, one need to focus on separably isogeny only) | |
| Jul 4, 2011 at 10:14 | comment | added | naf | If the characteristic of the base field divides $\deg(\phi)$ then you need to work with group schemes even if the field is perfect. In general, one can define the quotient of an abelian variety by any finite subgroup scheme, so the same proof works over any field. See for example, SGA 3, Expose VI_A. (As pointed out by Francois Brunault, the map you want is not what is usually called the dual isogeny (in higher dimensions) but it does exist.) | |
| Jul 4, 2011 at 10:06 | comment | added | Jason Starr | If the isogeny is separable, then the kernel of $\phi$ is split after passing to the separable closure of $k$. Now you can form the group quotient, etc., over the separable closure. And you can perform Galois descent for the extension of $k$ to the separable closure to get $\phi^\vee$ defined over $k$. Possibly the references are imposing "perfect" as a hypothesis because they want a result that applies to all isogenies, not just separable isogenies. | |
| Jul 4, 2011 at 9:58 | comment | added | François Brunault | The dual isogeny should be $\vee{\phi} : B^{\vee} \to A^{\vee}$. I think the map $\phi \mapsto \hat{\phi}$ you want to define will not be well-behaved, even in characteristic 0. For example it won't be involutive if the dim is >1 (look at the degrees of the isogenies which are involved). | |
| Jul 4, 2011 at 9:40 | history | asked | user565739 | CC BY-SA 3.0 |