Timeline for Do isogenies between AVs over finite fields separate finite subgroups?
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2011 at 16:38 | comment | added | Tommaso Centeleghe | I think I was looking at a false generalization of what W. uses. I included a proof of his statement below. Thanks for your time. | |
| Mar 2, 2011 at 16:29 | vote | accept | Tommaso Centeleghe | ||
| Mar 2, 2011 at 16:28 | answer | added | Tommaso Centeleghe | timeline score: 0 | |
| Mar 2, 2011 at 14:20 | comment | added | Chris Wuthrich | No, the point is that in your EDIT 2, you have $A = B$ as you start with a subset of endomorphisms. | |
| Mar 2, 2011 at 13:54 | answer | added | Yuri Zarhin | timeline score: 7 | |
| Mar 2, 2011 at 13:41 | comment | added | Tommaso Centeleghe | @Chris. You are right, in general it should not be true. However if the endomorphism ring of $E$ is a maximal order, then it should hold. As W. explains. | |
| Mar 2, 2011 at 13:31 | history | edited | Tommaso Centeleghe | CC BY-SA 2.5 |
added 725 characters in body
|
| Mar 2, 2011 at 13:17 | comment | added | Chris Wuthrich | They have different $j$-invariant. Yes, as right modules. I don't see why the fact that this ideal is principal will imply that they should be isomorphic. | |
| Mar 2, 2011 at 13:11 | comment | added | Chandan Singh Dalawat | numdam.org/numdam-bin/fitem?id=ASENS_1969_4_2_4_521_0 | |
| Mar 2, 2011 at 13:09 | comment | added | Tommaso Centeleghe | Thanks for your comment. I am not fully able to verify your counterexample. You are saying that the map from Hom(E,E') to End(E) sending x to $\hat\varphi$x gives an iso between Hom(E,E') and the ideal (2) as right End(E) modules, right? If this is the case then shouldn't E be isomorphic to E', because of the principality of that ideal? Anyway I am going to edit my question and post what Waterhouse actually says... | |
| Mar 2, 2011 at 12:54 | comment | added | Chris Wuthrich | I took $k=\mathbf{F}_5$. Am I doing something wrong, here ? Sorry I do not have Waterhouse at hand. | |
| Mar 2, 2011 at 12:53 | comment | added | Chris Wuthrich | I do not see why your statement is true in the following example. Take $\psi$ to be the $2$-isogeny from $E : y^2=x^3+x$ to the non-isomorphic $E': y^2 = x^3+x+3$ killing the point $(2,0)$. Now the image under the dual $\hat\psi$ of $I(E,E')$ in $End(E)$ is the ideal generated by $[2]$. I believe that one concludes from this that any isogeny $\varphi:E\to E'$ has to factor through $\psi$. So $H=<(2,0)>$ vanishes for all $\varphi$ | |
| Mar 2, 2011 at 12:52 | history | edited | Tommaso Centeleghe | CC BY-SA 2.5 |
added 196 characters in body
|
| Mar 2, 2011 at 11:07 | history | asked | Tommaso Centeleghe | CC BY-SA 2.5 |