Timeline for Isogeny of abelian varieties
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 10, 2013 at 23:14 | vote | accept | CommunityBot | ||
| Nov 10, 2013 at 22:23 | comment | added | Will Sawin | If $\alpha_i$ is greater than one, this implies that there are many different covering maps from $X$ to $E_i$. | |
| Nov 10, 2013 at 19:13 | answer | added | abx | timeline score: 4 | |
| Nov 10, 2013 at 18:28 | comment | added | user42616 | Fair enough. Suppose then that the $\alpha_i =1$ for all $i$. Does $\phi$ always define an isogeny? Can we deduce the degree directly? (is it $\prod_i d_i$?) | |
| Nov 10, 2013 at 18:24 | comment | added | abx | Obviously no, unless you take $\alpha_i=1$ for all $i$. | |
| Nov 10, 2013 at 18:21 | comment | added | user42616 | You're right - I should have thought of that. Is there any way, then, to construct an isogeny $A \to J$ by "using only" the maps $\phi_i$? Thanks! | |
| Nov 10, 2013 at 18:15 | comment | added | abx | But then it cannot be surjective! $\phi^*$ is a homomorphism, so the image of $\phi^*+\ldots+\phi^*:E^\alpha \rightarrow J$ is $\phi^*(E)$. | |
| Nov 10, 2013 at 18:05 | history | edited | user42616 | CC BY-SA 3.0 |
Ambiguous notation
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| Nov 10, 2013 at 18:02 | comment | added | user42616 | Good point, my notation is ambiguous -- what I meant with $\alpha \phi^\ast$ is $\phi^\ast + ... + \phi^\ast$ (there are $\alpha$ summands). | |
| Nov 10, 2013 at 17:45 | review | First posts | |||
| Nov 10, 2013 at 17:55 | |||||
| Nov 10, 2013 at 17:38 | comment | added | abx | There is something wrong with your formulation : what is the map from $E_i^{\alpha_i} $ to $J$? $\alpha_1 \phi_1^\ast$ is a map from $E_i$ to $J$, not from $E_i^{\alpha_i} $ to $J$. | |
| Nov 10, 2013 at 17:29 | history | asked | user42616 | CC BY-SA 3.0 |