Suppose we have a curve $X$ (of genus $\geq 3$), and we know that $\{\phi_i : X \to E_i\ \textrm{ for } i = 1, ..., r\}$ are covers of degrees $d_i$ (with the $d_i$'s not necessarily all equal), where the $E_i$ are (genuinely) distinct elliptic curves. Suppose further that we know, for example, that $J = Jac(X)$ is isogenous to $A := E_1^{\alpha_1} \times ... \times E_r^{\alpha_r}$. My question is: is the map $\alpha_1 \phi_1^\ast + ... + \alpha_r \phi_r^\ast$ an isogeny from $A$ to $J$, where $\alpha \phi^\ast$ is short-hand notation for $\phi^\ast + ... + \phi^\ast$ ($\alpha$ times)? If so, how would one show surjectivity? Again, if so, is it obvious what the degree of such an isogeny is? If, on the other hand, $\phi$ is not an isogeny, is there some standard way to construct an isogeny $A \to J$ by using the maps $\phi_i$?

  • $\begingroup$ 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 $J$. $\endgroup$
    – abx
    Nov 10 '13 at 17:38
  • $\begingroup$ Good point, my notation is ambiguous -- what I meant with $\alpha \phi^\ast$ is $\phi^\ast + ... + \phi^\ast$ (there are $\alpha$ summands). $\endgroup$
    – user42616
    Nov 10 '13 at 18:02
  • $\begingroup$ But then it cannot be surjective! $\phi^*$ is a homomorphism, so the image of $\phi^*+\ldots+\phi^*:E^\alpha \rightarrow J$ is $\phi^*(E)$. $\endgroup$
    – abx
    Nov 10 '13 at 18:15
  • $\begingroup$ 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! $\endgroup$
    – user42616
    Nov 10 '13 at 18:21
  • 1
    $\begingroup$ If $\alpha_i$ is greater than one, this implies that there are many different covering maps from $X$ to $E_i$. $\endgroup$
    – Will Sawin
    Nov 10 '13 at 22:23

OK, assume $\alpha _i=1$ for all $i$ (this is the only case where the question makes sense). You are given an isogeny $JC\sim A=E_1\times \ldots \times E_g$, with the $E_i$ in different isogeny classes. Poincaré complete reducibility theorem tells you that this decomposition is unique (up to isogeny), so any nontrivial map from $E_i$ to $A$ maps $E_i$ onto the corresponding summand of $A$. Therefore any family of maps $\phi_i:C\rightarrow E_i$ $(i=1,\ldots ,g)$ produces an isogeny $\ (\phi_i^*): \prod E_i\rightarrow JC$.


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