Let $E$ be an elliptic curve in characteristic zero with $Aut(E)=\pm$. Is it possible to have two different isogenies to another elliptic curve $E'$ with distinct cyclic kernels? This isn't possible if $End(E)\cong\mathbb{Z}$ (see for example Lang's Elliptic functions Chapter 2, $\S 2$), but how about if $E$ does have CM?
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asked Aug 14, 2012 at 11:09
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It is possible. Suppose $E$ has complex multiplication defined over a number field $F$ by the ring of integers $\mathcal{O}_K$ in a quadratic imaginary field $K$ (and we take $K$ that only has $\pm 1$ as roots of unity, so that your $Aut(E) = \pm 1$ condition is satisfied). Then one can take $E^{\prime} = E$ and the following for kernels of (infinitely many) distinct cyclic isogenies $E \rightarrow E$: for each rational prime $p$ split in $K$ take a prime $\mathfrak{p}|p$ of $K$ and let the kernel of the isogeny be $E[\mathfrak{p}] := \{ P \in E(\bar{F}): \alpha P = 0, \text{ for all } \alpha \in \mathfrak{p}\}$. By the theory of complex multiplication, $E[\mathfrak{p}] \cong \mathcal{O}_K/\mathfrak{p}$ as $End_F E = \mathcal{O}_K$-modules, so the kernels of the isogenies are indeed cyclic.
answered Aug 14, 2012 at 11:37
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$\begingroup$ @Kestutis: Thanks - don't you need $\mathfrak{p}$ to be principal for this to be an endomorphism? $\endgroup$ Commented Sep 3, 2012 at 5:28
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$\begingroup$ No, you don't. Even if $\mathfrak{p}$ is not principal, $E[\mathfrak{p}]$ is Galois invariant of order $\#(\mathcal{O}_K/\mathfrak{p}) = p$. Hence it is the kernel of some $p$-isogeny $E \rightarrow E$. $\endgroup$ Commented Sep 3, 2012 at 12:33