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Let $E_1$ and $E_2$ be elliptic curves defined over a field $\mathbb{K}$. An isogeny is a finite morphism $$\varphi:E_1\to E_2$$ such that $\varphi(\mathcal{O})=\mathcal{O}$. It is possible to show that an isogeny induces a morphisms of the groups of $E_1$ and $E_2$. It is clear that isogeny is an equivalence relation. Furthermore, following Hartshorne's Algebraic Geometry exercise 4.9.b, for a fixed elliptic curve $E$, the set elliptic curves of isogenous to $E$ is countable.

We are looking for a way to compute explicitly this set of isogenous curves for a fixed curve $E$. Our ultimate goal would be to give a set of curves $\lbrace E_i\rbrace_{i\in I}$ such that if $\varphi: E\to E'$ is an isogeny, then there exists $i\in I$ with $E_i\simeq E'$, each $E_i$ being expressed explicitely in the Weirstrass form (assuming $\mathrm{car}(\mathbb{K})\neq 2,3$).

Many theorems & propositions in Silverman's The Arithmetic of Elliptic Curves suggest a classification using subgroups of the fixed curve $E$. For example, proposition 4.12 tells us: "For an elliptic curve $E$ and $\Phi$ a finite subgroup of $E$, there is a unique elliptic curve $E'$ and a separable isogeny $\varphi:E\to E'$ such that $\ker \varphi=\Phi$." This approach did not show much progress yet unfortunatly.

I hope you find this question as interesting as I do. Any progress on the question, as tiny as it may seem, is more than welcome. If anything is unclear, please let me know.


Edit: Thanks to @Jyrki Lahtonen for pointing out this trivial case. Using Silverman's proposition cited above, if we have $\Phi=E[n]$, then the curve and the isogeny is $[n]:E\to E$.

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    $\begingroup$ Can you elaborate on your high lighted question? Compute explicitly a set means what exactly? $\endgroup$ – Mohan Jun 6 '17 at 21:04
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    $\begingroup$ A thing to keep in mind is that if $\Phi$ is the full $n$-torsion subgroup of $E$, then $E'\simeq E$ and the isogeny is just to multiple points by $n$. $\endgroup$ – Jyrki Lahtonen Jun 7 '17 at 10:55
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    $\begingroup$ $[n]:E\to E$ is a finite morphism such that the point at infinity is mapped to itself. Its kernel is $E[n]$. QED. If $n1_K=0_K$ then there are issues with separability. $\endgroup$ – Jyrki Lahtonen Jun 7 '17 at 16:08
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    $\begingroup$ Alexis, I have a vague recollection of having seen somebody claim that the method of constructing $E/\Phi$ that I used in this answer works quite generally. In other words, you get the function field $K(E/\Phi)$ as the subfield of $K(E)$ generated by $P\mapsto \sum_{Q\in\Phi}x(P+Q)$ and $P\mapsto \sum_{Q\in\Phi}y(P+Q)$. I was searching for it but couldn't find it. Unfortunately I'm not an expert. $\endgroup$ – Jyrki Lahtonen Jun 7 '17 at 16:20
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    $\begingroup$ Ok. I think that the result I recall is called Velu's formula. I'm not at all certain that it is exactly what I think it is, but the Google hits do suggest that it is a method for constructing the curve $E/\Phi$ given $E$ and $\Phi$. Do look it up! $\endgroup$ – Jyrki Lahtonen Jun 7 '17 at 16:32

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