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Suppose there are two isogenies $\phi:E \rightarrow E^\prime$ and $\psi:E \rightarrow E^\prime$ two isogenies between the same two elliptic curves. Can the isogenies be different? How can you tell?

AHusain
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José
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    You should look at the kernels. If same, the isogenies should be equal or equivalent in some strong sense (though you need to be careful in characteristic $p$). If the kernels are of different order, this will be reflected in the inequality between the degrees of $\phi$ and $\psi$, As a too-easy example, the degree of $[n]_E$ is $n^2$—this isogeny has the same domain and codomain, $E$. – Lubin May 15 '21 at 03:44
  • Ok, so different kernel orders means different degrees, which implies different isogenies. But what if we have two different kernels with the same order? – José May 15 '21 at 19:23
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    If the kernels have same order, you need to look more closely. For instance, if one kernel is $C_p\oplus C_p$ and the other is $C_{p^2}$, then they are certainly different. One really needs to look more closely. (Notation: “$C_n$” for cyclic group of order $n$.) – Lubin May 15 '21 at 21:36
  • So there is no way to know other than explicitely computing the abelian structure of the kernel? I wish there was a way to know as I'm studying isogeny volcanoes and each non-equivalent isogeny with a fixed degree is an edge on the graph, so knowing when the graph is a multigraph matters. – José May 15 '21 at 23:45
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    I think you need to be more specific about what the base is. A finite field? An algebraically closed field? Beyond that, I suppose that in the algebraically closed situation, the lattice of isogenies is the same as the lattice of subgroups of $\Bbb Z\oplus\Bbb Z$ that are of rank two. (Think of the complex case.) – Lubin May 16 '21 at 13:43
  • José, I think the question you just deleted can be answered with the techniques described here. Did you figure it out yourself? – Jyrki Lahtonen Jun 25 '21 at 19:47
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    I've figured it with the help of a friend taking the same class. You got to define a expression $\sum_{x \in \mathbb{F}p} \left[\left(1 + \left(\frac{x+1}{p}\right)\right)\cdot \dots \cdot \left(1 + \left(\frac{x+r}{p}\right)\right)\right]/2^r$ whose value is equal to the amount of sets of $r$ consecutive squares. Then you got to expand this to get $\frac{p}{2^r} +\frac{1}{2^r} \sum{x \in \mathbb{F}_p} \dots$. You can use the Hasse-Weil bound on the summatory to get the bound $O(\sqrt{p})$. – José Jun 28 '21 at 22:52
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    Great! That's what I would have suggested. I was looking for a suitable link as I have answered one with the same technique, and therefore I was slow to answer/comment. The question is settled anyway. No harm done. – Jyrki Lahtonen Jun 30 '21 at 06:51

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