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Do there exist non-isomorphic elliptic curves $(E_1, E_2)$ defined over $\mathbb{Q}$ that are connected by two or more isogenies defined over $\mathbb{Q}$ of different coprime degrees?

My thought : In my opinion, such curves do not exist in the case of coprime degrees. The existence of two isogenies of coprime degrees induces isomorphisms of Tate modules at every prime number. This leads to an isomorphism $E_2 \cong \text{Hom}(E_1, E_2) \otimes_{\text{End } E_1} E_1$ (see the answer at https://mathoverflow.net/questions/41931/about-isogeny-theorem-for-elliptic-curves), which recovers a $\mathbb{Q}$-isomorphism $E_1 \cong E_2$. Indeed, even if $E_1, E_2$ do not have CM, class number of $\text{EndE}\otimes_{\Bbb{Z}}\Bbb{Q}$ is $1$.

Would an explanation using Tate modules or Faltings' theorem be standard (unavoidable)?

Is there a simpler approach to the original problem? Even if you don't see one, I would be grateful if anyone knows of references to such problems and methods.

Poitou-Tate
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No, they do not exist but only because CM over $\mathbb{Q}$ means class number $1$ (see the last paragraph of my answer to see why you have CM in this case). Over number fields, the answer is as follows.

Let $\mathfrak{a}$ and $\mathfrak{b}$ be any two coprime integral ideals in an imaginary quadratic order $\mathcal{O}$. Suppose that $\mathfrak{a}$ and $\mathfrak{b}$ have the same class in the class group of $\mathcal{O}$ and further suppose that the class is nontrivial (so that in particular the class group of $\mathcal{O}$ is nontrivial). Such ideals always exist when the class group of $\mathcal{O}$ is nontrivial.

By CM theory the ideals $\mathfrak{a}$ and $\mathfrak{b}$ determine isogenies $\phi_{\mathfrak{a}}, \phi_{\mathfrak{b}} : E \to E'$ where $E/H$ is any elliptic curve with CM by $\mathcal{O}$ (here $H$ is some number field -- the ring class field of $\mathcal{O}$). The isogenies above are uniquely determined by the property that $\mathfrak{a}$ (resp. $\mathfrak{b}$) annihilate the kernel. But now the degrees are $\mathrm{Nm}(\mathfrak{a})$ and $\mathrm{Nm}(\mathfrak{b})$ which are coprime by assumption.

Conversely, if $\phi, \psi : E \to E'$ are coprime degree isogenies, then $\hat{\psi}\phi$ is an endomorphism of $E$ (here $\hat{\psi}$ is the dual) which is not multiplication by an integer, so $E$ (and thus $E'$) must have CM.

Mummy the turkey
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  • Sorry for bothering you, I think this is basic, but I've become confused about how the answer to the question follows from having CM by an order in a class number 1 field. If $\text{End}(E_1) \cong \text{Hom}(E_1, E_2)$, then I can say it without problem, but why does this (this isomorphism) hold? – Poitou-Tate Jun 03 '25 at 12:32
  • Off the top of my head I don't think that isomorphism does hold. The point is that non-principal ideals also determine isogenies – Mummy the turkey Jun 03 '25 at 15:05
  • @Mummy the turkey Could you tell me why 'CM over $\mathbb{Q}$ means class number $1$ ' implies 'There is no pair of elliptic curves over $\Bbb{Q}$ with isogenies of caprice degrees '? – Poitou-Tate Jun 03 '25 at 17:42
  • When $φ:E → E'$ is an isogeny where $E/\Bbb{Q}$ and $E'/\Bbb{Q}$ have CM by different orders $\text{End}E$ and $\text{End}E'$ respectively, it can happen that $j(E) ≠ j(E')$, and I think $φ(E)$ does not become isomorphic to any twist of $E$. – Poitou-Tate Jun 12 '25 at 16:15
  • I doubt the titled proposition of my question indeed holds. – Poitou-Tate Jun 12 '25 at 22:48
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    The easiest way to see any of this is the way hinted in the (now deleted) chain of comments under my post. There exists no elliptic curve $E/\mathbb{Q}$ with a non-integer endomorphism $\eta \colon E \to E$ defined over $\mathbb{Q}$ (i.e., $\textrm{End}_{\mathbb{Q}}(E) = \mathbb{Z}$ for all $E$ (regardless of CM). The argument above then shows that if $\phi \colon E \to E'$ and $\psi \colon E \to E'$ are isogenies we must have $\phi = [n] \circ \psi$ (or visa versa) since $\phi^{\vee} \circ \psi = [n]$ for some $n$. – Mummy the turkey Jun 13 '25 at 10:40
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    The more involved way (which also works over number fields) is to use the fact that isogenous CM curves have CM by the same number field and the degree of the isogeny connecting them must be related to the relative conductor – Mummy the turkey Jun 13 '25 at 10:42
  • Thank you very much. This indeed seems correct. Also, $\text{Hom}(E_1, E_2)\cong \text{End}E_1$ seems correct as well. I'm a bit confused as I don't know what the deleted comment was about. If possible (since I've made the answer editable now), could you edit your answer to include the content of that comment? – Poitou-Tate Jun 14 '25 at 12:45