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Let $E$ and $E'$ be isogenous elliptic curves and $K=\text{end}(E) \otimes \mathbb(Q) $

Is it true that $\text{end}(E') $ is a subring of $K$?

The only thing I thought is that the isogeny between $E$ and $E'$ and its dual give a map between the endomorphism rings, which as far as i know needs not to be surjective or injective.

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    $\begingroup$ It's true. Maybe a good way to see it is to look at lattices in $\Bbb C$. $\endgroup$
    – Lubin
    Commented Aug 27, 2017 at 14:58
  • $\begingroup$ I'll think about the lattices. Honestly I didn't think about them because this question rose while studying the section on finite fields of "arithmetic of elliptic curves" $\endgroup$
    – karmalu
    Commented Aug 27, 2017 at 15:15
  • $\begingroup$ Afterwards, I realized that that suggestion might not have been too helpful. Can’t you show directly that if $E$ and $\mathscr E$ are isogenous, then $\text{End}(E)\otimes\Bbb Q$ and $\text{End}(\mathscr E)\otimes\Bbb Q$ are essentially equal? $\endgroup$
    – Lubin
    Commented Aug 27, 2017 at 17:35
  • $\begingroup$ How would you suggest to do that? $\endgroup$
    – karmalu
    Commented Aug 28, 2017 at 12:31

2 Answers 2

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Try this:
Let $\varphi:E\to E'$ and $\psi:E'\to E$ be the pair of isogenies, $\deg\varphi=\deg\psi=n$ and $\psi\circ\varphi=[n]_E$, while $\varphi\circ\psi=[n]_{E'}$. I’m going to map $\,f\in\text{End}(E)$ to an element $\,\tilde f\in\text{End}(E')\otimes\Bbb Q$, and we’ll see that the map is injective.

Simply set $\,f'=\frac1n\varphi\circ f\circ\psi$. You see immediately that it’s a ring homomorphism, and certainly injective, since the composition of (nonzero) isogenies is still nonzero.

I think this gets you to come down where you want to be.

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I think it is not true at least over $\mathbb{C}$ Let us consider the elliptic curve $E$ whose lattice is given by $\langle 1,\frac{i}{2} \rangle$ which has not the CM Now let us consider the isogeny given by the quotient for the translation of the point $\frac{1}{2}$ then you get an elliptic curve $E’$ whose lattice is $\langle\frac{1}{2},\frac{i}{2}\rangle$ which is isomorphic to $\langle 1,i\rangle$. Thus $E’\simeq E_i$ which has the CM given by $i$.

This i think that it may happen then after an isogeny you get an automorphism group bigger or not depending on the isogeny you are considering.

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  • $\begingroup$ Well, your first lattice has the endomorphism $z\mapsto2iz$, doesn’t it? $\endgroup$
    – Lubin
    Commented May 30, 2022 at 19:51
  • $\begingroup$ Sorry I correct my post since I wrote that the endomorphism ring change up to isogeny (which is false) what change is the automorphism group. So here the point is that when you consider an isogeny of elliptic curve (or abelian variety) you can loose or find some transformation of your lattice, right? But are you sure that is it an endomorphism of $E$? since I don't see why the lattice of $E$ is preserved by this map.. I mean an endomorphism must be a trasnformation of $E$ so it must preserve its lattice . $\endgroup$ Commented May 31, 2022 at 12:51
  • $\begingroup$ The map $z\mapsto2iz$ doesn’t preserve the lattice$L=\langle1,i/2\rangle$, it maps $L$ into itself. And this induces a map $E=\Bbb C/L$ to $E$. And $E$ has the nonintegral endomorphism coming from $z\mapsto2iz$. So you see that the endomorphism ring of $E$ is $\Bbb Z[2i]$, not all tlhe Gaussian integers, but of index $2$ therein. $\endgroup$
    – Lubin
    Commented May 31, 2022 at 20:10

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