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Q: Show that any two CM elliptic curves with the same endomorphism algebra (say $K$) are isogenous.

I was thinking of the following:

Let $E_1\simeq \mathbb{C}/L_1$ and $E_2\simeq \mathbb{C}/L_2$, then since both elliptic curves have CM by $K$, we know $L_1\otimes \mathbb{Q} = L_2\otimes \mathbb{Q}=K$. WLOG we may assume $L_1\subseteq L_2$, then $L_1\cap L_2$ is also a lattice and has finite index in both $L_1$ and $L_2$ (this is where we probably use CM but I am unsure of the details, in fact is it even okay to assume the containment?). From here we conclude that there is an isogeny $\varphi: \mathbb{C}/L_1 \rightarrow \mathbb{C}/L_2$; indeed, since the map is surjective and kernel=$L_2/L_1$ is finite.

What happens for the case of non-CM elliptic curves?

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  • $\begingroup$ If you assume one lattice contains the other, then the intersection is simply the latter. $\endgroup$ – Mariano Suárez-Álvarez Jan 17 '18 at 4:48
  • $\begingroup$ Yes, I see I was being dumb there! But I am not convinced I can assume that one contains the other $\endgroup$ – debanjana Jan 17 '18 at 4:59
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Non-CM elliptic curves over the complexes all have endomorphism ring $\Bbb Z$. Up to isomorphism, there are only countably many curves isogenous to a given curve, but there are uncountably many curves (consider the $j$-invariants) so there are uncountably man isogeny classes of non-CM elliptic curves.

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