# Elliptic curves over $\mathbb C$ have same endomorphism ring but not isomorphic

I am finding elliptic curves over $\mathbb C$ have same endomorphism ring but not isomorphic.

Elliptic curves over $\mathbb C$ can be identified with $\mathbb C/\land$ for some lattice $\land$. And endomorphism ring is just a set of z s.t. z$\land \subseteq \land$.

But I am stuck here, the lattice in my mind is too simple, I guess I need to see more example. Can somebody help me?

What about lattice generated by {$1$,$\sqrt{2} i$} and {$i$,$\sqrt{2}$}? Is this an example?

• The vast majority of elliptic curve over $\Bbb C$ have $\Bbb Z$ as their endomorphism ring... – mercio Dec 6 '15 at 18:33
• @mercio You are absolutely right. So the point here is to find example with endomorphism ring larger than $\mathbb Z$ – user198206 Dec 6 '15 at 23:15

Your two lattices actually give isomorphic elliptic curves, because one can be rotated into the other via multiplication by $i$. So it doesn't work.
It is actually slightly tricky to find an example. You need two lattices which are both closed under multiplication by some subring $R \subseteq \mathbb{C}$ strictly containing $\mathbb{Z}$, but are not multiples of each other. Note that "multiple" can be any complex number.
One way to find an example is to fix a subring $R$, then look for sublattices within $R$ which are closed under multiplication by $R$. By definition, these sublattices are the ideals of $R$. So we are looking for 2 ideals of $R$ that are not multiples of each other by any complex number.
Now we enter the land of algebraic number theory: Two ideals of $R$ are complex multiples of each other iff they belong to the same ideal class. So to find our example, we need a ring $R \subseteq \mathbb{C}$ with more than one ideal class. This is equivalent to the statement that $R$ does not have unique factorization into elements.
The standard example is $R = \mathbb{Z}[\sqrt{-5}]$. It has 2 ideal classes. There is the trivial ideal class represented by $R$ itself, and the non-trivial ideal class represented by the ideal $J = (2, 1 + \sqrt{-5})$. So $R$ and $J$ represent two elliptic curves which are non-isomorphic, but both have endomorphism ring $R$. For some more details on the ideal classes in $\mathbb{Z}[\sqrt{-5}]$, see here.
• I still think that $\Bbb Q(\sqrt{-6}\,)$ gives a clearer example. Then the lattices $\Bbb Z+\sqrt{-6}\Bbb Z$, generated by $1$ and $\sqrt{-6}$; and $2\Bbb Z+\sqrt{-6}\Bbb Z$, generated by $2$ and $\sqrt{-6}$, work just as well, and are more transparent. – Lubin Dec 7 '15 at 23:51