Let $E$ be an elliptic curve over $\mathbb C$ with CM by ring of integers $O\_K$ of an imaginary quadratic number field $K$. Let $O$ be an order of $O\_K$.


Is there a number field $L$ such that $E$ has a model $E\_L$ over $L$ with $\mathrm{End}\_L(E\_L) = O$?


Of course, if $O = O\_K$, the answer is positive. But what if $O$ has a non-trivial conductor?


Concretely: Let $E : y^2= x^3 +x$ over $\mathbb Q$, and let $f\geq 2$ be an integer. Is there a number field $L$ such that $\mathrm{End}\_L(E\_L) = \mathbb Z[fi]$?



Endomorphisms of elliptic curves with CM; can we have an order?
No. If $\alpha$ is an endomorphism over $\bar L$ such that $n\alpha$
is defined over $L$, then so is $\alpha$. Proof: suppose some element
of Gal$(\bar L / L)$ takes $\alpha$ to an endomorphism $\beta$.
We shall prove that $\beta=\alpha$. Indeed by hypothesis $n\beta = n\alpha$.
But then $n(\beta-\alpha) = 0$, so the endomorphism $\beta-\alpha$ 
takes $E$ to the $n$-torsion group $E[n]$, which is finite.
Hence $\beta-\alpha$ is the zero endomorphism; that is, $\beta=\alpha$.
Therefore $\alpha$ is Galois-invariant,
and thus defined over $L$, as claimed. $\Box$


This proof works more generally with the elliptic curve $E$ 
replaced by any abelian variety.


