TITLE: Endomorphisms of elliptic curves with CM; can we have an order?
QUESTION [6 upvotes]: 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 $\mathrm{End}_L(E_L) = \mathbb Z[fi]$?

REPLY [13 votes]: 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.