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 $\mathrm{End}_L(E_L) = \mathbb Z[fi]$?


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.


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