# Is any imaginary quadratic field equal to $\mathrm{End}(E) \otimes \Bbb Q$ for some $E / \Bbb F_q$?

Let $$K$$ be an imaginary quadratic field. Is there always a finite field $$k$$ and an elliptic curve $$E$$ such that the endomorphism algebra $$\mathrm{End}_{\overline k}(E) \otimes \Bbb Q$$ is isomorphic to $$K$$ ?

A refined question would be whether any order $$O \subset K$$ is isomorphic to $$End(E)$$ of some $$E$$ ?

(The analogue for $$K=\Bbb Q$$ is discussed here:Is the ring of integers of any imaginary quadratic field equal to $\mathrm{End}(E)$ for some $E / \Bbb Q$?)

I know that $$E$$ should be ordinary in that case (the supersingular case is known due to Deuring correspondence).

One idea would be to take some $$E / \Bbb C$$ with CM by $$K$$, it is then defined over some number field $$L$$, and then find a prime ideal $$Q$$ of $$O_L$$ so that $$End(E) \cong End(E \times_{O_L} O_L / Q)$$...

Yes, from Serge Lang's book Elliptic Functions theorem 13.4.12 (Deuring) says the following:

So it suffices to take a prime integer $$q$$ that splits in $$O_K$$ (not necessarily in $$O_L$$ !) such that $$q$$ doesn't divide the conductor of the order $$End(E) \subset O_K$$ (there exist infinitely many such $$q$$ by Cebotarev, or Frobenius theorem... !), then the reduction of the curve $$E / L$$ mod $$Q$$ (where $$Q$$ is any prime above $$q$$) has the same endomorphism ring as $$E$$.

I guess that any order $$O \subset O_K$$ can be $$End(E)$$ for some $$E$$ over a number field (take a suitable complex torus... then CM theory tells us it is defined over $$\overline{\Bbb Q}$$). But of course, over a given finite field $$k$$, there are only finitely many possible $$End(E)$$ for $$E/k$$. But what about $$E / \overline k$$ (e.g. this includes all maximal orders in the quaternion algebra $$Q_{p, \infty}$$ by Deuring)?

Then I guess we have the following: an order $$O$$ in an imaginary quadratic field $$K$$ is the endomorphism ring of some $$E / \overline{\Bbb F_p}$$ iff the conductor of $$O$$ is coprime to $$p$$ (i.e. $$O = \Bbb Z + f O_K$$ for $$gcd(f,p)=1$$) and $$p$$ splits in $$K$$.

One direction was just proved. The other uses Deuring lifting theorem (13.5.14 in Lang), and combined with part i) of thm 12 above.

• See Theorem 9.10.18 in math.auckland.ac.nz/~sgal018/crypto-book/ch9.pdf Commented May 10, 2021 at 16:35
• See also mat.uniroma2.it/~schoof/cubiccurves.pdf thm 4.3 (refers to Waterhouse 1969 thesis) Commented May 10, 2021 at 16:40
• Notice that for ordinary E / F_q, we have $End(E) \otimes Q = Q(\sqrt{D})$ where $D= a_q^2 - 4q < 0$ (by Hasse and $D=0$ implies $E$ supersingular). We have $D \geq -4q$, so there are finitely many imag. quad. fields that can occur (and also fin. many orders therein, since there are fin. many E / F_q, for fixed q !). Commented May 21, 2021 at 8:33