Let's say $p$ is a prime and $t\neq 0$ is a trace of Frobenius that occurs over $\mathbb{F}\_p$. The discriminant of the Frobenius polynomial is
$\Delta:=t^2-4p.$
So we obtain $4p=t^2-\Delta.$ If $E$ is an elliptic curve over $\mathbb{F}\_p$ with trace of Frobenius $t$, then the Frobenius, call it $\sigma$, generates an order in the endomorphism ring of $E$, $End(E)$. Symbollically, $\mathbb{Z}[\sigma]\subseteq End(E).$ Put another way, the discriminant of $End(E)$ divides $\Delta$.


Now $\Delta$ may not be the discriminant of a maximal order. We write $\Delta=B\Delta\_{\mathcal{O}\_K}$ where $\Delta\_{\mathcal{O}\_K}$ is the discriminant of the ring of integers of some imaginary quadratic field $K$ and $B\geq 1$. Suppose there is an integer $b>1$ with $b^2\mid B$ and $b$ prime to $p$. Then $b$, the *conductor*, is the index of the order of discriminant $b\Delta\_{\mathcal{O}\_K}$ in $\mathcal{O}\_K$. There could be an elliptic curve over $\mathbb{F}\_p$ whose endomorphism ring is the order of discriminant $b\Delta\_{\mathcal{O}\_K}$.


My question is, in general, does every possible order occur as an endomorphism ring of an elliptic curve over $\mathbb{F}\_p$? Kohel seems to indicate this is the case in his thesis, but I can't find a proof. By work of Deuring and because the $j$-invariants in question are algebraic, we know that over some finite extension of $\mathbb{F}\_p$ there is an elliptic curve with an endomorphism ring that is isomorphic to each possible order. For my applications though, I'd like to have this elliptic curve be defined over $\mathbb{F}\_p.$



Endomorphism rings of ordinary elliptic curves
The paper "Abelian varieties over finite fields" of Waterhouse (Annales scientifiques de l'École Normale Supérieure, Série 4, Volume 2 (1969) no. 4, p. 521-560), available at


<http://www.numdam.org/item/?id=ASENS_1969_4_2_4_521_0>


has, I think, relevant information (see Section 4). There might be more recent works in this direction also.


