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 $\mathbb{F}_p.$