**Conjecture:** If I have an elliptic curve with j-invariant 0 of the form $y^2 = x^3 + b$ over some prime-order field $\mathbb{F}\_p$ (where $p$ is not 2 or 3), and the group of rational curvepoints has prime order $q$ (which is not 2 or 3), then there is some curve of the form $y^2 = x^3 + b'$ over $\mathbb{F}\_q$ with prime order $p$.


I'm not sure about the characteristic 2-or-3 case, I guess I could check that exhaustively, but it's less interesting to me and I haven't taken the time to do it yet. For larger primes I've found a few dozen cases with small primes, a couple with large primes, and no counterexamples.


I have 3 questions:


1. Is this conjecture true?
2. What does it mean? This seems to give an equivalence relation between ground fields of different characteristic, which seems really weird to me... what am I looking at here? (Sorry this is so vague.)
3. What is the natural way to generalize this to curves whose groups of rational points has composite order, and/or fields with non-prime order?



Elliptic-curve related equivalence between fields of different characteristic?
If $\# E(\mathbb{F}\_p) = q$ and $j=0$, then the endomorphism ring is an order in the field of third roots of unity so $(p+1-q)^2 - 4p = -3u^2$ for some integer $u$. Now note that $(p+1-q)^2 - 4p$ is symmetric in $p$ and $q$. Hence, if there is an elliptic curve at all over $\mathbb{F}\_q$ with $p$ points, then it automatically has endomorphism ring by an order in the field of third roots of unity and thus has $j=0$ and is of the required form. So the remaining issue is whether $p$ lands in the Hasse interval for $q$. But, let's say $p < q$, then $q < (\sqrt{p}+1)^2$ so $\sqrt{p} > \sqrt{q}-1$ and $p$ is in the Hasse interval for $q$. The case $q<p$ is similar. This answers 1. I don't know about 2. I didn't really use the primality of $p,q$ in the proof except to ensure there is a finite field of that order. So, I guess 3. should go the same way. The only thing to watch out is that I am assuming $p,q$ are $1$ modulo $3$ for the curves to be ordinary. As supersingular curves over prime fields of cardinality $p>3$ have order $p+1$, hence not prime, this is not an issue but can become an issue when considering non-prime fields. 


Edit: To get $j=0$ the endomorphism ring has to be the maximal order. I think my proof is incomplete but hopefully is OK.


