Let $E$ be a supersingular elliptic curve which is defined over field $\mathbb{F}_{p^2}$ and $l$ be a prime such that $gcd(l,p)=1$. Is there an endomorphism $\phi\in End(E)$ such that $deg(\phi)=l$? Suppose that $H$ is a cyclic subgroup of $E$ and $|H|=l$. I also want to know if I construct an isogeny using velo formula, is it possible to have $j(E/H)=j(E)$?
$\begingroup$
$\endgroup$
4
-
2$\begingroup$ Regarding your first question, having a degree $l$ endomorphism means that you have an element in your endomorphism ring of norm $l$. Since $E$ is supersingular, its endomorphism ring is an order in the quaternion algebra ramified at $p$ and $\infty$. Then the question is whether the norm form on such order represents $l$ or not. For example the maximal order in the quaternion algebra ramified at $11$ and $\infty$ represents all primes congruent to $1$ modulo $4$ (since it contains the Gaussian integers) but does not represent $3$. Clearly they will always represents all big enough primes $\endgroup$– A. PacettiCommented Mar 5, 2016 at 11:51
-
$\begingroup$ It is clear that the answer to the second questions is true, since up to isomorphism there are only finitely many supersingular elliptic curves, and isogeny curves are supersingular (and there are infinitely many primes $l$ satisfying your first question) $\endgroup$– A. PacettiCommented Mar 5, 2016 at 12:30
-
$\begingroup$ @A.Pacetti I think for some supersingular elliptic curves over $\mathbb F_{p^2}$, the endomorphism ring is lower because not all the endomorphisms are defined over $\mathbb F_{p^2}$. Indeed by Honda-Tate there are elliptic curves with characteristic polynomial of Frobenius $T^2 - a p T + p^2$ for $a \in \{-1,0,1\}$ which provide examples. $\endgroup$– Will SawinCommented Mar 5, 2016 at 17:40
-
$\begingroup$ The question does not ask the isogeny to be defined over $F_{p^2}$, but the isogenous elliptic curve will be defined over $F_{p^2}$ (since they are supersingular). $\endgroup$– A. PacettiCommented Mar 5, 2016 at 17:46
Add a comment
|