# Supersingular elliptic curves with automorphisms of order 6

Let $E$ be the elliptic curve over $\mathbb{Q}$ defined by $$y^2=x^3-1.$$ Let $p$ be an odd prime congruent $-1$ modulo $3$. Then, this curve becomes supersingular after ''reduction'' modulo $p$, also denoted by $E$. Note that the group of automorphisms of $E$, $\text{Aut} (E)$, is cyclic of order $6$. Let $\rho$ be a generator of $\text{Aut} (E)$.

Let $q=3k+1$ be another odd prime (for some $k\geq 2$). Then, $E$ has $(q+1)=(3k+2)$ distinct cyclic subgroups of order $q$. Let $\{C_1, \dots, C_{3k+2}\}$ be the complete list of these cyclic subgroups. Then, there are exactly two $i$ such that $$\rho C_i = C_i.$$ Let $D_1$, $D_2$ be two such subgroups. Then, for $C_i \not\in \{D_1, D_2\}$, $\rho C_i \neq C_i$. Thus, $(E, C_i), (E, \rho(C_{i}))$ and $(E, \rho^2(C_{i}))$ are isomorphic pairs. (Here, two pairs $(E, C)$ and $(E, D)$ are isomorphic if there is an automorphism sending $C$ to $D$.) By reordering them, we may assume that $$\rho C_i = C_{i+k} \text{ for } 1\leq i \leq 2k, \quad C_{3k+1}=D_1 \quad\text{ and } \quad C_{3k+2}=D_2.$$ Then, the complete list of the isomorphism classes of pairs $(E, C)$ (with same $E$) is $$S:=\{[(E, D_1)], [(E, D_2)], [(E, C_i)] : 1 \leq i \leq k\}.$$ For $1\leq i \leq k$, $\text{Aut}([(E, C_i)]) \simeq \{ \pm 1 \}$ and $\text{Aut}([(E, D_1)]) \simeq \text{Aut}([(E, D_2)]) \simeq \langle \rho \rangle$.

Any element of $S$ can be regarded as a point of $X_0(q)(\overline{\mathbb{F}}_p)$, where $\overline{\mathbb{F}}_p$ is an algebraic closure of $\mathbb{F}_p$, the finite field of $p$ elements. Some of them can be defined over $\mathbb{F}_p$ and the others are not.

Now I am asking the following: As a point of $X_0(q)(\overline{\mathbb{F}}_p)$, is $[(E, D_1)]$ defined over $\mathbb{F}_p$? i.e., $[(E, D_1)] \in X_0(q)(\mathbb{F}_p)$? (namely, $\text{Frob}_p(D_1)=D_1$? Here $\text{Frob}_p$ is the Frobenius endomorphism in characteristic $p$.) Similarly, is $[(E, D_2)]$ defined over $\mathbb{F}_p$?

To prove this, consider the eigenvalue of $\rho$ on each of those subgroups, which is one of the two sixth roots of unity in $\mathbb F_q$. Because $\operatorname{Frob}_p(\zeta)=\zeta^{-1}$ for $\zeta$ a third root of unity, we have $\operatorname{Frob}_p (\rho)=\rho^{-1}$, and so Frobenius does not preserve this invariant and hence does not preserve the subgroup.
• How to compare elements in $\mathbb{F}_q$ and $\mathbb{F}_p$? Are you lifting the eigenvalue in $\mathbb{F}_q$ to a sixth root of unity in $\mathbb{C}$? Commented Aug 24, 2017 at 12:11
• And why does Frobenius preserve this invariant if $\text{Frob}_p(D_1)=D_1$? Commented Aug 24, 2017 at 12:24
• @user1225 If $Frob_p(D_1) = D_1$, then $Frob_p$ and $\rho$ both act by scalar multiplication on $D_1$, so $Frob_p(\rho) = Frob_p \circ \rho \circ Frob_p^{-1}$ acts by the same eigenvalue as $\rho$. Commented Aug 24, 2017 at 12:26
• @user1225 I don't compare $\mathbb F_q$ and $\mathbb F_p$. The eigenvalue in $\mathbb F_q$ is just a number from $0$ to $q-1$ for me. Basically, this means that if I have a subgroup $D_1$ defined over some field, I can distinguish $\rho$ and $\rho^{-1}$ over that field (using their eigenvalues on $D_1$), so the third roots of unity must lie in that field. Commented Aug 24, 2017 at 12:28