When is a $j$-invariant supersingular?

Question

Assume $j \in \mathbb{F}_p$ is the $j$-invariant of a curve $E/\mathbb{F}_p$. The modular polynomial $\Phi_l(T,S)$ is such that the roots of $\Phi_l(T,j)$ are the $j$-invariants of curves $l$-isogenous to $E$, but apparently there are cases where this doesn't work, like if the $j$-invariant turns out to be $0$ or $1728$, or if the $j$-invariant comes from a supersingular curve. How do I know if the a $j$-invariant is supersingular?

Answer 1

Too long for a comment, I think I have a way to make sense to OP's question:

Let $E/\Bbb{F}_p$ be non-supersingular such that $E$ is not isogeneous to a curve with $j\in 0,1728$ (ie. $(End(E)\otimes_\Bbb{Z}\Bbb{Q})_{tors}^\times = \pm 1$ and $End(E)$ is an order in an imaginary quadratic field), together with an isogeny $f:E\to E_2$, of degree $l$, with $E_2/\Bbb{F}_p$,

where $f$ is defined over an extension $\Bbb{F}_{p^n}$ with possibly $n> 1$.

From the assumptions the Frobenius and the dual isogeny satisfiy $$\phi_{E_2}^n f=f\phi_{E}^n,\qquad f^* f=[l]$$

Observe that $f^*\phi_{E_2} f\in End(E)$ satisfies $$(f^*\phi_{E_2} f)^n=[l^{n-1}]f^*\phi_{E_2}^n f= [l^{n-1}]f^* f \phi_{E}^n= [l^n]\phi_{E}^n=([l] \phi_{E})^n$$

Whence $$f^*\phi_{E_2} f= u [l] \phi_{E}= u f^* f\phi_E$$ for some $u\in (End(E)\otimes_\Bbb{Z}\Bbb{Q})_{tors}^\times$, which must be $u=\pm 1$.

• If $u=1$ then "the" Frobenius commutes with $f$ so that $f$ is defined over $\Bbb{F}_p$.

• If $u=-1$ then let $h$ be the isomorphism (defined over $\Bbb{F}_{p^2}$) from $E_2$ to its quadratic twist $E_3/\Bbb{F}_p$ (so that $j(E_2)=j(E_3)$).

  We get that $$ \phi_{E_3}hf=-h\phi_{E_2}f=hf\phi_{E}$$ so that "the" Frobenius commutes with $hf$ which is defined over $\Bbb{F}_p$.

Therefore, when $p\nmid l$ the roots $\in \Bbb{F}_p$ of $\Phi_l(T,j(E))\in \Bbb{F}_p[T]$ are exactly the $j(C)$ such that there is an isogeny $E\to C$ of degree $l$ defined over $\Bbb{F}_p$.