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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?

José
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    At least for $p\nmid l$ I don't see where supersingular comes into play. $\Phi_l(T,j(E))=\prod_{H\subset E,|H|=l} (T-j(E/H))$. Taking $E$ such that its reduction is $\widetilde{E}$ it remains to check that $j(E/H)\equiv j(\widetilde{E}/H)\bmod \mathfrak{p}$. – reuns Apr 10 '21 at 18:47
  • Maybe you meant the isogeny has degree $l$ and is defined over $\Bbb{F}_p$? – reuns Apr 10 '21 at 19:01
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    Is your question about supersingular $j$-invariants, or about the modular polynomial? Much of what you want is in Silverman. Presumably, you know that $y^2=x^3+b$ is supersingular for $p\equiv5\pmod6$ And $y^2=x^3+ax$ is supersingular for $p\equiv3\pmod4$. There are jolly formulas of Deuring telling you how many supersingular values of $j$ there are, depending on the congruence of $p$ modulo $12$. The largest prime with only one supersingular $j$ is $13$, where $y^2=x^3+x +4$ gives the one. – Lubin Apr 10 '21 at 20:44
  • @reuns It doesn't come into play if $p \nmid l$? My mistake. I will look into it. Lubin I think I expressed myself badly. What I want to know is if there is a way to know if a $j$-invariant is supersingular without computing the formula of a curve with that $j$-invariant. – José Apr 10 '21 at 21:17

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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 $\Bbb{F}_p$.

reuns
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