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Let E be an elliptic curve over $F_p$. Suppose that its j invarient is not supersingular and that $j\neq 0 $ or 1728.

Then the modular polynomial $\Phi_l(j,T)$ has a zero $\tilde{\jmath} \in \mathbb{F}_{p^r}$ if and only if the kernel $C$ of the corresponding isogeny $E \mapsto E/C$ is a one-dimensional eigenspace of $\phi^r_p$ in $E[l]$, with $\phi_p$ the Frobenius endomorphism of $E$.

In the proof: Counting points on eliptic curve over finite field page no:236. Conversly, if $\Phi_l(j,\tilde{\jmath} )=0$, then there exists a cyclic subgroup $C$ of $E[l]$ such that the $j$ invariat of $E/C$ is equal to $\tilde{\jmath}\in \mathbb{F}_{p^r} $. Let $E^1$ be an elliptic curve over $\mathbb{F}_{p^r}$ with $j$ invariant equal to $\tilde{\jmath}$. Let $E/C \mapsto E^1$ be an $\bar{\mathbb{F}}_p$ isomorphism and let $f: E \mapsto E/C \mapsto E^1$ be the composite isogeny. It has kernel $C$.

I can see $f$ is defined over $\mathbb{F}_{p^r}$, so this implies existence of an isogeny $E \mapsto E^1$ over $\mathbb{F}_{p^r}$.

How this will imple $C$ is an eigenspace of $\phi^r_p$. How the statement that,"Frobenius endomorphisms over $\mathbb{F}_{p^r}$ satisfies the same characteristic equation" will help to see that $C$ is an eigenspace of $\phi^r_p$.

Please help me to understand this.

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The key point is to see that, because $f$ is defined over $\mathbb F_{p^r}$, $f \circ \phi_p^r = \phi_p^r \circ f$.

This immediately implies, because $\phi_{p^r}$ is injective, that kenel of $f$ is stable under $\phi_{p^r}$ and hence, because it is one-dimensional, is an eigenspace for $\phi_{p^r}$.

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