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 $E$.
I am reading the proof given by schoof, (1995)Counting points on eliptic curve over finite field page no:236-237. 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 $\mathbb{F}_{p^r}$.
Now, when it shows $E^1$ can be chosen to be $\mathbb{F}_{p^r}$ isogenenous to $E$. Let $K=\mathbb{Q}(i)$ or $\mathbb{Q}{(\zeta)}$ where $\zeta $ is primitive cube root of unity.
The endomorphism rings of $E$ and $E^1$ are orders of conductor $f$ and $f^1$ respectively in $K$. Consider the proof for, The quotient $f/f^1$ is equal to $l, 1$ or $1/l$:
Let $R_C=\{f\in End(E): f(C)\subset C\} \subset End(E)$
How the index of this ring in $End(E)$ divides $l$?
Please help me to understand this.