Using the definitions that an elliptic curve $E$ over a finite field $K$ of characteristics $p$ is supersingular if dual of $p^r$-Frobenius map $F:E\to E^{(p^r)}$ is purely inseparable for all $r≧1$, how can I show that $E$ is supersingular if and only if dual of $p^r$ -Frobenius map $F:E\to E^{(p^r)}$ is purely inseparable?
dual of $F:E\to E^{(p^r)}$ is purely inseparable for all $r≧1$⇨ dual of $F:E\to E^{(p^r)}$ is inseparable is obvious, but the inverse direction is unclear..
dual of $F:E\to E^{(p^r)}$ is inseparable ⇨dual of $F:E\to E^{(p^r)}$ is purely inseparable does not hold because dual of $F$ has inseparable degree devides $p^r$ does not imply dual of $F$ has inseparable degree $p^r$..
$\phi_{p^d}^* \phi_{p^d} = [p^d]$, if $\phi_{p^d}^*$ is purely inseparable then $[p^d]$ is purely inseparable, ie. $\ker([p^d])=\{O\}$, this implies that $\ker([p])=\{O\}$ so that $\ker([p^r])=\{O\}$ and hence $\ker(\phi_{p^r}^*)=\{O\}$ ie. $\phi_{p^r}^*$ is purely inseparable.