Why is $\deg_s(\hat\phi_r)=\deg_s[p^r]$?

I'm having trouble with a part of the proof of V$.3.1$ in Silverman's Arithmetic of Elliptic Curves. In this part of the proof he's trying to prove the following equivalence:

Let $K$ be a perfect field of characteristic $p$, and $E/K$ an elliptic curve. For $r\ge1$, let $\phi_r:E\to E^{(p^r)}$ be the $p^r$-power Frobenius map and $\hat\phi_r$ its dual. The following are equivalent:

(i) $E[p^r]=0$

(ii) $\hat\phi_r$ is purely inseparable

Here is how the proof goes:

Since the Frobenius map is purely inseparable, we have $$\deg_s(\hat\phi_r)=\deg_s[p^r]=(\deg_s[p])^r=\deg_s(\hat\phi_1)^r.$$

From this it follows that $$\# E[p^r]=\deg_s(\hat\phi_r)=\deg_s(\hat\phi)^r,$$ from which the equivalence follows immediately.

Here, $\deg_s(\psi)$ denotes the separable degree of $\psi$, $E[m]$ denotes the $m$-torsion subgroup of $E$ and $[m]$ denotes the $m$-times multplication map, $P\mapsto m P$.

The author seems to be implicitly using the fact that $\deg_s(\hat\phi_r)=\deg_s[p^r]$ for all $r$, which is where my confusion lies. Why is this the case? Does this follow from a more general fact?

In particular, the author says, "Since the Frobenius map is purely inseparable, we have...", but where is the inseparability of $\phi_r$ being used?

• By the way, this is Theorem V.3.1, isn't it? Or maybe I have an older edition... – André 3000 Jul 2 '17 at 18:00
• @Quasicoherent good catch, that was a typo – Alex Mathers Jul 2 '17 at 18:01

Let's just consider the $p^\text{th}$-power Frobenius map $\phi$. Since $\deg(\phi) = p$ and $\phi$ is purely inseparable, then $\deg_i(\phi) = p$, $\deg_s(\phi) = 1$. By Theorem III.6.1 we have $\widehat{\phi} \circ \phi = [p]$, so since separable degree is multiplicative, we find $$\require{cancel} \deg_s([p]) = \deg_s(\widehat{\phi} \circ \phi) = \deg_s(\widehat{\phi}) \cancelto{1}{\deg_s(\phi)} = \deg_s(\widehat{\phi}) \, .$$