Let $\phi: E_1\to E_2$ be an isogeny of elliptic curves over a field $K$ of characteristic $p>0$. Suppose that $\phi$ is separable and let $\hat{\phi}: E_2\to E_1$ denote the dual isogeny. Then $\hat{\phi}$ satisfies several nice properties. Two of these that I am interested in are:
- $\deg\hat{\phi} = \deg\phi$;
- $\hat\phi\circ\phi = [\deg\phi]$ on $E_1$ and $\phi\circ\hat\phi = [\deg\phi]$ on $E_2$.
Note that if $\deg\phi$ is divisible by $p$ then both compositions $\hat\phi\circ\phi$ and $\phi\circ\hat\phi$ are inseparable.
But is it possible for $\phi$ to be separable while $\hat\phi$ is inseparable? Or, by the above remark, can $\phi$ be separable whilst having degree divisible by the characteristic $p$? These properties, plus others in Silverman's book, don't tell me how the inseparability degrees of an isogeny and its dual are related. My intuition (and hope!) is that these two things are impossible, but I am wary of strange counterexamples in characteristic $p$.
Just so this has an answer, let me expand upon what I said in the comments slightly.
So, let us begin with the following trivial observation:
Indeed,this follows immediately from the follow facts: in a tower $E/K/F$ the extension $E/F$ is separable if and only if $E/K$ and $E/F$ is separable, $[n]$ is separable if and only if $p\nmid n$, and $f\circ\widehat{f}=[n]$.
Thus, your question comes down to whether or not there even exist separable isogenies $f:E\to E'$ with $p\mid \deg(f)$. The answer is yes if and only if $E$ is ordinary.
To see this most clearly, note that every isogeny $f:E\to E'$ is of the form $E\to E/K$ (quotient map) for $K\subseteq E$ a finite subgroup (scheme) and that $\ker f=K$ so that $\deg(f)=|K|$. Moreover, $f$ is separable if and only if it's étale (separable=generically étale=étale for group schemes since one can translate the generic étaleness everywhere) if and only if $K$ is étale (over $\text{Spec}(k)$).
Now, we may as well assume that $\deg(f)=p^r$ for some $r$ (since this is the only case of real interest to us) in which case we see that $K\subseteq E[p^r]$. Thus, your whole question comes down to whether or not $E[p^r]$ has an étale subgroup scheme. Moreover, it's easy to deduce from the fact that $E[p^r]$ is an iterative extension of $E[p]$ that this is true for some $r$ if and only if it's true for $r=1$. Thus, the existence of an isogeny $f$ which is separable but for which $\widehat{f}$ is inseparable is equivalent to the question of whether or not $E[p]$ has an étale subgroup scheme.
But, this is precisely equivalent to whether or not $E$ is ordinary or supersingular. Namely, depending on your definitions, $E$ is defined to be ordinary if $E[p]$ has a non-trivial étale subgroup (and supersingular otherwise). If you, instead, use the definition that $E$ is ordinary if and only if $|E[p](\overline{k})|>1$ note merely that a finite scheme $X/k$ is étale if and only if $\#(X_{\overline{k}})=\dim(\mathcal{O}_X(X))$. Using this it's easy to see that this definition of ordinary agrees with the one I said above.
To summarize, the only interesting isogenies in this question are of the form $f:E\to E'$ with $\deg(f)=p^r$ for some $r\geqslant 1$, and are of the form $E/\to E/K$ (quotient map). Moreover, this $f$ is separable with inseparable dual if and only if $K$ is an étale subgroup scheme of $E[p^r]$, which exist if and only if $E$ is ordinary.