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 $p$.

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    $\begingroup$ Dear Alex: Let $E$ be an abelian variety over $\overline{\mathbb{F}_p}$ which is ordinary. Then, there exists a copy of $\mathbb{Z}/p\mathbb{Z}\subseteq E$. The isogeny $f:E\to E/(\mathbb{Z}/p\mathbb{Z})$ is separable, in fact it's étale, since its kernel is $\mathbb{Z}/p\mathbb{Z}$. That said, if one looks at the dual isogeny the kernel is the Cartier dual of the kernel of the original map. In particular, $\ker\widehat{f}=\mu_p$ which is non-étale. $\endgroup$ Apr 27, 2016 at 22:05
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    $\begingroup$ So, $\widehat{f}$ is not étale and so not separable (since separable morphisms are generically étale and this is equivalent to étale for group schemes since one can translate the generic étaleness everywhere). $\endgroup$ Apr 27, 2016 at 22:06
  • $\begingroup$ Thanks for this nice example @AlexYoucis! $\endgroup$
    – Alex Saad
    Apr 27, 2016 at 22:13
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    $\begingroup$ It should be noted that this is, in a rigorizable sense, the only example over $\overline{\mathbb{F}_p}$. Namely, as you pointed out, we really should only be caring about isogenies of prime power order. Then, the kernel is going to be of the form $\mu_{p^m}\times(\mathbb{Z}/p^n\mathbb{Z})$ in the ordinary case and a connected-connected group scheme in the supersingular case. The latter can't lead to anything interesting (since it's connected connected) and so we focus on the former, which produces examples as above. $\endgroup$ Apr 27, 2016 at 22:26
  • $\begingroup$ @AlexYoucis I'm still out of my depth with the group scheme stuff but it's reassuring to know this is basically the only counterexample of this behaviour (at least over $\bar{\mathbb{F}_p}$)! Thank you for clarifying this! $\endgroup$
    – Alex Saad
    Apr 27, 2016 at 23:16

1 Answer 1


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:

Observation: If $f:E\to E'$ is an isogeny of elliptic curves over $k$, with $\mathrm{char}(k)=p>0$, then $f$ and $\widehat{f}$ are separable if and only if $p\nmid\deg(f)$.

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.

  • $\begingroup$ Thanks for writing this up and simplifying the discussion a bit from the comments above. As usual it's an excellent answer and is so helpful to me for learning this subject and understanding some of the intricate things that can happen! $\endgroup$
    – Alex Saad
    Apr 29, 2016 at 16:34
  • $\begingroup$ Excellent answer $\endgroup$
    – Hanno
    Jan 2, 2018 at 15:30

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