**Question:** Let $k$ be an algebraically closed field, and $A,B$ abelian varieties over $k$. Suppose there exists an isogeny $A\to B$. Does this imply existence of a *separable* isogeny $A\to B$?




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**Known Cases:** The answer is yes when $k$ has characteristic zero (because every isogeny is separable). More interestingly, the answer is also yes for elliptic curves over $\overline{\mathbb{F}\_p}$: Since every isogeny factors into a Frobenius map and a separable map, it suffices to prove the result for $B=A^{(p)}$.


* If $A$ is ordinary, then the Verschiebung map (dual to Frobenius) is separable, and this is true for all isogenous curves as well. So if the field of definition of $A$ is $\mathbb{F}\_{p^k}$, then the composition of Verschiebung maps
$$A\simeq A^{(p^k)}\to A^{(p^{k-1})}\to\cdots\to A^{(p^2)}\to A^{(p)} $$
is separable. (This argument applies more generally to any ordinary abelian variety over $\overline{\mathbb{F}\_p}$.)
* If $A$ is supersingular, we can take any prime $\ell\neq p$ and use the fact that the $\ell$-isogeny graph of supersingular curves is connected.




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**Further thoughts:**
I don't think the second bullet point above is the right place to look for a generalization; connectedness of $\ell$-isogeny graphs of abelian varieties is a very difficult problem in general, and it's also much stronger than we need. I also don't know how to approach the problem when $k\neq\overline{\mathbb{F}\_p}$ has characteristic $p$.



Is there a separable isogeny between any two isogenous abelian varieties?
The answer is no. I think Asvin's example in the comments is correct but there may be a few things to check. I will give a different example that is easier to check, using Moret-Bailly's famous example of a nonisotrivial supersingular abelian surface.


Let $E$ be a supersingular elliptic curve over $F$, the algebraic closure of a finite field of characteristic $p$ and $A = E \times E$. The kernel of Frobenius on $A$ is isomorphic to $\alpha\_p \times \alpha\_p$ and we take $B = A/G$ where $G$ is the subgroup scheme of kernel of Frobenius given by $\{y=tx\} \subset \alpha\_p \times \alpha\_p$ where $t$ is transcendental over $F$. Then $k$ will be the algebraic closure of $F(t)$, $B$ is defined over $k$ but does not descend to $F$.


Now, any separable isogeny with source $A$ is $A \to A/H$, where $H$ is a finite etale subgroup of $A$, but such a subgroup is defined over $F$ so $A/H$ is defined over $F$ and can't be isomorphic to $B$.


