**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 $A\to B$?

**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.

**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 $p$.