# Is there a separable isogeny between any two isogenous abelian varieties?

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 $$p$$.

• Hi Jonathan! If there were a separable isogeny, the p(-power)-torsion group schemes would be isomorphic, right? Is it known that p-power isogenous abelian varieties have isomorphic p-divisible groups? Commented Mar 15, 2023 at 1:32
• Oh hi Asvin! Thanks for the comment. That's a very good question but I don't know the answer - it somehow seems more unlikely when phrased that way, but I don't immediately see a reason it would be false. (I also just have no idea how one would go about showing that two abelian varieities have non-isomorphic p-divisible groups without using isogeny invariants.) Commented Mar 15, 2023 at 2:42
• But do let me know if you have any ideas, partial or otherwise! Commented Mar 15, 2023 at 2:43
• I think the answer should also be yes for abelian varieties over $\overline{\mathbb F}_p$ because we can lift them to CM AVs in char 0 with the same rational ring of endomorphisms along with a lift of the isogeny. The given isogeny will then correspond to an ideal $I$ that reduces to the inseparable isogeny mod $p$ and we can pick a different ideal $J$ in the same ideal class but now co-prime to $p$. Does this work? Commented Mar 15, 2023 at 4:32
• Finally, I think this should be false for any non-isotrivial elliptic curve $E_t/ \overline{\mathbb F}_p(t)$ and the Frobenius isogeny. Suppose there were a $N$-power isogeny between $E_t$ and $E_t^{(p)}$ and let's assume we set things up so the parameter $t$ is the $j$-invariant. Then, we would have an equation of the form $\Phi_N(t,t^p) = 0$ wher $\Phi_N$ is the appropriate modular polynomial. But this is impossible for degree considerations! $\Phi_N(x,y)$ cuts out a $(\varphi(n),\varphi(n))$ cycle in $\mathbb P^1\times \mathbb P^1$ and it's intersection with $x = y^p$ is finite. Commented Mar 15, 2023 at 4:41

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 $$B$$.