Endomorphism of abelian variety which kills $p$-torsion is divisible by $p$ Let $A/\mathbb{Q}$ be an abelian variety, and suppose $T \in \operatorname{End}_{\mathbb{Q}}(A)$ is such that $T(A[p]) = 0$.  I'd like to show that there is a $T' \in \operatorname{End}_{\mathbb{Q}}(A)$ such that $T = [p] \circ T'$.
It seems to me that such a $T'$ has to be defined in the following way: given $P \in A$, choose $Q \in A$ such that $[p] Q = P$, and set $T'(P) := T(Q)$.  This is well-defined because of the condition $T(A[p]) = 0$.  My question is, why should such a $T'$ be a morphism of varieties, and moreover why is it defined over $\mathbb{Q}$?
 A: I've figured out a more general statement, which is: Let $A, B$ be abelian varieties over $\mathbb{Q}$, and let $S, T \in \operatorname{Hom}_{\mathbb{Q}}(A, B)$ be isogenies from $A$ to $B$ defined over $\mathbb{Q}$.  Suppose furthermore that $T$ annihilates (the $\overline{\mathbb{Q}}$-points of) $\ker(S)$.  Then there is a $T' \in \operatorname{End}_{\mathbb{Q}}(B)$ such that $T = T' \circ S$.
To see this, note that $\ker(S)(\overline{\mathbb{Q}}) \subset A$ is a finite subgroup of $A(\overline{\mathbb{Q}})$ which acts freely on $A(\overline{\mathbb{Q}})$, and $B$ is the quotient variety $A/\ker(S)$; once we make this identification the map $S$ is just the projection $A \to A/\ker(S)$.  In particular, $B$ satisfies the universal property of the quotient, so if $T \in \operatorname{Hom}_{\mathbb{Q}}(A, B)$ is an isogeny such that $T(\ker(S)(\overline{\mathbb{Q}})) = 0$, there is a unique $T' \in \operatorname{End}_{\overline{\mathbb{Q}}}(B)$ such that $T = T' \circ S$.
Now $T$ and $S$ are defined over $\mathbb{Q}$, and moreover $S$ is an isogeny, hence surjective on $\overline{\mathbb{Q}}$-points, so one can check by hand that $\sigma(T'(Q)) = T'(\sigma(Q))$ for any $Q \in B(\overline{\mathbb{Q}})$ and $\sigma \in \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. It follows that $T'$ is defined over $\mathbb{Q}$, as desired.
In characteristic $p$, the same argument will break down because we can't argue on geometric points, but I suspect an analogous statement will be true if we use the group scheme kernel instead.
