Let $k$ be an algebraically closed field of characteristic $p > 0$ and let $A$ be an abelian variety over $k$ such that $A[p](k) = 0$, i.e., such that $A$ has $p$-rank $0$. If I am not mistaken, this implies that $\mathrm{Ker}(F) \subset A[p]$, where $\mathrm{Ker}(F)$ is the Frobenius kernel of $A$. Is it true that $A[p] = \mathrm{Ker}(F^2)$, i.e., that $A[p]$ is the $p^2$-Frobenius kernel of $A$? Equivalently, is $I^{p^2} = 0$, where $I$ denotes the augmentation ideal of $A[p]$?


I know that the claim is true for elliptic curves but am confused about the higher dimensional case...



$p$-torsion of an abelian variety of $p$-rank $0$
No, this is another entry in the list of ways in which elliptic curves can be a poor guide to the higher-dimensional case. The kernel of $F\_{A/k}:A \rightarrow A^{(p)}$ is *always* contained in $A[p]$ since $\ker F\_{A/k}$ is an infinitesimal commutative group scheme whose own Frobenius morphism vanishes (and all such are killed by $p$). The $p$-rank being 0 expresses that the slopes of the Dieudonne module of the $p$-divisible group are all positive.


In general $F\_{A/k}$ is an isogeny of degree $p^g$ for $g = \dim A$, so $F^2$ is an isogeny of degree $p^{2g}$. Strictly speaking, we really mean the 2-fold Frobenius isogeny $F\_{A/k,2}$, since it doesn't literally make sense to compose $F\_{A/k}$ with itself (as its source and target are not "the same"). Thus, $A[p] = \ker(F\_{A/k,2})$ if and only if either of these is contained in the other, or equivalently $[p]$ factors through $F\_{A/k,2}$ via an isomorphism, or vice-versa, so it amounts to saying that the Frobenius slopes of the Dieudonne module of $A[p^{\infty}]$ are *all* equal to 1/2; i.e., it is "isoclinic" of slope 1/2 (and rank $2g$). 


For elliptic curves, if both slopes are positive then they are forced to be 1/2 and the elliptic curve is forced to be supersingular. But in higher dimensions Honda-Tate theory provides many absolutely simple abelian varieties over finite fields whose slopes are positive but not all equal to 1/2 (by constructing appropriate Weil numbers), and then passing up to $k$ gives counterexamples. See 2.3.5 in the book "Complex multiplication and lifting problems" for one concrete supply of such examples (which have some additional interesting features), but there is a vast array of many more. 


