TITLE: $p$-torsion of an abelian variety of $p$-rank $0$
QUESTION [7 upvotes]: 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...

REPLY [2 votes]: An explicit example is the Jacobian of the hyperelliptic genus-$3$ curve in characteristic two, $Y^2+Y=X^7$. The vertices of the polygon are $(0,0)$, $(3,1)$, and $(6,3)$, giving slopes of $1/3$ and $2/3$.