Let $E$ be an ordinary elliptic curve defined over a non-perfect field $K$ of characteristic $p$. If $P \in E(K)$ satisfies $P \not\in [p]E(K)$, is it true that its $p^m$-division points of $P$ are defined in inseparable $p^m$-degree extensions? Say, if $[p^m]P_m = P$, then $[K(P_m):K]_i \geq p^m$? Thanks in advance.

**EDIT**: As Lubin pointed out, I didn't consider the case when the coordinates of $P$ are defined in the prime field, thus the inverse images of $P$ by $[p]$ lie in separable extensions of $K$. My attempt now is to consider only the case when the coordinates of $P$ are not in contained in the image of the Frobenius of $K$. If I assume that the $p$-torsion is rational, the projection $f: E \to E/E[p]$ is separable, defined over $K$ and has kernel $E[p] = \ker[p]$, so its dual is the Frobenius morphism. In this situation, if the coordinates of $P$ are not in $K^p$ and $[p]P_1 = P$, then the coordinates of $f(P_1)$ are $p$-th roots of elements not in $K^p$, so they are defined in a inseparable extensions of $P$. However, I cannot see how to extend this argument to $m>1$ and I can't keep assuming that the $p^m$-torsion is rational.

For some context: I guess this can be thought of as an analog of the corollary of Kummer theory for abelian varieties which asserts that $[k({1 \over n} P):k]\geq cn^\alpha$, for some constants $c,\alpha$ not depending on $n$ ($k$ number field)

realexperts can see it and maybe offer help. $\endgroup$