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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)

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    $\begingroup$ Do you mean $m$-division points or $p^m$-division points? You say the former, but your definition of $P_m$ seems to imply you mean the latter. The multiplication-by-$p$ map factors as $p=F\circ G$ with $F$ the Frobenius map and $G$ separable (since you specify that $E$is ordinary). So doesn't your question just come down to the kernel of $F^m$? $\endgroup$ – Joe Silverman Apr 3 '16 at 12:23
  • $\begingroup$ @JoeSilverman, Sorry, I edit the question, I was thinking about $p^m$-division points of $P$ (I don't know if this is the standard name for $P_m$, but they are well defined up to $p^m$-torsion, right?) $\endgroup$ – Vinicius M. Apr 3 '16 at 12:59
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    $\begingroup$ I think your conjecture needs refinement. Let $k$ be a finite field, and $K=k(t)$. Let $E$ be defined over $k$ and thus as well over $K$. Let $P$ be a $k$-point not in $[p]E(k)$ and thus not in $[p]E(K)$. But the $p$-divisions of $P$ are defined over a finite extension of $k$, so that $[K(P'):K]_i=1$. The only refinement of your conjecture that I can think of will still be false (I think), but I’ll leave that refinement to you. $\endgroup$ – Lubin Apr 4 '16 at 17:57
  • $\begingroup$ @Lubin, Indeed, I didn't consider the case when $P$ has coordinates in the prime field. Maybe I should consider only points $P$ such that its coord. are not contained in the image of the Frobenius of $K$. Assuming that the $p$-torsion is rational, the projection $f:E \to E/E[p]$ is separable with kernel $E[p]$ and its dual is the Frobenius, so, if the coord. of $P$ are not $p$-th powers in $K$ and $[p]P_1 = P$, the coord. of $f(P_1)$ are $p$-th roots of elements not in $K^p$, so they should be defined over an inseparable extensions of $P$. Do you think it's still false for $m>1$. $\endgroup$ – Vinicius M. Apr 5 '16 at 13:59
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    $\begingroup$ This looks as if it might be right. Your restatement of the conjecture is better than what I had in mind. Maybe if you delete this question and instead put forward the reformulated conjecture? Then the real experts can see it and maybe offer help. $\endgroup$ – Lubin Apr 5 '16 at 17:28

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