Timeline for Field of definition of a point in $[p]^{-1}E(K)$
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| S Apr 21, 2016 at 5:55 | history | bounty ended | CommunityBot | ||
| S Apr 21, 2016 at 5:55 | history | notice removed | CommunityBot | ||
| S Apr 13, 2016 at 4:28 | history | bounty started | Vinicius M. | ||
| S Apr 13, 2016 at 4:28 | history | notice added | Vinicius M. | Draw attention | |
| Apr 6, 2016 at 8:14 | history | edited | Vinicius M. | CC BY-SA 3.0 |
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| Apr 5, 2016 at 17:28 | comment | added | Lubin | 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. | |
| Apr 5, 2016 at 13:59 | comment | added | Vinicius M. | @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 $m>1$. | |
| Apr 4, 2016 at 17:57 | comment | added | Lubin | 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. | |
| Apr 3, 2016 at 12:59 | comment | added | Vinicius M. | @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?) | |
| Apr 3, 2016 at 12:53 | history | edited | Vinicius M. | CC BY-SA 3.0 |
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| Apr 3, 2016 at 12:23 | comment | added | Joe Silverman | 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 $F^m$? | |
| Apr 3, 2016 at 9:40 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Added top-level tag; fixed a typo.
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| Apr 3, 2016 at 9:31 | review | First posts | |||
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| Apr 3, 2016 at 9:25 | history | asked | Vinicius M. | CC BY-SA 3.0 |