Timeline for Multiplication-by-m map as an isogeny
Current License: CC BY-SA 4.0
14 events
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| Oct 15 at 20:45 | history | edited | samething | CC BY-SA 4.0 |
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| Oct 15 at 19:10 | vote | accept | samething | ||
| Oct 15 at 19:01 | history | edited | samething | CC BY-SA 4.0 |
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| Oct 14 at 22:35 | vote | accept | samething | ||
| Oct 15 at 19:10 | |||||
| Oct 14 at 22:09 | answer | added | Joe Silverman | timeline score: 2 | |
| Oct 14 at 21:23 | comment | added | Will Sawin | But when $p$ divides $m$, the isogeny is indeed inseparable, which can be argued by exactly the reasoning you give: The kernel has fewer than $m^2$ points since there are at most $p$ points of order $p$, hence the map must be inseparable. | |
| Oct 14 at 21:23 | comment | added | Will Sawin | Your argument seems to prove the converse of what you claim: $mP = \mathcal O$, so the order of every multiple of $P$ divides $m$, and a point is in the kernel if and only if its order divides $m$. It follows that every multiple of $P$ is in the kernel but it does not follow that every element of the kernel is a multiple of $P$. If you look at slide 8 of the slides you posted you will see that the kernel has order $m^2$ as long as the characteristic $p$ does not divide $m$. So the isogeny is separable in this case. | |
| Oct 14 at 20:49 | review | Close votes | |||
| Oct 20 at 3:08 | |||||
| Oct 14 at 20:36 | comment | added | samething | @ChrisWuthrich thanks a lot for the example, I'll read Silverman's chapter. | |
| Oct 14 at 20:33 | comment | added | Chris Wuthrich | $E[4](\mathbb{F}_5) = \{O\}$ but $E[4]$ as a group scheme has $16$ points over $\mathbb{F}_{5^6}$. This is a basic question about isogenies, please read chapter III.4 in Silverman's "The arithmetic of elliptic curves". | |
| Oct 14 at 20:25 | comment | added | samething | Because $mP = \mathcal{O}$, the subgroup is an additive group and the order of each point in kernel divides $m$. Am I wrong? Can you give an example please? | |
| Oct 14 at 20:19 | comment | added | Will Sawin | Why do you assume a point of order $m$ generates all points in the kernel? | |
| S Oct 14 at 20:12 | review | First questions | |||
| Oct 14 at 20:42 | |||||
| S Oct 14 at 20:12 | history | asked | samething | CC BY-SA 4.0 |