Timeline for All rational periodic points
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 4, 2020 at 11:14 | history | edited | Joe Silverman | CC BY-SA 4.0 |
added 80 characters in body
|
| Nov 4, 2020 at 11:01 | history | edited | Joe Silverman | CC BY-SA 4.0 |
Added an alternative way to rule out period 4
|
| Nov 4, 2020 at 9:43 | comment | added | nomadd | Thank you so much for your answers Mr. Silverman, Mr. Elkies and Mr. Sawin. So with this technique we examine finitely many cases, in fact if we are lucky it is 2 or 3. Yes it is PCF as you said. | |
| Nov 3, 2020 at 14:33 | history | edited | Joe Silverman | CC BY-SA 4.0 |
Added further information about the problem
|
| Nov 1, 2020 at 13:17 | comment | added | Joe Silverman | @NoamD.Elkies Good point, that would be another way to do it, although working out the explicit $O(1)$ constant can be a bit annoying. But as with elliptic curves, if there are some small primes of good reduction, that usually lets one pin things down to the extent that it can almost be done by hand. | |
| Nov 1, 2020 at 3:16 | comment | added | Noam D. Elkies | One can also figure out an explicit $c>0$ such that the height $H(f(x))$ is always at least $c H(x)^d$ where $d = \deg f$ (which is $3$ in our case). Then any periodic point must have height at most $\root d-1 \of {1/c}$, reducing the problem to a finite and probably quick search. (That's what I expected to read when I saw Joe Silverman's name on the answer . . .) | |
| Nov 1, 2020 at 3:01 | history | edited | Bombyx mori | CC BY-SA 4.0 |
small typo fixed
|
| Nov 1, 2020 at 2:25 | history | edited | Joe Silverman | CC BY-SA 4.0 |
Completed the proof
|
| Nov 1, 2020 at 2:20 | history | edited | Joe Silverman | CC BY-SA 4.0 |
added 1 character in body
|
| Nov 1, 2020 at 2:15 | history | edited | Joe Silverman | CC BY-SA 4.0 |
added 66 characters in body
|
| Nov 1, 2020 at 2:11 | comment | added | Joe Silverman | @WillSawin Oops, that will teach me to try to answer a MO question while also participating in a Zoom conference call. Thanks. I had also done the mod $5$ calculation, but didn't like the fact that it gave an $8$. But you're right, mod $3$ and mod $5$ get one down to periods $1$, $2$, and $4$. I'll fix my answer. | |
| Nov 1, 2020 at 1:48 | comment | added | Will Sawin | mod $5$ I get the orbits are $\{0,3\}$ and $\{2\}$, so $m_5$ is $1$ or $2$ and $r_5$ divides $4$ and thus the period divides $8$ times a power of $5$, which combined with mod $3$ gets the same $1, 2$, or $4$ conclusion. | |
| Nov 1, 2020 at 1:44 | comment | added | Will Sawin | Doesn't the mod $3$ calculation in fact show the period divides $4$ times a power of $3$, since $p=3$? | |
| Nov 1, 2020 at 1:23 | history | edited | Joe Silverman | CC BY-SA 4.0 |
Gave more details of how to solve the problem
|
| Nov 1, 2020 at 1:14 | history | edited | Joe Silverman | CC BY-SA 4.0 |
added 561 characters in body
|
| Oct 31, 2020 at 23:06 | history | answered | Joe Silverman | CC BY-SA 4.0 |