Timeline for Is $\mathbb{Q}$ the orbit of a rational function under iteration?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 7, 2022 at 9:34 | vote | accept | Ivan Meir | ||
| Jan 5, 2022 at 3:52 | comment | added | alpoge | Now time to learn formatting: this is the promised link (if it works)! (Oy, I meant to also say $\deg{g} > 0$ too above but there's no room left to edit it in! Otherwise you should look at $1/p$ instead as in the link.) | |
| Jan 5, 2022 at 3:50 | comment | added | alpoge | Just like (link after), you don't need Hilbert for nonsurj.! Wlog $f,g\in\mathbb{Z}[x]$ are coprime in $\mathbb{Q}[x]$, let $0\neq R:=\mathrm{Res}_x(f,g)\in\mathbb{Z}$, $k:=\deg{f}-\deg{g}$, which wlog (via $f/g\mapsto g/f$ or $f/g + c$) is $>0$, and let $c_0:=$ leading coeff. of $f$ (it could be made $1$ too). Let $p$ be large. If $f(a/b)/g(a/b)=p$ with $a,b\in\mathbb{Z}$ coprime, then $F(a,b)/(b^k G(a,b)) = p$ with $F,G$ the homogenizations of $f,g$. Hence $b\vert F(a,b)$, so $b\vert c_0$, and also $G(a,b)\vert F(a,b)$ so $G(a,b)\vert R b^{\deg{f}}$ which now bounds $a$ and thus $p$, contra. | |
| Jan 4, 2022 at 20:16 | history | edited | LSpice | CC BY-SA 4.0 |
Links; attribution of comment to pregunton
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| Jan 4, 2022 at 19:20 | comment | added | Wojowu | @SaúlRodríguezMartín I indeed am using the fact that $f,g$ have rational coefficients. This follows from the fact that it takes infinitely many rational values on rational numbers, see e.g. here | |
| Jan 4, 2022 at 19:15 | comment | added | Saúl RM | I think you are supposing that $f,g$ have rational coefficients. I don´t know if that is what Ivan Meir means with the question but he doesn´t make it explicit | |
| Jan 4, 2022 at 16:51 | comment | added | Joe Silverman | Nice application of Hilbert irreducibility for degree $\ge2$. But as I mentioned in the comments, there's a more elementary proof using growth of height functions that shows that the points in the forward orbit are extremely sparse within the full set of rationals. Indeed, among the $O(B^2)$ rational numbers $p/q$ with $\max\{|p|,|q|\}\le B$, the number of them in a forward orbit is $O(\log\log B)$ as $B\to\infty$. | |
| Jan 4, 2022 at 11:33 | history | answered | Wojowu | CC BY-SA 4.0 |