Timeline for Are there unconditional results for boundedness of finitely many rational points on $f(x,y)=n$ for all $n$?
Current License: CC BY-SA 3.0
11 events
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| Nov 25, 2015 at 18:40 | comment | added | Terry Tao | Conditionally on the Bombieri-Lang conjecture, this should be true for all $f$ that are not degenerate in the sense that the curves $f(x,y)=n$ are reducible or have genus at most 1, from the work of Caporaso, Harris, and Mazur: ams.org/mathscinet-getitem?mr=1325796 . But this type of result is unlikely to be made unconditional in the near future. | |
| Nov 25, 2015 at 9:39 | history | edited | joro | CC BY-SA 3.0 |
major rewrite due to comments
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| Nov 23, 2015 at 17:43 | comment | added | joro | @eric Yes, this is the question. | |
| Nov 23, 2015 at 17:11 | comment | added | eric | I see. So here's the question: Does there exist $f$ in $Q[x,y]$ such that there's a uniform bound independent of $n$ for number of solutions to $f(x,y)=n$ in rationals $x$ and $y$? That's a nice question because it's weaker than the question which is apparently probably true but open (same question but with bound 1). | |
| Nov 21, 2015 at 16:06 | comment | added | joro | @eric Injection Q^2 to Q indeed will answer the question with bound one, but AFAICT it is not known. | |
| Nov 21, 2015 at 10:54 | comment | added | joro | @eric Maybe will edit, thanks. $f$ is polynomial with rational coefficients and the rational points must be finite. I am asking are the bounded by $\deg{f}$ for all rational $n$ (again, when they are finite). Maybe this is open, since it is not known for Thue equations (in this case they depend on $n$). I know several conjectures imply even constant bound, but am asking about unconditional results. | |
| Nov 21, 2015 at 9:51 | comment | added | eric | joro -- your question is uncharacteristically imprecise. What order do the quantifiers go in? And where does f live? I'm assuming the answer isn't "Faltings' theorem". Note also Tom Leinsters comment on mathoverflow.net/questions/21003/… , so one answer to one interpretation of your question might be "bound = 1, answer is unknown" | |
| Nov 20, 2015 at 15:19 | history | edited | joro | CC BY-SA 3.0 |
added 14 characters in body; edited title
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| Nov 20, 2015 at 15:18 | comment | added | joro | @JasonStarr Over the rationals. Your example has infinitely many points, but I am asking about bounded finitely many. Maybe will edit, thanks. | |
| Nov 20, 2015 at 15:05 | comment | added | Jason Starr | Could you please make your question more precise? For every integer $d$, for every rational number $n$, the polynomial $x^d + y = n$ has infinitely many rational solutions. Are you asking about rational points on irrational curves? | |
| Nov 20, 2015 at 14:53 | history | asked | joro | CC BY-SA 3.0 |