Timeline for On the distribution of roots modulo primes of an integral polynomial
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 22, 2016 at 7:59 | history | edited | user98708 | CC BY-SA 3.0 |
math error
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| Sep 22, 2016 at 7:58 | vote | accept | user98708 | ||
| Sep 20, 2016 at 22:54 | answer | added | Gerry Myerson | timeline score: 9 | |
| Sep 20, 2016 at 13:25 | comment | added | user98708 | @GerryMyerson — Fantastic. That answers my first question. If you post it as an answer, I will mark it as accepted. | |
| Sep 20, 2016 at 12:42 | comment | added | Gerry Myerson | Related problems are discussed in my paper, Polynomial congruences and density, Mathematics Magazine 80 (2007) 299-302, maa.org/sites/default/files/Myerson10-200716952.pdf | |
| Sep 20, 2016 at 10:10 | comment | added | user98708 | @DenisChaperondeLauzières — Thanks for your comment. That probably gives the state of the art for question 1. I am curious if someone has any comments about question 2 (maybe only the quadratic case); or if the link with Sato–Tate is contrived or not… | |
| Sep 20, 2016 at 9:49 | comment | added | Denis Chaperon de Lauzières | It's conjectured (though I'm not sure by whom originally) that if $f$ is irreducible of degree $\geq 2$, then one gets equidistribution . The only non-trivial case known is degree $2$, by Duke-Friedlander-Iwaniec (negative discriminant, "Equidistribution of roots of a quadratic congruence to prime moduli", Annals of Math., 1995) and Toth (positive discriminant, "Roots of quadratic congruences", IMRN, 2000). | |
| Sep 20, 2016 at 9:37 | comment | added | Uri Bader | I tried to imply two things: (1) Seems natural that you have different limits along "arithmetically define" subsequences, such as 1 mod 4, 3 mod 4, and (2) I would start by asking for a solution for this specific case, which might be well known or a not-so-hard interpretation of a well known theorem. | |
| Sep 20, 2016 at 9:28 | comment | added | user98708 | @UriBader — Yes. In your example $f$ has no roots if $p \cong 3 \pmod{4}$. But if you lump together all the roots that it has modulo all the primes, then you get a picture that looks very much like a uniform distribution. All those roots will be of the form $a \pmod{p}$ with $p \cong 1 \pmod{4}$, but that does not matter; I think. (Also see my remark about Chebotarev's density theorem, at the end of my question.) | |
| Sep 20, 2016 at 9:23 | comment | added | Uri Bader | Did you look closely at $f=x^2+1$? | |
| Sep 20, 2016 at 9:20 | history | edited | user98708 | CC BY-SA 3.0 |
added 2 characters in body
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| Sep 20, 2016 at 9:12 | review | First posts | |||
| Sep 20, 2016 at 9:54 | |||||
| Sep 20, 2016 at 9:07 | history | asked | user98708 | CC BY-SA 3.0 |