Timeline for Primes mod 4 and integer polynomials
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2020 at 2:49 | vote | accept | CommunityBot | moved from User.Id=6976 by developer User.Id=69903 | |
| Feb 17, 2020 at 2:36 | comment | added | user6976 | @GHfromMO: Thank you! I understand now. | |
| Feb 17, 2020 at 2:27 | comment | added | GH from MO | @MarkSapir: You are right, I overlooked this. At any rate, my comments about splitting prime sets yield that, for any $f(x)\in\mathbb{Z}[x]$ and any positive integer $d$, there exists a prime $p\equiv 1\pmod{d}$ such that $f(x)$ mod $p$ splits into linear factors. This yields a negative answer to your second question. | |
| Feb 17, 2020 at 1:51 | comment | added | user6976 | @GHfromMO: in your definition of "splitting set of primes" you want splitting into linear factors? If so, it is not the same as "reducible sets of primes" in the OP. I think the new version of SashaP's answer deals precisely with that issue. | |
| Feb 17, 2020 at 1:34 | comment | added | GH from MO | @MarkSapir: Let us call a set of primes $S$ splitting if there exists a polynomial $f(x)\in\mathbb{Z}[x]$ such that $p\in S$ if and only if $f(x)$ splits into linear factors modulo $p$. It is known that $S$ is infinite, and its relative density among primes is $1/|G|$, where $G$ is the Galois group of the splitting field of $f(x)$. The intersection of two splitting sets is a splitting set. It is also known that the primes $p\equiv 1\pmod{d}$ plus certain prime divisors $p\mid d$ form a splitting set. Hence a set of primes $p\equiv a\pmod{d}$ can only be splitting when $a\equiv 1\pmod{d}$. | |
| Feb 17, 2020 at 1:00 | comment | added | user6976 | @GHfromMO: I do not understand neither the part before "so" nor the part after "so". Is this a justification for $4k+1$ or an answer for $4k+3$? | |
| Feb 16, 2020 at 19:16 | comment | added | user6976 | @Seva: $5, 13$ are random prime numbers =1 mod 4 . I do not know the answer for any finite set of exceptions, including the empty set and sets consisting of one prime. As for $4k+1$, it is also random arithmetic progression (loosely associated with the polynomial $x^2+1$). | |
| Feb 16, 2020 at 17:23 | comment | added | Seva | Exactly what property of $5$ and $13$ makes them exceptional? Interestingly, I also have an open problem which deals with primes $p\equiv1\pmod 4$, $p\notin\{5,13\}$, and I wonder whether there is any relation. | |
| Feb 16, 2020 at 13:14 | history | became hot network question | |||
| Feb 16, 2020 at 6:24 | comment | added | user6976 | @KConrad: The motivation is the question linked to in the OP. That question had a trivial and implici (my) answer. So a concrete example would be good. $4k+1$ can be replaced by any arithmetic progression containing infinitely many primes, of course (these sets have positive density and I do not know the answer for any of these sets of primes ). But this snswer (if correct) shows that these sets are indeed, the required explicit examples. | |
| Feb 16, 2020 at 5:52 | comment | added | KConrad | Please give some context for these questions. In particular, what kind of mathematical problem leads you to be interested in all primes of the form $4k+1$ except 5 and 13? | |
| Feb 16, 2020 at 4:48 | answer | added | SashaP | timeline score: 5 | |
| Feb 16, 2020 at 4:14 | comment | added | user6976 | $x^2+1$ was suggested as a comment in mathoverflow.net/questions/352105/… . After it was noticed there that it is reducible $\mod 2$, I suggested $4x^2+1$ but it is not monic, unfortunately. | |
| Feb 16, 2020 at 4:03 | comment | added | YCor | $x^2+1$ is reducible mod $p$ iff $p=2$ or $4$ divides the order of $(Z/p)^*$, iff $p=2$ or has the form $4k+1$. This is fairly close to (1), but I understand from (3) that you really mean the given subset and not the subset modulo finitely many exceptions. | |
| Feb 16, 2020 at 3:45 | history | asked | user6976 | CC BY-SA 4.0 |