Timeline for A sufficient condition for a set of primes to be the set of reducibility of an integer polynomial
Current License: CC BY-SA 4.0
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| Feb 9, 2020 at 18:36 | history | edited | user145520 | CC BY-SA 4.0 |
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| Feb 9, 2020 at 11:17 | comment | added | user6976 | The word "monic" I did not notice. So $4x^2+1$ and $16900x^2+1$ are not good and my question above is still open. | |
| Feb 9, 2020 at 11:14 | comment | added | user6976 | There is no word "Chebotarev" anywhere in the question. | |
| Feb 9, 2020 at 6:30 | comment | added | Pasten | @MarkSapir The OP does in fact require this in the question! (and it is not Chebyshev's theorem, it's Cebotarev's). If the required monic polynomial exists, then S is a Cebotarev set of primes and the density is a rational number. Of course this is only a necessary condition, while the title asks for a sufficient condition. | |
| Feb 8, 2020 at 20:23 | comment | added | Pasten | @MarkSapir well, if a set of primes has natural density $\delta$, then it also has Dirichlet density and it is equal to $\delta$. In any case, when people refer to the Cebotarev density theorem they usually mean the version about natural density. | |
| Feb 8, 2020 at 14:22 | comment | added | user6976 | @Pasten: The Chebotarev density theorem is about Dirichlet density of some sets of primes (and it can be an arbitrary nonnegative number). If the OP meant something else, (s)he should have said it. | |
| Feb 8, 2020 at 6:15 | comment | added | Pasten | The Cebotarev (or Frobenius) density theorem shows that the density is a rational number. The set of primes considered in this question is an example of a Cebotarev set. | |
| Feb 8, 2020 at 2:09 | comment | added | user6976 | In general if $S$ is a proper set of primes, and $f(x)$ is a polynomial reducible mod $p$ iff $p\in S$, and $p_1,..., p_n$ are in $S$, then let $a=p_1\cdot...\cdot p_n$ and let $g(x)=f(ax)$. Then $g$ is reducible modulo prime $p$ iff $p$ is in $S$ and does not divide $a$. | |
| Feb 7, 2020 at 22:45 | comment | added | user6976 | I guess that $16900x^2+1$ answers my question above. | |
| Feb 7, 2020 at 22:24 | comment | added | user6976 | I forgot to include iff as in OP. | |
| Feb 7, 2020 at 21:50 | comment | added | Robert Israel | Yes, but you asked about primes of the form $4k+1$. | |
| Feb 7, 2020 at 21:30 | comment | added | user6976 | Your polynomial is reducible mod 2 and 3. | |
| Feb 7, 2020 at 20:27 | comment | added | Robert Israel | $(65 x + 1)(x + 1)$? | |
| Feb 7, 2020 at 15:00 | review | Close votes | |||
| Feb 12, 2020 at 3:04 | |||||
| Feb 7, 2020 at 6:16 | answer | added | user6976 | timeline score: 1 | |
| Feb 7, 2020 at 2:40 | comment | added | user6976 | Yes, $5, 13$. Using a phone to type leads to misprints. | |
| Feb 7, 2020 at 2:38 | comment | added | user44191 | @MarkSapir so, just except 5? (maybe $5$ and $13$?) | |
| Feb 7, 2020 at 2:34 | comment | added | user6976 | Is there a polynomial which is reducible modulo every prime of the form $4k+1$ except $5$ and $11$? | |
| Feb 7, 2020 at 2:13 | comment | added | user145520 | related: math.stackexchange.com/a/3534817/700841 mathoverflow.net/a/352098/145520 | |
| Feb 7, 2020 at 2:12 | history | asked | user145520 | CC BY-SA 4.0 |