Timeline for Primes mod 4 and integer polynomials
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
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| Feb 17, 2020 at 23:33 | comment | added | user6976 | Vefa Goksel informed me that the group theory problem about a permutation group where 1/2 of the elements are big cycles has been solved in "Goksel, V. Markov processes and some PCF quadratic polynomials. Res. number theory 5, 29 (2019)" (see Theorem A in section 4 of the paper). | |
| Feb 17, 2020 at 20:20 | comment | added | user6976 | @Lucia: Thank you! | |
| Feb 17, 2020 at 19:28 | comment | added | Lucia | @MarkSapir: If the polynomial factors in $F_p[x]$ as $f_1 ... f_r$ with $f_i$ of degree $d_i$, then this corresponds to a conjugacy class in the Galois group with cycles of length $d_1$, $\ldots$, $d_r$. In your problem you had $f$ being irreducible for a bunch of primes; the Frobenius for these primes is therefore an $n$-cycle. | |
| Feb 17, 2020 at 18:29 | comment | added | user6976 | @Lucia: Now that you understood, could you explain it to me? Why does the Galois group contain a big cycle? | |
| Feb 17, 2020 at 16:18 | comment | added | Lucia | Ah ok -- I misread the problem thinking that $f$ is assumed to have no roots mod p for $p$ not in $S$. Instead the problem gives you that $f$ is irreducible for these primes. So what you have is fine of course. | |
| Feb 17, 2020 at 14:01 | comment | added | user6976 | @SashaP: Can you answer Lucia's question? | |
| Feb 17, 2020 at 13:56 | comment | added | user6976 | If the set is of density 1/2 there should be a cycle type occurring 1/2 of the time. It is in the Wikipedia article on Chebotarev theorem. | |
| Feb 17, 2020 at 13:49 | comment | added | Lucia | @MarkSapir: I only see that half the time there must be a fixed point, and half the time not. Why must there be a cycle type occurring 1/2 the time? | |
| Feb 17, 2020 at 13:47 | comment | added | user6976 | @Lucia: By Chebotarev theorem there must be a cycle type that occurs 1/2 of the time. So it is stronger than what you wrote. But, indeed, it is not immediately clear why this type is the cycle of max length. | |
| Feb 17, 2020 at 8:22 | comment | added | Lucia | Why must exactly $1/2$ the elements of $G$ correspond to cycles of length $n$? I would have thought that the condition is that $1/2$ the elements have at least one fixed point, and $1/2$ the elements have none. | |
| Feb 17, 2020 at 2:52 | comment | added | user6976 | Why is the answer "Community Wiki"? | |
| Feb 17, 2020 at 2:49 | vote | accept | CommunityBot | moved from User.Id=6976 by developer User.Id=69903 | |
| Feb 17, 2020 at 0:27 | comment | added | user6976 | @SashaP: looks good now. I need to educate myself about Frobenius elements. Then I will accept the answer. | |
| Feb 17, 2020 at 0:27 | comment | added | SashaP | @YCor Sure, corrected | |
| Feb 17, 2020 at 0:27 | history | edited | SashaP | CC BY-SA 4.0 |
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| Feb 17, 2020 at 0:20 | history | edited | YCor | CC BY-SA 4.0 |
fixed typos (spacing missing before opening parentheses). Emphasized "some $S'$"
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| Feb 16, 2020 at 23:52 | comment | added | SashaP | @MarkSapir The existence of such subgroup forces $n$ to be a power of two. Either way, there is an argument around it for your original question, see the edits. | |
| Feb 16, 2020 at 23:51 | history | edited | SashaP | CC BY-SA 4.0 |
added 1159 characters in body; Post Made Community Wiki
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| Feb 16, 2020 at 21:44 | comment | added | user6976 | @SashaP: I thought a little about your problem. The answer is probably unknown but can be checked for small $n$ using GAP or Magma. Say, for $n\le 7$. | |
| Feb 16, 2020 at 17:30 | comment | added | user6976 | @SashaP: It is probably known. You can ask a separate question on MO. | |
| Feb 16, 2020 at 15:56 | comment | added | SashaP | @MarkSapir To show that the existence of a subgroup in $S_n$ with precisely half of its elements being cycles of length $n$ forces $n=2$, as stated in the second paragraph of the edited answer. | |
| Feb 16, 2020 at 15:43 | comment | added | user6976 | @SashaP: What problem? | |
| Feb 16, 2020 at 14:57 | comment | added | SashaP | Yes, you're right. This comes down to a problem in group theory that I'm not sure how to solve. | |
| Feb 16, 2020 at 14:56 | history | edited | SashaP | CC BY-SA 4.0 |
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| Feb 16, 2020 at 11:58 | comment | added | user6976 | @SashaP: Your answer only proves that quadratic polynomials won't work? | |
| Feb 16, 2020 at 5:16 | comment | added | user44191 | @MarkSapir For primes in $S$, $D$ is a square residue, and $-1$ is a square residue, while for primes not in $S$, $D$ is not a square residue, and $-1$ is not a square residue (with finitely many exceptions). Therefore, for all primes, $-D = -1*D$ is a square residue (with finitely many exceptions). | |
| Feb 16, 2020 at 4:59 | comment | added | user6976 | 1) Why does the Chebotarev density theorem implies that the degree should be 2? 2) In the last paragraph, what are $a,b,c$? 3) Also in the last paragraph, why $-D$ is a square mod all but finitely many primes if $S=\{p| p=4k+1\}$? | |
| Feb 16, 2020 at 4:57 | history | edited | SashaP | CC BY-SA 4.0 |
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| Feb 16, 2020 at 4:48 | history | answered | SashaP | CC BY-SA 4.0 |