Timeline for nontrivial cube root of unity [closed]
Current License: CC BY-SA 2.5
18 events
| when toggle format | what | by | license | comment | |
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| Mar 29, 2011 at 15:08 | history | edited | Niti | CC BY-SA 2.5 |
added 2 characters in body; edited title
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| Mar 26, 2011 at 16:48 | vote | accept | Niti | ||
| Mar 25, 2011 at 20:19 | history | closed |
Franz Lemmermeyer Zev Chonoles Mariano Suárez-Álvarez Daniel Litt Felipe Voloch |
off topic | |
| Mar 25, 2011 at 19:56 | answer | added | Charles Matthews | timeline score: 0 | |
| Mar 25, 2011 at 19:54 | history | edited | Charles Matthews | CC BY-SA 2.5 |
formula
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| Mar 25, 2011 at 19:30 | comment | added | Charles Matthews | Still a problem, though. There is already a square root of -1 in the quadratic extension of the prime field (p is congruent to 3 mod 4, so its square is congruent to 1 mod 4, and therefore this extension has a cyclic multiplicative group of order divisible by 4). So the first suggestion is probably wrong, and instead the quadratic extension is supposed to be generated over the prime field by the square root of -1? That at least makes sense ... | |
| Mar 25, 2011 at 19:29 | comment | added | Zev Chonoles | But even though it is irreducible over $F_p$ when $p\equiv 3\bmod 4$, it will still split in $F_{p^2}$. | |
| Mar 25, 2011 at 19:28 | comment | added | Niti | no actually (x^2+1) is irreducible in F_p for p = 3 mod(4). | |
| Mar 25, 2011 at 19:27 | comment | added | Mariano Suárez-Álvarez | @Niti: do ask in math.stackexchange.com | |
| Mar 25, 2011 at 19:27 | comment | added | Zev Chonoles | Right. However, the polynomial $x^2+1$ already has its roots in $F_{p^2}$, so $F_{p^2}[x]/(x^2+1)$ cannot be a finite field. | |
| Mar 25, 2011 at 19:26 | comment | added | Niti | Actually, I am stuck with this for weeks and I would appreciate a help! Or maybe some reference at least. | |
| Mar 25, 2011 at 19:24 | comment | added | Niti | probably you are right. I am not a mathematician so am not quite familiar with the terminology. So I need sigma where sigma^3=1 in Fp^2[x]/(x^2+1). | |
| Mar 25, 2011 at 19:23 | comment | added | Zev Chonoles | Also, I'm afraid that your question is not at the right level for MO, but there are many other sites where your question would be better suited. See mathoverflow.net/faq#whatnot. | |
| Mar 25, 2011 at 19:22 | comment | added | Zev Chonoles | You can edit your question - the "edit" button is right underneath the tags. Also, you mean "nontrivial cube root of unity", not "third nontrivial root of unity", because every element of a finite field other than 0 and 1 is a nontrivial root of unity. | |
| Mar 25, 2011 at 19:22 | history | edited | Niti | CC BY-SA 2.5 |
added 15 characters in body
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| Mar 25, 2011 at 19:19 | comment | added | Niti | Yes. I meant that! | |
| Mar 25, 2011 at 19:17 | comment | added | Zev Chonoles | Presumably you mean $F_{p^2}[x]/(x^2+1)$? | |
| Mar 25, 2011 at 19:11 | history | asked | Niti | CC BY-SA 2.5 |