Timeline for Finding the n-th root of unity in a finite field
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2018 at 9:57 | comment | added | Ruggero | @poncho Ops. I confess my mind is often limited to prime fields. | |
| Nov 2, 2018 at 21:11 | vote | accept | fraiser | ||
| Oct 31, 2018 at 21:00 | history | tweeted | twitter.com/StackCrypto/status/1057739137749581828 | ||
| Oct 31, 2018 at 18:41 | answer | added | Thomas Pornin | timeline score: 18 | |
| Oct 31, 2018 at 18:29 | comment | added | poncho♦ | If the multiplicative group you're working in has prime order and n is a power of 2, then the answer is easy; either you're working in $GF(3)$ (and so the answer is 1 and 2), or you're working in a larger group, in which case the only n-th root of 1 is 1, that is $x^n = 1$ only if $x=1$ | |
| Oct 31, 2018 at 16:35 | comment | added | fraiser | @Ruggero The group I'm working with is certainly prime order. | |
| Oct 31, 2018 at 16:32 | comment | added | poncho♦ | @Ruggero: actually, there do exist finite fields with prime-sized multiplicative groups; for example, $GF(2^{127})$. | |
| Oct 31, 2018 at 16:01 | comment | added | Ruggero | But you are working in the multiplicative group, which has not prime order. See also this answer | |
| Oct 31, 2018 at 15:44 | history | asked | fraiser | CC BY-SA 4.0 |