The definition of quadratic residue codes involves finding a primitive p-th root of unity in some finite extension field of $GF(2)$. 2 is a quadratic residue of prime $p$. By brute force search I found solutions for $p=7, \zeta=Z(2^3)$ and $p=23,\zeta=Z(2^{11})^{89}$; $Z(2^m)$ is a primitive element in $GF(2^m)$. My question is how is this solved in general? In particular, what is the solution for $p=47$?
Asked
Active
Viewed 145 times
2
-
1In every group, if an element $a$ has order $n$, then $a^k$ has order $n/\gcd(n,k)$. If $\zeta$ is a primitive element of the field $GF(2^{23})$ it has order $2^{23}-1$. As $$\frac{2^{23}-1}{47}=178481,$$ we can conclude that $\zeta^{178481}$ is a root of unity of order $47$. – Jyrki Lahtonen Oct 21 '23 at 06:56
-
See for example this old thread and others linked to it. – Jyrki Lahtonen Oct 21 '23 at 06:56
-
1Compare: The exponent $89$ is similarly gotten as $(2^{11}-1)/23$. – Jyrki Lahtonen Oct 21 '23 at 06:57
-
thanks...this is what I was looking for. – unknown Oct 21 '23 at 15:22
1 Answers
2
This question was answered to some satisfaction in the comment by Jyrki Lahtonen, as written by the poster in another comment.
From the comments:
"In every group, if an element $a$ has order $n$, then $a^k$ has order $n / \text{gcd}(n,k)$. If $\zeta$ is a primitive element of the field $GF(2^{23})$ it has order $2^{23}-1$. As $$\frac{2^{23}-1}{47} = 178481,$$ we conclude that $\zeta^{178481}$ is a root of unity of oder 47."
"See for example this old thread and others linked to it."
"Compare: The exponent $89$ is similarly gotten as $(2^{11}-1)/23$."
Aidan W. Murphy
- 1,317