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I'm trying to understand how to compute the generator of a finite field.

$\alpha \in GF(q)$

so for example if I was working with $GF(2^8)$

I think I would need to find

$\alpha ^ {256-1} \equiv 1 (\mod 256)$

How do I calculate that?

sav
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    How do you realize the field $\operatorname{GF}(2^8)$? – MooS Nov 03 '15 at 06:46
  • @Moos I'm not sure what you mean by realize. I'm reading about Reed-Solomon error detection and GF(2^8) seems to be used in some of the documents I've found which describe the algorithm. – sav Nov 03 '15 at 07:17
  • http://www.cs.duke.edu/courses/spring10/cps296.3/finitefields.pdf – sav Nov 03 '15 at 07:18
  • So your $GF(2^8)$ is $\Bbb F_2[X]/(p(X))$ for a suitable polynomial $p$ ... ?[ – Hagen von Eitzen Nov 03 '15 at 07:50
  • @Hagen von Eitzen maybe I am missunderstanding galois fields. But I just thought that a galois field was a field with a finite number of elements. In my case, the integers from 0 to 255, with + and × in (mod 256). I'm not sure how polynomials come into this. – sav Nov 03 '15 at 13:10
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    $GF(2^8)$ is not " the integers from 0 to 255, with + and × in (mod 256)" which is $\mathbb Z_{256}$ or $\mathbb Z/256\mathbb Z$. I doubt that a CS course at Duke University claims what you say it does. – Dilip Sarwate Nov 03 '15 at 17:30
  • Yea I've most likely misunderstood something. What is $GF(2^8)$ then ? – sav Nov 04 '15 at 01:41
  • The others got it right - the arithmetic of the field $GF(256)$ is not that of integers modulo $256$. That model only works for $GF(p)$, where $p$ is a prime number. At the risk of blowing my own trumpet... in this answer I seek to describe what these fields and their arithmetic look like. – Jyrki Lahtonen Nov 04 '15 at 08:00

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