Bill Allombert on Mon, 09 Jan 2012 23:02:12 +0100

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Re: composite field GF(q^mn) as vector space over GF(q^m)

On Sat, Jan 07, 2012 at 11:46:21AM +0400, ÐÐÐÐÐÐÐ ÐÐÑÑ wrote:
> Hello, all!
> I have a problem:
> 1. I have a composite field GF(q^{m*n}) and let A be its element;
> 2. I want to get the n-dimensional vector over GF(2^m) corresponding to A.
> How can I get a minimal/irreducible polynomial f(x) from GF(q^m)[x] to
> represent GF(q^{m*n})
> as GF(q^m)[x]/<f(x)>?

There are various ways:
- If A is a primitive root of GF(q^{m*n}), then obviously B=A^((q^(n*m)-1)/(q^m-1)) will be a
primitive root of GF(q^m). You can then compute minpoly(B) to get f.

- If not, you can still use this formula. If minpoly(B) does not have the right
degree, try a different element A (e.g. random(A)).

- Alternatively you can try B=sum(k=0,n-1,A^(q^(m*k)))

- or you can find a root of ffinit(q,m) in GF(q^{m*n}) using factorff.