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. Cheers, Bill.