Loic Grenie on Wed, 31 Jan 2007 16:52:20 +0100 |
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Re: Construct a "complex" finite field? |
> Hi everybody. > > > Is it possible to define the following using pari? > > ((Z/2Z)[a] / a2 + a + 1) [x] / g(x) I don't think there is a real notion of finite field in pari. You can compute in finite fields but not define them. It depends on what you want to do. You can compute in the double extension: gp> g = Mod(1,2)*x^3 + Mod(Mod(1,2)*a,Mod(1,2)*(a^2+a+1)) %1 = Mod(1,2)*x^3 + Mod(Mod(1, 2)*a, Mod(1, 2)*a^2 + Mod(1, 2)*a + Mod(1, 2)) gp> Mod(Mod(1,2)*x, g)^3 %2 = Mod(Mod(Mod(1, 2)*a, Mod(1, 2)*a^2 + Mod(1, 2)*a + Mod(1, 2)), Mod(1, 2)*x^3 + Mod(Mod(1, 2)*a, Mod(1, 2)*a^2 + Mod(1, 2)*a + Mod(1, 2))) (here g(x)=x^3+[a], and I've computed the cube of [x] in the field you wanted). Notice that this is not very readble and error prone. gp > Mod(Mod(1,2)*(x+1), g)^3 %3 = Mod(Mod(1, 2)*x^2 + Mod(1, 2)*x + Mod(Mod(1, 2)*a + Mod(1, 2), Mod(1, 2)*a^2 + Mod(1, 2)*a + Mod(1, 2)), Mod(1, 2)*x^3 + Mod(Mod(1, 2)*a, Mod(1, 2)*a^2 + Mod(1, 2)*a + Mod(1, 2))) > where g(x) is an irreducible polynom of degree n. With coefficients in F_4 ? > I have a finite field Z/2Z > Now i want to extend it by some irreducible polynome of degree m (in this > example, m was 2) > Now i obtain a finite field again. > Finally, i want to extend this field again by some g(x). > > Is this possible? Ive tried Singular and some others, but i was only able > to do one single extension of (Z/pZ). > > By the way: is it possible to obtain an irreducible polynome of degree n > using pari? gp> P2_6 = ffinit(2, 6) %4 = Mod(1, 2)*x^6 + Mod(1, 2)*x^5 + Mod(1, 2)*x^3 + Mod(1, 2)*x^2 + Mod(1, 2) It is probably much easier to use this polynomial instead of the couple of polynomials above. The class of a modulo a^2+a+1 above is the class of x^21 modulo P2_6: gp> a = Mod(Mod(1, 2)*x, P2_6)^21 %5 = Mod(Mod(1, 2)*x^4 + Mod(1, 2)*x^2 + Mod(1, 2)*x + Mod(1, 2), Mod(1, 2)*x^6 + Mod(1, 2)*x^5 + Mod(1, 2)*x^3 + Mod(1, 2)*x^2 + Mod(1, 2)) gp> a^2+a+1 %6 = 0 Hope this helps, Loïc