brilliant zeno from efnet #math
09:48 <+andrewl> in GF(2^3), F(x)=x^4 is a linear map, so component-wise, I should be able to
write y1=F1(x1,x2,x3), y2=F2(x1,x2,x3) and y3=F3(x1,x2,x3) where each Fi is
first-degree... how can the Fi be calculated?
09:50 <+andrewl> looking at F(x) for each of the 8 field elements, I can infer that F1=x1,
F2=x2+x3, and F3=x2, but I have no idea what to do if the field were much larger
09:51 <+zeno> Once you pick any particular irreducible polynomial p(t) of degree three over K =
GF(2), you can use that to do the calculations in K[t]/(p(t))
09:55 <+zeno> your x1,x2,x3 are really the 0th, 1st, and 2nd coefficients of each polynomial
coset in K[x]/(p(x))
09:55 <+zeno> call those a,b,c instead
09:56 <+zeno> so GF(2^3) is really { a + bx + cx^2 : a,b,c in GF(2) } where things in there add
like ordinarly polynomials, and they multiply the ordinary by then followed by
reduction mod p(x)
09:56 <+zeno> I mean "the ordinary way but then"
09:59 <+zeno> so once you have the specific p(x) chosen, you can find out what x^4 and (x^2)^4
come out to, and then that will let you calculate the general (a + bx + cx^2)^4 = a
+ bx^4 + cx^8
09:59 <+zeno> you just need to know the remainder after dividing that last thing by p(x)
10:00 <+zeno> (I also just there used the fact that a^n = a for any a in GF(2), any natural n)