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)