[gckdtyeg] cubic field extension

complex numbers a+bi are a field extension of the reals using the quadratic reduction polynomial i^2+1.

we can similarly define a field a + b*i + c*i^2 with the reduction polynomial i^3+1.

if we use the notation (a,b,c) = a + b*i + c*i^2 , then multiplication is

(a,b,c) * (d,e,f) = (ad-ce-bf , bd+ae-cf , cd+be+af)

reciprocal is

1/(a,b,c) = ( (a^2 + bc) , (-c^2 - ab) , (b^2 - ac) ) / (a^3 - b^3 + c^3 + 3abc)

(scalar division distributes over comma.)

future work jjuenpnj: 3D Mandelbrot set

UPDATE/CORRECTION: the above structure is a commutative ring, not a field.  using i^3+1 as a reduction polynomial does not result in a field because i^3+1 is not irreducible over the reals.  i = -1 is a real root, so the polynomial factors (i + 1) * (i^2 - i + 1).  because of the factorization, (through a process which I don't understand), there will exist nonzero elements which doesn't have reciprocals (multiplicative inverses), violating the field axioms.  for example, (a,b,c) = (1,1,0) will make the denominator of the reciprocal above equal to zero.  (a,b,c) = (1,1,0) is equal to 1 + i, which is one of the factors of the reduction polynomial.  another element with no reciprocal is (a,b,c) = (1,-1,1) which is the other factor.  any polynomial multiples of the factors will also have no reciprocal.