Suppose we have polynomials over $GF (2^8)$.
$a(x) = x^8$.
A textbook says that.
$x^8 \bmod m(x) = [m(x) - x^8]=(x^4+x^3+x+1)$
So I would like to understand, how do we reach this answer?
It's unclear to me how we perform division here when $m(x) > a(x)$. All the examples in the book show division or finding $a(x) \bmod m(x)$ when $m(x) < a(x)$.
If I understand the comments correctly, $x^8+x^8 = x^8-x^8=0 $ since this is GF$(2)$. So then $ -x^8 + -(x^4+x^3+x^1+1) = -m(x) $, so if $r(x) =(x^4+x^3+x^1+1)$, then $a(x)-r(x) \equiv 0 \bmod m(x)$. Do I have this correct?