A Question about Irreducible Polynomials
In the page $104$ you can see:
$$F_{2^8}=\frac{Z_2[x]}{x^8+x^4+x^3+x+1}.$$
this means that irreducible polynomial that generate the $F_{2^8}$ is $x^8+x^4+x^3+x+1$. so, in multiplications, you should note that $x^8=-x^4-x^3-x-1=x^4+x^3+x+1$. Therefore $x^{8+i}=x^8\cdot x^i=x^{4+i}+x^{3+i}+x^{1+i}+x^i$.