How to do multiplication in $GF(2^8)$? I am taking an Internet Security Class and we received some practice problems and answers, but I do not know how to do these problems , an explanation would be greatly appreciated
Try to compute the following value:(the number is in hexadecimal and each represents a polynomial in GF(2^8)
1) {02} * {87}
answer: 
{02} = {0000 0010} = x
{87} = {1000 0111} = x
-> {02} * {87} = (000 1110) XOR (0001 1011) = 0001 0101 = {15}

To carry out the operation, we need to know the irreducible polynomial that is being used in this representation. By reverse-engineering the answer, I can see that the irreducible polynomial must be $x^8+x^4+x^3+x+1$ (Rijndael's finite field). To carry out a product of any two polynomials then, what you want to do is multiply them and then use the relation $x^8+x^4+x^3+x+1\equiv 0$, or in other words $x^8\equiv x^4+x^3+x+1$, to eliminate any terms $x^k$ where $k\geq 8$, reducing modulo 2 as you go along.
The binary {0000 0010} corresponds to the polynomial $x$ (i.e., $0x^7+0x^6+0x^5+0x^4+0x^3+0x^2+1x^1+0x^0$), while the binary {1000 0111} corresponds to $x^7+x^2+x+1$ (i.e., $1x^7+0x^6+0x^5+0x^4+0x^3+1x^2+1x^1+1x^0$). So to do the multiplication, we calculate
\begin{align*}
x*(x^7+x^2+x+1) &= x^8 +x^3+x^2+x \\
&\equiv (x^4+x^3+x+1) + x^3+x^2+x \\
&= x^4+2x^3+x^2+2x+1 \\
&\equiv x^4+x^2+1
\end{align*}
which is represented in binary as {0001 0101}.
