# simple multiplication in GF(8)

I am trying to do multiplication in the GF($2^3$) defined by the irreducible minimum binary polynomial $X^3+X^2+1$. I want to multiply $A(x) * B(x)$ where $A(x) = x$ and $B(x) = x^2$. The multiplication is easy, and I get the result: $C'(x) = x^3$ However, I am confused on how to do the modulo in $X^3 + X^2 +1$. I am convinced that it can be done by XOR'ing in long division: $1000 \oplus 1101 = 0101$. Is this the correct result then: $101$?

-

$A \equiv A + B\ \ (\bmod\ B)$
Thus, if $A = X^3$ and $B = X^3+X^2+1$, that implies that $X^3$ is equivalent to the sum of $X^3$ and $X^3+x^2+1$, which is $X^2+1$, which is represented by $101$.