I am still trying to understand polynomial arithmetic with Galois Field. Can someone explain to me what this question is asking to be constructed. I have searched all over google for GF(4) with modulo and I am not getting anywhere.
I have read over a few of the other posts here on the forums but none seem to apply to my specific situation. Can anyone help me understand GF with modulo of a polynomial?
Thanks in advance.
What you're asking saying $GF(4)$ modulo $x^2+x+1$ is equivalent to do:
$$GF(2)\cong \frac{\mathbb{F}_2[x]}{x^2+x+1}$$
So the sum of $GF(2)$ is the same as in $\mathbb{F}_2[x]$ and then doing the modulo $x^2+x+1$: Since all your elements of degree $2$ or greater can be reduced considering $x^n=x^2x^{n-2}=(x+1)x^{n-2}$, you only will have elements of degree $1$ or $0$, and this will be the table:
$$ \begin{array}{ | c | c | c |c| c|} \hline & \color{green}{0} & \color{green}{1} & \color{green}{x} & \color{green}{x+1}\\ \hline \color{green}{0} & 0 & 1 & x & x+1 \\ \hline \color{green}{1} & 1 & 0 & x+1 & x \\ \hline \color{green}{x} & x & x & 0 & 1\\ \hline \color{green}{x+1} & x+1 & x & 1 & 0\\ \hline \hline \end{array}$$
For the case of the multiplication, you can do it
$$ \begin{array}{ | c | c | c |c| c|} \hline & \color{green}{0} & \color{green}{1} & \color{green}{x} & \color{green}{x+1}\\ \hline \color{green}{0} & 0 & 0 & 0 & 0 \\ \hline \color{green}{1} & 0 & 1 & x & x+1 \\ \hline \color{green}{x} & 0 & x & x+1 & 1\\ \hline \color{green}{x+1} & 0 & x+1 & 1 & x\\ \hline \hline \end{array}$$
Why? It's clear that $1\cdot a=a$ for any element or why $0\cdot a=0$, so let's see the case with $x+1$. Look that $(x+1)x=x^2+x$, but by your quotient, $x^2+x+1\equiv 0$, so $x^2+x\equiv 1$ (we're modulo 2), and by the same reason, $(x+1)^2=x^2+2x+1\equiv x^2+1\equiv x$ And that's the way the multiplication happens.
