Representations of the formulas with the elements of the linear representations of $GF(2^2)$

Question

The problem statement is as below.

Represent the following formulas with the elements which are of the linear representations of $GF(2^2)$ where the root $\alpha$ satisfies $x^2+x+1=0$ .

$(1)$ $\alpha^4$

$(2)$ $\alpha^2-(\alpha-1)$

$(3)$ $\frac{\alpha}{\alpha^4}$

I have the following formula but I don't know how should I use it.

$GF(2^2)=\{0,1,\alpha,\alpha+1\}$ $,(\alpha^2+\alpha+1)$

Can someone let me know the solutions or some website that explain about the solutions?

Answer 1

I will do one example, that isn't on your list, and then that should show you what is meant. I choose $\alpha^2+1$.

Now the elements of the field are $0,1,\alpha,\alpha+1$, so this is equal to one of those four elements. We have the equation $\alpha^2+\alpha+1=0$, so (as we are over a field of characteristic $2$) $\alpha^2=\alpha+1$.

Then $\alpha^2+1=(\alpha+1)+1=\alpha$.

Answer 2

Well, by hypothesis, $\alpha^2+\alpha+1=0$ and so $\alpha^2=\alpha+1$. From here you can easily answer the questions.

First, $\alpha^4 = \alpha^2\alpha^2 = (\alpha+1)^2 = \alpha^2+1 = (\alpha+1)+1 = \alpha.$

Second, $\alpha^2 - (\alpha-1) = \alpha^2 - \alpha+1$. But $-1=+1$ in the coefficient field $GF(2)$, and so you get $\alpha^2+\alpha+1=0$.

Third, $\alpha/\alpha^4 = \alpha^{-3}$. But $\alpha^3=1$ since $\alpha$ is 3rd root of unity, and so $\alpha^{-3}=1$.