Construction of addition and multiplication table for GF(4) I am dealing with finite fields and somehow got stuck. The construction of a prime field $GF(p),  p \in \mathbb{P}$ is pretty easy because every operation is modulo p. In other words $GF(p)$ contains all integers ranging from 0 to $p-1$.
However, non prime fields are a bit trickier. Given the power $q = p^n$ with $p \in \mathbb{P}$ and $n \in \mathbb{N}$, one has to find an irreducable polynom g(x) of degree n. Then the construction of $GF(p^n)$ is the following:
$GF(p^n) = \frac{GF(p)[x]}{g}$
Theoretically I get along with this definition. Unfortunately I fail to construct addition and multiplication tables for any GF(q). Though I can easily find the wanted table on the internet, I have not found an explication yet that really made me understand.
I would like to know how to create the addition and multiplication table for $GF(2^2)$ with the above knowledge. $GF(2^2)$ contains four elements. Let's call them $\{0,1, \alpha, \beta \}$. $g$ must be $x^2 + x + 1$ as there no other irreducable polynom of degree 2. So far I am able to construct the addition table partly (question marks indicating despair...):
| + | 0  1  $\alpha$  $\beta$ |
0 | 0  1  $\alpha$  $\beta$ |
1 | 1  0  ?  ? |
$\alpha$ | $\alpha$ ?  ?  ? |
$\beta$ | $\beta$  ? ? ? |
I don't understand how to calculate for example $1+\alpha$. The result is $\beta$, but I don't know why. Concerning the above short explanation, I have to divide $1+\alpha$ by $g$. But how can I do this? 
Thanks for your help

This case is fairly easy because all the calculations are modulo 2 and the field has only 4 elements. Let $\alpha$ be a root of $g$, i.e, $\alpha^2 + \alpha + 1 = 0$. This immediately implies that $1+\alpha = -\alpha^2 = \alpha^2$, which you are calling $\beta$. I would suggest to try a polynomial of higher degree so that the field has more elements. 
