Sunday, September 16, 2018

finite fields - 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

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