In our exam we were supposed to find the normal basis of $GF(3^6)$ taken as polynomial space over $GF(3^2)$. Our second task was to make the multiplication matrix for the chosen element.
Dimension of $GF(3^2)$ is 2, dimension of the extension and length of the basis is 3.
I know that $GF(3^2)$ has 9 elements, so I claim that normal basis should be in the form of $(\alpha,\alpha^9,\alpha^{81})$, $\alpha\in GF(3^6)$.
But I don't fully understand how should I make the multiplication matrix for the chosen element. Can you please show me a short example?
Over $\mathbb{F}_3$ : $x^2-2$ has no root so it is irreducible and $\mathbb{F}_{9} = \mathbb{F}_3[a]/(a^2-2)$.
Over $\mathbb{F}_{9}$ : $X^3+X+a\ $ has no root. So it is irreducible and $\mathbb{F}_{9^3} = \mathbb{F}_9[X]/(X^3+X+a)=\mathbb{F}_3[a,X]/(a^2-2,X^3+X+a)$.
In $\mathbb{F}_{9^3}$ the addition table is the same as in $\mathbb{F}_3[a,X]$, and the multiplication is given by $a^2 = 2$ and $X^3 = -a-X$