If I work over $GF(2^4)$, then the condition for the type I optimal normal basis holds, in fact $4+1=5$ is prime and $2$ is a primitive element in $\mathbb{Z}_5$. My question is about the existence of an optimal normal basis in case that one starts from $GF(2^2)$ and construct $GF(2^4)$ as quotient ring $\dfrac{GF(2^2)[x]}{<q(x)>}$ with $q(x)$ an irreducible polynomial of degree $2$.
For example, starting from $GF(2)$, one chooses $x^2+x+1$ as irreducible polynomial so that $GF(2^2)=\dfrac{GF(2)[x]}{<x^2+x+1>}$. Then, using polynomial basis one has $GF(2^2)=\{0,1,\alpha,1+\alpha\}$ with $\alpha^2=\alpha+1$. In this case, $\alpha$ and $1 + \alpha$ are optimal normal basis elements, in fact $\alpha^3=(1+\alpha)^3=1$ so that the normal basis $\{\alpha, 1+\alpha\}$ is also optimal. Then, if one chooses $x^2+x+\alpha^2$ as irreducible polynomial, one has $GF(2^4)=\dfrac{GF(2^2)[x]}{<x^2+x+\alpha^2>}$. I don't represent all the elements for simplicity: they are linear combination of, e.g., polynomial basis $\{1, \beta\}$ with $\beta^2=\beta+\alpha^2=\beta+\alpha+1$ with coefficients in $GF(2^2).$ In this case, is it possible to find an optimal normal basis? If yes, can you explain me how? I tried to compute all $16$ elements to the power $3$, but anyone returns $1$, except for $1$, $\alpha$ and $1+\alpha$. Maybe I'm doing something wrong, but I don't know where.
Thank you!
EDIT:
Thanks to Jyrki Lahtonen
The third roots of unit are $1, \alpha, \alpha+1$. In this case it is not possible to find an optimal normal basis due to the fact that the condition about primitivity of $4$ in $Z_3$ falls.