Finding normal basis for GF(q^m) over GF(q)

Question

Could you kindly explain, how can one find a normal basis for GF$(3^6)$ over the GF$(3^2)$?

As I understood, I should start with finding the polynomial in a form $$a(x^2) + (a^9)x + a^{81},$$ which is relatively prime to polynomial $x^3 - 1$. If such exists, normal basis should be $(a, a^9, a^{81})$. Am I right?

I would appreciate any help. Thanks in advance.

Answer 1

Since the degree of $GF(3^6)$ over $GF(3^2)$ is a power of the characteristic, an element $a$ generates a normal basis if and only if its trace to $GF(3^2)$ is not zero. You can for instance take for $a$ a zero of the polynomial $X^3-X^2+1$.