Show that GF(81) is an $x^{26}+x^{8}+x^{2}+1$ decomposition field

Question

I tried decomposing the polynomial, but after taking out $(x^{2}+1)$ you have to break the remainder into polynomials of degree 4, which is manually hard. Perhaps this is solved by using Frobenius automorphism, I have never been able to find a way to use it

Answer 1

If $p(x) = x^{26}+x^8+x^2+1$, then $q(x) = xp(x) = x^{27}+x^9+x^3+x$ is invariant under the Frobenius $\sigma(t)=t^3$ (good instincts Alvaro!) for every $x\in GF(81)$, since \begin{multline*} \sigma(q(x)) = \sigma(x^{27}+x^9+x^3+x) = \sigma(x^{27})+\sigma(x^9)+\sigma(x^3)+\sigma(x) \\ = x^{81}+x^{27}+x^9+x^3 = x+x^{27}+x^9+x^3 = q(x), \end{multline*} where we have used the fact that every $x\in GF(81)$ is a root of $x^{81}-x$. In particular, $q(x)\in GF(3)$ (the fixed field of $\sigma$) for all $x\in GF(81)$. But each of $q(x)=0$ and $q(x)=1$ and $q(x)=2$ can have at most $27$ solutions due to the degree of $q$; it follows that each of them has exactly $27$ solutions. In particular, $q(x) = xp(x)$ has $27$ roots in $GF(81)$ and thus splits completely, which implies that $p(x)$ itself splits completely in $GF(81)$.