Suppose we have the field of rational functions $GF(7)(t)$ in the indeterminate $t$ over $GF(7)$. Define $σ,τ ∈ Aut(E)$ by $σ(t) = 2t$ and $τ(t) = 1 /t$. Set $G = <σ,τ>$ and $F = Fix(G)$.
There was a question in a past exam paper I was looking at that said considering the minimal polynomial of $t$ over $F$ describe $F$ as a simple extension of $GF(7)$. I want to make sure I understand this correctly
Firstly we note that $\sigma(t^3)=t^3$ so $t^3 \in Fix(<\sigma>)$ , also we have that $\tau(t+1/t)=t+1/t $ so $(t+1/t)\in Fix(<\tau>)$.
So then we have that $(t^3+1/t^3)\in Fix(G)$.
Next we note that $t^6-t^3(t^3+1/t^3)+1=0, $so a minimal polynomial for $t$ over $GF(7)$ is $x^6-x^3(t^3+1/t^3)+1$.
So then as a simple extension of $GF(7)$ is $F$ just $GF(7)(t^3+1/t^3)$ ?