splitting field of $x^8-1$ over $\mathbb F_3$

Question

Suppose $F=\mathbb F_3$ and $f(x)=x^8-1$ in $F[x]$.
I tried finding the Galois group of the splitting field of $f(x)$ over $F$ and I'm not so sure if what I did was correct.
I began by looking at \begin{equation*}g(x)=x^{3^2}-x=xf(x)\end{equation*} $g(x)$ and $f(x)$ have the same roots besides $0$, but since $0\in F$ their splitting field is the same.
It is known that the splitting field of $g(x)$ is isomorphic to the field with $3^2$ elements and hence the splitting field of $f(x)$ is $\mathbb F_{3^2}$. Another common fact is that for a prime $p$ and $n\in \mathbb N$ \begin{equation*}\textrm{Gal}(\mathbb F_{p^n}/\mathbb F_p)\cong C_n\end{equation*} where the Frobenius homomorphism generates the group.
the conclusion is that \begin{equation*}\textrm{Gal}(\mathbb F_{3^2}/\mathbb F_3)\cong C_2\end{equation*}The part that bothers me the most is that I examined the splitting field of $g(x)$ instead of constructing the splitting field of $f(x)$ by adding $8$-th roots of the identity to $F$.

Answer 1

There's nothing wrong, because the splitting field of $x^8-1$ is the same as the splitting field of $x^9-x$: we're adding a root that's already in the field. This is known to be the field with nine elements.

You can factor the polynomial as $$ x^8-1=(x^4-1)(x^4+1)=(x-1)(x+1)(x^2+1)(x^4+1) $$ and note that $$ x^4+1=x^4-2x^2+1-x^2=(x^2-1)^2-x^2=(x^2-x-1)(x^2+x-1) $$ If we add a root $i$ of $x^2+1$, so $i^2=2=-1$, we also have roots of $x^2-x-1$ and $x^2+x-1$, as it's easily seen. So the splitting field is $\mathbb{F}_3[i]$.