What is the Galois Group of $x^4+1$ over $\mathbb{F}_3$ and describe the action of the group on the roots of it's polynomial

Question

I know the Galois Group is cyclic of order 2 and I got the splitting field to be $\mathbb{F}_9$ but I don't understand how to write the frobenius automorphism that describes the action

Answer 1

The Froebenius automorphism $\sigma : x \mapsto x^3$ swaps the roots since it transforms a root into a root and $\sigma(x) \neq x$ for all $x \notin \mathbb{F}_3$.

And of course $\sigma^2$ is the identity on $\mathbb{F}_9$.

And that's it, the galaois group is $\{\sigma, Id\}$.