Consider the finite field $\mathbb F_9 = \mathbb F_3[x]/ \langle x^2 + 1\rangle$, and recall the Frobenius isomorphism $Frob_3 : \mathbb F_3 → \mathbb F_3$, given by $Frob_3(x) = x^3$ Find all the elements of $\mathbb F_9$ that are fixed by $Frob_3$, and use this to see which elements are fixed by $(Frob_3)^2 = Frob_3 ◦ Frob_3$.
What does it mean by find elements that are fixed by $Frob_3$? Can't find it anywhere. Please help.
An element of $\mathbb{F}_9 = \mathbb{F}_3[x]/(x^2 + 1)$ is of the form $(x^2 + 1) + f(x)$, where $f(x) \in \mathbb{F}_3[x]$.
Let $\sigma : \mathbb{F}_9 \to \mathbb{F}_9$ be the Frobenius automorphism $\alpha \mapsto \alpha^3$. To say that an element $\alpha = (x^2 + 1) + f(x) \in \mathbb{F}_9$ is fixed by $\sigma$ means that $(x^2 + 1) + (f(x))^3 = (x^2 + 1) + f(x)$. For example, $(x^2 + 1) + 1$ is fixed by $\sigma$ because $1^3 = 1$.