Let $i \in \mathbb{F}_3$ be a zero of $x^2+ 1$. Let $\mathbb{F} = \mathbb{F}_3(i)$, and determine the minimal polynomial of $\alpha\in \mathbb{F}$ over $\mathbb{F}_3$.
I'm trying to solve the question above. I showed that $\mathbb{F}$ is a field extension with 9 elements. Thus $\mathbb{F}\cong \mathbb{F}_9$. It's clear that we shouldn't bother about the elements contained in $\mathbb{F}_3$, nor $\pm i$, since they're trivial for the former and we already know them for the latter. But there are 4 more elements to go. I don't know how they look like, so it's hard for me to imagine how I could even get their respective minimal polynomials.
I would really appreciate any help possible.
As lulu mentioned, the elements of $\mathbb{F}_{3}(i)$ are precisely those of the form $a + bi$ where $a, b \in \mathbb{F}_{3}$
Consider $\alpha = a + bi \in \mathbb{F}_{3}(i)$. Now let $x = a + bi$, where $a, b \in \mathbb{F}_{3}$. Then
$$x - a = bi \iff (x-a)^{2} = -b^{2} \iff x^{2} - 2ax + a^{2} + b^{2} = 0.$$
Hence $h = x^{2} - 2ax + a^{2} + b^{2} \in \mathbb{F}_{3}[x]$ is possibly the desired minimal polynomial. We have that $\alpha$ is a root of $h$, the other root of $h$ is its conjugate $\overline{\alpha} = a - bi$. Now either $b = 0$ or $b \neq 0$.
If $b = 0$: we have that $\alpha = a \in \mathbb{F}_{3}$, and the linear (thus irreducible) polynomial $X - a \in \mathbb{F}_{3}[x]$ is the minimal polynomial of $\alpha$ over $\mathbb{F}_{3}$
If $b \neq 0$: we have that $\alpha = a + bi \in \mathbb{F}_{3}(i)$. Note that $h$ is irreducible over $\mathbb{F}_{3}$ since $i \notin \mathbb{F}_{3}$. Then $h$ is the minimal polynomial of $\alpha$ over $\mathbb{F}_{3}$