Does $\mathbb{F}_9$ contain a 4th root of unity? I realised that I don't know how to construct $\mathbb{F}_9$. I'm guessing that $\mathbb{F}_9 = \mathbb{F_3(\theta)}$, where $\theta$ is the root of some irreducible polynomial over $\mathbb{F}_3[x]$ of degree two? 
Must I even construct $\mathbb{F}_9$ in order to determine whether it contains a 4th root of unity or is there some other simpler way I'm missing?

A non-trivial $4$th root of unity is a square root of $-1$, since $(x^4-1)=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)$, and $x^2+1$ is irreducible over $\mathbf F_3$. So the answer is yes:
$$\mathbf F_9\simeq \mathbf F_3[x]/(x^2+1),$$
and if you denote $\omega$ the congruence class of $x$, the non-trivial $4$th roots of unity  are $\omega$ and $-\omega$.
