Field $F$ with $\operatorname{char}F=3$ and algebraic over $\mathbb{F}_3$ has a primitive root of unity.

Question

Suppose that $F$ is a field with $\operatorname{char}F=3$ and $F$ is an algebraic extension of $\mathbb{F}_3$. Prove that $F$ contains a primitive $n$th root of unity for some $n>2$.

Answer 1

Assuming that $F \neq \mathbb{F}_3$, let $x \in F \setminus \mathbb{F}_3$. By assumption $x$ is algebraic, so $[\mathbb{F}_3(x):\mathbb{F}_3]<+\infty$. This means that the unit group $\mathbb{F}_3(x)^\times$ is a finite group. Let $n = \vert \mathbb{F}_3(x)^\times \vert$ be the order of the unit group. Then by Lagrange's theorem $x^n = 1$, so $x$ is a root of unity. It now suffices to show that $x\neq 1$ and $x^2 \neq 1$. The first is clear from $x \notin \mathbb{F}_3$. The second equality would imply $x = \pm 1$ which is also wrong (you use that $F$ is a field here, i.e. that $X^2-1$ has only two roots, namely $\pm 1$!). This shows the statement.

Answer 2

Note that any finite extension $F$ of $\mathbb F_3$ with $[F:\mathbb F_3] = n$ is isomorphic to the splitting field of the polynomial $$X^{3^n}-X,$$ which clearly contains the roots of $$X^{3^{n}-1}-1,$$

ie: the $3^{n}-1$-th roots of unity.

Since any nontrivial algebraic extension of $\mathbb F_3$ contains such a subfield, we are done.