Let $F$ be a finite field.
If the characteristic of $F$ doesn’t divide $n$, then $F$ contains a primitive $n^{th}$ root of unity.
I believe the converse is true, too, but I can’t prove either direction?
I know that $F^{*}$ is cyclic and of order $p^n -1$, where $p$ is the characteristic of $F$ and $n >0$. So it contains a generator, $\alpha$ with $\alpha^{p^n -1} = 1$.
But how does this relate to the characteristic not dividing $n$?
fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.
Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.
If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.
The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.
