TITLE: Characteristic and primitive roots of unity
QUESTION [5 upvotes]: 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 $n$?

REPLY [3 votes]: 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 $n$.