# Characteristic and primitive roots of unity

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 $$n$$?

• Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity. Commented Jan 18, 2019 at 10:23
• More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $n\mid q^k-1$. This is implicit in Servaes' answer. Commented Jan 18, 2019 at 10:25

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 $$n$$.