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 $n$?

  • $\begingroup$ 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. $\endgroup$ Jan 18, 2019 at 10:23
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 18, 2019 at 10:25

3 Answers 3


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 $n$.


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