Are the $n$-th roots of unity over an arbitrary field generated by a single element?

Question

Let $\mathbb F$ be a field. If we take the set of numbers such that $x^n=1$ (for a fixed $n$), is it true that said group is finitely generated?

What I mean is, let $n\in\mathbb Z^+$, $G_n(\mathbb F):=\{x\in\mathbb F| x^n=1\}$. Is there any $y\in G_n(\mathbb F)$ such that $G_n(\mathbb F)=\langle y^m\rangle$?

I found this earlier question: Do the $n$-th roots of unity of an *arbitrary* field form a cyclic group? but there was no proof provided for what I'm asking.

For what is worth, I know that they form a cyclic group, but I'm failing to find if they're generated by a single element.

Answer 1

It has been answered in the comments but the explicit solution looks like a good exercice :

With $p=char(F)$ and $n=p^k m,p\nmid m$ then the polynomial $x^n-1\in F[x]$ splits completely in some finite extension $E/F$ and its roots are the cyclic group generated by any root of the cyclotomic polynomial $\Phi_m(x)\in E[x]$

Answer 2

For any field $K$, any finite subgroup G of the multiplicative group $K^*$ is cyclic.

Proof. From the structure theorem for finite abelian groups, $G\cong Z/a_1 \times...\times Z/a_n$ (in additive notation), wih $a_1\mid a_2\mid ...\mid a_n$. Denote by $y$ the class of $1$ in $Z/a_n$, then $x=(0,0,...,y)$ has obviously order $a_n$. For any $z=(z_1,...,z_n)\in G$, because $a_i$ divides $a_n$, one has $a_n z=0$, so the order of $z$ divides $a_n$. Thus the order $m=a_n$ of $x$ is the l.c.m. of the orders of the elements of $G$, hence any $g\in G$ is a root of the polynomial $X^m -1$. As $X^m -1$ has at most $m$ roots in the field $K$, the order of $G$ is at most $m$. But, as $x$ has order $m$, the powers $x, x^2,..., x^m=1$ are distinct. It follows that $G$ is cyclic of order $m$.