Char(F)=0, K/F is abelian and F contains a primitive root of unity

Question

$\mathrm{Char}(F)=0, K/F$ is abelian . Let $n$ be a positive integer such that $f^n=1$ for every $f$ in $G(K/F)$ and $F$ contains a primitive $n^{th}$ root of unity. Prove there exist $x_1, x_2, …, x_r$ in $K$ such that $K=F(x_1, x_2, …, x_r)$ with ${x_i}^{n_i}$ in F some $n_i | n$ for $i = 1,\ldots, r$. Can anyone give me some hints ?

Answer 1

As pointed out by LUXun this is basic Kummer theory. A sequence of steps that takes you there that first came to mind is to:

1. Write $G$ as a direct sum of cyclic groups, $G=\bigoplus_{i=1}^k G_i$, with $G_i$ generated by $\sigma_i, i=1,\ldots,k$.
2. Let $H_i$, $i=1,2,\ldots,k,$ be the subgroup $H_i=\bigoplus_{j=1,j\neq i}^kG_j\le G$ generated by $\sigma_j, j\neq i$. Let $K_i$ be the fixed field of $H_i$ for all $i$.
3. Show that the restriction of $\sigma_i$ to $K_i$ generates $\mathrm{Gal}(K_i/F)$. Let $m$ be the order of this restriction. Then $m\mid n$.
4. Let $\zeta\in K$ be a primitive root of unity of order $m$. Using the independence of characters theorem (IIRC credited to Artin or Dedekind, depending on the source) to prove that there exists an element $z\in K_i$ such that the sum $$ x_i=z+\zeta\sigma_i(z)+\zeta^2\sigma_i^2(z)+\cdots\zeta^{m-1}\sigma_i^{m-1}(z)\neq0. $$
5. Show that $\sigma_i(x_i)=\zeta^{-1}x_i$. Show that this implies that $K_i=F(x_i)$, and that $x_i^m\in F$.
6. Show that $K$ is the compositum of the fields $K_i,i=1,2,\ldots,k$.