Finite Galois Field extension of a field $F$ containing all roots of unity Let $F$ be a field that contains all roots of unity. Furthermore, let $K$ be a finite algebraic extension of $F$ with abelian Galois group . Then 
$$K= F(z_1,\ldots , z_n)$$
for some $z_i \in K$ $(i=1,\ldots, n)$ and for each $i$ there is a non-zero integer $n_i$ such that $z_i^{n_i}\in F$. 
I came across this while reading a paper but I am not able to realise this. Please help. Any hint will be appreciated.
Thanks! 
 A: First, you can argue that any abelian extension of $F$ is a composite of cyclic extensions.  Next, it suffices to show that any such cyclic extension $E/F$ is of the form $F(\sqrt[m]{a})$ for some $m$ and some $a \in F$.  Let $m = [E : F]$, so the Galois group of $E/F$ is isomorphic to $\mathbb{Z}/m\mathbb{Z}$.  Let $\sigma$ generate this Galois group, and let $\zeta \in F$ be a primitive $m$th root of unity in $F$.  
We have $N_{E/F}(\zeta) = \zeta^m = 1$, and so by Hilbert's Theorem 90, there must exist an element $c \in E$ such that $\zeta = \sigma(c)/c$.  Then $$\sigma(c) = \zeta c $$ $$\sigma^2(c) = \zeta \sigma(c) = \zeta^2 c $$ $$\vdots $$ $$\sigma^{m-1}(c) = \zeta^{m-1}c$$ In other words, the $m$ conjugates of $c$ in $E$ are distinct.  This shows that the minimal polynomial of $c$ over $F$ has $m$ distinct elements.  So $[F(c) : F] = m$, which implies $F(c) = E$.  
Finally, we claim that $c^m \in F$:  we have $1 = \sigma(c^m)/c^m$, so $\sigma(c^m) = c^m$.  Hence $\sigma^i(c^m) = c^m$ for all $i$, so $c^m$ must be in $F$.
