Is the field generated by adding several p-th roots a Kummer extension Let $F$ be a field with $(char(F),p)=1$ or $char(F)=0$ containing the roots of unity of order $p$. Consider the following extension of $F$: $$F[a_1 ^{1/p},...,a_k ^{1/p}]$$ and assume $a_i\notin(F^*)^p$. One can prove that the polynomials $f_i(x)=x^p-a_i$ are either irreducible or they split. Indeed, every extension in the tower is galois both over $F$ and over the previous extension. My question is whether the final extension is a Kummer (abelian) extension over $F$? 
It is abelian over the $(k-1)$ extension, and from the fundamental theorem of Galois extensions, for every $i=1,..,k-1$ $Gal(F[a_1 ^{1/p},...,a_{i+1} ^{1/p}]/F)$ is normal in $Gal(F[a_1 ^{1/p},...,a_{k} ^{1/p}]/F)$. However this a group $G$ can have a normal abelian subgroup $N$ and an abelian quotient $G/N$ without being abelian. I would like to understand why (and if) is $F[a_1 ^{1/p},...,a_k ^{1/p}]/F$ abelian.
 A: Yes it's an Abelian extension. Each $\sigma$ in the Galois group maps
$a_i^{1/p}$ to $\omega_i a_i^{1/p}$ where $\omega_i$ s a $p$-th root of unity. The map $\sigma\mapsto(\omega_1,\ldots,\omega_k)$ is a group homomorphism embedding the Galois group in an elementary Abelian group.
A: Nothing wrong with Lord Shark's answer, just adding a different way of arriving at the conclusion.
Because the $a_i$s are all in $F$ you can adjoin their $p$th roots in any which order you want. This freedom forces the big extension to be Abelian. We can do it by induction $k$:


*

*You know that $K_1:=F[a_1^{1/p}]$ is an Abelian extension of $F$.

*The induction hypothesis tells us that $K_2:=F[a_2^{1/p},\ldots,a_k^{1/p}]$ is also an abelian extension of $F$.

*So if $K:=F[a_1^{1/p},a_2^{1/p},\ldots,a_k^{1/p}]$, then we know that $Gal(K/K_1)$ and $Gal(K/K_2)$ are both normal subgroups of $Gal(K/F)$. Furthermore, basic Galois correspondence tells us that those two subgroups intersect trivially. Therefore the big Galois group is the direct product of those two subgroups. Consequently 
$Gal(K/K_1)\simeq Gal(K_2/F)$ and $Gal(K/K_2)\simeq Gal(K_1/F)$ so the factor groups, hence also their direct product, are Abelian.

