Galois group of $ x^n - a $ over a field containing $ \zeta_n $ I'm having trouble solving an exercise regarding Galois theory.
Suppose $ n > 0 $ and $\zeta_n \in F \subset \mathbb{C} $, where $ \zeta_n $ denotes the primitive root of unity of degree $ n $.
Let $ f(x) = x^n - a \in F[x] $ and $ E $ be the splitting field of $ F $. 
Prove that the Galois group of $ E/F $ is abelian and that if $ f $ is irreducible over $ F $ then for every natural $ m $ dividing $ n $ there exists precisely one subextension $ F \subset L \subset E $ such that $ [L:F] = m $.
My attempt: It appears that the crucial step here is knowing the fundamental theorem of Galois theory. We know that the roots of $ g $ are $ \sqrt[n]{a}\zeta_n^j $ for $ j = 0, 1,\dots, n-1 $, so if $ \sqrt[n]{a} \in F $, then there's nothing to prove - the Galois group is trivial. 
Otherwise any automorphism if defined by its action on $ \sqrt[n]{a} $. So the Galois group $ G $ acts faithfully on the set $ \{\sqrt[n]{a}\zeta_n^j, ~j = 0\dots n-1\} $, so it's a subgroup of $ S_n $ (is this right?). Moreover, $ \sigma_k: \sqrt[n]{a} \rightarrow \sqrt[n]{a}\zeta_n $ generates all automorphisms $ \sigma_j : \sqrt[n]{a} \rightarrow \sqrt[n]{a}\zeta_n^j$, so the Galois group is cyclic (I can't see any errors so far, but I'm suspicious), hence it's abelian.
Is it true that $ \sqrt[n]{a} \notin F \iff  f $ is irreducible?
About the other one - if it is true that the Galois group is cyclic, then the one-to-one correspondence between subgroups of $ G $ and intermediate fields makes it obvious.
Am I getting it right?
 A: Consider $x^4-4$ over $F=\Bbb{Q}[i=\zeta_4]$. What goes wrong here?
Over this field, $x^4-4$ factors as $(x^2-2)(x^2+2)$, which factors over $\Bbb{C}$ as $(x-\sqrt{2})(x+\sqrt{2})(x-\sqrt{2}i)(x+\sqrt{2}i)$. The splitting field is clearly $F[\sqrt{2}]$, which has degree 2 over $F$, and the the nontrivial automorphism is $\sqrt{2}\mapsto-\sqrt{2}$. There is no automorphism that maps $\sqrt{2}$ to $\sqrt{2}i$ since these have different minimal polynomials over $F$ even though $i=\zeta_4$.
Your error is that what you have claimed is an automorphism isn't always an automorphism.
Also this answers your other question, $\sqrt[4]{4}\not\in F$, but $x^4-4$ is not irreducible over $F$.
However your claim is true when $f(x)$ is irreducible, since then the Galois group acts transitively on the roots so the desired automorphism exists, then the Galois group is cyclic and the correspondence is true.
When $f$ is not irreducible however, again the automorphisms are $\sigma_k : \sqrt[n]{a}\mapsto \sqrt[n]{a}\zeta_n^k$ however only for certain $k$. What's important to note is that $(\sigma_k\circ\sigma_j)(\sqrt[n]{a})=\sigma_k(\sqrt[n]{a}\zeta^j)=\sqrt[n]{a}\zeta^{k+j}=\sigma_j(\sqrt[n]{a}\zeta^k)=(\sigma_j\circ\sigma_k)(\sqrt[n]{a})$. Then since the splitting field is $E=F[\sqrt[n]{a}]$, and $\sigma_j$ and $\sigma_k$ fix $F$, $\sigma_j\sigma_k x=\sigma_k\sigma_j x$ for all $x$ in $E$. Therefore $\sigma_j\sigma_k=\sigma_k\sigma_j$, so the Galois group is abelian.
