Understanding Galois theory, swapping roots I am trying to understand the rudiments of galois theory but I have a hard time answering the following question: Suppose we have $f(X)=x^5-2$ over $\mathbb{Q}$. What is $Gal(E_f: \mathbb{Q})$?
After seeing similar questions here I am more confused: Isn't it true that since $f$ is irreducible, every possible permutation of its roots can be extended linearly to a endomorphism $E_f \rightarrow E_f$ while holding $\mathbb{Q}$ constant? If that is so then shouldn't the Galois Group be $S_5$? 
I understand that this isn't the case, but can't really understand why...

First note that the splitting field of $f$ is given by $L = \mathbb{Q}(\sqrt[5]{2}, \zeta_5)$, where $\zeta_5$ is the primitive root of unity. 
Now we know that any automorphism on $L$ is uniquely determined by the action on the adjoined elements $\sqrt[5]{2}$ and $\zeta_{5}$. Additionally we know that if $\alpha_1$ and $\alpha_2$ are roots of the same irreducible polynomial over $\mathbb{Q}$ then the identity automorphism on $\mathbb{Q}$ can be extended to a automorphism $\sigma$ of the splitting field s.t. $\sigma(\alpha_1) = \alpha_2$. In particular in our case we can have automorphism of $L$ s.t. $\sigma(\sqrt[5]{2}) = \zeta_5^2\sqrt[5]{2}$. Now we can do the same for $\zeta_5$ and for example get an automorphism $\tau$ on $L$, s.t. $\tau(\zeta_5) = \zeta_5^3$ and it is identity on $\mathbb{Q}$. 
However note that this doesn't give you every possible permutation of roots, as it's not possible to permute the first three roots, while to keep the other two fixed. So to summarize there is an element of the Galois group of $f$ (assuming $f$ is irreducible) s.t. that it send any root to any other root of $f$, but it's not necessarily keeping the other roots fixed or permuting them in arbitrary manner.
