I'll be learning Galois theory for the first time later this year and wanted to clear up something that was puzzling me.
If f(x) = $\sum {a_ix^{i}}$ is a polynomial which is irreducible over $\mathbb{Q}$ and K is the splitting field of f (so it is the intersection of all fields which contain $\mathbb{Q}$ and all the roots of f), it is easy to see that any automorphism $\phi$ of K which fixes elements of $\mathbb{Q}$ will permute the roots of f. This is because if $r$ is a root of f, then $0$ = $\sum {a_ir^{i}}$ $\Rightarrow$ $0$ = $\phi (0)$ = $\phi(\sum {a_ir^{i}})$ = $\sum {a_i\phi(r)^{i}}$, whence $\phi(r)$ is also a root of f.
My question is: how does the other direction work? How does a permutation of the roots $r_1, r_2 $ etc. of f induce an automorphism on K (which fixes elements of $\mathbb{Q}$)? The impression I have gotten from reading guides online is that there is a very obvious, natural way to build an automorphism from a permutation of roots, but it does not appear to me to be at all obvious.
EDIT: okay, so there is no guarantee that there exists an automorphism which permutes the roots in a predetermined manner. But how do you know that two different automorphisms cannot permute the roots in the same way? In other words if there is an automorphism $\phi$ which DOES give some permutation $\sigma$ on the roots, how can I use the roots alone to uniquely determine $\phi$?