Let $f(x)\in\mathbb Q[X]$ be a polynomial of degree $n$, and let $K$ be a splitting field of $f$ over $\mathbb Q$. Suppose that $\mathrm{Gal}(K/\mathbb Q)$ is the symmetric group $S_n$ with $n>2$.
(a) Show that $f$ is irreducible over $\mathbb Q$.
(b) If $\alpha$ is a root of $f$, show that the only automorphism of $\mathbb Q(\alpha)$ is the identity.
I am beginning to learn Galois Theory, but I feel this question is a little hard for me. Can someone tell me how to solve this problem?
a) If $f$ is reducible then the action of the Galois group on the set of roots is not transitive. Indeed the Galois group never maps a root of one factor of $f$ into a root of another factor. The permutations generating the Galois group generate a group that is a proper subgroup of $S_n$, otherwise the action would be transitive.
b) The action of the Galois group is transitive so the set of roots has only one orbit. By the orbit-stabilizer theorem the stabilizer of a root $\alpha$ is the trivial group.