Counter example to Automorphism group

Question

I am working on a problem that starts off with the given: Let $f(x) \in k[x]$ be a separable irreducible polynomial of degree n over a field k, and let F be its splitting field. Assume $Aut_k(F) \cong S_n$, and let a be a root of $f(x)$ in F.

My question is: Doesn't separable irreducible of degree n imply that the Automorphism group is isomorphic to symmetric group?

Answer 1

No. For instance, if $f(x)\in\mathbb{Q}[x]$ is an irreducible cubic polynomial with splitting field $E_f$, then $\mathrm{Gal}(E_f/\mathbb{Q})\simeq C_3$ if the discriminant of $f$ is a square in $\mathbb{Q}$, and otherwise $\mathrm{Gal}(E_f/\mathbb{Q})\simeq S_3$.

One example is $f(x)=x^3-x^2-4x-1$, which has discriminant $169$.

Answer 2

No it doesn't. Consider $f(x) = x^4 + 1$ over $\mathbb{Q}$.