This is a question from Lang's Algebra Chapter $VI$ $Q10.
Let $f(x)\in Q[x]$ be a polynomial of degree n, and let $K$ be a splitting field of $f$ over $Q$. Suppose that $Gal(K/Q)$ is the symmetric group $S_{n}$ with $n>2$.
$(a)$ show that $f$ is irreducible over $Q$
$(b)$ If $a$ is a root of $f$, show that the only automorphism of $Q(a)$ is the idenity.
$(c)$ If $n\geq 4$, show that $a^{n} \notin Q$.
I have shown the part $(b)$, but I am having trouble in $(a)$ and $(c)$. For $(a)$, I tried to assume that $f$ is reducible but I failed.
Any hints and detailed explanations are really really appreciated!!!!
$A)$ Since $K$ is galois over $Q$, $[K:Q] = |Gal(K/Q)| = n!$
Since $f(x)$ is of degree $n$; there are obviously $n$ roots, $a_1, ... , a_n$. Suppose $f$ is reducible. Then $[Q(a_1) :Q] < n$. Therefore [$K:Q] < n!$ and this is a contradiction with the statement above.
$C)$ Since $f$ is irreducible $[Q(a):Q] = n$. If $a^n \in Q$, then $x^n - a^n$ would be a degree $n$ polynomial that contains $a$ as a root. This implies $f(x) = x^n - a^n$. You don't have to search too deeply in chapter $6$ of Lang to see that the roots of $x^n - a^n$ are $\rho^i a$ for $1 \leq i \leq n$ where $\rho$ is a primitive $n^{th}$ root of unity. Hence the splitting field of $x^n - a^n$ is $Q(\rho, a)$. You also don't need to search to deeply in Lang to see that $[Q(\rho):Q] = \phi(n)$ where $\phi$ is the Euler function. Therefore, $[Q(\rho,a):Q] =[Q(\rho,a):Q(\rho)] [Q(\rho):Q] \leq n \phi(n)$.
It's trivial to see that $n \phi(n) < n!$ and this is a contradiction.