Is it true that if $ f \in \mathbb{Q}[X] $ is a polynomial of degree n and $ L $ its splitting field over $ \mathbb{Q} $ such that $ [L:\mathbb{Q}]=n! $ then $ f $ is irreducible?
If $ f $ is separable, then $ L/\mathbb{Q} $ is Galois which, given the hypothesis, implies that $ Gal(L/\mathbb{Q}) \simeq S_{n} $. But how can I deduce from this whether or not $ f $ is irreducible?
Also, what about the case when $ f $ is not separable?
I would appreciate any hints. Thank you!
First: the splitting field of a polynomial over $\mathbb{Q}$ (or any perfect field) is always Galois, because if the polynomial has repeated roots, it must be divisible by powers of the corresponding minimal polynomial.
If $f$ is not irreducible, it can be written as a product $f=f_1f_2$. Then any automorphism of $L$ must induce a permutation of the roots of $f_1$ and of those of $f_2$, but it can never send a root of $f_1$ to a root of $f_2$. Thus the group of permutations of the roots of $f$ induced by automorphism of $L$ must be a proper subgroup of $S_n$, which contradicts the fact that it has order $[L:\mathbb{Q}]=n!$. Incidentally, the same argument shows that $f$ must in fact be separable : if it has repeated roots, then it has less than $n$ roots and the group of permutations must be a subgroup of $S_{n-1}$.