Let $f(x) \in \mathbb{Q}[x]$ be a polynomial of degree 4 with distinct roots $r_{1}, r_{2}, r_{3}, r_{4} \in \mathbb{C}$, such that:
$$ \prod_{i<j}\left(r_{i}-r_{j}\right)=9 $$
Let $E$ be the splitting field of $f$. Can the Galois group of $E/\mathbb{Q}$ be $S_{4}$?
I am supposed to answer this without any other information, like the resolvent cubic. Also, we have not introduced in class the concept of the discriminant for quartics, so maybe I should not use that as well. I cannot come up with a solution.
Thank you in advance for any help.
Let $D=\prod_{i<j}\left(r_{i}-r_{j}\right)$. For any permutation $\sigma\in S_{\{r_1,...,r_4\}}$ we have:
$\sigma.D=sgn(\sigma)\times D$.
This follows from the definition of sign of a permutation. It means that $\sigma.D=D$ if and only if $\sigma\in A_{\{r_1,...,r_4\}}$. Now, $D=9\in\mathbb{Q}$ and hence $D$ belongs to the fixed field of the group $Gal(E/\mathbb{Q})$. But only permutations in the alternating group are fixing $D$, hence there can be only even permutations in $Gal(E/\mathbb{Q})$. So $Gal(E/\mathbb{Q})\leq A_4$.