TITLE: Is there any irreducible polynomial over $\mathbb{Q}$ whose Galois group is $S_4$?
QUESTION [5 upvotes]: As simple as that: Can we find an irreducible polynomial in $\mathbb{Q}$ such that, if $K$ is its splitting field over $\mathbb{Q}$, $\operatorname{Gal}(K|\mathbb{Q})\cong S_4$?
I've thought a lot and I have not reached any. Can you give me a hand please?

REPLY [4 votes]: Consider the polynomial
$$
p(x) \;=\; x^4 - 6x^2 +16x - 3
$$
whose roots are
$$
-\sqrt{1+\sqrt[3]{3}} \,\pm\, \sqrt{2-\sqrt[3]{3}+\frac{4}{\sqrt{1+\sqrt[3]{3}}}},\qquad
\sqrt{1+\sqrt[3]{3}} \,\pm\, i\sqrt{-2+\sqrt[3]{3}+\frac{4}{\sqrt{1+\sqrt[3]{3}}}}.
$$
It should come as no surprise that $p$ is irreducible. Indeed, since $\sqrt{1+\sqrt[3]{3}}$ and hence $\sqrt[3]{3}$ lies in the splitting field of $p$, the degree of the splitting field is a multiple of $3$, so $p$ does not factor into quadratics over $\mathbb{Q}$.
Since $p$ is irreducible, the Galois group acts transitively on the roots, so the only possibilities are $V$, $\mathbb{Z}_4$, $D_4$, $A_4$, and $S_4$.  Since the degree of the extension is a multiple of $3$, it must be either $A_4$ or $S_4$.  But complex conjugation is an automorphism of the splitting field that acts as a transposition of the two complex roots, and therefore the Galois group is $S_4$.
In some sense, this example is fairly typical, if somewhat simpler than usual.  Any irreducible quartic with two real and two complex roots has a transposition in its Galois group from complex conjugation, which means that the Galois group is either $D_4$ or $S_4$.  Those with Galois group $S_4$ also have the roots of some irreducible cubic in their splitting field -- in this case the cubic $x^3 - 3$.