Finding a quartic polynomial in $\mathbb{Q}[X]$ with four real roots such that Galois group is ${S_4}$. Is there a quartic polynomial $p(x)\in\mathbb{Q}[x]$ irreducible with four real roots such that Galois group is ${S_4}$?
If it really exists, can someone give me a example?  
 A: As hinted in the comments to a related question
and as shown in Keith Conrad's notes,
a degree-$4$ polynomial from $K[X]$, where $K$ is a field with $\operatorname{char}K\not\in\{2,3\}$,
has Galois group $S_4$ if


*

*it is irreducible over $K$
(thus its Galois group has order divisible by $4$), and

*its resolvent cubic
is irreducible over $K$
(thus the Galois group has order divisible by $3$), and

*its discriminant
is not a square in $K$ (thus the Galois group is not a subgroup of $A_4$).


Now consider
$$f(X) = X^4 - 6 X^2 + 2 X + 2$$
which changes sign between $-\infty,-1,0,+1,+\infty$
and therefore has four real roots.
$f(X)$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion
with $p=2$.
Therefore its Galois group has order divisible by $4$.
The resolvent cubic of $f(X)$ is
$$\begin{align}
g(X) &= X^3 + 6 X^2 - 8 X - 52 &
g(X-2) &= X^3 - 20 X - 20
\end{align}$$
so $g(X-2)$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion
with $p=5$, thus $g(X)$ is irreducible over $\mathbb{Q}$ as well.
The extension degree of the splitting field of $g(X)$ over $\mathbb{Q}$
is therefore divisible by $3$.
By construction of the resolvent cubic,
the splitting field of $g(X)$ is contained in the splitting field of $f(X)$,
so the order of the Galois group of $f(X)$ must be divisible by $3$.
Again by construction, the discriminant of $f(X)$
equals the discriminant of $g(X)$ and of $g(X-2)$, which is
$$\Delta=21200$$
You easily recognize the square $100$ as a factor,
but the cofactor $212\equiv2\pmod{5}$ is not a square.
Therefore $\Delta$ is not a square in $\mathbb{Q}$.
Consequently, the Galois group of $f(X)$ is not a subgroup of $A_4$.
Summarizing the above results, the Galois group of $f(X)$ has order
divisible by $3$ and $4$, yet is not a subgroup of $A_4$.
This leaves $S_4$ as only possible Galois group for $f(X)$.
