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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?

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Following @user64687’s hint in the comments,

thus according to 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 $A_4$).

In example 3.2, Keith Conrad shows that the Galois group of

$f(X) = X^4 – X – 1$ over $\mathbb{Q}$ is $S_4$.

Let me demonstrate the very similar case

$$f(X) = X^4 + X + 1$$

instead.

Since $f(X) – (X^{2^2}-X) = 1$ in $\mathbb{F}_2[X]$,

$f(X)$ is coprime to every irreducible polynomial over $\mathbb{F}_2$

of degree dividing $2$, this implies that $f(X)$ is irreducible $\bmod{2}$,

thus $f(X)$ is irreducible over $\mathbb{Q}$.

Therefore its Galois group has order divisible by $4$.

$f(X)$ has the resolvent cubic

$$g(X) = X^3 – 4 X – 1$$

which, by its low degree and the rational root theorem,

would need to have an integer root dividing $1$ if it were reducible

over $\mathbb{Q}$, but it hasn’t, so $g(X)$ is irreducible over $\mathbb{Q}$.

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 $g(X)$, which is

$$\Delta=229$$

and is thus easily recognizable as a nonsquare in $\mathbb{Q}$.

In particular, it’s congruent to the nonsquare $5\pmod{8}$.

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 $f(X)$.

If you are looking for a quartic polynomial with only real roots,

see here;

the argumentation there is almost the same.

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 $x^3 – 3$.

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