Galois group of $x^4+x^3+1\in\mathbb Q[x]$

Question

I am interested in the Galois group of $f(x)=x^4+x^3+1\in\mathbb Q[x]$. We have that $f(x)>0$ for all $x\in\mathbb Q$, so the roots of $f$ are two pairs of conjugates that all lie in $\mathbb C\backslash\mathbb Q$, namely $\alpha,\overline\alpha$ and $\beta,\overline\beta$. But what is the splitting field? And of which degree is it? I basically don't know anything about the roots to form any kind of statement here.

Answer 1

Consider the resolvent cubic, the monic polynomial with roots $\alpha\bar{\alpha} + \beta\bar{\beta}, \alpha\bar{\beta} + \beta\bar{\alpha}, \alpha\beta + \bar{\alpha}\bar{\beta}$. In general, according to Wikipedia, the resolvent cubic of $x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$ is $$y^3-a_2y^2+(a_1a_3-4a_0)y+4a_0a_2-{a_1}^2-a_0{a_3}^2.$$ (n.b. the symmetric functions theorem guarantees a priori that the coefficients of the resolvent cubic can be written in terms of the coefficients of its quartic.)

The resolvent cubic always splits over the splitting field of its quartic, so determining its (ir)reducibility gives information about the splitting field. In this case your resolvent cubic is $y^3 -4y - 1$, which is irreducible over $\mathbb{Q}$ since it has no rational roots. Hence, the degree of your splitting field is either $12$ or $24$, corresponding to Galois groups of $A_4$ or $S_4$.

Now recall that the Galois group of an irreducible rational polynomial is contained in the alternating permutation subgroup on the roots if and only if the discriminant is a square. The discriminant of $x^4 + bx^3 + e$ is (again according to Wikipedia) $$ 256e^3 -27b^4e^2,$$ so in your case the discriminant is $256-27 = 229$, which is not a square. Hence, the Galois group of $x^4 + x^3 + 1$ over $\mathbb{Q}$ is $S_4$, and the splitting field over $\mathbb{Q}$ has degree 24.