Bourbaki's Algebra Chapter V exercise $1$ of paragraph 2 asks to show that there are infinitely many non-zero integers $a$ such that the field $\mathbb{Q}(\alpha)$ has no intermediate subfield, where $\alpha$ is any root of $T^4 – aT – 1$.
I can factor $T^4 – aT – 1$ to $(T – \alpha)(T^3 + \alpha T^2 + \alpha^2 T + \alpha^3 – a)$ in $\mathbb{Q}(\alpha)[T]$, and I know that there are no intermediate subfields if the polynomial $P(T) = T^3 + \alpha T^2 + \alpha^2 T + \alpha^3 – a$ has no roots in $\mathbb{Q}(\alpha)$. So I want to show there are infinitely many $a$ such that this polynomial has no roots.
I'm stuck at this point. I know no easy criterion to ensure that a cubic has no roots. I've tried messing around a bit with $P$ with the hope to find something more tractable. After a change of variable, $P$ can be put to normal form $T^3 + \frac{2}{3}\alpha^2T + \frac{20}{27}\alpha^3 – a$, which has discriminant $24a\alpha^3 – 16\alpha^2 – 27a^2$. Unfortunately, I do not kow how to link the discriminant to the existence of at least one root (I know that if it is a square, then $P$ may have roots, but I do not know if the converse holds).
Any hint or help would be appreciated. The exercise being in the early part of the chapter, the solution is supposed to be quite elementary, but I'm fine with more involved methods if that's what it takes to prove this.
Edit:
I had Sage brutally compute the number of subfields of $\mathbb{Q}(\alpha)$ for a few thousands of values of $a$ and it seems that the only one for which $T^3 + \alpha T^2 + \alpha^2T + \alpha^3 – a$ splits in $\mathbb{Q}(\alpha)$ is $a = \pm 4$. This gives at least a hint of what the answer should be, yet I still don't see how to attain this answer.
Best Answer
A quick look told me that this happens at least whenever $a\equiv1\pmod{14}$. I am using Dedekind's theorem relating factorization modulo primes to cycle structure of permutations in the Galois group $G$ (here a subgroup of $S_4$)