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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.

user26857
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    I think you are doing things backwards. An intermediate field would be a field $E$, $\mathbb{Q}\subseteq E\subseteq \mathbb{Q}(\alpha)$, they are not exclusively determined by other roots of $T^4-aT-1$ in $E$. (For instance, if you had the polynomial $x^4-2$, with $F=\mathbb{Q}(\sqrt[4]{2})$, then the only intermediate field is $\mathbb{Q}(\sqrt{2})$, and $\sqrt{2}$ is not a root of the original polynomial). You look more like you are trying to find fields intermediate between $F$ and the splitting field of $T^4-aT-1$... – Arturo Magidin May 02 '20 at 21:43
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    @ArturoMagidin My reasoning is that, if I have an intermediate field extension $\mathbb{Q} \subset E \subset \mathbb{Q}(\alpha)$, then I have two choices: Either $T^4 - aT - 1$ remains irreducible in $E[T]$, or either it splits into smaller degree polynomials in $E[T]$. The first case is impossible since in that case, $\alpha$ would be a root of a degree $4$ polynomial of $E$, and thus the degree of $[\mathbb{Q}(\alpha) : E]$ would be $4$, contradicting the fact that $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is already $4$. 1/2 –  May 02 '20 at 21:48
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    Now if $T$ splits into two degree $2$ polynomials, then $\alpha$ must be a root of one of them, and so the two degree $2$ polynomials will be completely split. –  May 02 '20 at 21:50
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    Okay, I see what you are trying to do. Fair enough... – Arturo Magidin May 02 '20 at 21:50
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    An entreaty from an aging mathematician with failing vision: Please, please, never use “$\alpha$” and “$a$” anywhere near each other. Preferably not in the same formula, and maybe not even in the same paragraph. There are so many letters to choose from! – Lubin Aug 30 '22 at 18:44

1 Answers1

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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 $S_4$)

  • Then your polynomial is $\equiv T^4+T+1\pmod2$. This is irreducible over $\Bbb{F}_2$, so $G$ contains a 4-cycle.
  • Modulo $p=7$ we get $\equiv x^4-x-1\equiv(x+4)(x^3+3x^2+2x+5)$ with the cubic factor irreducible. Dedekind tells us that $G$ contains a 3-cycle.
  • A subgroup of $S_4$ containing a 4-cycle and a 3-cycle is easily seen to be all of $S_4$.
  • The key is that a point stabilizer $H$ of $\alpha$ (= a copy of $S_3$) is a maximal subgroup of $G=S_4$. By Galois correspondence this implies that there are no intermediate fields between $\operatorname{Inv}(H)=\Bbb{Q}(\alpha)$ and $\operatorname{Inv}(G)=\Bbb{Q}.$
Jyrki Lahtonen
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  • Sorry about slapping off a quick answer rather than delving into your approach (which would be pedagogically more productive). Thankfully Arturo seems to be covering that side. – Jyrki Lahtonen May 02 '20 at 21:51
  • Thank you for the answer! I was not aware of Dedekind's theorem, so it gives an opportunity to learn something. I'll leave the question open in case some magical elementary answer exists, and will probably accept your answer after some time in the case I can't find anything elementary. –  May 02 '20 at 21:56
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    @RobinCarlier I wouldn't say Jyrki answer is based on a specific theorem but instead on the full Galois theory of number fields including the part about the surjection from a subgroup of $G$ to the Galois group of $O_K/\mathfrak{p}$ for a prime ideal (here $K$ is the splitting field of your polynomial with $a\equiv 1\bmod 14$) – reuns May 02 '20 at 23:43
  • But can't we get the fixed field of $A_3$, which will be a non-trivial intermediate field extension? – permutation_matrix Aug 30 '22 at 16:36
  • @permutation_matrix No. $\Bbb{Q}(\alpha)$ is the fixed field of (one copy of) $S_3$. The fixed field of $A_3$ would be a quadratic extension of $\Bbb{Q}(\alpha)$. There are no intermediate groups between $S_3$ and $S_4$ so there cannot be intermediate fields between $\Bbb{Q}$ and $\Bbb{Q}(\alpha)$. – Jyrki Lahtonen Aug 30 '22 at 17:30