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I suspect that the field $\mathbb{Q}(\sqrt{1+\sqrt[3]{2}})$ does not contain any subfield $K$ with $[K:\mathbb{Q}]=2$, but I'm not sure how to prove it.

More generally, if $L$ denotes the splitting field of $x^6 -3x^4 + 3x^2 -3$ (the minimal polynomial of $\sqrt{1+\sqrt[3]{2}}$), can we identify $\text{Gal}(L/\mathbb{Q})$?

In response to some of the comments: The roots of $x^6 -3x^4 + 3x^2 -3$ are $\pm \sqrt{1+\zeta_3^i \sqrt[3]{2}}$, where $\zeta_3$ is a primitive cube root of unity and $0\leq i \leq 2$. Using this it is easy to show that $[L:\mathbb{Q}]$ is either 12, 24, or 48.

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  • $\begingroup$ Are you saying that $\mathbb{Q}(\sqrt{1+\sqrt[3]{2}})$ is the splitting field of that polynomial? Seems like this field is real and the polynomial has only 2 real roots . $\endgroup$
    – Myself
    Mar 15, 2017 at 18:57
  • $\begingroup$ @Myself No, I am certainly not suggesting that. $\endgroup$
    – Mr. Frog
    Mar 15, 2017 at 18:58
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    $\begingroup$ And do you know the degree of the splitting field? That seems like a good start to determine the Galois group! $\endgroup$
    – Myself
    Mar 15, 2017 at 19:00
  • $\begingroup$ @Myself No, I do not know the degree. It is divisible by 12, and at most 48. $\endgroup$
    – Mr. Frog
    Mar 15, 2017 at 19:07

3 Answers 3

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With the aid of Dedekind's theorem we can conclude that $[L:\Bbb{Q}]=48$. The idea is to view the Galois group as acting on the set $X$ of the six zeros of $$ f(x)=x^6-3x^4+3x^2-3. $$ Factorization of $f(x)$ modulo distinct primes (excluding those that give rise to factors of multiplicity $>1$, among others the ramified primes) then gives cycle structures of Frobenius elements of $G=Gal(L/\Bbb{Q})$.

  • Modulo $5$ we get $f(x)=(x+2)(x-2)(x^4+x^2+2)$. Therefore $G$ contains a 4-cycle.
  • Modulo $31$ we get $f(x)=(x+6)(x-6)(x+15)(x-15)(x^2+10)$. Therefore $G$ contains a 2-cycle.
  • $G$ acts transitively on $S$. By the first bullet the stabilizer of at least one of the zeros contains a 4-cycle. All the point stabilizers are conjugate, so we can deduce that $Stab_G(a),a=\sqrt{1+\root3\of2},$ contains a 4-cycle also. Call it $\sigma$. Similarly, the second bullet implies that $Stab_G(a)$ contains a 2-cycle, call it $\alpha$. Because $\sigma^2$ is the product of two disjoint 2-cycles we know that $\alpha\neq\sigma^2$.
  • The previous bullet implies that $|Stab_G(a)|\ge8$. By the orbit-stabilizer theorem $|G|\ge48$.
  • Others already concluded that $|G|\le48$ (this follows from the fact that the six zeros come in three pairs that are negatives of each other). Therefore $|G|=48$.

The consideration in the previous bullet now shows that $G\simeq C_2\wr S_3$, or if you are unfamiliar with wreath products, $$ G\simeq C_2^3\rtimes S_3 $$ with $S_3$ permuting the three components of $C_2^3$ (one for each pair of zeros). I think that the only subgroups of index two are the subgroup $H_1=C_2^3\rtimes A_3$ and the subgroup $H_2=V\rtimes S_3$, where $V$ is the zero sum subgroup of $C_2^3$ (this is stable under the action of $S_3$). Because neither contains the point stabilizer ($Stab_G(x)$ looks like $(1\times C_2\times C_2)\rtimes C_2\simeq D_4$), there won't be a quadratic subfield of $K$.


How to? Confession time! I just crunched the primes up to $p=31$ with Mathematica. Only then it dawned on me that $p=31$ is an obviously good candidate for finding low degree factors. After all, $2^{10}-1=(2^5-1)(2^5+1)$ is divisible $31$, so $2$ has a cubic root modulo $31$. Also, $6\mid (31-1)$, so the sixth roots of unity will be there. With a bit of luck some of the numbers $1+\zeta_3^i\root3\of2$ turn out to be quadratic residues as well!

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  • $\begingroup$ It may be possible to figure this out without Dedekind's theorem. It is just my go to -technique nowadays. $\endgroup$ Mar 15, 2017 at 20:15
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The given extension is only ramified at $ 2 $ and $ 3 $, and it is totally ramified at $ 3 $ (for example, by Eisenstein). Thus, the only possible candidates for quadratic subfields are $ \mathbf Q(\sqrt{3}) $ and $ \mathbf Q(\sqrt{6}) $. However, the prime $ 17 $ splits as $ \mathfrak p_1 \mathfrak p_2 \mathfrak p_3 $, with inertia degrees $ f_1 = f_2 = 1 $ and $ f_3 = 4 $. Neither $ 3 $ nor $ 6 $ are quadratic residues modulo $ 17 $, thus $ 17 $ remains inert in both of these quadratic extensions; therefore the prime factors of $ 17 $ should all have even inertia degree. This is a contradiction, thus the extension has no quadratic subfield.

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  • $\begingroup$ Nice! ${}{}{}{}$ $\endgroup$ Mar 15, 2017 at 20:20
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An idea for you to think:

Put

$$x=\sqrt{1+\sqrt[3]2}\implies x^2-1=\sqrt[3]2\implies f(x):=x^6-3x^4+3x^2-3=0$$

Using Eisenstein's Test we can see the above polynomial is irreducible over the rationals, and thus $\;[\Bbb Q(\sqrt{1+\sqrt[3]2}):\Bbb Q]=6\;$ .

Observe $\;\pm\sqrt{1+\sqrt[3]2}\;$ are roots of $\;f(x)\;$ so you can divide by $\;(x-\sqrt{1+\sqrt[3]2})(x+\sqrt{1+\sqrt[3]2})\;$ and get a quartic.

Anyway, since we have at least two roots in the same extension, the splitting field has degree at most $\;\frac{6!}2\;$ ... But the quartic seems to be also an even one (check this!), so we can divide furthermore the degree (again, two roots ($\;\pm\,$ something) in the same extension), and it seems to be reasonable to expect the degree of the splitting field will something like $\;24\;$ or perhaps $\;48\;$ . Try to do something with this (disclaimer: I don't know the final answer) .

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