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Find the Galois Group of $\mathbb{Q}(2^{1/3},5^{1/3},\zeta_{3})/\mathbb{Q}$, for $\zeta_{3}$ being a third primitive root of unity.


It's easy to show this is a Galois extension since it will be the splitting field of some separable polynomial (and one will compute that the degree of the extension is 18 as well). We know as well that $\mathbb{Q}(2^{1/3},\zeta_{3})$ and $\mathbb{Q}(5^{1/3},\zeta_{3})$ are Galois extensions over $\mathbb{Q}$, where the intersection is $\mathbb{Q}(\zeta_{3})$

Then, a characterization of the Galois group of the compositum would be:

$Gal(\mathbb{Q}(2^{1/3},5^{1/3},\zeta_{3})/\mathbb{Q})=\{(\tau,\sigma) \in Gal(\mathbb{Q}(5^{1/3},\zeta_{3})) \times Gal(\mathbb{Q}(5^{1/3},\zeta_{3})): \tau_{\mathbb{Q}(\zeta_{3})} =\sigma_{\mathbb{Q}(\zeta_{3})} \}$

However, I have no idea on how to use the characterization of the compositum in this manner to find explicitly what's the Galois group. It seems like it would be very computational (taking isomorphisms and seeing if they satisy the restriction conditions). What would be a more elegant or simpler way to solve this problem? I'm very curious.

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  • $\begingroup$ $P(x) = (1+x+x^2)(x^3-5)(x^3-2)$ and trying to make a minimal polynomial ? $\endgroup$ – reuns Apr 14 '16 at 20:39
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If you look at the subfield $\mathbb{Q}(\zeta_3,2^{1/3})$, this is Galois over $\mathbb{Q}$ with Galois group $S_3$ generated by $\sigma: \zeta_3 \mapsto \zeta_3^2$ and $\tau: 2^{1/3} \mapsto \zeta 2^{1/3}$ (I've omitted the elements that they fix).

Similarly, we have the subfield $\mathbb{Q}(\zeta_3,5^{1/3})$ which comes with a Galois element $\rho: 5^{1/3} \mapsto \zeta 5^{1/3}$.

Taking the compositum, we find the Galois group is generated by these three elements as this gives a group of order 18, where we take $\tau$ to fix $5^{1/3}$ and $\rho$ to fix $2^{1/3}$; these are all certainly automorphisms.

Checking their actions, we find that $\tau$ and $\rho$ commute giving us a presentation \begin{equation*} G= \langle \sigma, \tau, \rho | \sigma^2, \tau^3, \rho^3, \sigma\tau\sigma^{-1}=\tau^{-1}, \sigma\rho\sigma^{-1}=\rho^{-1}, \rho\tau=\tau\rho \rangle. \end{equation*}

As a final check, we can check that $|G|=18$ and therefore we have the full Galois group.

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Allowing a bit more theory, one could give a more « sleeky » proof in a more general situation. For an odd prime $p$ , let $K = \mathbf Q(\zeta_p)$, which is Galois over $\mathbf Q$ with group $\mathbf Z/(p-1)$. Consider a subgroup $A$ of $\mathbf Q^*/ \mathbf Q^{*p}$ generated by $n$ independent classes $a_i$ mod $\mathbf Q^{*p}$, $a_i \in \mathbf Q^*$ ; in additive notation, this means that $A$ is a $\mathbf Z/p$ - sub-vector space of $\mathbf Q^*/ \mathbf Q^{*p}$ with a basis consisting of the classes $a_i$ mod $\mathbf Q^{*p}$. Because $K$ has degree $(p-1)$ over $\mathbf Q$ and $p$ and $(p-1)$ are coprime, the natural map $\mathbf Q^*/ \mathbf Q^{*p}$ --> $ K^*/K^{*p}$ is injective, and $A$ can also be viewed as a sub-vector space of $ K^*/K^{*p}$ of dimension $n$. Let $L = K (… \sqrt [p] {a_i}…)$, which is obviously the normal closure of $k =\mathbf Q (…\sqrt [p] {a_i}…)$.

Write $G = Gal (L/\mathbf Q)$ and $H = Gal(L/K)$. The problem is to describe $G$ as an extension of $G/H$ by $H$. By Kummer theory, $H \cong Hom (A , <\zeta_p >)$, hence is isomorphic to $(\mathbf Z/p)^{n}$ as a group. Since $p$ and $(p-1)$ are coprime, the above extension of groups is split (this is a theorem of Frobenius, I believe), i.e. $G$ is the semi-direct product of $\mathbf Z/(p-1)$ (identified with $Gal(L/k)$) and $(\mathbf Z/p)^{n}$. In particular, this answers your question for $p = 3$ and $n=2$ .

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