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Given the polynomial $f(T) = (T^4-3)(T^6-3)$, I would like to calculate the Galois group of $f$. What I've done is the following: setting $\alpha = 3^{1/4}$ and $\beta= 3^{1/6}$, $\zeta_k = e^{2\pi i / k}$, since the decomposition field of $f$ over $\mathbb Q$ is $L := \mathbb Q(\alpha,...,\alpha \zeta_4^3,\beta,...,\beta\zeta_6^5) = \mathbb Q(\alpha,\beta,\zeta_4,\zeta_6) = \mathbb Q(3^{1/3},3^{1/4},i)$ since $\zeta_6 = (1+i(3^{1/4})^2)/2$ and $3^{1/6}=\frac{1}{3^4}(3^{1/3}(3^{1/4})^2)^5$ and then: $$ \mathbb Q \subset \mathbb Q(3^{1/4}) \subset \mathbb Q(3^{1/4},3^{1/3})\subset \mathbb Q(3^{1/4},3^{1/3},i) = L $$ with $$ [L:\mathbb Q(3^{1/3},3^{1/4})] = 2,\quad [\mathbb Q(3^{1/3},3^{1/4}):\mathbb Q(3^{1/4})] = 3, \quad [\mathbb Q(3^{1/4}):\mathbb Q]=4. $$ The first is clear to me, since the first field contains complex numbers, the second is the complicated one and the last is because of $\deg(T^4-3)$, which is the minimum polynomial $\mathbb Q$. So, $[L:\mathbb Q] = 24$. But I don't know how to calculate the Galois group. Any hint? Thanks in advance

Edit: I'll try the following. If $\phi\in Gal(L:\mathbb Q)$, $\phi$ is completely determined by $\phi(i)$, $\phi(3^{1/3})$ and $\phi(3^{1/4})$. For the first, $(\phi(i))^2=\phi(i^2) = \phi(-1) = -1$, so $\phi(i) = \pm i$. For the second, $3 = \phi((3^{1/3})^3) = \phi(3^{1/3})^3$ and necessarily $\phi(3^{1/3}) = \zeta_3^{k} 3^{k/3}$ for some $k \in \{0,...,2\}$. This gives $Gal(L:\mathbb Q)\cong \mathbb Z_2\times \mathbb Z_4\times \mathbb Z_3$?

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I think your answer for the Galois group is correct.

Note that we can also express $L = \mathbb{Q}(\sqrt[12]{3}, \zeta_{12})$ where $\zeta_{12}$ is a primitive $12$th root of unity. enter image description here

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  • $\begingroup$ Thanks, but I would appreciate a hint in order to determine the Galois group $\endgroup$ – user55268 May 19 '15 at 22:19
  • $\begingroup$ I've updated my answer. $\endgroup$ – André 3000 May 19 '15 at 23:29

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