Let \[ L=\QQ(\sqrt[3]2,\sqrt3,\zeta_3), \] where \(\zeta_3\) is a primitive cube root of unity. Compute \(\operatorname{Gal}(L/\QQ)\) and describe the intermediate fields, at least up to conjugacy.
Solution
Set \[ a=\sqrt[3]2, \qquad b=\sqrt3, \qquad \omega=\zeta_3. \] The splitting field of \(x^3-2\) is \(\QQ(a,\omega)\) and has degree \(6\) over \(\QQ\). The element \(b\) does not lie in this field: the only quadratic subfield of the \(S_3\)-extension \(\QQ(a,\omega)/\QQ\) is \(\QQ(\sqrt{-3})\), whereas \(\QQ(\sqrt3)\) is different. Hence \[ [L:\QQ]=12. \]
Define automorphisms \[ \sigma:a\mapsto\omega a,\quad \omega\mapsto\omega,\quad b\mapsto b, \] \[ \tau:a\mapsto a,\quad \omega\mapsto\omega^2,\quad b\mapsto b, \] and \[ \gamma:a\mapsto a,\quad \omega\mapsto\omega,\quad b\mapsto-b. \] Then \[ \sigma^3=\tau^2=\gamma^2=1, \qquad \tau\sigma\tau=\sigma^{-1}, \] and \(\gamma\) commutes with both \(\sigma\) and \(\tau\). Therefore \[ \operatorname{Gal}(L/\QQ) \cong S_3\times C_2, \] which is also the dihedral group of order \(12\) (often denoted \(D_6\)).
The subgroup/fixed-field correspondence gives the intermediate fields. Representatives of the conjugacy classes of proper nontrivial subgroups are:
\[ \begin{array}{c|c} H & L^H\\ \hline \langle\sigma\rangle & \QQ(\omega,b)\\ \langle\tau\rangle & \QQ(a,b)\\ \langle\gamma\rangle & \QQ(a,\omega)\\ \langle\tau\gamma\rangle & \QQ(a,i)\\ \langle\tau,\gamma\rangle & \QQ(a)\\ \langle\sigma,\tau\rangle & \QQ(b)\\ \langle\sigma,\gamma\rangle & \QQ(\omega)\\ \langle\sigma,\tau\gamma\rangle & \QQ(i) \end{array} \]
Here \[ i=\frac{\omega-\omega^2}{b}, \] which is fixed by \(\tau\gamma\).
The conjugates of \(\langle\tau\rangle\), \(\langle\tau\gamma\rangle\), and \(\langle\tau,\gamma\rangle\) give the remaining conjugate intermediate fields: \[ \QQ(\omega^j a,b), \qquad \QQ(\omega^j a,i), \qquad \QQ(\omega^j a) \quad(j=0,1,2). \] Together with \(\QQ\) and \(L\), these account for the full subgroup lattice under the Galois correspondence.