Determine all subgroups of the Galois group and all intermediate fields of the splitting (over \(\mathbb{Q}\)) of the polynomial \((x^{3}-2)(x^{2}-3)\in \mathbb{Q}[x]\).
Solution
Put \[ a=\sqrt[3]{2},\qquad \zeta=e^{2\pi i/3},\qquad b=\sqrt3. \] The splitting field is \[ L=\QQ(a,\zeta,b). \]
1. One has\[ \operatorname{Gal}(L/\QQ)\cong S_3\times C_2. \]
Let \[ M=\QQ(a,\zeta), \qquad N=\QQ(b). \] The field \(M\) is the splitting field of \(x^3-2\) and \[ \operatorname{Gal}(M/\QQ)\cong S_3. \] Its unique quadratic subfield is the fixed field of \(A_3\), namely \[ \QQ(\sqrt{-3})=\QQ(\zeta). \] Since \(N=\QQ(\sqrt3)\) is a different quadratic field, \[ M\cap N=\QQ. \] Both extensions are Galois, so their compositum satisfies \[ \operatorname{Gal}(L/\QQ) \cong\operatorname{Gal}(M/\QQ)\times\operatorname{Gal}(N/\QQ) \cong S_3\times C_2. \]
Choose generators \(\rho,\tau,\epsilon\) by \[ \rho(a)=\zeta a,\quad \rho(\zeta)=\zeta,\quad \rho(b)=b, \] \[ \tau(a)=a,\quad \tau(\zeta)=\zeta^2,\quad \tau(b)=b, \] and \[ \epsilon(a)=a,\quad \epsilon(\zeta)=\zeta,\quad \epsilon(b)=-b. \] Then \[ \rho^3=\tau^2=\epsilon^2=1, \qquad \tau\rho\tau=\rho^{-1}, \qquad \epsilon\rho=\rho\epsilon, \qquad \epsilon\tau=\tau\epsilon. \] For \(j=0,1,2\), set \[ a_j=\zeta^j a, \qquad \tau_j=\rho^j\tau\rho^{-j}. \] Thus \(\tau_j\) is the transposition fixing \(a_j\).
Also put \[ i=\frac{\sqrt{-3}}{\sqrt3}\in L. \]
2. The complete subgroup list of \(G=S_3\times C_2\) is\[ \begin{array}{c|c|c} |H|&H&L^H\\ \hline 1&\{1\}&L\\[2mm] 2&\langle\epsilon\rangle&\QQ(a,\zeta)\\ 2&\langle\tau_j\rangle&\QQ(a_j,b)\quad(j=0,1,2)\\ 2&\langle\tau_j\epsilon\rangle&\QQ(a_j,i)\quad(j=0,1,2)\\[1mm] 3&\langle\rho\rangle&\QQ(\zeta,b)\\[1mm] 4&\langle\tau_j,\epsilon\rangle&\QQ(a_j)\quad(j=0,1,2)\\[1mm] 6&\langle\rho,\tau\rangle&\QQ(b)\\ 6&\langle\rho,\epsilon\rangle&\QQ(\zeta)\\ 6&\langle\rho,\tau\epsilon\rangle&\QQ(i)\\[1mm] 12&G&\QQ. \end{array} \]
We first verify that no subgroups are missing.
The involutions of \(G\) are \[ \epsilon,\quad \tau_0,\tau_1,\tau_2, \quad \tau_0\epsilon,\tau_1\epsilon,\tau_2\epsilon, \] so there are exactly seven subgroups of order \(2\).
The only elements of order \(3\) are \(\rho\) and \(\rho^2\), giving the unique subgroup \(\langle\rho\rangle\) of order \(3\).
If \(H\) has order \(4\), its projection to \(S_3\) has order \(2\), so the kernel of that projection on \(H\) has order \(2\) and equals \(\langle\epsilon\rangle\). Therefore \[ H=\langle\tau_j,\epsilon\rangle \] for exactly one \(j\), giving the three subgroups of order \(4\).
Every subgroup of order \(6\) has index \(2\), hence is the kernel of a nontrivial homomorphism \(G\to C_2\). Since \[ G_{\mathrm{ab}}\cong C_2\times C_2, \] there are exactly three such kernels, namely \[ \langle\rho,\tau\rangle, \qquad \langle\rho,\epsilon\rangle, \qquad \langle\rho,\tau\epsilon\rangle. \] Together with the trivial subgroup and \(G\), this is the full subgroup list by Lagrange’s theorem.
It remains to verify the fixed fields in the table. The subgroup \(\langle\epsilon\rangle\) changes only \(b\), so its fixed field is \(M\). The involution \(\tau_j\) fixes \(a_j\) and \(b\), and the field \(\QQ(a_j,b)\) has degree \(3\cdot2=6=[G:\langle\tau_j\rangle]\), hence is its full fixed field.
Every transposition \(\tau_j\) changes \(\sqrt{-3}\) to its negative, while \(\epsilon\) changes \(b=\sqrt3\) to its negative. Consequently \(\tau_j\epsilon\) fixes both \(a_j\) and \[ i=\frac{\sqrt{-3}}{\sqrt3}. \] The field \(\QQ(a_j,i)\) has degree \(6\), so it is exactly \(L^{\langle\tau_j\epsilon\rangle}\).
The subgroup \(\langle\rho\rangle=A_3\) fixes \(\zeta\) and \(b\), and \(\QQ(\zeta,b)\) has degree \(4\), so this is its fixed field. The subgroup \(\langle\tau_j,\epsilon\rangle\) fixes \(a_j\), and \([\QQ(a_j):\QQ]=3\) is its index, so \(\QQ(a_j)\) is its fixed field.
Finally, the three index-\(2\) subgroups correspond to the three quadratic characters of \(S_3\times C_2\). Their fixed fields are respectively \[ \QQ(b)=\QQ(\sqrt3), \qquad \QQ(\zeta)=\QQ(\sqrt{-3}), \qquad \QQ(i), \] the last because \(\tau\epsilon\) fixes the quotient \(\sqrt{-3}/\sqrt3=i\). The degree of each field equals the subgroup index, so each displayed inclusion is the full fixed field.
3. The table in 2 gives all intermediate fields of \(L/\QQ\).
1. The fundamental theorem ofGalois theory gives a bijection between its subgroups and its intermediatefields via \(H\mapsto L^H\). Since 2 lists every subgroup and identifies itsfixed field, it lists every intermediate field exactly once.