Let \(K\) be the splitting field of \(x^3 + 2\) over \(\mathbb{Q}\).
What is the dimension of \(K\) over \(\mathbb{Q}\)?
What is the Galois group of \(K\) over \(\mathbb{Q}\)? (Identify the isomorphism type of the group, and also indicate how its elements act as permutations of the roots of \(x^3 + 2\).)
Show that \(K\) contains a primitive third root of unity \(\zeta_3\). What is the Galois group of \(K\) over \(\mathbb{Q}(\zeta_3)\)? What is the Galois group of \(\mathbb{Q}(\zeta_3)\) over \(\mathbb{Q}\)? Again identify these groups as permutation groups and indicate their isomorphism type.
How are the three groups discussed in this problem related?
Solution
Let \(a=-\sqrt[3]{2}\) be the real root of \(x^3+2\), and choose \(\zeta=e^{2\pi i/3}\). Label the three roots by \(a_j=\zeta^ja\), with indices in \(\mathbb Z/3\mathbb Z\).
1. The splitting field is \(K=\mathbb Q(a,\zeta)\), and\([K:\mathbb Q]=6\).
The roots \(a,\zeta a,\zeta^2a\) lie in \(\mathbb Q(a,\zeta)\). Conversely, the field generated by them contains \(a\) and \(\zeta=(\zeta a)/a\), so it is precisely \(K\). Eisenstein’s criterion at \(2\) makes \(x^3+2\) irreducible over \(\mathbb Q\), giving \([\mathbb Q(a):\mathbb Q]=3\) [@DF04]. The field \(\mathbb Q(a)\) is real, whereas \(\zeta\) is nonreal and satisfies \(\zeta^2+\zeta+1=0\). Thus adjoining \(\zeta\) has degree \(2\), and the tower law gives \([K:\mathbb Q]=6\). As a splitting field in characteristic zero, this is a Galois extension [@DF04].
2. The full Galois group is \(S_3\) acting on the three roots,with generators\[ r(a)=\zeta a,\quad r(\zeta)=\zeta, \qquad s(a)=a,\quad s(\zeta)=\zeta^{-1}. \]
The polynomial \(x^2+x+1\) is the minimal polynomial of \(\zeta\) over \(\mathbb Q\), since its roots are nonreal. The tower law therefore gives \([K:\mathbb Q(\zeta)]=3\). Hence \(x^3+2\) remains the minimal polynomial of \(a\) over \(\mathbb Q(\zeta)\). Sending \(a\) to \(\zeta a\) defines an embedding of \(K\) into itself fixing \(\zeta\). Its image contains \(a\) as well, so it is the automorphism \(r\). Complex conjugation restricts to \(s\).
Their actions on the root indices are \[ r:j\longmapsto j+1,\qquad s:j\longmapsto-j. \] Thus \(r\) is the cycle \((0\,1\,2)\) and \(s\) the transposition \((1\,2)\). Their six products \(r^us^v\), with \(0\leq u<3\) and \(v\in\{0,1\}\), act by \(j\mapsto u+(-1)^vj\) and are distinct permutations. They exhaust the Galois group, whose order is \(6\). Equivalently, these are all permutations of the three roots, since the roots generate \(K\) and the action is faithful. The relations are \(r^3=s^2=1\) and \(srs=r^{-1}\).
3. The other two groups are \(C_3\) and \(C_2\), linked to thefull group by a split exact sequence.
An automorphism \(r^us^v\) fixes \(\zeta\) exactly when \(v=0\). Consequently \[ \operatorname{Gal}(K/\mathbb Q(\zeta))=\langle r\rangle \cong C_3, \] acting on \(a_0,a_1,a_2\) by cyclic rotations. The extension \(\mathbb Q(\zeta)/\mathbb Q\) is the splitting field of \(x^2+x+1\) and has degree \(2\). Its nonidentity automorphism \(\bar s\) interchanges the roots \(\zeta,\zeta^2\). Thus its Galois group is the full two-letter permutation group \(S_2\cong C_2\).
Restriction to \(\mathbb Q(\zeta)\) sends \(r\) to the identity and \(s\) to \(\bar s\), so is onto and has kernel \(\langle r\rangle\). This gives \[ 1\longrightarrow\operatorname{Gal}(K/\mathbb Q(\zeta)) \longrightarrow\operatorname{Gal}(K/\mathbb Q) \xrightarrow{\mathrm{res}} \operatorname{Gal}(\mathbb Q(\zeta)/\mathbb Q) \longrightarrow1. \] The map sending \(\bar s\) to \(s\) is a homomorphic section, because both have order \(2\). Thus \(G\cong C_3\rtimes C_2\), where the nonidentity element of \(C_2\) acts on \(C_3\) by inversion, as the relation \(srs=r^{-1}\) explicitly shows. This is not a direct product, since \(r\ne r^{-1}\).