Let \(K\) be the splitting field of \(f(x)=x^3 - 5\) 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 Galois group, and also indicate how its elements act as permutations of the roots of \(f\).)
What are all the subfields of \(K\)? Be sure to explain how you know you have included all of them.
Which of these subfields are Galois over \(\mathbb{Q}\)?
Solution
Put \(a=\sqrt[3]{5}>0\), choose \(\zeta=e^{2\pi i/3}\), and write \(a_j=\zeta^ja\) for \(j\in\mathbb Z/3\mathbb Z\).
1. One has \(K=\mathbb Q(a,\zeta)\) and \([K:\mathbb Q]=6\).
The roots are \(a_0,a_1,a_2\). The field they generate contains \(a=a_0\) and \(\zeta=a_1/a_0\), and conversely all the roots belong to \(\mathbb Q(a,\zeta)\). This proves the field identity. The polynomial \(x^3-5\) is Eisenstein at \(5\), hence irreducible over \(\mathbb Q\) [@DF04]. Thus \([\mathbb Q(a):\mathbb Q]=3\). Since \(\mathbb Q(a)\subset\mathbb R\) and \(\zeta\) is nonreal with \(\zeta^2+\zeta+1=0\), adjoining \(\zeta\) has degree two. The tower law gives the asserted degree six. The polynomial has distinct roots, and its splitting field is a finite Galois extension [@DF04].
2. The Galois group is \(S_3\), acting as all permutationsof \(a_0,a_1,a_2\).
An automorphism permutes the roots; that action is faithful because they generate \(K\). Since the Galois group has order \([K:\mathbb Q]=6=|S_3|\), its image is all of \(S_3\). Explicit generators are \[ r(a)=\zeta a,\quad r(\zeta)=\zeta, \qquad s(a)=a,\quad s(\zeta)=\zeta^{-1}. \] To verify existence of \(r\), first note that \([\mathbb Q(\zeta):\mathbb Q]=2\), since \(x^2+x+1\) has no rational root. Therefore \([K:\mathbb Q(\zeta)]=3\), and \(x^3-5\) is the minimal polynomial of \(a\) over \(\mathbb Q(\zeta)\). Sending \(a\) to its root \(\zeta a\) defines an embedding fixing that field; its image contains \(a=\zeta^{-1}(\zeta a)\), so the embedding is onto. Complex conjugation restricts to the automorphism \(s\).
On indices, \(r\) is \(j\mapsto j+1\) and \(s\) is \(j\mapsto-j\). Thus \(r=(0\,1\,2)\) and \(s=(1\,2)\), and the six maps \(r^us^v\) act by \(j\mapsto u+(-1)^v j\), with \(u\in\mathbb Z/3\mathbb Z\) and \(v\in\{0,1\}\). These specify every automorphism and its root permutation.
3. The complete subfield list is\[ \mathbb Q,\qquad \mathbb Q(\zeta)=\mathbb Q(\sqrt{-3}), \qquad \mathbb Q(a_0),\quad\mathbb Q(a_1),\quad\mathbb Q(a_2), \qquad K. \]Their degrees over \(\mathbb Q\) are respectively \(1,2,3,3,3,6\).
By the Galois correspondence, intermediate fields correspond bijectively to subgroups \(H\leq S_3\), and the fixed field has degree \([S_3:H]\) [@DF04]. Lagrange’s theorem restricts subgroup orders to \(1,2,3,6\). There is just the identity subgroup of order one and the whole group of order six. A subgroup of order two is generated by an involution; the three transpositions are exactly the involutions of \(S_3\). There is exactly one subgroup of order three, namely \(\langle r\rangle=A_3\), because the only elements of order three are \(r\) and \(r^2\), which generate the same subgroup. This proves that the six subgroups just described are exhaustive.
The whole group and identity subgroup fix \(\mathbb Q\) and \(K\). The subgroup \(\langle r\rangle\) fixes \(\zeta\), and its fixed field has degree two, so it is precisely \(\mathbb Q(\zeta)\). The formula \(\zeta=(-1+\sqrt{-3})/2\) identifies that field. For each \(j\), the stabilizer of \(a_j\) is the order-two subgroup generated by the transposition of the other two roots. Its fixed field contains \(\mathbb Q(a_j)\). Each \(a_j\) has minimal polynomial \(x^3-5\), and both this field and the fixed field have degree three. They are therefore equal. The three stabilizers are different, so the three cubic fields are different. Every subfield of \(K\) contains its prime field \(\mathbb Q\), so no additional subfields lie outside this correspondence.
4. Exactly \(\mathbb Q\), \(\mathbb Q(\zeta)\), and \(K\)are Galois over \(\mathbb Q\).
A fixed field \(K^H\) is Galois over \(\mathbb Q\) exactly when \(H\) is normal in \(S_3\) [@DF04]. The identity subgroup and the whole group are normal. The subgroup \(A_3\) has index two and is normal, giving the quadratic Galois field. Conjugation by \(r\) cyclically permutes the three transpositions, so none of their order-two subgroups is normal. Thus none of the three cubic fields is Galois, and the exhaustive list in step 3 proves that these decisions cover every subfield.