Let \(K\) be the splitting field of \(x^3 - 7\) over \(\mathbb{Q}\).
Find the Galois group of \(K\) over \(\mathbb{Q}\).
Find all of the subfields of \(K\).
Which of the subfields you found above are Galois over \(\mathbb{Q}\)?
Solution
Let \(\alpha=\sqrt[3]{7}\) be the real cube root and let \(\zeta=(-1+i\sqrt3)/2\), a primitive cube root of unity. The answers are \(\operatorname{Gal}(K/\mathbb Q)\cong S_3\) and the following six fields:
| Subfield \(F\) | \([F:\mathbb Q]\) | Galois over \(\mathbb Q\)? |
|---|---|---|
| \(\mathbb Q\) | \(1\) | Yes |
| \(\mathbb Q(\zeta)\) | \(2\) | Yes |
| \(\mathbb Q(\alpha)\) | \(3\) | No |
| \(\mathbb Q(\zeta\alpha)\) | \(3\) | No |
| \(\mathbb Q(\zeta^2\alpha)\) | \(3\) | No |
| \(K=\mathbb Q(\alpha,\zeta)\) | \(6\) | Yes |
1. The splitting field has degree \(6\), with Galois group \(S_3\).
The three roots are \(\alpha,\zeta\alpha,\zeta^2\alpha\). Their generated field contains \(\zeta=(\zeta\alpha)/\alpha\), and \(\mathbb Q(\alpha,\zeta)\) contains all of them. Thus this is \(K\). The polynomial \(x^3-7\) is irreducible over \(\mathbb Q\) by Eisenstein at \(7\) [@DF04], so \([\mathbb Q(\alpha):\mathbb Q]=3\). This cubic field lies in \(\mathbb R\), whereas \(\zeta\) does not. Since \(\zeta\) satisfies \(x^2+x+1\), its degree over \(\mathbb Q(\alpha)\) is therefore exactly \(2\). The tower law gives \[ [K:\mathbb Q]=2\cdot3=6. \] The polynomial is separable in characteristic zero, so its splitting field is Galois and its automorphism group has order \(6\) [@DF04]. Each automorphism permutes the three roots, and one fixing all of them fixes their generated field \(K\). Hence this permutation action embeds the group into \(S_3\). Both groups have order \(6\), so the embedding is an isomorphism.
2. The table lists every subfield, without repetition.
Every subfield of the characteristic-zero field \(K\) contains its prime field \(\mathbb Q\). The Galois correspondence therefore gives a bijection between these subfields and the subgroups of \(S_3\) [@DF04]. By Lagrange’s theorem a subgroup of \(S_3\) has order \(1\), \(2\), \(3\), or \(6\). There is one trivial subgroup, three subgroups of order \(2\) generated by the three transpositions, one subgroup of order \(3\) generated by either of the two inverse \(3\)-cycles, and one subgroup of order \(6\), namely \(S_3\) itself. This is the complete list of six subgroups.
The field \(\mathbb Q(\zeta)\) has degree \(2\), since the quadratic \(x^2+x+1\) has no rational root. For \(j=0,1,2\), the element \(\zeta^j\alpha\) is a root of the irreducible polynomial \(x^3-7\), so \(\mathbb Q(\zeta^j\alpha)\) has degree \(3\). These three cubic fields are distinct. Indeed, if two coincided, their common field would contain the ratio of two different roots, which is \(\zeta\) or \(\zeta^2\), and hence would contain \(\zeta\). The tower law would then force \(2\mid3\), a contradiction.
Consequently the table contains six different intermediate fields: one of degree \(1\), one of degree \(2\), three of degree \(3\), and one of degree \(6\). The subgroup count proves exhaustiveness.
3. Exactly \(\mathbb Q\), \(\mathbb Q(\zeta)\), and \(K\) are Galoisover \(\mathbb Q\).
The identity extension is Galois. Both roots \(\zeta,\zeta^2\) of \(x^2+x+1\) lie in \(\mathbb Q(\zeta)\), so that field is a splitting field of a separable polynomial and is Galois. The same property for \(K\) was proved in step 1.
For any of the cubic fields in the table, the ratio argument in step 2 shows that it cannot contain two distinct roots of \(x^3-7\). It contains one root of that irreducible polynomial, but the polynomial does not split there. Thus the extension is not normal and cannot be Galois. This covers all the remaining fields.