- Characterize the splitting field and all intermediate fields of the polynomial \(x^{4}-3\) over \(Q\)
Solution
1. Splitting field and Galois group of \(f(x) = x^4 - 3\) over \(\mathbb{Q}\):
1.1. The four complex roots of \(x^4 - 3\) are \(\{\alpha, i\alpha, -\alpha, -i\alpha\}\) where \(\alpha = \sqrt[4]{3} \in \mathbb{R}_{>0}\).Thus the splitting field is \(K = \mathbb{Q}(\alpha, i)\).
roots of \(x^4 - 3 = 0\).
1.2. The polynomial \(x^4 - 3 \in \mathbb{Q}[x]\) is irreducible by Eisenstein’s criterion at \(p = 3\).Hence \([\mathbb{Q}(\alpha) : \mathbb{Q}] = 4\).
Eisenstein’s criterion.
1.3. Since \(\mathbb{Q}(\alpha) \subset \mathbb{R}\) and \(i \notin \mathbb{R}\), \([K : \mathbb{Q}(\alpha)] = 2\), so \([K : \mathbb{Q}] = 8\).
degree tower \([K : \mathbb{Q}] = [K : \mathbb{Q}(\alpha)][\mathbb{Q}(\alpha) : \mathbb{Q}] = 2 \times 4 = 8\).
1.4. The Galois group \(G = \operatorname{Gal}(K/\mathbb{Q})\) is generated by:\[ \sigma: \begin{cases} \alpha \mapsto i\alpha \\ i \mapsto i \end{cases} \quad (\text{order } 4), \qquad \tau: \begin{cases} \alpha \mapsto \alpha \\ i \mapsto -i \end{cases} \quad (\text{order } 2). \]They satisfy \(\tau \sigma \tau = \sigma^{-1}\), so \(G \cong D_4\), the dihedral group of order 8.
action on generators \(\alpha\) and \(i\).
2. Classification of all intermediate fields via the Galois correspondence:
2.1. Quadratic extensions of \(\mathbb{Q}\) (fixed fields of subgroups of order 4 / index 2):- \(H_1 = \langle \sigma \rangle \cong \mathbb{Z}_4 \implies K^{H_1} = \mathbb{Q}(i)\).- \(H_2 = \langle \sigma^2, \tau \rangle \cong V_4 \implies K^{H_2} = \mathbb{Q}(\alpha^2) = \mathbb{Q}(\sqrt{3})\).- \(H_3 = \langle \sigma^2, \sigma\tau \rangle \cong V_4 \implies K^{H_3} = \mathbb{Q}(i\alpha^2) = \mathbb{Q}(i\sqrt{3})\).
fixed fields of the three index-2 normal subgroups of \(D_4\).
2.2. Quartic extensions of \(\mathbb{Q}\) (fixed fields of subgroups of order 2 / index 4):- Non-normal extensions:- \(J_1 = \langle \tau \rangle \implies K^{J_1} = \mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt[4]{3})\).- \(J_2 = \langle \sigma^2\tau \rangle \implies K^{J_2} = \mathbb{Q}(i\alpha) = \mathbb{Q}(i\sqrt[4]{3})\).- \(J_3 = \langle \sigma\tau \rangle \implies K^{J_3} = \mathbb{Q}((1+i)\alpha)\).- \(J_4 = \langle \sigma^3\tau \rangle \implies K^{J_4} = \mathbb{Q}((1-i)\alpha)\).- Normal extension:- \(J_5 = \langle \sigma^2 \rangle = Z(D_4) \implies K^{J_5} = \mathbb{Q}(\alpha^2, i) = \mathbb{Q}(\sqrt{3}, i)\).
fixed fields of the five order-2 subgroups of \(D_4\).
2.3. Extreme fields:- Fixed field of \(\{1\}\) is \(K = \mathbb{Q}(\sqrt[4]{3}, i)\) (degree 8).- Fixed field of \(D_4\) is \(\mathbb{Q}\) (degree 1).
Galois correspondence.
3. Conclusion:The splitting field is \(\mathbb{Q}(\sqrt[4]{3}, i)\) with Galois group \(D_4\), and there are exactly 10 intermediate fields: \(\mathbb{Q}\), three quadratic fields \(\mathbb{Q}(i), \mathbb{Q}(\sqrt{3}), \mathbb{Q}(i\sqrt{3})\), five quartic fields \(\mathbb{Q}(\sqrt[4]{3}), \mathbb{Q}(i\sqrt[4]{3}), \mathbb{Q}((1\pm i)\sqrt[4]{3}), \mathbb{Q}(\sqrt{3}, i)\), and the splitting field \(\mathbb{Q}(\sqrt[4]{3}, i)\). Q.E.D.
1 and 2.