Calculate the Galois group of \(x^5 - 2\).
Solution
Goal: Determine the Galois group \(\operatorname{Gal}(K/\mathbb{Q})\) of the polynomial \(f(x) = x^5 - 2\) over \(\mathbb{Q}\).
1. Irreducibility and root generation:
1.1. The polynomial \(f(x) = x^5 - 2 \in \mathbb{Q}[x]\) is monic and Eisenstein with respect to the prime \(p = 2\).
1.2. By Eisenstein’s Criterion and Gauss’s Lemma, \(f(x)\) is irreducible over \(\mathbb{Q}\).
1.3. Let \(\alpha = \sqrt[5]{2} \in \mathbb{R}\) be the real fifth root of 2. Then \([\mathbb{Q}(\alpha) : \mathbb{Q}] = \deg f = 5\).
1.4. The five complex roots of \(f(x)\) are \(\alpha_k = \zeta^k \alpha\) for \(k = 0, 1, 2, 3, 4\), where \(\zeta = e^{2\pi i/5}\) is a primitive fifth root of unity.
2. Degree of the splitting field \(K\):
2.1. The splitting field of \(f(x)\) over \(\mathbb{Q}\) is \(K = \mathbb{Q}(\alpha, \zeta\alpha, \dots, \zeta^4\alpha) = \mathbb{Q}(\alpha, \zeta)\).
2.2. The minimal polynomial of \(\zeta\) over \(\mathbb{Q}\) is the fifth cyclotomic polynomial\[\Phi_5(x) = x^4 + x^3 + x^2 + x + 1,\]which is irreducible over \(\mathbb{Q}\) with degree \([\mathbb{Q}(\zeta) : \mathbb{Q}] = 4\).
2.3. Because \([\mathbb{Q}(\alpha) : \mathbb{Q}] = 5\) and \([\mathbb{Q}(\zeta) : \mathbb{Q}] = 4\) are coprime (\(\gcd(5, 4) = 1\)), the tower law implies\[[K : \mathbb{Q}] = [\mathbb{Q}(\alpha, \zeta) : \mathbb{Q}] = [\mathbb{Q}(\alpha) : \mathbb{Q}] \cdot [\mathbb{Q}(\zeta) : \mathbb{Q}] = 5 \times 4 = 20.\]
2.4. In characteristic 0, every algebraic extension is separable, so \(K/\mathbb{Q}\) is a normal and separable Galois extension with \(|\operatorname{Gal}(K/\mathbb{Q})| = [K : \mathbb{Q}] = 20\).
3. Construction of Galois automorphisms:
3.1. Since \([K : \mathbb{Q}(\zeta)] = 5\) and the minimal polynomial of \(\alpha\) over \(\mathbb{Q}(\zeta)\) is \(x^5 - 2\), there exists an automorphism \(\sigma \in \operatorname{Gal}(K/\mathbb{Q}(\zeta)) \subset \operatorname{Gal}(K/\mathbb{Q})\) satisfying\[\sigma(\alpha) = \zeta \alpha, \qquad \sigma(\zeta) = \zeta.\]Iterating gives \(\sigma^k(\alpha) = \zeta^k \alpha\), so \(\sigma\) has order 5 and generates a cyclic subgroup \(N = \langle \sigma \rangle \cong C_5\).
3.2. Since \([K : \mathbb{Q}(\alpha)] = 4\) and the minimal polynomial of \(\zeta\) over \(\mathbb{Q}(\alpha)\) is \(\Phi_5(x)\), there exists an automorphism \(\tau \in \operatorname{Gal}(K/\mathbb{Q}(\alpha)) \subset \operatorname{Gal}(K/\mathbb{Q})\) satisfying\[\tau(\alpha) = \alpha, \qquad \tau(\zeta) = \zeta^2.\]Because 2 is a primitive root modulo 5 (\(2^1 \equiv 2, 2^2 \equiv 4, 2^3 \equiv 3, 2^4 \equiv 1 \pmod 5\)), \(\tau\) has order 4 and generates a cyclic subgroup \(H = \langle \tau \rangle \cong C_4\).
4. Group structure and semidirect product presentation:
4.1. The subgroup \(N = \operatorname{Gal}(K/\mathbb{Q}(\zeta))\) is the kernel of the restriction homomorphism \(\operatorname{Gal}(K/\mathbb{Q}) \to \operatorname{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})\), so \(N \trianglelefteq \operatorname{Gal}(K/\mathbb{Q})\).
4.2. Since \(\gcd(|N|, |H|) = \gcd(5, 4) = 1\), \(N \cap H = \{e\}\).
4.3. By the product formula, \(|NH| = |N| \cdot |H| = 20 = |\operatorname{Gal}(K/\mathbb{Q})|\), so \(\operatorname{Gal}(K/\mathbb{Q}) = NH = N \rtimes H\).
4.4. Compute the conjugation action of \(\tau\) on \(\sigma\):\[(\tau \sigma \tau^{-1})(\alpha) = \tau(\sigma(\tau^{-1}(\alpha))) = \tau(\sigma(\alpha)) = \tau(\zeta \alpha) = \tau(\zeta) \tau(\alpha) = \zeta^2 \alpha = \sigma^2(\alpha),\]\[(\tau \sigma \tau^{-1})(\zeta) = \tau(\sigma(\tau^{-1}(\zeta))) = \tau(\sigma(\zeta^3)) = \tau(\zeta^3) = (\zeta^2)^3 = \zeta^6 = \zeta = \sigma^2(\zeta).\]Since \(\tau \sigma \tau^{-1}\) and \(\sigma^2\) agree on the generators \(\alpha\) and \(\zeta\), \(\tau \sigma \tau^{-1} = \sigma^2\).
5. Conclusion:
The Galois group of \(x^5 - 2\) over \(\mathbb{Q}\) is the non-abelian Frobenius group of order 20 (isomorphic to the affine linear group \(\operatorname{AGL}_1(\mathbb{F}_5)\)), with presentation \[\operatorname{Gal}(K/\mathbb{Q}) \cong C_5 \rtimes C_4 \cong \langle \sigma, \tau \mid \sigma^5 = e, \; \tau^4 = e, \; \tau \sigma \tau^{-1} = \sigma^2 \rangle.\]