Let \(\zeta_n \in \mathbb{C}\) denote a primitive \(n\)-th root of unity, and let \(p_n(x) \in \mathbb{Q}[x]\) be its minimal polynomial over \(\mathbb{Q}\). You may assume that the roots of \(p_n(x)\) are precisely the primitive \(n\)-th roots of unity in \(\mathbb{C}\).
Show that the field extension \(\mathbb{Q}(\zeta_n)/\mathbb{Q}\) is Galois, and prove that its Galois group is isomorphic to \((\mathbb{Z}/n\mathbb{Z})^\times\).
How many subfields are there of \(\mathbb{Q}(\zeta_{20})\)?
Solution
Goal: Prove that cyclotomic extensions are Galois with abelian Galois group \((\mathbb{Z}/n\mathbb{Z})^\times\) in (a), and count the subfields of \(\mathbb{Q}(\zeta_{20})\) via the Galois correspondence in (b).
1. Part (a): \(\mathbb{Q}(\zeta_n)/\mathbb{Q}\) is Galois.
1.1. Splitting field: The polynomial \(x^n - 1 \in \mathbb{Q}[x]\) has roots \(\{e^{2\pi i k / n} : 0 \le k < n\} = \{\zeta_n^k : 0 \le k < n\}\). All of these roots belong to \(\mathbb{Q}(\zeta_n)\).
1.2. Thus \(\mathbb{Q}(\zeta_n)\) is the splitting field of \(x^n - 1\) over \(\mathbb{Q}\).
1.3. Since \(\mathbb{Q}\) has characteristic 0, every algebraic extension over \(\mathbb{Q}\) is separable.
1.4. Because \(\mathbb{Q}(\zeta_n)/\mathbb{Q}\) is a splitting field of a separable polynomial, it is normal and separable, hence Galois.
2. Part (a): \(\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\times\).
2.1. Field degree: By hypothesis, the roots of the minimal polynomial \(p_n(x)\) of \(\zeta_n\) over \(\mathbb{Q}\) are precisely the primitive \(n\)-th roots of unity. The number of primitive \(n\)-th roots of unity is \(\varphi(n)\), so \([\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \deg p_n(x) = \varphi(n)\).
2.2. Action on generators: Any automorphism \(\sigma \in \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\) is uniquely determined by its value \(\sigma(\zeta_n)\), since \(\mathbb{Q}(\zeta_n) = \mathbb{Q}[\zeta_n]\).
2.3. Roots of minimal polynomial: \(\sigma(\zeta_n)\) must be a root of \(p_n(x)\), which means \(\sigma(\zeta_n) = \zeta_n^a\) for some integer \(a\) with \(1 \le a \le n\) and \(\gcd(a, n) = 1\).
2.4. Group homomorphism: Define the map\[\Psi: \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \to (\mathbb{Z}/n\mathbb{Z})^\times, \quad \sigma \mapsto [a]_n \text{ where } \sigma(\zeta_n) = \zeta_n^a.\]For any \(\sigma, \tau \in \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\) with \(\sigma(\zeta_n) = \zeta_n^a\) and \(\tau(\zeta_n) = \zeta_n^b\):\[(\sigma \circ \tau)(\zeta_n) = \sigma(\zeta_n^b) = (\sigma(\zeta_n))^b = (\zeta_n^a)^b = \zeta_n^{ab}.\]Thus \(\Psi(\sigma \circ \tau) = [ab]_n = [a]_n [b]_n = \Psi(\sigma) \Psi(\tau)\), so \(\Psi\) is a group homomorphism.
2.5. Injectivity: If \(\Psi(\sigma) = [1]_n\), then \(\sigma(\zeta_n) = \zeta_n^1 = \zeta_n\). Since \(\sigma\) fixes \(\mathbb{Q}\) and \(\zeta_n\), \(\sigma = \operatorname{id}\). Thus \(\ker \Psi = \{\operatorname{id}\}\), so \(\Psi\) is injective.
2.6. Isomorphism: Since \(|\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})| = [\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n) = |(\mathbb{Z}/n\mathbb{Z})^\times|\), the injective homomorphism \(\Psi\) is a bijection, and hence an isomorphism:\[\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\times.\]
3. Part (b): Number of subfields of \(\mathbb{Q}(\zeta_{20})\).
3.1. Galois group structure:For \(n = 20 = 2^2 \cdot 5\), by the Chinese Remainder Theorem:\[G = \operatorname{Gal}(\mathbb{Q}(\zeta_{20})/\mathbb{Q}) \cong (\mathbb{Z}/20\mathbb{Z})^\times \cong (\mathbb{Z}/4\mathbb{Z})^\times \times (\mathbb{Z}/5\mathbb{Z})^\times \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}.\]
3.2. Galois Correspondence: By the Fundamental Theorem of Galois Theory, the subfields \(E\) with \(\mathbb{Q} \subseteq E \subseteq \mathbb{Q}(\zeta_{20})\) correspond bijectively to the subgroups of \(G \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}\).
3.3. Classification of subgroups of \(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}\):- Order 1 (index 8): Exactly 1 subgroup: \(\{(0, 0)\}\), corresponding to \(\mathbb{Q}(\zeta_{20})\).- Order 2 (index 4): The elements of order 2 in \(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}\) are \((1, 0)\), \((0, 2)\), and \((1, 2)\). Each generates a distinct cyclic subgroup of order 2:\[H_1 = \langle (1, 0) \rangle, \quad H_2 = \langle (0, 2) \rangle, \quad H_3 = \langle (1, 2) \rangle.\]There are exactly 3 subgroups of order 2.- Order 4 (index 2):- Cyclic subgroups of order 4: The elements of order 4 are \((0, 1)\), \((0, 3)\), \((1, 1)\), and \((1, 3)\). They generate 2 cyclic subgroups:\[K_1 = \langle (0, 1) \rangle = \langle (0, 3) \rangle, \quad K_2 = \langle (1, 1) \rangle = \langle (1, 3) \rangle.\]- Non-cyclic subgroup (\(\cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\)): Generated by the elements of order 2:\[K_3 = \{(0, 0), (1, 0), (0, 2), (1, 2)\}.\]There are exactly \(2 + 1 = 3\) subgroups of order 4.- Order 8 (index 1): Exactly 1 subgroup: \(G\) itself, corresponding to \(\mathbb{Q}\).
3.4. Total count:\[1 + 3 + 3 + 1 = 8 \text{ subgroups}.\]
3.5. Therefore \(\mathbb{Q}(\zeta_{20})\) has exactly 8 subfields.
4. Conclusion:
$\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois with group $(\mathbb{Z}/n\mathbb{Z})^\times$, and $\mathbb{Q}(\zeta_{20})$ has 8 subfields.