分圆扩张与 Q 上的阿贝尔扩张
我们已经确定，Q ( ζ n ) \mathbb{Q}\left( {\zeta }_{n}\right) Q ( ζ n ) n th {n}^{\text{th }} n th Q \mathbb{Q} Q φ ( n ) \varphi \left( n\right) φ ( n ) φ \varphi φ n th {n}^{\text{th }} n th ζ n {\zeta }_{n} ζ n n th {n}^{\text{th }} n th Φ n ( x ) {\Phi }_{n}\left( x\right) Φ n ( x ) σ ( ζ n ) = ζ n a \sigma \left( {\zeta }_{n}\right) = {\zeta }_{n}^{a} σ ( ζ n ) = ζ n a a , 1 ≤ a < n a,1 \leq a < n a , 1 ≤ a < n n n n φ ( n ) \varphi \left( n\right) φ ( n ) a a a Q ( ζ n ) \mathbb{Q}\left( {\zeta }_{n}\right) Q ( ζ n ) σ a {\sigma }_{a} σ a n n n a a a σ a {\sigma }_{a} σ a a a a n n n
定理 26。单位根 Q ( ζ n ) \mathbb{Q}\left( {\zeta }_{n}\right) Q ( ζ n ) ( Z / n Z ) × {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } ( Z / n Z ) ×
( Z / n Z ) × → ∼ Gal ( Q ( ζ n ) / Q ) {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times }\overset{ \sim }{ \rightarrow }\operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\right) ( Z / n Z ) × → ∼ Gal ( Q ( ζ n ) / Q ) a ( mod n ) ↦ σ a a\left( {\;\operatorname{mod}\;n}\right) \mapsto {\sigma }_{a} a ( mod n ) ↦ σ a 其中 σ a {\sigma }_{a} σ a
σ a ( ζ n ) = ζ n a {\sigma }_{a}\left( {\zeta }_{n}\right) = {\zeta }_{n}^{a} σ a ( ζ n ) = ζ n a 证明：上述讨论表明对于任何 a ( mod n ) a\left( {\;\operatorname{mod}\;n}\right) a ( mod n ) σ a {\sigma }_{a} σ a
( σ a σ b ) ( ζ n ) = σ a ( ζ n b ) = ( ζ n b ) a \left( {{\sigma }_{a}{\sigma }_{b}}\right) \left( {\zeta }_{n}\right) = {\sigma }_{a}\left( {\zeta }_{n}^{b}\right) = {\left( {\zeta }_{n}^{b}\right) }^{a} ( σ a σ b ) ( ζ n ) = σ a ( ζ n b ) = ( ζ n b ) a = ζ n a b = {\zeta }_{n}^{ab} = ζ n ab 这表明 σ a σ b = σ a b {\sigma }_{a}{\sigma }_{b} = {\sigma }_{ab} σ a σ b = σ ab σ a {\sigma }_{a} σ a a a a n n n
示例
(1) 该领域 Q ( ζ 5 ) \mathbb{Q}\left( {\zeta }_{5}\right) Q ( ζ 5 ) Q \mathbb{Q} Q ( Z / 5 Z ) × ≅ Z / 4 Z {\left( \mathbb{Z}/5\mathbb{Z}\right) }^{ \times } \cong \mathbb{Z}/4\mathbb{Z} ( Z /5 Z ) × ≅ Z /4 Z Q \mathbb{Q} Q { σ 1 = 1 , σ 2 , σ 3 , σ 4 } \left\{ {{\sigma }_{1} = 1,{\sigma }_{2},{\sigma }_{3},{\sigma }_{4}}\right\} { σ 1 = 1 , σ 2 , σ 3 , σ 4 } σ 2 : ζ 5 ↦ ζ 5 2 {\sigma }_{2} : {\zeta }_{5} \mapsto {\zeta }_{5}^{2} σ 2 : ζ 5 ↦ ζ 5 2 ( Z / 5 Z ) × {\left( \mathbb{Z}/5\mathbb{Z}\right) }^{ \times } ( Z /5 Z ) ×
正好存在一个非平凡子域，即 Q \mathbb{Q} Q { 1 , σ 4 = σ − 1 } \left\{ {1,{\sigma }_{4} = {\sigma }_{-1}}\right\} { 1 , σ 4 = σ − 1 }
α = ζ 5 + σ − 1 ζ 5 = ζ 5 + ζ 5 − 1 \alpha = {\zeta }_{5} + {\sigma }_{-1}{\zeta }_{5} = {\zeta }_{5} + {\zeta }_{5}^{-1} α = ζ 5 + σ − 1 ζ 5 = ζ 5 + ζ 5 − 1 因为这个元素显然被 σ − 1 {\sigma }_{-1} σ − 1 ζ 5 {\zeta }_{5} ζ 5
ζ 5 4 + ζ 5 3 + ζ 5 2 + ζ 5 + 1 = 0 {\zeta }_{5}^{4} + {\zeta }_{5}^{3} + {\zeta }_{5}^{2} + {\zeta }_{5} + 1 = 0 ζ 5 4 + ζ 5 3 + ζ 5 2 + ζ 5 + 1 = 0 注意接着有
α 2 + α − 1 = ( ζ 5 2 + 2 + ζ 5 − 2 ) + ( ζ 5 + ζ 5 − 1 ) − 1 {\alpha }^{2} + \alpha - 1 = \left( {{\zeta }_{5}^{2} + 2 + {\zeta }_{5}^{-2}}\right) + \left( {{\zeta }_{5} + {\zeta }_{5}^{-1}}\right) - 1 α 2 + α − 1 = ( ζ 5 2 + 2 + ζ 5 − 2 ) + ( ζ 5 + ζ 5 − 1 ) − 1 = ζ 5 2 + 2 + ζ 5 3 + ζ 5 + ζ 5 4 − 1 = 0. = {\zeta }_{5}^{2} + 2 + {\zeta }_{5}^{3} + {\zeta }_{5} + {\zeta }_{5}^{4} - 1 = 0. = ζ 5 2 + 2 + ζ 5 3 + ζ 5 + ζ 5 4 − 1 = 0. 明确求解 α \alpha α α \alpha α Q \mathbb{Q} Q Q ( 5 ) \mathbb{Q}\left( \sqrt{5}\right) Q ( 5 )
Q ( ζ 5 + ζ 5 − 1 ) = Q ( 5 ) \mathbb{Q}\left( {{\zeta }_{5} + {\zeta }_{5}^{-1}}\right) = \mathbb{Q}\left( \sqrt{5}\right) Q ( ζ 5 + ζ 5 − 1 ) = Q ( 5 ) 一般情况下可以证明（这并非完全平凡），对于奇素数 p p p Q ( ζ p ) \mathbb{Q}\left( {\zeta }_{p}\right) Q ( ζ p ) Q ( ± p ) \mathbb{Q}\left( \sqrt{\pm p}\right) Q ( ± p ) p ≡ 1 mod 4 p \equiv 1{\;\operatorname{mod}\;4} p ≡ 1 mod 4 p ≡ 3 mod 4 p \equiv 3{\;\operatorname{mod}\;4} p ≡ 3 mod 4
(2) Q ( ζ 13 ) \mathbb{Q}\left( {\zeta }_{13}\right) Q ( ζ 13 ) p p p Q ( ζ p ) \mathbb{Q}\left( {\zeta }_{p}\right) Q ( ζ p ) Q \mathbb{Q} Q
1 , ζ p , ζ p 2 , … , ζ p p − 2 . 1,{\zeta }_{p},{\zeta }_{p}^{2},\ldots ,{\zeta }_{p}^{p - 2}. 1 , ζ p , ζ p 2 , … , ζ p p − 2 . 因为
ζ p p − 1 + ζ p p − 2 + ⋯ + ζ p + 1 = 0 {\zeta }_{p}^{p - 1} + {\zeta }_{p}^{p - 2} + \cdots + {\zeta }_{p} + 1 = 0 ζ p p − 1 + ζ p p − 2 + ⋯ + ζ p + 1 = 0 我们看到元素
ζ p , ζ p 2 , … , ζ p p − 2 , ζ p p − 1 {\zeta }_{p},{\zeta }_{p}^{2},\ldots ,{\zeta }_{p}^{p - 2},{\zeta }_{p}^{p - 1} ζ p , ζ p 2 , … , ζ p p − 2 , ζ p p − 1 构成基础。选择这个基础的原因是，在伽罗瓦群 σ \sigma σ Gal ( Q ( ζ p ) / Q ) \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{p}\right) /\mathbb{Q}}\right) Gal ( Q ( ζ p ) / Q ) p th {p}^{\text{th }} p th p p p n th {n}^{\text{th }} n th n th {n}^{\text{th }} n th Q \mathbb{Q} Q 4 th {4}^{\text{th }} 4 th
设 H H H Q ( ζ p ) \mathbb{Q}\left( {\zeta }_{p}\right) Q ( ζ p ) Q \mathbb{Q} Q
α H = ∑ σ ∈ H σ ζ p (14.10) {\alpha }_{H} = \mathop{\sum }\limits_{{\sigma \in H}}\sigma {\zeta }_{p} \tag{14.10} α H = σ ∈ H ∑ σ ζ p ( 14.10 ) 是 ζ p {\zeta }_{p} ζ p H H H τ ∈ H \tau \in H τ ∈ H τ σ {\tau \sigma } τ σ σ \sigma σ H H H H H H τ α = α {\tau \alpha } = \alpha τα = α α \alpha α H H H τ \tau τ H H H τ α {\tau \alpha } τα τ ( ζ p ) \tau \left( {\zeta }_{p}\right) τ ( ζ p ) τ α = α {\tau \alpha } = \alpha τα = α τ ( ζ p ) = σ ( ζ p ) \tau \left( {\zeta }_{p}\right) = \sigma \left( {\zeta }_{p}\right) τ ( ζ p ) = σ ( ζ p ) σ ζ p \sigma {\zeta }_{p} σ ζ p τ σ − 1 = 1 \tau {\sigma }^{-1} = 1 τ σ − 1 = 1 ζ p {\zeta }_{p} ζ p τ = σ ∈ H \tau = \sigma \in H τ = σ ∈ H α \alpha α H H H Q ( α ) \mathbb{Q}\left( \alpha \right) Q ( α ) H H H
以一个特定例子来说，考虑 Q ( ζ 13 ) \mathbb{Q}\left( {\zeta }_{13}\right) Q ( ζ 13 ) ( Z / 13 Z ) × ≅ Z / 12 Z {\left( \mathbb{Z}/{13}\mathbb{Z}\right) }^{ \times } \cong \mathbb{Z}/{12}\mathbb{Z} ( Z / 13 Z ) × ≅ Z / 12 Z σ = σ 2 \sigma = {\sigma }_{2} σ = σ 2 ζ 13 {\zeta }_{13} ζ 13 ζ 13 2 {\zeta }_{13}^{2} ζ 13 2 σ 6 , σ 4 , σ 3 {\sigma }^{6},{\sigma }^{4},{\sigma }^{3} σ 6 , σ 4 , σ 3 σ 2 {\sigma }^{2} σ 2 Q \mathbb{Q} Q ( ζ = ζ 13 ) \left( {\zeta = {\zeta }_{13}}\right) ( ζ = ζ 13 )
ζ + σ 6 ζ = ζ + ζ 2 6 = ζ + ζ − 1 \zeta + {\sigma }^{6}\zeta = \zeta + {\zeta }^{{2}^{6}} = \zeta + {\zeta }^{-1} ζ + σ 6 ζ = ζ + ζ 2 6 = ζ + ζ − 1 ζ + σ 4 ζ + σ 8 ζ = ζ + ζ 2 4 + ζ 2 8 = ζ + ζ 3 + ζ 9 \zeta + {\sigma }^{4}\zeta + {\sigma }^{8}\zeta = \zeta + {\zeta }^{{2}^{4}} + {\zeta }^{{2}^{8}} = \zeta + {\zeta }^{3} + {\zeta }^{9} ζ + σ 4 ζ + σ 8 ζ = ζ + ζ 2 4 + ζ 2 8 = ζ + ζ 3 + ζ 9 ζ + σ 3 ζ + σ 6 ζ + σ 9 ζ = ζ + ζ 8 + ζ 12 + ζ 5 \zeta + {\sigma }^{3}\zeta + {\sigma }^{6}\zeta + {\sigma }^{9}\zeta = \zeta + {\zeta }^{8} + {\zeta }^{12} + {\zeta }^{5} ζ + σ 3 ζ + σ 6 ζ + σ 9 ζ = ζ + ζ 8 + ζ 12 + ζ 5 ζ + σ 2 ζ + σ 4 ζ + σ 6 ζ + σ 8 ζ + σ 10 ζ = ζ + ζ 4 + ζ 3 + ζ 12 + ζ 9 + ζ 10 . \zeta + {\sigma }^{2}\zeta + {\sigma }^{4}\zeta + {\sigma }^{6}\zeta + {\sigma }^{8}\zeta + {\sigma }^{10}\zeta = \zeta + {\zeta }^{4} + {\zeta }^{3} + {\zeta }^{12} + {\zeta }^{9} + {\zeta }^{10}. ζ + σ 2 ζ + σ 4 ζ + σ 6 ζ + σ 8 ζ + σ 10 ζ = ζ + ζ 4 + ζ 3 + ζ 12 + ζ 9 + ζ 10 . 这个扩展的子域格如下所示：
方程（10）中构造的元素及其共轭被称为 ζ \zeta ζ
假设 n = p 1 a 1 p 2 a 2 ⋯ p k a k n = {p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\cdots {p}_{k}^{{a}_{k}} n = p 1 a 1 p 2 a 2 ⋯ p k a k n n n ζ n p 2 a 2 ⋯ p k a k {\zeta }_{n}^{{p}_{2}^{{a}_{2}}\cdots {p}_{k}^{{a}_{k}}} ζ n p 2 a 2 ⋯ p k a k p 1 a 1 {p}_{1}^{{a}_{1}} p 1 a 1 K 1 = Q ( ζ p 1 a 1 ) {K}_{1} = \mathbb{Q}\left( {\zeta }_{{p}_{1}^{{a}_{1}}}\right) K 1 = Q ( ζ p 1 a 1 ) Q ( ζ n ) \mathbb{Q}\left( {\zeta }_{n}\right) Q ( ζ n ) K i = Q ( ζ p i a i ) , i = 1 , 2 , … , k {K}_{i} = \mathbb{Q}\left( {\zeta }_{{p}_{i}^{{a}_{i}}}\right) ,i = 1,2,\ldots ,k K i = Q ( ζ p i a i ) , i = 1 , 2 , … , k Q ( ζ n ) \mathbb{Q}\left( {\zeta }_{n}\right) Q ( ζ n ) ζ p 1 a 1 ζ p 2 a 2 ⋯ ζ p k a k {\zeta }_{{p}_{1}^{{a}_{1}}}{\zeta }_{{p}_{2}^{{a}_{2}}}\cdots {\zeta }_{{p}_{k}^{{a}_{k}}} ζ p 1 a 1 ζ p 2 a 2 ⋯ ζ p k a k n th {n}^{\text{th }} n th Q ( ζ n ) \mathbb{Q}\left( {\zeta }_{n}\right) Q ( ζ n ) [ K i : Q ] \left\lbrack {{K}_{i} : \mathbb{Q}}\right\rbrack [ K i : Q ] φ ( p i a i ) , i = 1 , 2 , … , k \varphi \left( {p}_{i}^{{a}_{i}}\right) ,i = 1,2,\ldots ,k φ ( p i a i ) , i = 1 , 2 , … , k φ ( n ) = φ ( p 1 a 1 ) φ ( p 2 a 2 ) ⋯ φ ( p k a k ) \varphi \left( n\right) = \varphi \left( {p}_{1}^{{a}_{1}}\right) \varphi \left( {p}_{2}^{{a}_{2}}\right) \cdots \varphi \left( {p}_{k}^{{a}_{k}}\right) φ ( n ) = φ ( p 1 a 1 ) φ ( p 2 a 2 ) ⋯ φ ( p k a k ) K i {K}_{i} K i k k k Q \mathbb{Q} Q Q ( ζ n ) \mathbb{Q}\left( {\zeta }_{n}\right) Q ( ζ n ) Q \mathbb{Q} Q K i {K}_{i} K i
推论27。设 n = p 1 a 1 p 2 a 2 ⋯ p k a k n = {p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\cdots {p}_{k}^{{a}_{k}} n = p 1 a 1 p 2 a 2 ⋯ p k a k n n n Q ( ζ p i a i ) , i = 1 , 2 , … , k \mathbb{Q}\left( {\zeta }_{{p}_{i}}^{{a}_{i}}\right) ,i = 1,2,\ldots ,k Q ( ζ p i a i ) , i = 1 , 2 , … , k Q \mathbb{Q} Q Q ( ζ n ) \mathbb{Q}\left( {\zeta }_{n}\right) Q ( ζ n )
Gal ( Q ( ζ n ) / Q ) ≅ Gal ( Q ( ζ p 1 a 1 ) / Q ) × Gal ( Q ( ζ p 2 a 2 ) / Q ) × ⋯ × Gal ( Q ( ζ p k a k ) / Q ) \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\right) \cong \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{{p}_{1}^{{a}_{1}}}\right) /\mathbb{Q}}\right) \times \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{{p}_{2}^{{a}_{2}}}\right) /\mathbb{Q}}\right) \times \cdots \times \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{{p}_{k}^{{a}_{k}}}\right) /\mathbb{Q}}\right) Gal ( Q ( ζ n ) / Q ) ≅ Gal ( Q ( ζ p 1 a 1 ) / Q ) × Gal ( Q ( ζ p 2 a 2 ) / Q ) × ⋯ × Gal ( Q ( ζ p k a k ) / Q ) 在定理26的同构下，这是中国剩余定理：
( Z / n Z ) × ≅ ( Z / p 1 a 1 Z ) × × ( Z / p 2 a 2 Z ) × × ⋯ × ( Z / p k a k Z ) × . {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \cong {\left( \mathbb{Z}/{p}_{1}^{{a}_{1}}\mathbb{Z}\right) }^{ \times } \times {\left( \mathbb{Z}/{p}_{2}^{{a}_{2}}\mathbb{Z}\right) }^{ \times } \times \cdots \times {\left( \mathbb{Z}/{p}_{k}^{{a}_{k}}\mathbb{Z}\right) }^{ \times }. ( Z / n Z ) × ≅ ( Z / p 1 a 1 Z ) × × ( Z / p 2 a 2 Z ) × × ⋯ × ( Z / p k a k Z ) × . 证明：唯一尚未证明的陈述是伽罗瓦群的同构与关于群 ( Z / n Z ) × {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } ( Z / n Z ) ×
由定理26可知，Q ( ζ n ) / Q \mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q} Q ( ζ n ) / Q
定义。如果 K / F K/F K / F Gal ( K / F ) \operatorname{Gal}\left( {K/F}\right) Gal ( K / F ) K / F K/F K / F
由于阿贝尔群的所有子群和商群都是阿贝尔群，根据伽罗瓦理论的基本定理，我们得知每个包含 F F F F F F F F F
确定哪些群可以作为 Q \mathbb{Q} Q Q \mathbb{Q} Q
设 n = p 1 p 2 ⋯ p k n = {p}_{1}{p}_{2}\cdots {p}_{k} n = p 1 p 2 ⋯ p k
( Z / n Z ) × ≅ ( Z / p 1 Z ) × × ( Z / p 2 Z ) × × ⋯ × ( Z / p k Z ) × {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \cong {\left( \mathbb{Z}/{p}_{1}\mathbb{Z}\right) }^{ \times } \times {\left( \mathbb{Z}/{p}_{2}\mathbb{Z}\right) }^{ \times } \times \cdots \times {\left( \mathbb{Z}/{p}_{k}\mathbb{Z}\right) }^{ \times } ( Z / n Z ) × ≅ ( Z / p 1 Z ) × × ( Z / p 2 Z ) × × ⋯ × ( Z / p k Z ) × ≅ Z p 1 − 1 × Z p 2 − 1 × ⋯ × Z p k − 1 . (14.11) \cong {\mathbb{Z}}_{{p}_{1} - 1} \times {\mathbb{Z}}_{{p}_{2} - 1} \times \cdots \times {\mathbb{Z}}_{{p}_{k} - 1}. \tag{14.11} ≅ Z p 1 − 1 × Z p 2 − 1 × ⋯ × Z p k − 1 . ( 14.11 ) 现在，假设 G G G
G ≅ Z n 1 × Z n 2 × ⋯ × Z n k G \cong {\mathbb{Z}}_{{n}_{1}} \times {\mathbb{Z}}_{{n}_{2}} \times \cdots \times {\mathbb{Z}}_{{n}_{k}} G ≅ Z n 1 × Z n 2 × ⋯ × Z n k 对于某些整数 n 1 , n 2 , … , n k {n}_{1},{n}_{2},\ldots ,{n}_{k} n 1 , n 2 , … , n k m m m p p p p ≡ 1 mod m p \equiv 1{\;\operatorname{mod}\;m} p ≡ 1 mod m
根据这个结果，选择不同的素数 p 1 , p 2 , … , p k {p}_{1},{p}_{2},\ldots ,{p}_{k} p 1 , p 2 , … , p k
p 1 ≡ 1 mod n 1 {p}_{1} \equiv 1{\;\operatorname{mod}\;{n}_{1}} p 1 ≡ 1 mod n 1 p 2 ≡ 1 mod n 2 {p}_{2} \equiv 1{\;\operatorname{mod}\;{n}_{2}} p 2 ≡ 1 mod n 2 p k ≡ 1 mod n k {p}_{k} \equiv 1{\;\operatorname{mod}\;{n}_{k}} p k ≡ 1 mod n k 并令 n = p 1 p 2 ⋯ p k n = {p}_{1}{p}_{2}\cdots {p}_{k} n = p 1 p 2 ⋯ p k
根据构造，n i {n}_{i} n i p i − 1 {p}_{i} - 1 p i − 1 i = 1 , 2 , … , k i = 1,2,\ldots ,k i = 1 , 2 , … , k Z p i − 1 {Z}_{{p}_{i} - 1} Z p i − 1 p i − 1 n i \frac{{p}_{i} - 1}{{n}_{i}} n i p i − 1 H i {H}_{i} H i i = 1 , 2 , … , k i = 1,2,\ldots ,k i = 1 , 2 , … , k n i {n}_{i} n i ( Z / n Z ) × {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } ( Z / n Z ) × H 1 × H 2 × ⋯ × H k {H}_{1} \times {H}_{2} \times \cdots \times {H}_{k} H 1 × H 2 × ⋯ × H k G G G
根据定理26和伽罗瓦理论的基本定理，我们发现存在一个 Q ( ζ p 1 p 2 ⋯ p k ) \mathbb{Q}\left( {\zeta }_{{p}_{1}{p}_{2}\cdots {p}_{k}}\right) Q ( ζ p 1 p 2 ⋯ p k ) Q \mathbb{Q} Q G G G
推论28。设 G G G Gal ( K / Q ) ≅ G \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \cong G Gal ( K / Q ) ≅ G K K K
这个结果有一个逆命题（其证明超出了我们的范围），著名的克罗内克-韦伯定理：
定理（克罗内克-韦伯）设 K K K Q \mathbb{Q} Q K K K Q \mathbb{Q} Q
"Q \mathbb{Q} Q Q \mathbb{Q} Q Q \mathbb{Q} Q F F F Q \mathbb{Q} Q F F F F F F Q \mathbb{Q} Q Q ( − D ) \mathbb{Q}\left( \sqrt{-D}\right) Q ( − D ) D D D e x {e}^{x} e x x x x
"我们以正 n n n
"回顾（参看第13.3节）一个元素 α \alpha α Q \mathbb{Q} Q Q ( α ) \mathbb{Q}\left( \alpha \right) Q ( α ) K K K
Q = K 0 ⊂ K 1 ⊂ ⋯ ⊂ K i ⊂ K i + 1 ⊂ ⋯ ⊂ K m = K (14.12) \mathbb{Q} = {K}_{0} \subset {K}_{1} \subset \cdots \subset {K}_{i} \subset {K}_{i + 1} \subset \cdots \subset {K}_{m} = K \tag{14.12} Q = K 0 ⊂ K 1 ⊂ ⋯ ⊂ K i ⊂ K i + 1 ⊂ ⋯ ⊂ K m = K ( 14.12 ) "with"
[ K i + 1 : K i ] = 2 , i = 0 , 1 , … , m − 1. \left\lbrack {{K}_{i + 1} : {K}_{i}}\right\rbrack = 2,\;i = 0,1,\ldots ,m - 1. [ K i + 1 : K i ] = 2 , i = 0 , 1 , … , m − 1. 正则n n n R 2 {\mathbb{R}}^{2} R 2 n th {n}^{\text{th }} n th n th {n}^{\text{th }} n th C \mathbb{C} C n n n
ζ n {\zeta }_{n} ζ n ζ n {\zeta }_{n} ζ n x x x R 2 {\mathbb{R}}^{2} R 2 ζ n {\zeta }_{n} ζ n ζ n {\zeta }_{n} ζ n ζ n − 1 {\zeta }_{n}^{-1} ζ n − 1 ζ n {\zeta }_{n} ζ n x = 1 2 ( ζ n + ζ n − 1 ) x = \frac{1}{2}\left( {{\zeta }_{n} + {\zeta }_{n}^{-1}}\right) x = 2 1 ( ζ n + ζ n − 1 ) ζ n {\zeta }_{n} ζ n ζ n 2 − 2 x ζ n + 1 = 0 {\zeta }_{n}^{2} - {2x}{\zeta }_{n} + 1 = 0 ζ n 2 − 2 x ζ n + 1 = 0 Q ( x ) \mathbb{Q}\left( x\right) Q ( x ) [ Q ( ζ n ) : Q ( x ) ] = 2 \left\lbrack {\mathbb{Q}\left( {\zeta }_{n}\right) : \mathbb{Q}\left( x\right) }\right\rbrack = 2 [ Q ( ζ n ) : Q ( x ) ] = 2 Q ( x ) \mathbb{Q}\left( x\right) Q ( x ) Q \mathbb{Q} Q φ ( n ) / 2 \varphi \left( n\right) /2 φ ( n ) /2
因此，如果正则n n n φ ( n ) \varphi \left( n\right) φ ( n ) φ ( n ) = 2 m \varphi \left( n\right) = {2}^{m} φ ( n ) = 2 m Gal ( Q ( ζ n ) / Q ) \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\right) Gal ( Q ( ζ n ) / Q ) Gal ( Q ( x ) / Q ) \operatorname{Gal}\left( {\mathbb{Q}\left( x\right) /\mathbb{Q}}\right) Gal ( Q ( x ) / Q ) 2 m {2}^{m} 2 m G G G
G = G m > G m − 1 > ⋯ > G i + 1 > G i > ⋯ > G 0 = 1 G = {G}_{m} > {G}_{m - 1} > \cdots > {G}_{i + 1} > {G}_{i} > \cdots > {G}_{0} = 1 G = G m > G m − 1 > ⋯ > G i + 1 > G i > ⋯ > G 0 = 1 具有
[ G i + 1 : G i ] = 2 , i = 0 , 1 , 2 , … , m − 1. \left\lbrack {{G}_{i + 1} : {G}_{i}}\right\rbrack = 2,\;i = 0,1,2,\ldots ,m - 1. [ G i + 1 : G i ] = 2 , i = 0 , 1 , 2 , … , m − 1. 将其应用于群G = Gal ( Q ( x ) / Q ) G = \operatorname{Gal}\left( {\mathbb{Q}\left( x\right) /\mathbb{Q}}\right) G = Gal ( Q ( x ) / Q ) G i , i = 0 , 1 , … , m − 1 {G}_{i},i = 0,1,\ldots ,m - 1 G i , i = 0 , 1 , … , m − 1
我们得出结论，正 n n n φ ( n ) \varphi \left( n\right) φ ( n ) n n n φ ( n ) \varphi \left( n\right) φ ( n ) n = 2 k p 1 ⋯ p r n = {2}^{k}{p}_{1}\cdots {p}_{r} n = 2 k p 1 ⋯ p r p i {p}_{i} p i p i − 1 {p}_{i} - 1 p i − 1 p p p p − 1 p - 1 p − 1
p = 2 2 s + 1 p = {2}^{{2}^{s}} + 1 p = 2 2 s + 1 对于某个整数 s s s
3 = 2 1 + 1 3 = {2}^{1} + 1 3 = 2 1 + 1 5 = 2 2 + 1 5 = {2}^{2} + 1 5 = 2 2 + 1 17 = 2 4 + 1 {17} = {2}^{4} + 1 17 = 2 4 + 1 257 = 2 8 + 1 {257} = {2}^{8} + 1 257 = 2 8 + 1 65537 = 2 16 + 1 {65537} = {2}^{16} + 1 65537 = 2 16 + 1 （但 2 32 + 1 {2}^{32} + 1 2 32 + 1
命题29。正 n n n n = 2 k p 1 ⋯ p r n = {2}^{k}{p}_{1}\cdots {p}_{r} n = 2 k p 1 ⋯ p r
上面的证明实际上指出了构建正 n n n Q ( ζ 17 ) \mathbb{Q}\left( {\zeta }_{17}\right) Q ( ζ 17 ) ζ 17 {\zeta }_{17} ζ 17 13 th {13}^{\text{th }} 13 th Q ( ζ 17 ) \mathbb{Q}\left( {\zeta }_{17}\right) Q ( ζ 17 ) H 1 < H 2 {H}_{1} < {H}_{2} H 1 < H 2 [ H 2 : H 1 ] = 2 \left\lbrack {{H}_{2} : {H}_{1}}\right\rbrack = 2 [ H 2 : H 1 ] = 2 H 1 {H}_{1} H 1 H 2 {H}_{2} H 2 σ 2 {\sigma }_{2} σ 2 ( ζ = ζ 17 ) \left( {\zeta = {\zeta }_{17}}\right) ( ζ = ζ 17 )
η 1 = ζ + ζ 2 + ζ 4 + ζ 8 + ζ 9 + ζ 13 + ζ 15 + ζ 16 {\eta }_{1} = \zeta + {\zeta }^{2} + {\zeta }^{4} + {\zeta }^{8} + {\zeta }^{9} + {\zeta }^{13} + {\zeta }^{15} + {\zeta }^{16} η 1 = ζ + ζ 2 + ζ 4 + ζ 8 + ζ 9 + ζ 13 + ζ 15 + ζ 16 η 2 = ζ 3 + ζ 5 + ζ 6 + ζ 7 + ζ 10 + ζ 11 + ζ 12 + ζ 14 {\eta }_{2} = {\zeta }^{3} + {\zeta }^{5} + {\zeta }^{6} + {\zeta }^{7} + {\zeta }^{10} + {\zeta }^{11} + {\zeta }^{12} + {\zeta }^{14} η 2 = ζ 3 + ζ 5 + ζ 6 + ζ 7 + ζ 10 + ζ 11 + ζ 12 + ζ 14 这些周期“减半”了整个伽罗瓦群的周期，并满足
η 1 + η 2 = − 1 {\eta }_{1} + {\eta }_{2} = - 1 η 1 + η 2 = − 1 （由 ζ 17 {\zeta }_{17} ζ 17
η 1 η 2 = − 4 {\eta }_{1}{\eta }_{2} = - 4 η 1 η 2 = − 4 （这需要计算 - 我们知道它必须是理性的，因为该乘积由伽罗瓦群的所有元素固定）。因此这两个周期是二次方程的根
x 2 + x − 4 = 0 {x}^{2} + x - 4 = 0 x 2 + x − 4 = 0 我们可以显式地解这个方程。类似地，指数为4的子群（由 σ 4 {\sigma }_{4} σ 4 ζ 17 {\zeta }_{17} ζ 17 8 ( ζ + ζ − 1 ) = 16 cos ( 2 π 17 ) 8\left( {\zeta + {\zeta }^{-1}}\right) = {16}\cos \left( \frac{2\pi }{17}\right) 8 ( ζ + ζ − 1 ) = 16 cos ( 17 2 π )
− 1 + 17 + 2 ( 17 − 17 ) + 2 17 + 3 17 − 2 ( 17 − 17 ) − 2 2 ( 17 + 17 ) . - 1 + \sqrt{17} + \sqrt{2\left( {{17} - \sqrt{17}}\right) } + 2\sqrt{{17} + 3\sqrt{17} - \sqrt{2\left( {{17} - \sqrt{17}}\right) } - 2\sqrt{2\left( {{17} + \sqrt{17}}\right) }}. − 1 + 17 + 2 ( 17 − 17 ) + 2 17 + 3 17 − 2 ( 17 − 17 ) − 2 2 ( 17 + 17 ) . J.H. Conway 向我们展示了一种相对简单的正17边形构建方法，该方法在练习中有说明。
虽然我们已经看到，在一般情况下，仅使用连续的平方根求解 ζ n {\zeta }_{n} ζ n n th {n}^{\text{th }} n th ζ n {\zeta }_{n} ζ n n n n
