06.3.巡回拡大

準同型写像の一次独立性

Statement

$S$:Semigroup $\quad(\stackrel{\text{def}}{\Leftrightarrow} \forall a,b,c \in S, (a\cdot b)\cdot c = a \cdot (b \cdot c))$
$F$:Field
$(\phi_i)_{i=1\cdots n}$
s.t.
$\phi_i:S \rightarrow F^*$
$\phi_i$:Homomorphism
then
$(\forall i,j \text{ s.t. } i \neq j \Rightarrow \phi_i \neq \phi_j)$
$\Leftrightarrow$
$(\forall s \in S,a_1 \phi_1(s) + \cdots + a_n \phi_n(s) = 0 \quad (a_1,\cdots,a_n \in F)\Rightarrow a_1 = \cdots = a_n = 0)$
$\Leftrightarrow$
$(a_1,\cdots,a_n) \neq 0 \Rightarrow \exists s \in S, a_1 \phi_1(s) + \cdots + a_n \phi_n(s) \neq 0 \quad(a_1,\cdots,a_n) \in F$

Proof

$\Rightarrow$
Suppose $\exists (a_1, \cdots, a_n) \neq 0 \text{ s.t. } a_1 \phi_1(s) + \cdots + a_n \phi_n(s) = 0$
Without loss of generality, suppose $a_1, \cdots a_m \neq 0\quad(1 \le m \le n)$ and $m$ is smallest
(i) $m = 1$
$\phi(s) \neq 0, a_1 \neq 0 \therefore a_1 \phi(s) \neq 0$ leads to contradiction.

(ii) $m \ge 2$
for $s,t \in S$
$a_1 \phi_1(st) + \cdots + a_m \phi_m(st) = 0$
$\Leftrightarrow a_1 \phi_1(s)\phi_1(t) + \cdots + a_m \phi_m(s)\phi_m(t) = 0 \quad ---(\ast)$

$(a_1 \phi_1(s) + \cdots + a_m \phi_m(s))\phi_m(t) = 0$
$\Leftrightarrow a_1 \phi_1(s)\phi_m(t) + \cdots + a_m \phi_m(s)\phi_m(t) = 0---(\ast\ast)$
$(*) - (**)$
$a_1\phi_1(s)(\phi_1(t) - \phi_m(t)) + \cdots + a_{m-1}\phi_{m-1}(s)(\phi_{m-1}(t) - \phi_m(t)) = 0$
let $a_i(\phi_i(t) - \phi_m(t)) = b_i$
then
$\exists t \text{ s.t. } a_i(\phi_i(t) - \phi_m(t)) = b_i \neq 0$
$b_1 \phi(s) + \cdots + b_{m-1} \phi(s) = 0$
$m - 1 \lt m$ contradiction.
$\Leftarrow$
Suppose $\exists i,j \text{ s.t. } \phi_i = \phi_j\quad(i \neq j)$
then
$\phi_i(s) + (-1)\phi_j(s) = 0$
Proved by contraposition.
QED.

原始n乗根添加による拡大

Statement

$F$:Field
$p = \text{char} F$
if $p \gt 0$ then $p \nmid n$

$E$:Minimal Splitting Field of $x^n - 1$ over $F$
$\overset{\text{(i)}}{\Leftrightarrow}$
$E = F(\sqrt[n]{1})$
$\overset{\text{(ii)}}{\Rightarrow}$
$E/F$:Cyclic Extension

Proof

let $f(x) = x^n - 1$
then $f'(x) = n x^{n-1}$
$f'(x) = n x^{n-1} = 0 \Rightarrow x = 0 \quad(\because p \nmid n)$
$\sqrt[n]{1},\sqrt[n]{1}^2,\cdots,\sqrt[n]{1}^{n-1},\sqrt[n]{1}^{n} = \sqrt[n]{1}^{0} = 1 \neq 0$
$f(\sqrt[n]{1})=f(\sqrt[n]{1}^2)=\cdots=f(\sqrt[n]{1}^{n-1})=f(1) = 0$
$\therefore \sqrt[n]{1}^{i} \neq \sqrt[n]{1}^{j} \quad (0 \le i,j \lt n, i \neq j)\quad(\because\text{重根(があるとすれば)は}x=0\text{の時のみ})$

$\overset{\text{(i)}}{\Rightarrow}$
$\therefore f(x) = x^n - 1 = (x - 1)(x - \sqrt[n]{1})\cdots(x - \sqrt[n]{1}^{n-1})$:Separable
$\sqrt[n]{1},\sqrt[n]{1}^2,\cdots,\sqrt[n]{1}^{n-1} \in E \quad(\because E$:Splitting Field of $x^n - 1$)
$\therefore E \supset F(\sqrt[n]{1},\cdots,\sqrt[n]{1}^{n-1}) = F(\sqrt[n]{1})$
$\therefore E = F(\sqrt[n]{a}) \quad(\because E:\text{Minimal Splitting Field of }x^n - 1)$
$\overset{\text{(i)}}{\Leftarrow}$
$f(x) = x^n - 1 = (x - 1)(x - \sqrt[n]{1})(x - \sqrt[n]{1}^2)\cdots(x - \sqrt[n]{1}^{n-1})$
$\therefore F(\sqrt[n]{1})$:Minimal Splitting Field of $x^n - 1$
$\therefore E$:Minimal Splitting Field of $x^n - 1\quad(\because E = F(\sqrt[n]{1}))$
$\overset{\text{(ii)}}{\Rightarrow}$
$x^n - 1$:Separable
$E = F(\sqrt[n]{1})$:Minimal Splitting Field of Separable Polynomial $x^n - 1$
$\therefore E:\text{Galois Extension}\quad (\because\text{Prop.ガロア拡大と最小分解体})$
$\forall \sigma \in \text{Gal}(E/F), f(\sigma(\sqrt[n]{1})) = \sigma(f(\sqrt[n]{1})) = \sigma(0) = 0$
$\therefore \exists i \in 1 \cdots n, \sigma(\sqrt[n]{1}) = \sqrt[n]{1}^i$
Define $\phi$ as
$\phi:\text{Gal}(E/F) \rightarrow \mathbb{Z}/n\mathbb{Z}$
$\phi:\sigma \mapsto \bar i \quad(\sigma(\sqrt[n]{1}) = \sqrt[n]{1}^i)$
let
$\sigma,\tau \in \text{Gal}(E/F), \phi(\sigma) = i, \phi(\tau) = j$
then
$i = j\quad(\text{mod} n) \Leftrightarrow n \mid i - j \Leftrightarrow \sqrt[n]{1}^{(i-j)} = 1 \Leftrightarrow \sqrt[n]{1}^i = \sqrt[n]{1}^j$
$\Leftrightarrow \sigma(\sqrt[n]{1}) = \sqrt[n]{1}^i = \sqrt[n]{1}^j = \tau(\sqrt[n]{1}) \Leftrightarrow \sigma = \tau \quad(\because E = F(\sqrt[n]{1}), \forall x \in E, \sigma(x) = \tau(x))$
$\phi$:Injective Homomorphism
$\therefore \text{Gal}(E/F):\text{SubGroup of }\mathbb{Z}/n\mathbb{Z}$
$\therefore E/F:\text{Cyclic Extension}$

巡回拡大と冪根拡大

Statement

$F$:Field
$p = \text{char} F$
$\theta = \sqrt[n]{1}$:primitive n-th root of unity(原始n乗根)
if $p \gt 0$ then $p \nmid n$
$\theta \in F$

(i)
$a \in F, a \neq 0$
then
$E$:Minimal Splitting Field of $x^n - a$ over $F$
$\Leftrightarrow$
$E = F(\sqrt[n]{a})$
(ii)
$a \in F$
$E:\text{Minimal Splitting Field of }x^n - a\text{ over }F\quad(\Leftrightarrow E = F(\sqrt[n]{a}))$
$\Rightarrow$
$E/F$:Cyclic Extension (defined by $E/F$:Galois Extension and $G = \text{Gal}(E/F)$:Cyclic Group )
$[E:F] \mid n$
(iii)
$E/F$:Cyclic Extension (defined by $E/F$:Galois Extension and $G = \text{Gal}(E/F)$:Cyclic Group )
$\Rightarrow$
$\exists a \in F \text{ s.t. }$
$E$:Minimal Splitting Field of $x^n - a$ over $F\quad(\Leftrightarrow E = F(\sqrt[n]{a}))$
$$ \begin{xy} \xymatrix { & E & \exists a \in F_1, E = F_1(\sqrt[n]{a}) \\ & F_1 = F_0(\sqrt[n]{1}) \ar[u] & \text{Gal}(E/F_1):\text{Cyclic} \ar@{-}[u]_{\Leftrightarrow} \\ & F_0 \ar[u]_{(\text{Cyclic})}\\ } \end{xy} $$

Proof

(i)
let $f(x) = x^n - a$
then $f'(x) = n x^{n-1}$
$f'(x) = n x^{n-1} = 0 \Rightarrow x = 0 \quad(\because p \nmid n)$
$\sqrt[n]{a},\sqrt[n]{a}\theta,\cdots,\sqrt[n]{a}\theta^{n-1} \neq 0$
$f(\sqrt[n]{a})=f(\sqrt[n]{a}\theta)=\cdots=f(\sqrt[n]{a}\theta^{n-1}) = 0$
$\therefore \sqrt[n]{a}\theta^{i} \neq \sqrt[n]{a}\theta^{j} \quad (0 \le i,j \lt n, i \neq j)\quad(\because\text{重根(があるとすれば)は}x=0\text{の時のみ})$

$\Rightarrow$
$\therefore f(x) = x^n - a = (x - \sqrt[n]{a})(x - \sqrt[n]{a}\theta)\cdots(x - \sqrt[n]{a}\theta^{n-1})$:Separable
$\sqrt[n]{a},\sqrt[n]{a}\theta,\cdots,\sqrt[n]{a}\theta^{n-1} \in E \quad(\because E$:Splitting Field of $x^n - a$)
$\therefore E \supset F(\sqrt[n]{a}\theta,\cdots,\sqrt[n]{a}\theta^{n-1}) = F(\sqrt[n]{a}) \quad(\because \theta \in F)$
$\therefore E = F(\sqrt[n]{a}) \quad(\because E:\text{Minimal Splitting Field of }x^n - a)$
$\Leftarrow$
$f(x) = x^n - a = (x - \sqrt[n]{a})(x - \sqrt[n]{a}\theta)\cdots(x - \sqrt[n]{a}\theta^{n-1})$
$\therefore F(\sqrt[n]{a})$:Minimal Splitting Field of $x^n - a$
$\therefore E$:Minimal Splitting Field of $x^n - a\quad(\because E = F(\sqrt[n]{a}))$
(ii)
$x^n - a$:Separable
$E$:Minimal Splitting Field of Separable Polynomial $x^n - a$
$\therefore E:\text{Galois Extension}\quad (\because\text{Prop.ガロア拡大と最小分解体})$
$\forall \sigma \in \text{Gal}(E/F), f(\sigma(\sqrt[n]{a})) = \sigma(f(\sqrt[n]{a})) = \sigma(0) = 0$
$\therefore \exists i \in 1 \cdots n, \sigma(\sqrt[n]{a}) = \sqrt[n]{a}\theta^i$
Define $\phi$ as
$\phi:\text{Gal}(E/F) \rightarrow \mathbb{Z}/n\mathbb{Z}$
$\phi:\sigma \mapsto \bar i \quad(\sigma(\sqrt[n]{a}) = \sqrt[n]{a}\theta^i)$
Let
$\sigma,\tau \in \text{Gal}(E/F), \phi(\sigma) = i, \phi(\tau) = j$
then
$i = j\quad(\text{mod} n) \Leftrightarrow n \mid i - j \Leftrightarrow \theta^{i-j} = 1 \quad(\because \theta^n = 1) \Leftrightarrow \theta^i = \theta^j \Leftrightarrow \sqrt[n]{a}\theta^i = \sqrt[n]{a}\theta^j$
$\Leftrightarrow \sigma(\sqrt[n]{a}) = \sqrt[n]{a}\theta^i = \sqrt[n]{a}\theta^j = \tau(\sqrt[n]{a}) \Leftrightarrow \sigma = \tau \quad(\because E = F(\sqrt[n]{a}), \forall x \in E, \sigma(x) = \tau(x))$
$\phi$:Injective Homomorphism
$\therefore \text{Gal}(E/F)$:SubGroup of $\mathbb{Z}/n\mathbb{Z}\therefore |\text{Gal}(E/F)| \mid n$
$\therefore E/F:\text{Cyclic Extension}, [E:F] \mid n$
(iii)
Let $\text{Gal}(E/F) = G = \langle \sigma \rangle$
then
$\exists b \in E \text{ s.t. } 0 \neq b + \theta^{-1}\sigma(b) + \cdots + \theta^{-(n-1)}\sigma^{n-1}(b) \in E \quad(\because\text{Prop.準同型写像の一次独立性})$
Let $\alpha = \sum_{i = 0}^{n-1} \theta^{-i} \sigma^i(b) \in E$
then
$\sigma(\alpha) = \sigma(b) + \theta^{-1}\sigma^2(b) + \cdots + \theta^{-(n-1)}\sigma^{n}(b) \quad(\because \theta \in F)$
$= \theta \theta^{-1}(\sigma(b) + \theta^{-1}\sigma^2(b) + \cdots + \theta^{-(n-1)}\sigma^{n}(b))$
$= \theta (\theta^{-1}\sigma(b) + \theta^{-2}\sigma^2(b) + \cdots + \theta^{-(n)}\sigma^{n}(b))$
$= \theta (b + \sigma(b)\theta^{-1} + \theta^{-2}\sigma^2(b) + \cdots + \theta^{-(n-1)}\sigma^{n-1}(b))\quad(\because \theta^n = 1, \sigma^n = id\therefore \theta^{-(n)}\sigma^{n}(b) = b)$
$= \theta \alpha$
$\therefore \sigma^i(\alpha) = \theta^i \alpha$

$\sigma(\alpha^n) = \sigma(\alpha)^n = (\theta \alpha)^n = \theta^n \alpha^n = \alpha^n$
$\therefore \sigma^i(\alpha^n) = \alpha^n$
$\therefore \alpha^n \in E^{G} = F$
let $a = \alpha^n$
then
$a \in F, \alpha \in E, \sqrt[n]{a} = \alpha$
$F(\alpha) \subset E$
$\sigma^i(\alpha) = \theta^i \alpha \neq \alpha \quad(i = 1,\cdots,n-1)$
$(\because \theta:\text{primitive n-th root of unity})$
$\therefore \sigma^i \in G - \{id\}\Rightarrow \sigma^i(\alpha) \neq \alpha$
$\therefore \sigma^i \notin \text{Gal}(E/F(\alpha)) \quad(i = 1,\cdots,n-1)$
$\therefore \text{Gal}(E/F(\alpha)) = \{id\}$
$\therefore E = F(\alpha) = F(\sqrt[n]{a})$
$\therefore E$:Minimal Splitting Field of $x^n - a$ over $F$
$$ \begin{xy} \xymatrix { & E \ar@{-}[r] & \{id\} = \text{Gal}(E/E) \\ & F(\alpha) \ar@{-}[u] \ar@{-}[r] & \{id\} = \text{Gal}(E/F(\alpha)) \ar@{-}[u] & \\ & F \ar@{-}[u] \ar@{-}[r] & \text{Gal}(E/F) \ar@{-}[u] & \\ } \end{xy} $$

巡回拡大列と可解拡大

Statement

$E/F$:Galois Extension
$F = M_0 \subset M_1 \subset \cdots \subset M_n = E$
$M_{i+1}/M_i:\text{Cyclic Extension}\quad (\overset{\text{def}}\Leftrightarrow M_{i+1}/M_i:\text{Galois Extension},\text{Gal}(M_{i+1}/M_i):\text{Cyclic Group})$
$\Leftrightarrow$
$E/F$:Solvable Extension

Proof

$(\Rightarrow)$
$M_i \subset M_{i+1} \subset E$
$E/M_i:\text{Galois}\quad(\because E/F:\text{Galois})$
$E/M_{i+1}:\text{Galois}\quad(\because E/F:\text{Galois})$
$\text{Gal}(E/M_{i+1}) \triangleleft \text{Gal}(E/M_i)\quad(\because M_{i+1}/M_i:\text{Galois Extension and }\text{Prop.ガロアの基本定理(6-1)})$
$\text{Gal}(M_{i+1}/M_i) \simeq \text{Gal}(E/M_i)/\text{Gal}(E/M_{i+1}) \quad(\because\text{Prop.ガロアの基本定理(6-2)})$
$\therefore\text{Gal}(E/M_i)/\text{Gal}(E/M_{i+1})):\text{Cyclic Group}$

$\text{Gal}(E/M_0) \triangleright \text{Gal}(E/M_1) \triangleright \cdots \text{Gal}(E/M_n) = \{id\}$
$\therefore E/F = E/M_0:\text{Solvable Extension}\quad(\because\text{Prop.有限可解群の同値条件})$
$(\Leftarrow)$
$\exists (H_i)_{i=0\cdots n} \text{s.t.}$
$\text{Gal}(E/F) = G = H_0 \triangleright \cdots \triangleright H_n = \{id\}, \quad H_i / H_{i+1}:\text{Cyclic}$

let $M_i = E^{H_i}$ then
$M_{i+1}/M_i:\text{Galois} \quad(\because\text{Prop.ガロアの基本定理(6-1)}))$
$H_i/H_{i+1} \simeq \text{Gal}(M_{i+1}/M_i):\text{Cyclic} \quad(\because\text{Prop.ガロアの基本定理(6-2)})$
$\therefore M_{i+1}/M_i:\text{Cyclic Extension}$

$$ \begin{xy} \xymatrix { E & M_n \ar@{-}[r] & \{id\} = \text{Gal}(E/E) & & \\ & \vdots & \vdots & & \text{Gal}(E/M_{n}) / \text{Gal}(E/M_{n-1})\text{:Cyclic} \\ & M_{i+1} \ar@{-}[r] & \text{Gal}(E/M_{i+1}) & \{id\} & \vdots \\ & M_i \ar@{-}[u]^{\text{Cyclic}} \ar@{-}[r] & \text{Gal}(E/M_i) \ar@{-}[u]_{\triangleleft} & \text{Gal}(M_{i+1}/M_{i})\simeq \ar@{-}[u] & \text{Gal}(E/M_{i}) / \text{Gal}(E/M_{i+1})\text{:Cyclic} \\ & \vdots & \vdots & & \vdots\\ F \ar@{-}[uuuuu]^{\text{Galois}} & M_0 \ar@{-}[r] & \text{Gal}(E/F) & & \text{Gal}(E/M_{1}) / \text{Gal}(E/M_{0}) \text{:Cyclic} \\ } \end{xy} $$ QED.