05.5.ガロア拡大

ガロア拡大と自己同型写像

Statement

$E/F$:Galois Extension
then
$\{\sigma:E \rightarrow \bar F \mid \sigma_F = id \} = \text{Aut}(E/F)$
$|\text{Aut}(E/F)| = [E:F]$

Proof

obviously
$\{\sigma:E \rightarrow \bar F \mid \sigma_F = id \} \supset \text{Aut}(E/F)$
$\forall \sigma \in \{\sigma:E \rightarrow \bar F \mid \sigma_F = id \}\Rightarrow \sigma(E) \subset E \quad(\because\text{Prop.正規拡大の同値条件})$
$\therefore \sigma:E \rightarrow E$
$\sigma:\text{injection}\quad(\because E:\text{Field})$
$\forall \alpha \in E - F$
$\exists p(x) \in F[x]:\text{Minimal Polynomial s.t. } p(\alpha)= 0$
$p(x) = (x - \alpha_1)\cdots(x - \alpha_n) \in E[x]\quad(\because E/F:\text{Normal Extension})$
$\alpha \in \{\alpha_1, \cdots, \alpha_n\} = \{\sigma(\alpha_1),\cdots,\sigma(\alpha_n)\}$
$(\because \forall i, p(\sigma(\alpha_i)) = \sigma(p)(\sigma(\alpha_i)) = \sigma(p(\alpha_i)) = \sigma(0) = 0, E/F:\text{Separable},\sigma(\alpha_i) \neq \sigma(\alpha_j))$
$\therefore \exists \alpha_i \text{ s.t. } \sigma(\alpha_i) = \alpha$
$\therefore\sigma:\text{surjection}$
$\therefore \sigma \in \text{Aut}(E/F)$
$\therefore \{\sigma:E \rightarrow \bar F \mid \sigma_F = id \} \subset \text{Aut}(E/F)$
$\therefore\{\sigma:E \rightarrow \bar F \mid \sigma_F = id \} = \text{Aut}(E/F)$
$|\{\sigma:E \rightarrow \bar F \mid \sigma_F = id \}| = [E:F] \quad(\because E/F:\text{Separable})$
$\therefore|\text{Aut}(E/F)| = |\{\sigma:E \rightarrow \bar F \mid \sigma_F = id \}| = [E:F]$
QED.

ガロア拡大と最小分解体

Statement

$E/F$:Finite Field Extension
then
$E/F$:Galois Extension
$\Leftrightarrow$
$\exists f(x) \in F[x] \text{ s.t. } f(x): \text{Separable and } E:\text{Minimal Splitting Field of }f(x)$

Proof

$\Rightarrow$

$E/F$:Finite Separable Extension $\Rightarrow \exists \theta \text{ s.t. } E = F(\theta) \quad(\because\text{Prop.有限次分離拡大は単拡大-2})$

let $f(x) \in F[x]$:Minimal Polynomial of $\theta$
then $f(x)$:Separable ($\because E/F$:Separable Extension)

let $f(x) = (x - \alpha_1)\cdots(x - \alpha_n)$
then
$\alpha_1, \cdots, \alpha_n \in E \quad(\because E/F:\text{Normal Extension})$
$\theta = \exists \alpha_i \in F(\alpha_1, \cdots, \alpha_n)$
$\therefore F(\theta) \subset F(\alpha_1, \cdots, \alpha_n)$

$\alpha_i$:Conjugate of $\theta$
$\therefore \alpha_i \in E = F(\theta) \quad(\because\text{Prop.正規拡大の同値条件2})$
$\therefore F(\alpha_1, \cdots, \alpha_n) \subset F(\theta)$

$\therefore E = F(\theta) = F(\alpha_1, \cdots, \alpha_n)$

$\Leftarrow$
$E$:Minimal Splitting Field of $f(x)$
$\therefore \exists (\alpha_1, \cdots, \alpha_n) \in E \text{ s.t. } f(x) = (x - \alpha_1)\cdots(x - \alpha_n) \text{ and } E = F(\alpha_1, \cdots, \alpha_n)$
$\forall \alpha_i \in \{\alpha_1,\cdots,\alpha_n\}$
let $p_i(x) \in F[x]$:Minimal Polynomial of $\alpha_i$
then $p_i(x) | f(x)$
$\therefore p_i(x):\text{Separable}\quad(\because f(x):\text{Separable})$
$\therefore \alpha_i$:Separable
$\therefore F(\alpha_1, \cdots, \alpha_n)/F:\text{Separable}\quad(\because\text{Prop.分離的元の添加})$
$\therefore E/F$:Separable

$E/F$:Splitting Field of $f(x)$
$\therefore E/F$:Normal Extension $\quad(\because\text{Prop.正規拡大の同値条件3})$

$\therefore E/F$:Galois Extension
QED.

ガロア拡大と固定体

Statement

$E/F$:Finite Field Extension
$G = \text{Aut}(E/F)$
then
$E/F:\text{Galois Extension} \Leftrightarrow E^G = F$

Proof

$\Rightarrow$
let $x \in F$
$\forall g \in G, g(x) = g|_F(x) = id(x) = x$
$\therefore x \in E^G$
$\therefore F \subset E^G$

let
$\alpha \in E^G$
$p(x) \in F[x]$:Minimal Polynomial of $\alpha$
$p(x) = (x - \alpha_1)\cdots(x - \alpha_n)$
then
$\exists \sigma_i \in G \text{ s.t. } \sigma_i(\alpha) = \alpha_i \quad(\text{Prop.単拡大の同型})$
$\forall \sigma_i, \sigma_i(\alpha) = \alpha\quad(\because \alpha \in E^G)$
$\therefore \forall \alpha_i = \alpha$
$\therefore p(x) = (x - \alpha)^n$
$n = 1 \quad(\because E/F:\text{Galois Extension})$
$p(x) = (x - \alpha) \in F[x]$
$\therefore \alpha \in F$
$\therefore F \supset E^G$

$\therefore F = E^G$
$\Leftarrow$
$|G| \le [E:F] \quad(\because\text{Prop.有限次拡大と同型写像})$
$[E:F] \lt \infty$
$\therefore |G| \lt \infty$
$\forall \alpha \in E$
let
$p(x) \lt F[x]$:Minimal Polynomial of $\alpha$
$A = \{\alpha^\sigma \mid \sigma \in G\} \subset E$
$|A| = n \le |G| \lt \infty$
$A = \{\alpha^{\sigma_1}, \cdots, \alpha^{\sigma_n}\}\quad(\alpha^{\sigma_i} \neq \alpha^{\sigma_j}(i \neq j)),(\sigma_1 = id, \alpha^{\sigma_1} = \alpha)$
$f(x) = \prod_{i = 1}^{n} (x - \alpha^{\sigma_i}) \in E[x]$
then
$f(x)$:Separable Polynomial
$\forall \alpha^{\sigma_i} \in A,p(\alpha^{\sigma_i}) = p(\sigma_i(\alpha)) = \sigma_i(p)(\sigma_i(\alpha)) = \sigma_i(p(\alpha)) = \sigma_i(0) = 0$
$\therefore f(x) | p(x)$
$\forall \sigma,\sigma\sigma_i \in G\therefore \alpha^{\sigma\sigma_i} = \sigma\sigma_i(\alpha) \in A\therefore \sigma(A) = A$
$\forall \sigma \in G, \sigma(f)(x) = \sigma(\prod_{i = 1}^{n} (x - \alpha^{\sigma_i})) = \prod_{i = 1}^{n} (x - \alpha^{\sigma\sigma_i}) = f(x)$
$\therefore f(x) \in F[x]$
$\therefore f(x) = p(x) \quad(\because p(x):\text{Irreducible Polynomial})$
$f(x)$:Separable Polynomial and (all roots of $f(x)) \in E$
$\therefore p(x)$:Separable Polynomial and (all roots of $f(x)) \in E$
$\therefore E/F$:Galois Extension
QED.