05.6.アルティンの定理

アルティンの定理の補題

Statement

$E/F$:Algebraic Separable Extension
if $\exists n \ge 1, \forall \alpha \in E, [F(\alpha):F] \le n$
then $[E:F] \le n$

Proof

Select $\alpha \in E \text{ s.t. } [F(\alpha): F] = m$ is Maximum.
Assume $E \neq F(\alpha)$
then $\exists \beta \in E - F(\alpha)$
$F(\alpha) \subsetneq F(\alpha, \beta)$
Since $F(\alpha,\beta)/F$:Finite Separable Extension
(
$\because [F(\alpha,\beta):F] = [F(\alpha,\beta):F(\alpha)][F(\alpha):F] \le [F(\beta):F][F(\alpha):F] \le m m \lt \infty$
$\because [F(\alpha,\beta):F(\alpha)] \le [F(\beta):F] \le [F(\alpha):F] = m$
$\because$ if
$f(x) \in F[x]$:Minimal Polynomial of $\beta$
$g(x) \in F(\alpha)[x]$:Minimal Polynomial of $\beta$
then
$f(x) \in F(\alpha)[x] \therefore g(x)|f(x), \text{deg}g(x) \le \text{deg}f(x)$
$\therefore [F(\alpha,\beta):F(\alpha)] \le [F(\beta):F]$
)
$\therefore \exists \gamma \in E \text{ s.t. } F(\gamma) = F(\alpha, \beta) \quad(\because\text{Prop.有限次分離拡大は単拡大-2})$
$\therefore [F(\alpha):F] \lt [F(\alpha,\beta):F] = [F(\gamma),F]$
Contradiction of the maximality of $[F(\alpha):F]$
$\therefore E = F(\alpha), [E:F] = [F(\alpha):F]$

アルティンの定理

Statement

$E$:Field
$H$:Finite Group, $H \subset \text{Aut}(E)$
$|H| = h \lt \infty$

then

(i) $E/E^H$:Galois Extension
(ii) $|H| = [E:E^H]$
(iii) $\text{Gal}(E/E^H) = H$

Proof

$\forall \alpha \in E$
$A = \{\sigma(\alpha)\mid \sigma \in H\}$
$A = \{\sigma_1(\alpha), \cdots, \sigma_n(\alpha)\}\quad(\because n \le h = |H| \lt \infty)$
$(i \neq j \Rightarrow \sigma_i(\alpha) \neq \sigma_j(\alpha))$
$A \subset E \quad(\because H \subset \text{Aut}(E))$
$\forall \sigma \in H,\sigma\sigma_i \in H\therefore \sigma H \subset H$
$\sigma\sigma_i = \sigma\sigma_j \Rightarrow \sigma_i = \sigma_j \therefore |\sigma H| = |H| \therefore \sigma H = H$
$\therefore \sigma(A) = \{\sigma(\sigma_1(\alpha)), \cdots, \sigma(\sigma_n(\alpha))\} = \{\sigma_1(\alpha), \cdots, \sigma_n(\alpha)\} = A$
let $f(x) = \prod_{i = 1}^{n} (x - \alpha^{\sigma_i}) \in E[x]$
then $\sigma(f)(x) = \prod_{i = 1}^{n} \sigma(x - \alpha^{\sigma_i}) = \prod_{i = 1}^{n} (x - \alpha^{\sigma\sigma_i}) = \prod_{i = 1}^{n} (x - \alpha^{\sigma_i}) = f(x)$
$\therefore$ coefficient of $f(x) \in E^H$
$\therefore f(x) \in E^H[x]$
$id \in H \therefore \alpha = id(\alpha) \in A\therefore f(\alpha) = 0$

let $p_\alpha(x) \in E^H[x]$:Minimal Polynomial of $\alpha$
then $p_\alpha(x)$:Separable $\quad(\because f(x):\text{Separable}, p_\alpha(x)|f(x))$
$\therefore E/E^H$:Separable Extension
$p_\alpha(x)|f(x) \Rightarrow \exists q(x) \in E^H[x] \text{ s.t. } f(x) = p_\alpha(x)q(x)$
$p_\alpha(\forall \beta) = 0 \Rightarrow f(\beta) = p_\alpha(\beta)q(\beta) = 0 \therefore \beta \in A \subset E\therefore p_\alpha(x)$
$\therefore E/E^H$:Normal Extension
$\therefore E/E^H$:Galois Extension---(i)

$[E^H(\alpha):E^H] = \text{deg}p_\alpha(x) \le \text{deg}f(x) = n \le h$
$\therefore [E:E^H] \le h \quad (\because\text{Prop.アルティンの定理の補題})$
$H \subset \text{Gal}(E/E^H) \quad(\because a \in E^H \Rightarrow \forall h \in H, h(a) = a )$
$\therefore h = |H| \le |\text{Gal}(E/E^H)| = [E:E^H]$
$\therefore |H| = h = [E:E^H] = |\text{Gal}(E/E^H)|$---(ii)
$\therefore H = \text{Gal}(E/E^H)$---(iii)
QED.