05.1.有限次分離拡大

Proposition

Statement

$F$:Field
$F^{\times} = F - \{0\}$
$G \subset F^{\times}$
$G$:Group
$|G| = p^n \quad(p:\text{Prime Number}, n \in \mathbb{N})$
then
$\exists a \in G, G = \langle a \rangle = \{1,a,a^2,\cdots,a^{p^n - 1}\}$
$G \simeq \mathbb{Z}/p^n\mathbb{Z}\quad$ (as a Group)

Proof

$a$ as $a\in G, \forall g \in G, |\langle a \rangle| \ge |\langle g \rangle|$
$s := |\langle a \rangle|$
then
$s = p^m \quad (0 \lt m \le n)\quad(\because \text{Prop.ラグランジェの定理})$
$a^s = 1$
$\langle a \rangle = \{1, a, \cdots, a^{s-1}\} \subset G$
$\forall a^i \in \langle a \rangle, (a^i)^s = (a^s)^i = 1^i = 1$
$\therefore x^s - 1 = (x - 1)(x - a)\cdots(x - a^{s-1})$
for $\forall g \in G$
let $t := |\langle g \rangle|$
then
$t = p^l\quad (0 \lt l \le m \le n)$
$g^t = 1$
$s = p^m = p^l p^{m-l} = tu \quad(u:=p^{m-l} \in \mathbb{N})$
$\therefore g^s = g^{tu} = (g^t)^u = 1$
$\therefore g$ is a root of $x^s - 1 = 0$
$\therefore g \in \langle a \rangle$
$\therefore G \subset \langle a \rangle$
$\therefore G = \langle a \rangle$
$\therefore G \simeq \mathbb{Z}/p^{n}\mathbb{Z}\quad(\because \text{Prop.巡回群と同型})$

Proposition

Statement

$F$:Field
$|F| = \infty$
then
I
$F(\alpha,\beta)/F$:Separable Extension
$\Rightarrow \exists \theta \in \bar F, F(\alpha,\beta) = F(\theta)$
II
$F(\alpha_1,\dots,\alpha_n)/F$:Separable Extension
$\Rightarrow \exists \theta \in \bar F, F(\alpha_1,\dots,\alpha_n) = F(\theta)$

Proof

I
$f(x) \in F[x]$:Separable over $\alpha$
$g(x) \in F[x]$:Separable over $\beta$
then
$f(x) = (x - \alpha_1)\cdots(x - \alpha_m)$
$g(x) = (x - \beta_1)\cdots(x - \beta_n)$
$(\alpha = \alpha_1, \beta = \beta_1, \alpha_i, \beta_j \in \bar F)$
$\exists c \in F \text{ s.t. } c \neq \frac{\alpha_i - \alpha_j}{\beta_k - \beta_l} \quad (i \neq j, k \neq l) \quad(\because |F| = \infty)$
let
$\theta = \alpha + c \beta$
then
$F(\theta) = F(\alpha + c\beta) \subset F(\alpha, \beta) \quad(\because \alpha, \beta \in F(\alpha, \beta), c \in F)$

$F(\theta) = F(\alpha + c\beta) \supset F(\alpha, \beta)$
$\because$
$g(\beta) = 0$
$f(\theta - cx) \in F(\theta)[x] \quad (\because \theta \in F(\theta), c \in F)$
$f(\theta - cx) = (\theta - cx - \alpha_1)(\theta - cx - \alpha_2)\cdots(\theta - cx - \alpha_m)$
$f(\theta - cx) = (\alpha + c\beta - cx - \alpha_1)(\alpha + c\beta - cx - \alpha_2)\cdots(\alpha + c\beta - cx - \alpha_m)$
$=-c(x - \beta)(\alpha - \alpha_2 - c(x - \beta)) \cdots (\alpha - \alpha_m - c(x - \beta))$
$\therefore$
$f(\theta - c\beta) = 0$
$\beta_i \neq \beta \Rightarrow f(\theta - c\beta_i) \neq 0$
$\therefore\beta$ is an only common root for $g(x), f(\theta - cx)$

let $p(x) \in F(\theta)[x]$:Minimal Polynomial of $\beta$ over $F(\theta)$
then $p(x) \mid g(x)$ and $p(x) \mid f(\theta - cx)$
$\text{deg }p(x) = 1$ ($\because \beta$ is an only common root of $g(x), f(\theta - cx)$)
$\therefore p(x) = x - \beta \in F(\theta)[x]$
$\therefore \beta \in F(\theta)$

$\alpha = (\alpha + c \beta) - c \beta = \theta - c \beta \in F(\theta)\quad(\because c \in F \subset F(\theta),\beta \in F(\theta))$
$\therefore F(\alpha,\beta) \subset F(\theta) \quad(\because \alpha, \beta \in F(\theta))$

$\therefore F(\theta) = F(\alpha, \beta)$

II
let $\theta_1 = \alpha_1$
then $\exists \theta_2, F(\theta_2) = F(\theta_1, \alpha_2) = F(\alpha_1, \alpha_2)$
then $\exists \theta_3, F(\theta_3) = F(\theta_2, \alpha_3) = F(\theta_2)(\alpha_3) = F(\alpha_1, \alpha_2)(\alpha_3) = F(\alpha_1, \alpha_2, \alpha_3)$
$\vdots$
then $\exists \theta_n, F(\theta_n) = F(\theta_{n-1}, \alpha_n) = F(\alpha_1, \cdots, \alpha_n) = E$

$\therefore E = F(\theta_n)$

QED.

Theorem

Statement

$E/F$:Finite Separable Extension
$\Rightarrow E/F$:Simple Extension

Proof

$[E:F] = n \lt \infty$
$\therefore \exists \theta_1, \cdots, \theta_n \in E, E = F(\theta_1, \cdots, \theta_n)$
Case $|F| \lt \infty$
$E = f_1 e_1 + \cdots + f_n e_n \quad(f_i \in F, e_i \in E)$
$\therefore |E| \lt \infty$
$\therefore \exists \alpha \in E, E^{\times} = \langle \alpha \rangle$
($\because$
Let
$|E^{\times}| = p_1^{a_1} \cdots p_n^{a_n}\quad(p_1,\cdots p_n \in {\text{Prime Number}}, a_1,\cdots,a_n \in \mathbb{N})$
then
$\exists H_i:p-\text{Sylow Subgroup} \quad (|H_i| = p_i^{a_i}) \quad (\because\text{Prop.シローの定理})$
$\therefore H_i \simeq \mathbb{Z}/p_i^{a_i}\mathbb{Z} \quad (\text{Prop.素数位数の群})$

$H_i$ is a commutative group
$\therefore H_i \triangleleft E^{\times}$
$i \neq j \Rightarrow \forall h_i \in H_i, |\langle h_i\rangle| = p_i^s, \forall h_j \in H_j, |\langle h_j\rangle| = p_j^t$
$\therefore \text{gcd}(|\langle h_i\rangle|,|\langle h_j\rangle|) = 1$
$\therefore H_i \cap H_j = \{e\}\quad(i \neq j)$
($\because x \in H_i \cap H_j \Rightarrow |\langle x \rangle| = p_i^s = p_j^t \Rightarrow |\langle x \rangle| = 1 \because \text{gcd}(p_i,p_j) = 1$)(４月セミナー宿題)
$E^{\times} \supset H_1 \cdots H_n = \{h_1\cdots h_n| h_i \in H_i, \cdots , h_n \in H_n\}$
$|H_1 \cdots H_n| = p_1^{a_1}\cdots p_n^{a_n}\quad(\because \text{Prop.部分群の積の位数(４月セミナー宿題)})$
$\therefore |H_1 \cdots H_n| = |E^{\times}|$
$E^{\times} = H_1 \cdots H_n$

$\therefore E^{\times} \simeq H_1 \times \cdot \times H_n\quad(\text{Prop.群と同型になる直積群})$
$\simeq \mathbb{Z}/p_1^{a_1}\mathbb{Z} \times \cdot \times \mathbb{Z}/p_n^{a_n}\mathbb{Z} \quad (\text{Prop.素数位数の群})$
$\simeq \mathbb{Z}/p_1^{a_1}\cdots p_n^{a_n}\mathbb{Z} \quad (\because \text{Prop.整数の剰余群と直積})$
$\therefore E^{\times}$:巡回群
)
$\therefore E = F(\alpha)$
Case $|F| = \infty$
$\because \text{Prop.有限次分離拡大は単拡大-1}$
QED.