05.4.正規拡大

共役と最小多項式の根

Statement

$E/F$:Algebraic Extension
$\alpha \in E - F$
$p(x) \in F[x]$:Minimal Polynomial of $\alpha$
$E$:Minimal Splitting Field of $p(x)$
then
$\alpha'$:Conjugate of $\alpha$ over $F\quad(\Leftrightarrow \exists \sigma \in \text{Aut}(E/F) \text{ s.t. }\sigma(\alpha) = \alpha')$
$\Leftrightarrow$
$p(\alpha') = 0$

Proof

$\Rightarrow$
$p(\alpha') = p(\sigma(\alpha)) = \sigma(p)(\sigma(\alpha)) = \sigma(p(\alpha)) = \sigma(0) = 0$
$\Leftarrow$
$\exists \mu \in \text{Hom}_F(F(\alpha),F(\alpha')) \text{ s.t. }\mu(\alpha) = \alpha' \quad(\because\text{Prop.単拡大の同型})$
$\exists \sigma:E \rightarrow E \text{ s.t. } \sigma|_{F(\alpha)} = \mu\quad(\because\text{Prop.分解体の同型})$
QED.

正規拡大の同値条件

Statement

$E/F$:Finite Extension
$\mathbb{S} = \{\sigma:E \rightarrow \bar F, \sigma \mid_F = id_F\}$
then
$E/F$:Normal Extension $\Leftrightarrow \forall \sigma \in \mathbb{S},\sigma(E) \subset E$

Proof

$\Rightarrow$
$\forall \alpha \in E$
$\exists p(x) \in F[x]$:Minimal Polynomial of $\alpha\quad(p(\alpha) = 0)$
then
$\exists \alpha_1, \cdots, \alpha_n \in E, \text{ s.t. } p(x) = (x - \alpha_1)\cdots(x - \alpha_n) \quad(\because E/F\text{:Normal Extension})$
$p(\sigma(\alpha)) = \sigma(p)(\sigma(\alpha)) = \sigma(p(\alpha)) = \sigma(0) = 0$
$\therefore \sigma(\alpha) = \alpha_i \quad(1 \le i \le n)$
$\therefore \sigma(\alpha) \in E$
$\therefore \sigma(E) \subset E$
$\Leftarrow$
$\forall \alpha \in E$
$\exists p(x) \in F[x]$:Minimal Polynomial of $\alpha\quad(p(\alpha) = 0)$
$\exists \alpha_1, \cdots, \alpha_n \in \bar F, \text{ s.t. } p(x) = (x - \alpha_1)\cdots(x - \alpha_n) \quad(\alpha_1 = \alpha)$
$\forall \alpha_i(1\le i \le n),\exists \mu_i:F(\alpha) \rightarrow \bar F \text{ s.t. } \mu_i(\alpha) = \alpha_i \quad(\because\text{Prop.単拡大の同型})$ $F(\alpha) \simeq F(\alpha_i)$
$\exists \sigma_i:E \rightarrow \bar F \text{ s.t. } \sigma_i \mid_M = \mu_i \quad(\because\text{Prop.分解体の同型})$
$\therefore \alpha_i = \mu_i(\alpha) = \sigma_i(\alpha) \in \sigma_i(E) \subset E$
$\therefore \alpha_i \in E$
$\therefore E/F$:Normal Extension
QED.

正規拡大の同値条件2

Statement

$F$:Field
$\alpha_1,\cdots,\alpha_n$:Algebraic over $F$
$\alpha'_i$:Conjugate of $\alpha_i$ over $F$
$E = F(\alpha_1,\cdots,\alpha_n)$
then
$E/F$:Normal Extension $\Leftrightarrow \forall \alpha'_i \in E$

Proof

$\Rightarrow$
$\forall \alpha_i, \exists \sigma_i:E \rightarrow \bar F \text{ s.t. } \sigma_i(\alpha_i) = \alpha'_i\quad(\because\alpha'_i$:Conjugate of $\alpha_i$ over $F)$
$\forall \sigma:E \rightarrow \bar F, \sigma(\alpha_i) \in \sigma(E) \subset E \quad(\because\text{Prop.正規拡大の同値条件})$
$\therefore \sigma_i(\alpha_i) = \alpha'_i \in E$
$\Leftarrow$
$\forall \sigma:F(\alpha_i,\cdots,\alpha_n) \rightarrow \bar F \text{ s.t. } \sigma\mid_F = id_F$
$\sigma(F(\alpha_1,\cdots,\alpha_n)) = F(\sigma(\alpha_1),\cdots,\sigma(\alpha_n))$
$\sigma(\alpha_i) \in F(\alpha_1,\cdots,\alpha_n) \quad(\because \sigma(\alpha_i) = \alpha'_i \in E = F(\alpha_1,\cdots,\alpha_n))$
$\therefore F(\sigma(\alpha_1),\cdots,\sigma(\alpha_n)) \subset F(\alpha_1,\cdots,\alpha_n)$
$\therefore \sigma(F(\alpha_1,\cdots,\alpha_n)) \subset F(\alpha_1,\cdots,\alpha_n)$
$\therefore \sigma(E) \subset E$
$\therefore E/F:\text{Normal Extension}\quad(\because\text{Prop.正規拡大の同値条件})$
QED.

Proposition

正規拡大の同値条件3

Statement

$E/F$:Finite Field Extension
then
$E/F$:Normal Extension $\Leftrightarrow \exists f(x) \in F[x]$ such that $E$:Minimal Splitting Field over $f(x)$

Proof

$\Rightarrow$
$\exists \alpha_1,\cdots,\alpha_n \in E \text{ s.t. } E = F(\alpha_1,\cdots,\alpha_n) \quad(\because E/F\text{:Finite Field Extension})$
let $q_i(x) \in F[x]$:Minimal Polynomial of $\alpha_i$
let $f(x) = q_1(x)\cdots q_n(x)$
then all roots of $q_i(x)$ is in $E \quad(\because E/F\text{:Normal Extension})$
$\therefore$ all roots of $f(x)$ is in $E$
$\therefore F(\text{all roots of }f(x)) \subset E$
obviously $E = F(\alpha_1,\cdots,\alpha_n) \subset F(\text{all roots of }f(x)) \quad (\because f(\alpha_i) = 0)$
$\therefore E = F(\text{all roots of }f(x))$
$F(\text{all roots of }f(x))$ is the minimal splitting field of $f(x)$
$\therefore E$:Minimal Splitting Field over $f(x)$

$\Leftarrow$
let
$E$:Minimal Splitting Field over $f(x)$
$f(x) = (x - \alpha_1)\cdots(x - \alpha_n) \quad (\alpha_1, \cdots \alpha_n \in E)$
then
$E = F(\alpha_1, \cdots, \alpha_n)$

let $q_i(x) \in F[x]$:Minimal Polynomial of $\alpha_i$
then $q_i(x) | f(x)$
$\therefore$ all roots of $q_i(x) \in \{\alpha_1,\cdots,\alpha_n\} \subset E$
$\therefore$ all conjugates of $\alpha_i \in E\quad(\because\text{Prop.共役と最小多項式の根})$
$\therefore E/F:\text{Normal Extension}\quad(\because\text{Prop.正規拡大の同値条件2})$
その２
let
$E$:Minimal Splitting Field over $f(x)\in F[x]$
$f(x) = (x - \alpha_1)\cdots(x - \alpha_n) \quad (\alpha_1, \cdots \alpha_n \in E)$
then
$E = F(\alpha_1, \cdots, \alpha_n)$

let
$\forall \beta_1 \in E$
$g(x) \in F[x] \subset E[x]$:Minimal Polynomial of $\beta_1$
$g(x) = (x - \beta_1)\cdots(x - \beta_m)$
then
let
$L$:Minimal Splitting Field of $g(x) \in E[x]$
then
$L = E(\beta_1, \cdots, \beta_m)$
$L = F(\alpha_1, \cdots, \alpha_n, \beta_1, \cdots, \beta_m)$
$L \supset E \supset F$
let
$M$:Minimal Splitting Field of $g(x) \in F[x]$
then
$M = F(\beta_1, \cdots, \beta_m)$
$L = F(\alpha_1,\cdots,\alpha_n,\beta_1,\cdots,\beta_m) = M(\alpha_1,\cdots,\alpha_n)$
$L \supset M \supset F$
$\forall \beta_j,\exists \mu:M\rightarrow M, \text{ s.t. } \mu|_F = id, \mu(\beta_1) = \beta_j \quad (\because\text{Prop.分解体の同型3})$
$L = M(\alpha_1,\cdots,\alpha_n)$ and $f(x) \in F[x] \subset M[x]$
$\therefore L$:Minimal Splitting Field of $f(x) \in M[x]$
$\therefore \exists \lambda:L\rightarrow L, \text{ s.t. } \lambda|_M = \mu \quad (\because\text{Prop.分解体の同型2})$
$\forall \alpha_i,f(\lambda(\alpha_i)) = \lambda(f)(\lambda(\alpha_i)) = \lambda(f(\alpha_i)) = \lambda(0) = 0$
$\therefore \lambda(\alpha_i) \in \{\alpha_1,\cdots,\alpha_n\}$
$\therefore \lambda(E) = \lambda(F(\alpha_1,\cdots,\alpha_n)) = F(\lambda(\alpha_1),\cdots,\lambda(\alpha_n)) \subset F(\alpha_1,\cdots,\alpha_n) = E$
$\therefore \beta_j = \mu(\beta_1) = \lambda(\beta_1) \in \lambda(E) \subset E$
$\therefore \text{ all roots of } g(x) \in E$
$\therefore \text{ all conjugates of } \beta_1 \in E\quad(\because\text{Prop.共役と最小多項式の根})$
$\therefore E/F:\text{Normal Extension}\quad(\because\text{Prop.正規拡大の同値条件2})$
QED.

Proposition

正規拡大の中間体

Statement

$E/F$:Normal Field Extension
$E \supset M \supset F$
then
$E/M$:Normal Extension

Proof

$\forall \alpha \in E$
let $p(x) \in F[x]$:Minimal Polynomial of $\alpha$
then $p(\alpha_i) = 0 \Rightarrow \alpha_i \in E \quad (\because E/F:\text{Normal Extension})$
let $q(x) \in M[x]$:Minimal Polynomial of $\alpha$
then $q(x) | p(x)\therefore q(\beta) = 0 \Rightarrow p(\beta) = 0 \therefore \beta \in E$
$\therefore E/M$:Normal Extension
QED.

Proposition

正規拡大列

Statement

$E/M$:Normal Extension
$M/F$:Normal Extension
then
$E/F$ is't necessarily Normal Extension

Proof

let
$F = \mathbb{Q}$
$M = \mathbb{Q}(\sqrt{2})$
$E = \mathbb{Q}(\sqrt{2},\sqrt[4]{2}) = \mathbb{Q}(\sqrt[4]{2})$
then
$x^2 - 2$:Minimal Polynomial of $\sqrt{2}$ over F
$x^2 - \sqrt{2}$:Minimal Polynomial of $\sqrt[4]{2}$ over M
$x^2 - 2 = (x - \sqrt{2})(x + \sqrt{2})$
$x^2 - \sqrt{2} = (x - \sqrt[4]{2})(x + \sqrt[4]{2})$
$\pm\sqrt{2} \in M$
$\pm\sqrt[4]{2} \in E$
$M/F$:Normal Extension
$E/M$:Normal Extension
$x^4 - 2$:Minimal Polynomial of $\sqrt[4]{2}$ over $F$
$x^4 - 2 = (x - \sqrt[4]{2})(x - \sqrt[4]{2}i)(x + \sqrt[4]{2})(x + \sqrt[4]{2}i)$
$i \notin E$
$\therefore E/F$ is not Normal Extension
QED.