# 02.8.代数閉包の存在証明

## Proposition

### Statement

$E,F$:Field
$E/F$:Algebraic Extension
$F$:Algebraic Closed Field
Then
$E = F$

### Proof

$\forall a \in E, \exists f(x) \in F[x], \text{ s.t. } f(a) = 0$ ($\because E/F$:Algebraic Extension)
$a \in F$ ($\because F$Algebraic Closed Field)
$\therefore E \subset F$
$\therefore E = F$ QED

## Example

$\mathbb{C,R,Q}$:Field
$\mathbb{C}$:Algebraic Closed Field
$\mathbb{C}$ is Algebraic Closure over $\mathbb{R}$
$\mathbb{C}$ is not Algebraic Closure over $\mathbb{Q} \quad(\because \sin \theta$ is not Algebraic Element over $\mathbb{Q})$

## Proposition

### Statement

$E/F$:Field Extension
$E$:Algebraic Closed Field
$K = \{a \in E|\exists f(x) \in F[x], f(a) = 0\} \quad (= \{a \in E| a \text{ is algebraic over } F\})$
Then
$K$ is an Algebraic Closure over $F$

### Proof

$(E\supset )K \supset F \quad (\because a \in F \Rightarrow (x - a) \in F[x])$
$K$:Field $\quad(\because \text{Prop.代数的元で構成する体})$
$\forall a \in K, \exists f(x) \in F[x] \subset K[x], \text{ s.t. } f(a) = 0 \therefore K/F$:Algebraic Extension---(1)
$\forall g(x) \in K[x] \subset E[x]$
$g(x) = (x - b_1)\cdots(x - b_n)\quad (b_1, \cdots , b_n \in E \because E\text{;Algebraic Closed Field})$
$\forall b_i(i=1,\cdots, n), g(b_i) = 0 \therefore b_i$:Algebraic over $K$
$[K(b_i)/K] \le n \therefore K(b_i)$ is an Algebraic Extension ---(2)$\quad (\because \text{Prop.有限次拡大})$
(1),(2) $\therefore K(b_i)/F$:Algebraic Extension $\quad (\because \text{Prop.代数拡大列})$
$\therefore \exists h_{b_i}(x) \in F[x] \text{ s.t. } h_{b_i}(b_i) = 0$
$\therefore b_i \in K$
$\therefore K$:Algebraic Closed Field
QED.

## Theorem

### Statement

$F$:Field
then
$\exists \bar F$:Algebraic Closure of $F$

### Proof

It is sufficient to prove $\exists E/F$:Algebraic Closed Field.
($\because \text{Prop.代数閉体から代数閉包を構築する}$)
Let
$S = \{f(x) \in F[x]|f(x) \text{ is irreducible}\} (\subset F[x])$
$A = F[x_f]_{f \in S}$ $\quad (\leftarrow F \text{係数、無限変数多項式環})$
$I = (\{f(x_f) | f \in S\}) \quad (\leftarrow \text{Ideal of } A)$
$= \{a_1 f_1(x_1) + \cdots + a_n f_n(x_n)| a_1, \cdots, a_n \in A, f_1(x_1), \cdots f_n(x_n) \in A, f_1, \cdots f_n \in S\}$

$A \neq I$
$\because$
Suppose $A = I$
Then $1 \in I \quad (\because 1 \in A)$
$\exists f(x_1,\cdots, x_m,\cdots) \in I, 1 = f(x_1,\cdots,x_m,\cdots) = a_1 f_1(x_1) + \cdots + a_m f_m(x_m) \quad (a_1, \cdots, a_n \in A, f_1, \cdots f_n \in S)$
Let $f'(x) = f_1(x)f_2(x)\cdots f_m(x) \quad (\subset F[x])$
then $\exists E / F:\text{Splitting Field of }f'(x) \quad (\because \text{Prop.分解体の存在})$
$\therefore \forall f_i(x)_{i = 1 \cdots m}, \exists \alpha_i \in E \text{ s.t. } f_i(\alpha_i) = 0$
then $f(\alpha_1,\cdots,\alpha_m,0,\cdots,0) \in E$ and $1 = f(\alpha_1,\cdots,\alpha_m,0,\cdots,0) = 0$ contradiction.
$\exists J \supset I$:Maximal Ideal of $A \quad (\because \text{Prop.極大イデアルの存在})$
Let
$\psi:A \twoheadrightarrow A/J\quad$($=E_1$:Field)
$\phi:F \hookrightarrow A \twoheadrightarrow A/J = E_1$
then
$\phi$:injective $(\because \text{Prop.体の準同型写像は単射})$
$E_1/ \bar F$:Field Extension $\quad (\bar F = \phi(F))$
$F \simeq \bar F \quad \therefore E_1/ F$:Field Extension
Let $\psi(x_f) = \alpha_f \in E_1 \quad (f \in S)$
then
$\forall f \in S$
$\psi(f(x_f)) = \bar f(\alpha_f) \quad \because \psi \text{ is a Homomorphism}$
$\psi(f(x_f)) = 0 \quad \because f(x_f) \in I \subset J$
$\therefore \bar f(\alpha_f) = 0 \quad (f \in S)$
$\therefore f(\alpha_f) = 0 \quad (f \in S)$

$\forall g(x) \in F[x], \exists \{f_1, \cdots f_l\} \subset S, g(x) = f_1(x)\cdots f_l(x)$
$\therefore \exists \alpha \in E_1, g(\alpha) = 0$

Let
$F \rightarrow E_1 \rightarrow E_2 \rightarrow \cdots$:Extend recuisively
$E_\infty = \cup_{i=1}^\infty E_i$
then
$E_\infty$ is a field extension of $F$
$\because$
$\forall a,b \in E_\infty, \exists E_j, a,b \in E_\infty$
$a + b, ab, a^{-1} \in E_j \subset E_\infty \quad (a \neq 0)$
$\forall f(x) \in E_\infty[x]$
Let $f(x) = a_n x^n + \cdots + a_0 \quad (a_n,\cdots,a_0 \in E_\infty)$
$(\text{deg }f = n)$
then $\exists E_m \text{ s.t. } a_n, \cdots a_0 \in E_m$
$\exists \alpha_1 \in E_{m+1} \text{ s.t. } f(\alpha_1) = 0 \therefore f(x) = (x - \alpha_1) f'(x) \quad (\text{deg }f' = n-1)$
$\therefore \alpha_1 \cdots \alpha_n \in E_{m+n} \subset E_\infty \text{ s.t. } f(x) = (x - \alpha_1)\cdots(x - \alpha_n)$
$\therefore E_\infty$ is an Algebraic Closed Field
QED

## Proposition

### Statement

$F_1, F_2$:Field
$F_1 \simeq F_2$
$\sigma_0:F_1 \rightarrow F_2$
$E_1$:Algebraic Closure of $F_1$
$E_2$:Algebraic Closure of $F_2$
then
$E_1 \simeq E_2$
$\exists \sigma:E_1 \rightarrow E_2 \text{ s.t. } \sigma|_{F_1} = \sigma_0$

### Proof

Let
$S = \{(K, \sigma)|E_1/K/F1, \sigma:K \rightarrow E_2, \sigma|_{F_1} = \sigma_0\}$
$(K_1, \sigma_1) \le (K_2, \sigma_2) \overset{\text{def}}{\Leftrightarrow} K_1 \subset K_2, \sigma_2|_{K_1} = \sigma_1$
then
$S \neq \emptyset \quad \because (F_1, \sigma_0) \in S$

Let
$T$:全順序部分集合 of $S \quad (\emptyset \neq T \subset S)$
$K_\infty = \cup_{(K,\sigma) \in T} K$
$\sigma_\infty(a) = \sigma_i(a) \text{ s.t. } a \in K_i \subset K_\infty, (K_i, \sigma_i) \in T$
then
$E_1 / K_\infty / F_1$:Field
$\sigma_\infty$ is well-defined
$(K_\infty, \sigma_\infty) \in S$
$\forall (K,\sigma) \in T, (K,\sigma) \le (K_\infty, \sigma_\infty) \therefore (K_\infty, \sigma_\infty)$ is upper bound of $T$
$\therefore \exists (\bar K, \bar \sigma)$:極大元 of $S \quad \because \text{(Zornの補題)}$

$\bar K = E_1$
$\because$
Suppose $\bar K \subsetneq E_1$
$\exists \alpha \in E_1, \alpha \notin \bar K$
$\exists p(x) \in \bar K[x]$:Minimal Polynomial over $\bar K \quad (\because E_1:\text{Algebraic Closure of }F_1, F_1[x] \subset \bar K[x])$
$\exists \bar K' / \bar K$ :Minimal Splitting Field of $p(x) \quad (\because \text{Prop.分解体の存在})$
$\exists \sigma':\bar K' \rightarrow E_2, \text{ s.t. } \sigma'|_{\bar K} = \bar \sigma \quad(\because \text{Prop.分解体の同型})$
$(\bar K',\sigma') \ge (\bar K, \bar \sigma)$ leads to contradiction of $(\bar K, \bar \sigma)$:極大元 of $S$
$\therefore \bar K = E_1$

$$\begin{xy} \xymatrix { & \bar K' \ar[r]^{\sigma' \simeq} & \sigma'(\bar K') & \\ & \bar K \ar[u] \ar[r]^{\sigma \simeq} & \sigma(\bar K) \ar[u] \\ & F_1 \ar[u] \ar[r]^{\sigma_0 \simeq} & F_2 \ar[u] & \\ } \end{xy}$$ $\bar \sigma$:injective $\quad (\because \text{Prop.体の準同型写像は単射})$
$\bar \sigma$:surjective $\quad(\bar \sigma(\bar K) = E_2)$
$\because$
obiously $\bar \sigma(\bar K) \subset E_2$
$\beta \in E_2$
$\exists q(x) \in F_2[x]$:Minimal Polynomial of $\beta$
$\sigma_0^{-1}(q)(x) \in F_1[x]$
$\sigma_0^{-1}(q)(x) = (x - \alpha_1)\cdots(x - \alpha_n) \in \bar K[x]$
$\bar \sigma(\sigma_0^{-1}(q)(x)) = (x - \bar \sigma(\alpha_1))\cdots(x - \bar \sigma(\alpha_n)) \in \bar \sigma (\bar K)[x]$

$\bar \sigma(\sigma_0^{-1}(q)(x)) = \sigma_0(\sigma_0^{-1}(q)(x))= q(x) \quad \because \sigma_0^{-1}(q)(x) \in F_1[x], \bar \sigma|_{F_1} = \sigma_0$
$\therefore q(x) = (x - \bar \sigma(\alpha_1))\cdots(x - \bar \sigma(\alpha_n))$
$q(\beta) = (\beta - \bar \sigma(\alpha_1))\cdots(\beta - \bar \sigma(\alpha_n)) = 0$
$\therefore \exists \alpha_i \in \bar K, \bar \sigma(\alpha_i) = \beta \quad (i = 1..n)$
$\therefore \bar \sigma(\bar K) \supset E_2$

$\therefore E_1 \simeq E_2$
QED