# 02.5.代数拡大

## Proposition

### Statement

$E/F$:Field Extension
$\alpha \in E, f(x) \in F[x], f(\alpha) = 0$
($\alpha$:Algebraic over $F$)
Then $f(x)$:Irreducible Polynomial over $F$
$\Leftrightarrow$
$f(x)$:Minimal Polynomial over $F$

### Proof

$\Leftarrow$
Suppose
$\exists g(x),h(x) \in F[x], f(x) = g(x)h(x), \text{deg }g(x),\text{deg }h(x) \gt 0$
then
$0 = f(\alpha) = g(\alpha)h(\alpha) \therefore g(\alpha) = 0 \text{ or } h(\alpha) = 0 \quad (\because g(\alpha), h(\alpha) \in E, \text{:Field, Prop.体は整域})$
contradict to $f(x)$:Minimal Polynomial
$\Rightarrow$
Suppose
$g(x) \in F[x]$:Minimal Polynomial
and $\text{deg }g(x) = m \lt n = \text{deg }f(x)$
Then
Let
$f(x) = a_n x^n + \cdots + a_0$
$g(x) = b_m x^m + \cdots + b_0$
then
$f_{1}(x) = f(x) - a_n/b_m x^{n-m}g(x) = a'_{n-1} x ^{n-1} + \cdots a'_0$
$f_{2}(x) = f_{1}(x) - a'_{n-1}/b_m x^{n-m-1}g(x) = a''_{n-2} x ^{n-2} + \cdots a''_0$
repeat this operation. then
$f_{n-m+1}(x) = f_{n-m}(x) - a^{(n-m+1)}_0/b_m g(x) = a^{(n-m+1)}_{m-1} x ^{m-1} + \cdots a^{(n-m+1)}_0$
$f_{n-m+1}(\alpha) = 0$ and deg $f_{n-m+1}(x) \lt m$ $\therefore f_{n-m+1}(x) = 0\quad$($\because g(x)$:Minimal Polynomial and deg $g(x) = m$)
$\therefore f(x) = (a_n/b_m x^{n-m} + a'_{n-1}/b_m x^{n-m-1} + \cdots + a^{(n-m+1)}_0/b_m )g(x)$
contradict to $f(x)$:Irreducible Polynomial
QED

## Proposition

### Statement

$F$:Field
$E=F(a)$:Simple Extension of $F$
$a$ is Algebraic over $F$
$p(x) \in F[x]$ is an irreducible polynomial
$p(a) = 0$
$\text{deg }p(x) = n$
then
$F[x]/(p(x)) \simeq F(a)$
$[E:F] = n$
$1, a, a^2, a^3, \cdots, a^{n-1}$ is base vectors of vector spase $E/F$

### Proof

Let $\phi:f(x) \in F[x] \mapsto f(a) \in F[a]$
then
$\phi$: Homomorphism
$(p(x)) = \text{ker }\phi$
$F[x]/(p(x)) = F[x]/\text{Ker }\phi \simeq \text{Img }\phi = F[a] \quad(\text{Prop.環準同型定理})$
$F[x]$:PID, $p(x)$:Irreducible Polynomial $\Rightarrow (p(x))$:Maximal Ideal
$\therefore F[x]/(p(x))$:Field $\quad(\text{Prop.極大イデアルと体})$
$\therefore F[a]$:Field
$F \subset F[a], a \in F[a]$ and $F[a]$:Field $\therefore F[a] = F(a)$
$\forall f(x) \in F[x], \exists g(x), r(x) \in F[x], f(x) = p(x)g(x) + r(x), \quad (\text{deg }r(x) \lt \text{deg }p(x) = n)$
$\therefore f(a) = p(a)g(a) + r(a)$
$= 0 g(a) + r(a) = r(a) \in F[a], r(a) = b_{n-1} a^{n-1} + \cdots + b_1 a^1 + a^0 \quad(b_{n-1}, \cdots, b_0 \in F)$
$b_{n-1} a^{n-1} + \cdots + b_1 a^1 + a^0 = 0 \Rightarrow b_{n-1},\cdots, b_0 = 0$
$\because$ if $\exists (b_{n-1} \cdots b_0) \neq 0, b_{n-1} a^{n-1} + \cdots + b_1 a^1 + a^0 = 0$
then $r(x) = b_{n-1} x^{n-1} + \cdots + b_1 x^1 + x^0 \neq 0$ and $r(a) = 0$
contradict to $p(x)$:Minimal Polynomial ($\because \text{Prop.既約多項式と最小多項式}$)
QED

## Proposition

### Statement

$E/F$:有限次拡大 $\Rightarrow E/F$:代数拡大

### Proof

Let $[E:F] = n \lt \infty$
Then
$\forall \alpha \in E, (\alpha^n, \alpha^{n-1}, \cdots \alpha, \alpha^0 = 1)$ are linearly dependent over $F$, $\therefore \exists a_n, a_{n-1}, \cdots, a_1, a_0 \in F, \text{s.t. }a_n \alpha^n + \cdots + a_0 = 0 \quad (a_n, \cdots a_0) \neq 0$
$\therefore f(x) = a_n x^n + \cdots + a_0 \in F[x]$ and $f(\alpha) = 0$
QED

## Proposition

### Statement

$E\supset K \supset F$:拡大体
$\{e_\lambda\}_\Lambda \subset E$:basis for $E$ over $K$---(1)
$\{k_\mu\}_M \subset K$:basis for $K$ over $F$---(2)
then
$\{e_\lambda k_\mu\}_{\Lambda, M}$:basis for $E$ over F

### Proof

$\forall \theta \in E, \exists \{a_\lambda\} \subset K, \theta = \sum_\Lambda a_\lambda e_\lambda$
$a_\lambda \in K \therefore \exists \{b^{(\lambda)}_\mu\} \subset F, a_\lambda = \sum_M b^{(\lambda)}_\mu k_\mu$
$\therefore \theta = \sum_{\Lambda, M} b^{(\lambda)}_\mu k_\mu e_\lambda$

Suppose $\sum_{\Lambda,M} c^{(\lambda)}_\mu k_\mu e_\lambda = 0$
$\forall \lambda \in \Lambda, \sum_M c^{(\lambda)}_\mu k_\mu = 0 \quad \because$ (1)
$\forall \mu \in M, c^{(\lambda)}_\mu = 0 \quad \because$ (2)
QED

## Proposition

### Statement

$E\supset K \supset F$:拡大体
$[E:F] = [E:K][K:F]$

### Proof

Prop.拡大体の基底 より明らか. QED

## Proposition

### Statement

$E/F$:Field Expansion
$E_0 = \{x \in F|x:\text{Algebraic over }F\}$
then
$E_0$ is a field.

### Proof

$\forall a, b \in E_0$
$[F(a,b):F] = [F(a,b):F(a)][F(a):F] \quad (\because \text{Prop.拡大体の基底(次元)})$ $a,b \in E_0 \therefore \exists f(x) \in F[x], \exists g(x) \in F[x] \subset F(a)[x] \text{ s.t. } f(a) = g(b) = 0$
$\therefore [F(a,b):F(a)] \lt \infty, [F(a):F] \lt \infty$
$\therefore [F(a,b):F] \lt \infty$ $\therefore F(a,b)$:Algebraic Expansion of $F$---(1)
$a+b, ab, a^{-1} \in F(a,b)$---(2)
(1),(2) leads to $a+b, ab, a^{-1} \in E_0$
$\therefore E_0$ is a field.
QED

## Proposition

### Statement

$E/K,K/F$:Field Expansion
Then
$E/F$:Algebraic $\Leftrightarrow E/K,K/F$:Algebraic

### Proof

$\Rightarrow$
$\forall a \in E, \exists f(x) \in F[x] \subset K[x] \text{ s.t. } f(a) = 0$
$\therefore E/K$:Algebraic
$\forall a \in K \subset E, \exists f(x) \in F[x] \text{ s.t. } f(a) = 0$
$\therefore K/F$:Algebraic
$\Leftarrow$
$\forall a \in E, \exists f(x)$:Minimal Polynomial over $K$
Let $f(x) = k_n x^n + \cdots + k_0 \quad (k_n,\cdots k_0 \in K)$
$[F(a,k_n,\cdots,k_0):F] = [F(a,k_n,\cdots,k_0):F(k_n,\cdots,k_0)][F(k_n,\cdots,k_0):F(k_{n-1},\cdots,k_0)]\cdots[F(k_1,k_0):F(k_0)][F(k_0):F]$---(1)
$[F(a,k_n,\cdots,k_0):F(k_n,\cdots,k_0)] = n \lt \infty$---(2)
$\forall k_i, \exists f_i(x)$:Minimal Polynomial over $F$
$\therefore [F(k_i,k_{i-1},\cdots,k_0):F(k_{i-1},\cdots,k_0)] = \text{ deg } f_i(x) \lt \infty$---(3)
(1),(2),(3) leads to $[F(a,k_n,\cdots,k_0):F] \lt \infty$
$\therefore F(a,k_n,\cdots,k_0)$:Algebraic over $F$
$\therefore a$:Algebraic over $F$
$\therefore E$:Algebraic over $F$
QED