# 05.2.体の同型の数

## Proposition

### Statement

$F$:Field
$\alpha$:Algebraic element over $F$
$p(x)$:Minimal Polynomial of $\alpha$
$\mathbb{S} = \{\sigma:F(\alpha) \rightarrow \bar F|\sigma |_F = id_F,\sigma:\text{Homomorphism}\}$
then
$|\mathbb{S}| \le [F(\alpha):F]\quad(=\text{deg}p(x))$
and
$|\mathbb{S}| = [F(\alpha):F] \Leftrightarrow \alpha$:Separable over $F$

### Proof

let
$n = [F(\alpha):F]\quad (\text{deg }p(x) = n)$
$p(x) =a_n x^n + \cdots + a_1 x + a_0\quad(a_0,\cdots,a_n \in F)$
then
$\exists \alpha_1,\cdots,\alpha_n \in \bar F\text{ such that } p(\alpha_i)= 0$
(one of $\alpha_i$ is $\alpha$. let $\alpha_1 = \alpha$)
$p(x) = (x - \alpha_1)\cdots(x - \alpha_n)$

then
$\exists\sigma_i:F(\alpha) \rightarrow \bar F\quad (i = 1 ,\cdots, n)$
such that
$\sigma_i(x) = x \quad (x \in F)$
$\sigma_i(\alpha) = \alpha_i$
$(\because \text{Prop.単拡大の同型})$

Let $\mathbb{T} = \{\sigma_1, \cdots \sigma_n\}$
then
each $\alpha_i \in \mathbb{S}$
$\therefore \mathbb{T} \subset \mathbb{S}$

On the other hand
$\forall \sigma \in \mathbb{S}$
$p(\sigma(\alpha)) = a_n (\sigma(\alpha))^n + \cdots + a_1 \sigma(\alpha) + a_0$
$= \sigma(a_n) (\sigma(\alpha))^n + \cdots + \sigma(a_1) \sigma(\alpha) + \sigma(a_0) = \sigma(p)(\sigma(\alpha))$
$= \sigma(a_n \alpha^n + \cdots + a_1 \alpha + a_0) = \sigma(p(\alpha))$
$=\sigma(0)=0$
$\therefore \exists \alpha_i \quad (i = 1 ,\cdots, n) \text{ s.t. } \sigma(\alpha) = \alpha_i$
$\therefore \sigma \in \mathbb{T}$
$\therefore \mathbb{S} \subset \mathbb{T}$

$\therefore \mathbb{S} = \mathbb{T} =\{\sigma_1, \cdots \sigma_n\}$
$\therefore |\mathbb{S}| \le n$ (if $\exists i,j \text{ s.t. } \alpha_i = \alpha_j$, then $|\mathbb{S}| \lt n$.)
$\therefore$
$|\mathbb{S}| = n \Leftrightarrow \sigma_i \neq \sigma_j \quad (i \neq j)$
$\Leftrightarrow \alpha_i \neq \alpha_j \quad(i \neq j)$
$\Leftrightarrow \alpha$:Separable

QED.

## Proposition

### Statement

$E/F$:Finite Field Extension
$\phi:F \simeq F'$
$\mathbb{S} = \{\sigma:E \rightarrow \bar F' \mid \sigma |_F = \phi,\sigma:\text{Homomorphism}\}$
then
$|\mathbb{S}| \le [E:F]$

### Proof

(i)$[E:F] = 1$ then
$\mathbb{S} = \{\phi\}$
$\therefore |\mathbb{S}| = 1 = [E:F]$
(ii)$[E:F] = n \gt 1$ and when extension degree is less than $n$, the statement is true. then
$\exists \alpha_1, \cdots, \alpha_t, \text{ s.t. } E = F(\alpha_1, \cdots, \alpha_t)$
let
$M = F(\alpha_1,\cdots,\alpha_{t-1})$
$\mathbb{M} = \{\mu:M \rightarrow \bar F' \mid \mu |_F = \phi,\mu:\text{Homomorphism}\}$
$m = |\mathbb{M}|$
then
$E = M(\alpha_t)$
$[E:M][M:F] = [E:F]\therefore [M:F] \lt [E:F] = n$
$\therefore m \le [M:F]$

define
$\mathbb{S}^{(i)} = \{\sigma \in \mathbb{S} \mid \sigma|_M = \mu_i\}\quad(i=1,\cdots,m)$
$s_i = |\mathbb{S}^{(i)}|$
then
$\mathbb{S} = \cup_{i=1\cdots m} \mathbb{S}^{(i)}$
$\mathbb{S}^{(i)} \cap \mathbb{S}^{(j)} = \emptyset \quad(i \neq j)$
$\therefore |\mathbb{S}| = \sum_{i=1\cdots m} s_i$

let
$f(x)$:Minimal Polynomial of $\alpha_t$ over $M$
fix $\mathbb{S}^{(i)}$
$\forall \sigma^{(i)}_k \in \mathbb{S}^{(i)}$
$f(\alpha_t) = 0$
$\therefore$
$\sigma^{(i)}_k(f(\alpha_t)) = \sigma^{(i)}_k(f)(\sigma^{(i)}_k(\alpha_t)) = \mu^{(i)}(f)(\sigma^{(i)}_k(\alpha_t))$
$\sigma^{(i)}_k(0) = 0$
$\therefore \mu^{(i)}(f)(\sigma^{(i)}_k(\alpha_t)) = 0$
$\therefore(x - \sigma^{(i)}_1(\alpha_t))\cdots (x - \sigma^{(i)}_{s_i}(\alpha_t)) \mid \mu^{(i)}(f)(x)$
$k \neq l \Rightarrow \sigma^{(i)}_k(\alpha_t) \neq \sigma^{(i)}_l(\alpha_t)$
$|\mathbb{S}^{(i)}| = s_i \le \text{deg}\mu^{(i)}(f)$
$\therefore\text{deg}f = \text{deg}\mu^{(i)}(f)$
$|\mathbb{S}| = \sum_{i=1\cdots m} s_i \le \sum_{i=1\cdots m} \text{deg}\mu^{(i)}(f) = m \cdot \text{deg}f = m \cdot [E:M] \le [M:F][E:M] = [E:F]$
$\therefore |\mathbb{S}| \le [E:F]$
$$\begin{xy} \xymatrix { & M(\alpha_t)= & E \ar[r]^{\sigma^{(i)}_k} & \sigma^{(i)}_k(E) & &\\ & F(\alpha_1,\cdots,\alpha_{t-1})= & M \ar[u]^{f(\alpha) = 0} \ar[r]^{\sigma^{(i)}_k = \mu^{(i)}} & \sigma^{(i)}_k(M) \ar[u]_{\mu_i(f)(\sigma^{(i)}_k(\alpha_t)) = 0} & &\\ & & F \ar[u] \ar[r]^{\phi} & F' \ar[u] & &\\ } \end{xy}$$ $$\begin{xy} \xymatrix { & \mu^{(1)} & \sigma^{(1)}_1,\cdots,\sigma^{(1)}_{s_1} & (x - \sigma^{(1)}_1(\alpha_t))\cdots(x-\sigma^{(1)}_{s_1}(\alpha_t)) \mid \mu^{(1)}(f)(x) & &\\ & \vdots & \vdots & \vdots & &\\ & \mu^{(m)} & \sigma^{(m)}_1,\cdots,\sigma^{(m)}_{s_m} & (x - \sigma^{(m)}_1(\alpha_t))\cdots(x-\sigma^{(m)}_{s_m}(\alpha_t)) \mid \mu^{(m)}(f)(x) & &\\ } \end{xy}$$ QED.

## Proposition

### Statement

$E/F$:Finite Extension
$\phi:F \rightarrow F'$:Isomorphism
$\sigma:E \rightarrow E',\quad(\sigma \mid_F = id)$:Isomorphism
then
$\sigma':E \rightarrow \sigma(E-F) \cup \phi(F), \quad(\sigma' \mid_{E-F} = \sigma, \sigma' \mid_F = \phi)$ is a Isomorphism

### Proof

case $a,b \in F$
then
$ab \in F$
$\therefore \sigma'(ab) = \phi(ab) = \phi(a)\phi(b) = \sigma'(a)\sigma'(b)$

case $a,b \in E-F$
$ab \in F$
$\therefore \sigma'(ab) = \phi(ab) \simeq ab = \sigma(ab) = \sigma(a)\sigma(b) = \sigma'(a)\sigma'(b)$
$ab \in E - F$
$\therefore \sigma'(ab) = \sigma(ab) = \sigma(a)\sigma(b) = \sigma'(a)\sigma'(b)$

case $a \in F,b \in E-F$
$ab \in E - F \quad(\because ab \in F \Rightarrow b = ab a^{-1} \in F\quad(\text{contradiction}))$
$\therefore \sigma'(ab) = \sigma(ab) = \sigma(a)\sigma(b) = a\sigma(b) \simeq \phi(a)\sigma(b) = \sigma'(a)\sigma'(b)$

$\therefore \sigma'$:Homomorphism
$\sigma,\phi$:injection.$\therefore \sigma'$:injection
obviously $\sigma'$:surjection
$\therefore \sigma'$:Isomorphism
$$\sigma':E\rightarrow \sigma'(E)\\ \\ \begin{xy} \xymatrix { & E \ar[r]^{\sigma} & E' \ar[r]^{id} & E' &\\ & F \ar[u] \ar[r]^{id} & F \ar[u] \ar[r]^{\phi} & \phi(F) \ar[u] &\\ } \end{xy}$$ QED.

## Proposition

### Statement

$E/F$:Finite Extension
$F \subset M \subset E$
$\mathbb{S} = \{\sigma:E \rightarrow \bar F \quad | \quad \sigma |_F = id_F\}$
$\mathbb{M} = \{\mu:M \rightarrow \bar F \quad | \quad \mu |_F = id_F\}$
$\mathbb{L} = \{\lambda:E \rightarrow \bar F \quad | \quad \lambda |_M = id_M\}$

then
$|\mathbb{S}| = |\mathbb{M}||\mathbb{L}|$

### Proof

その１
define $\Psi$ as
$\Psi:\mathbb{M} \times \mathbb{L} \rightarrow \mathbb{S}$
$(\mu, \lambda) \mapsto \sigma \quad(\text{ s.t. } \sigma\mid_M = \mu,\sigma\mid_{E-M} = \lambda)$
(1)
$\psi$ is well-defined.$(\because\text{Prop.同型写像の数})$

(2)
let
$\sigma \quad(\text{ s.t. } \sigma\mid_M = \mu,\sigma\mid_{E-M} = \lambda)$
$\sigma' \quad(\text{ s.t. } \sigma'\mid_M = \mu',\sigma'\mid_{E-M} = \lambda')$
then
if $\sigma = \sigma'$
$\mu = \sigma \mid_M = \sigma' \mid_M = \mu'$
$\lambda = \sigma \mid_{E-M} = \sigma' \mid_{E-M} = \lambda'$
$\therefore (\mu,\lambda) = (\mu',\lambda')$
$\therefore \psi$ is injective.

(3)
$\forall \sigma \in \mathbb{S} \Rightarrow \exists \mu \in \mathbb{M} \text{ s.t. } \sigma\mid_M = \mu$
$\forall \sigma \in \mathbb{S} \Rightarrow \exists \lambda \in \mathbb{L} \text{ s.t. } \sigma\mid_{E-M} = \lambda$
$\therefore \psi$ is surjective.

$\therefore |\mathbb{S}| = |\mathbb{M}||\mathbb{L}|$
QED.