# 02.7体の同型

## Proposition

### Statement

$F,F'$:Field
$\sigma_0:F \rightarrow F'$:Isomorphism
$p(x) \in F[x]$:Irreducible polynomial
($p(x) = a_0 + a_1 x + \cdots + a_n x^n$)
$E$:Minimal Splitting Field over $p(x)$
$E'$:Minimal Splitting Field over $\sigma_0 (p)(x)$
($\sigma_0(p)(x) = \sigma_0(a_0) + \sigma_0(a_1) x + \cdots + \sigma_0(a_n) x^n$)
$\alpha \in E, p(\alpha) = 0$
$\alpha' \in E', \sigma_0(p)(\alpha') = 0$
then
$\exists \sigma: F(\alpha) \rightarrow F'(\alpha')$:Isomorphism over $F$
such that
$\sigma(\alpha) = \alpha'$
$a \in F \Rightarrow \sigma(a) = \sigma_0(a)$

### Proof

$$\begin{xy} \xymatrix { & F \ar[d] \ar[r]^{\sigma_0 \simeq} & F' \ar[d] \\ & F[x] \ar[dl]_{\phi} \ar[d]^{\pi} \ar[r]^{\sigma_0 \simeq} \ar[rd]^{\psi} & F'[x] \ar[d]_{\pi'} \ar[dr]^{\phi'} \\ F(\alpha) & F[x] / (p(x)) \ar[l]_{\simeq} \ar[r]^{\sigma \simeq} & F'[x] / (\sigma_0(p)(x)) \ar[r]^{\simeq} & F'(\alpha')\\ } \end{xy}$$ Let
$\sigma_0(p(x)) = \sigma_0(p)(x)$
$\pi:F'[x] \rightarrow F'[x]/(\sigma_0(p)(x))$
$\psi:F[x] \rightarrow F'[x]/(\sigma_0(p)(x))$
$\psi:f(x) \mapsto \sigma_0(f)(x) \twoheadrightarrow \sigma_0(f)(x) + (\sigma_0(p)(x))$
$\psi(f(x)) = \pi(\sigma_0(f(x)))$
Then
$\psi$:Homomorphism $\quad \because \sigma_0$:Isomorphism, $\pi$:Homomorphism
$\text{Ker }\psi = (p(x)) \quad \because \text{Ker }\pi = (\sigma_0(p)(x)) \simeq (p(x))$
$F[x]/(p(x)) = F[x]/\text{Ker }\psi \simeq \text{Im }\psi = F'[x]/(\sigma_0(p)(x)) \quad \because \text{Prop.環準同型定理}$
$F[x]/(p(x)) \simeq F(\alpha) \quad \because \text{Prop.代数拡大}$
$F'[x]/(\sigma_0(p)(x)) \simeq F'(\alpha') \quad \because \text{Prop.代数拡大}$
$\therefore F(\alpha) \simeq F'(\alpha')$
$a \in F \Rightarrow$
$\sigma:a \mapsto \bar a = a + (p(x)) \mapsto \sigma_0(a) + (\sigma_0(p)(x)) \mapsto \sigma_0(a)$
$\therefore \sigma(a) = \sigma_0(a)$
$\sigma:\alpha \mapsto \bar x = x + (p(x)) \mapsto x + (\sigma_0(p)(x)) \mapsto \alpha'$
$\therefore \sigma(\alpha) = \alpha'$
QED

## Proposition

### Statement

$F,F'$:Field
$\sigma_0:F \rightarrow F'$:Isomorphism
$f(x) \in F[x], \text{deg }f(x) = n$
$E$:Minimal Splitting Field over $f(x)$
$E'$:Minimal Splitting Field over $\sigma_0 (f)(x)$
then
$\exists \sigma: E \rightarrow E'$:Isomorphism
such that
$a \in F \Rightarrow \sigma(a) = \sigma_0(a)$
(Expansion of $\sigma_0$)

### Proof

when $n=1$
$F = E, F' = E'$
$\therefore \sigma = \sigma_0$ fills conditions.
when $n \ge 2$
Suppose $n-1$ follows the statement --- (1)
$\exists \alpha \in E, f(\alpha) = 0$
$\exists \alpha' \in E', \sigma_0(f)(\alpha') = 0$
$\exists \sigma_\alpha:F(\alpha) \rightarrow F'(\alpha')$:Isomorphism and Extension of $\sigma_0 \quad (\because \text{Prop.単拡大の同型})$
$f(x) = (x - \alpha)f_1(x) \quad f_1(x) \in F(\alpha)[x]$
$\sigma_\alpha(f)(x) = (x - \sigma_{\alpha}(\alpha))\sigma_{\alpha}(f_1)(x) = (x - \alpha')f_1'(x) \quad (f_1'(x) = \sigma_{\alpha}(f_1)(x) \in F'(\alpha'))$
$E$:Minimal Splitting Field over $f_1(x)$
$E'$:Minimal Splitting Field over $f_1'(x)$
$\text{deg }f_1(x) = \text{deg }f_1'(x) =n - 1$
$\exists \sigma:E \rightarrow E'$:Isomorphism and Extension of $\sigma_\alpha$
$\therefore \sigma$ is a Isomorphism and Extension of $\sigma_0$
$$\require{AMScd} \begin{CD} E @>{\sigma}>> E' \\ @AAA @AAA \\ F(\alpha) @>{\sigma_\alpha}>> F'(\alpha') \\ @AAA @AAA \\ F @>{\sigma_0}>> F' \end{CD}$$ QED

## Proposition

### Statement

$F$:Field
$f(x) \in F[x], \text{deg }f(x) = n$
$E$:Minimal Splitting Field over $f(x)$
$E'$:Minimal Splitting Field over $f(x)$
then
$E \simeq E'$:Isomorphism

### Proof

In Prop.分解体の同型, Alter F' to F.
$$\require{AMScd} \begin{CD} E @>{\sigma}>> E' \\ @AAA @AAA \\ F @>{\sigma_0}>> F \end{CD}$$ QED