03.2.完全体

Proposition

整数から体への写像

Statement

$\mathbb{Z}$:Integer
$F$:Field
$f:\mathbb{Z}\longrightarrow F$:Homomorphism
then
(1) $f$ is identical
(2) $\text{Ker}f = (\alpha) \text{ s.t. } \alpha \in \mathbb{Z}$ is 0 or prime number
(3) $\alpha$:characteristic of $F$

Proof

(1)
$f(1) = f(1\cdot 1) = f(1) \cdot f(1)$
$f(1) \in F \therefore \exists f(1)^{-1} \in F$
$\therefore 1_F = f(1)f(1)^{-1} = f(1)f(1)f(1)^{-1} = f(1)$
$\therefore f(1) = 1_F$
$\forall n \in \mathbb{Z} > 0, f(n) = f(1 + \cdots + 1) = f(1) \cdots f(1) = n \cdot f(1) = n \cdot 1_F$.

(2)
$\mathbb{Z} / \text{Ker}f \simeq \text{Im}f \quad (\because \text{Prop.環準同型定理})$
$\text{Im}f$ is a partial field of $F$
$\therefore \mathbb{Z} / \text{Ker}f$:Field $\therefore$Domain
$\therefore \text{Ker}f$:Prime Ideal $\quad(\because \text{Prop.素イデアルと整域})$
$\mathbb{Z}$ is PID $\therefore \exists \alpha \in \mathbb{Z} \text{ s.t. }\text{Ker}f = (\alpha)$
$\alpha = 0$ or prime element of $\mathbb{Z} \quad(\because \text{Prop.素イデアルと素元})$

(3)
$\text{Ker}f = (0) \Leftrightarrow \forall n(\neq 0) \in \mathbb{Z},f(n) \neq 0 \Leftrightarrow \forall n(\neq 0) \in \mathbb{Z}, n\cdot f(1) = n\cdot 1_F \neq 0$
$\text{Ker}f = (p) \Leftrightarrow \forall m \in \mathbb{Z},f(mp) = 0 \Leftrightarrow f(p) = p \cdot f(1) = p \cdot 1_F = 0$

Proposition

標数Pの分離多項式の条件

Statement

$F$:Field
$\text{ch }F = p \gt 0$:prime number
$f(x) \in F[x]$:Irreducible Polynomial
then
$f(x)$:non-Separable Polynomial
$\Leftrightarrow$
$\exists g(x) \in F[x]$:Irreducible Separable Polynomial
$\exists n \gt 0 \text{ s.t. } f(x) = g(x^{p^n})$

Proof

$\Rightarrow$
$f(x)$:non-Separable Polynomial $\Leftrightarrow f'(x) = 0 (\in F[x]) \quad(\because \text{Prop.分離多項式の条件})$
Let $f(x) = a_n x^n + \cdots + a_1 x + a_0$
then $f'(x) = n a_n x^{n-1} + \cdots a_1 = 0$
$\therefore i a_i = 0_F \quad (0 \lt i \le n)$
$\therefore a_i = 0_F$ or $i = pm$
$\therefore f(x) = a_{pm} x^{pm} + a_{p(m-1)} x^{p(m-1)} + \cdots + a_p x^p + a_0$
let $g_1(x) = a_{pm} x^{m} + a_{p(m-1)} x^{(m-1)} + \cdots + a_p x + a_0$
then $f(x) = g_1(x^p)$
$g_1(x)$:Irreducible $\because g_1(x)$:Reducible $\Rightarrow f(x)$:Reducible lead to contradiction
if $g_1(x)$:non-Separable then $\exists g_2(x) \text{ s.t. } g_1(x) = g_2(x^p)$:Irreducible $\quad \because g_1(x)$ has a same feature of $f(x)$
finally $\exists k \text{ s.t. } g_{k+1}(x) = g_k(x)$:Irreducible Separable Polynomial
$f(x) = g_1(x^p) = g_2(x^{p^2}) = \cdots g_k(x^{p^k})$
$\Leftarrow$
$f'(x) = g'(x^{p^n}) \cdot p^n \cdot x^{p^n - 1} = 0 \quad \because p^n = 0_F$
$\therefore f(x)$:non-Separable Polynomial

Proposition

正標数Pの完全体の条件

Statement

$F$:Field
$\text{ch }F = p \gt 0$:prime number
then
$F$:Perfect $\Leftrightarrow \forall a \in F, \exists b \in F \text{ s.t. } a = b^p$

Proof

$\Rightarrow$
$\forall a \in F, \exists E$:Minimal Splitting Field of $x^p - a \in F[x] \quad(\because \text{Prop.分解体の存在})$
(Let $f(x) = x^p - a$)
$E$:Algebraic Extension
$\therefore E/F$:Separable Extension $\quad(\because F$:Perfect)

$\exists b \in E \text{ s.t. } f(b) = 0 \quad (\because E:\text{Splitting Field of} f(x))$
$f(b) = b^p - a = 0 \therefore a = b^p$
$\therefore f(x) = x^p - a = x^p - b^p = (x - b)^p$
$\because$
$(x - b)^p = x^p + _p C_1 x^{p-1}(-b) + \cdots + _p C_{p-1} x (-b)^{p-1} + (-b)^p$
$=x^p + (-b)^p \quad \because _p C_j = _p P_j / _j P_j = p \cdot(\cdots) = 0_F$
if $p = 2$ then $1 = -1 \therefore x^p + (-b)^p = x^p - b^p$
if $p \gt 2$ then $(-1)^p = -1 \therefore x^p + (-b)^p = x^p - b^p$

Let $p(x) \in F[x]$:Minimal Polynomial of $b$
then $p(x)|f(x)$
$\therefore p(x) = (x - b)^i$
$p(x)$:Separable $\therefore p(x) = (x - b)$
$\therefore x - b \in F[x] \therefore b \in F$
$\Leftarrow$
$\alpha \in \bar F$
$p(x) \in F[x]$:Minimal Polynomial of $\alpha$

Suppose $p(x)$:Inseparable
then
$p(x)$:Irreducible $\therefore \exists q(x) \in F[x] \text{ s.t. } p(x) = q(x^p) \quad (\text{Prop.標数Pの分離多項式の条件})$
Let $q(x) = a_n x^n + \cdots + a_1 x + a_0$
then $p(x) = q(x^p) = a_n x^{pn} + \cdots + a_1 x^p + a_0$
$\exists b_0, \cdots, b_n \in F \text{ s.t. } a_0 = b_0^p, \cdots a_n = a_n^p$
$\therefore p(x) = b_n^p x^{pn} + \cdots + b_1^p x^p + b_0^p$
$= (b_n x^n + \cdots + b_1 x + b_0)^p$
leads to contradiction.
$\therefore p(x)$:Separable $\forall E/F$:Algebraic $\Rightarrow E \subset \bar F$
$\therefore F$:Perfect
QED

Proposition

Fpは完全体

Statement

$\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ is Perfect.

Proof

$\forall a \in \mathbb{F}_p$
when $a = 0$ then $a^p = a = 0$ --- (i)
when $a \neq 0$ then
$(a)$ is a SubGroup of $\mathbb{F}_p^{\times}$
$|(a)| | |\mathbb{F}_p^{\times}|$
$|\mathbb{F}_p^{\times}| = p - 1$
$a^{|\{a\}|}$ = 1
$\therefore a^{p-1} = 1$ --- (ii)
(i),(ii)leads to $\forall a \in \mathbb{F}, a = a^p$
$\therefore \mathbb{F}_p$ is Perfect. $\quad (\text{Prop.正標数Pの完全体の条件})$
QED

Proposition

完全体の代数拡大

Statement

$F$:Perfect
$E/F$:Algebraic Extension
then $E$:Perfect

Proof

$E \subset \bar F$
$\forall a \in \bar F$
$p(x) \in F[x]$:Minimal Polynomial of $a$
$q(x) \in E[x]$:Minimal Polynomial of $a$
then
$q(x) | p(x)$
$\because p(x) \in E[x]$ and $p(a) = q(a) = 0$
$F$:Perfect $\therefore p(x)$:Separable
$\therefore q(x)$:Separable
QED

Proposition

有限体は完全体

Statement

$F$:Field
$|F| \lt \infty$
then $F$:Perfect

Proof

$|F| \lt \infty \Rightarrow \text{ch}F = p \gt 0$
$\mathbb{F}_p \subset F$
$[F:\mathbb{F}_p] \lt \infty$
$\therefore F/\mathbb{F}_p$:Algebraic Extension
$\mathbb{F}_p$:Perfect $\therefore F$:Perfect
QED