03.1.分離多項式

Proposition

重根の条件

Statement

$F$:Field
$\bar F$:Algebraic Closure of $F$
$f(x) \in F[x]$
$\alpha \in \bar F$
then
$\alpha$:Multiple Root of $f(x) \quad ((x-\alpha)^2 | f(x)) \Leftrightarrow f(\alpha)=0, f'(\alpha)=0$

Proof

($\Rightarrow$)
$f(x) = (x - \alpha)^2 g(x) \in \bar F[x]$
$f'(x) = 2(x - \alpha)g(x) + (x - \alpha)^2 g'(x)$
$\therefore f(\alpha) = f'(\alpha) = 0$
($\Leftarrow$)
$f(x) = (x - \alpha)^2 g(x) + px + q$
$f'(x) = 2(x - \alpha) g(x) + (x - \alpha)^2 g'(x) + p$
$0 = f'(\alpha) = p$
$0 = f(\alpha) = p\alpha + q$
$\therefore p = q = 0$
$\therefore f(x) = (x - \alpha)^2 g(x)$
QED

Proposition

既約分離多項式の微分

Statement

$F$:Field
$f(x) \in F[x]$:Irreducible Polynomial
then
$f(x)$:Separable Polynomial $\Leftrightarrow f'(x) \neq 0$

Proof

Prove by contraposition.
$f(x)$:non-Separable Polynomial $\Leftrightarrow f'(x) = 0$

$\Rightarrow$
$f(x) \in F[x] \Rightarrow f'(x) \in F[x]$
$\exists \alpha \in \bar F, f(x) = (x - \alpha)^2 g(x) \quad(g(x) \in \bar F[x])\quad(\because f(x):\text{non-Separable})$
$f'(x) = 2(x - \alpha)g(x) + (x - \alpha)^2 g'(x)$
$\therefore f(\alpha) = f'(\alpha) = 0$

Let $p(x) \in F[x]$:Minimal Polynomial of $\alpha$
then $p(x) | f(x)$ and $p(x) | f'(x)$

Suppose $f'(x) \neq 0$
Then
$\text{deg}p(x) \le \text{deg}f'(x) \lt \text{deg}f(x)$
$\therefore\text{deg}p(x) \lt \text{deg}f(x)$
$\therefore \exists q(x) \in F[x], f(x) = p(x)q(x) \quad (\text{deg}q(x) \gt 0)$
contradiction to $f(x)$:Irreducible

$\Leftarrow$
Let $\alpha \in \bar F, f(\alpha) = 0$
then $f'(\alpha) = 0 \because f'(x) = 0$
$\therefore \exists g(x) \in \bar F[x] \text{ s.t. } f(x) = (x - \alpha)^2 g(x) \quad(\because\text{Prop.重根の条件})$
$\therefore f(x)$:non-Separable Polynomial

QED.

標数0の体上の既約多項式は分離的

Statement

$F$:Field
$f(x) \in F[x]$:Irreducible Polynomial
$\text{char}(F) = 0$
then
$f(x)$:Separable Polynomial

Proof

Suppose
$f(x):\text{non-Separable}\quad(\Leftrightarrow \exists \alpha \in \bar F, g(x) \in \bar F [x] \text{ s.t. } f(x) = (x - \alpha)^2 g(x))$
then
Let
$f(x) = c_0 + c_1 x + \cdots + c_n x^n$
$f'(x) = c_1 + 2 c_2 x + \cdots + n c_nx^{n-1}$
$\therefore c_1 = 2c_2 = \cdots = nc_n = 0 \quad(\because\text{Prop.既約分離多項式の微分})$
$\therefore c_1 = c_2 = \cdots = c_n = 0 \quad(\because \text{char(F) = 0})$
$\therefore f(x) = c_0$
$\text{deg}f(x) = 0$ contradiction.
$\therefore f(x)$:Separable QED.

標数0の体の代数拡大は分離拡大

Statement

$E/F$:Algebraic Field Extension
$f(x) \in F[x]$:Irreducible Polynomial
$\text{char}(F) = 0$
then
$E/F$:Separable Extension

Proof

$\forall \alpha \in E$
$f(x)\in F[x]:\text{Minimal Polynomial of } \alpha \Rightarrow f(x):\text{Separable}\quad(\because\text{Prop.標数0の体上の既約多項式は分離的})$
$\therefore E/F$:Separable Extension
QED.

正標数の体上の既約非分離多項式

Statement

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

Proof

$\Rightarrow$
Let
$f(x) = c_0 + c_1 x + \cdots + c_n x^n$
then
$f'(x) = c_1 + 2 c_2 x + \cdots + n c_nx^{n-1}$
$\therefore c_1 = 2c_2 = \cdots = nc_n = 0 \quad(\mod p)\quad(\because\text{Prop.既約分離多項式の微分})$
$\therefore$
$i \neq 0 \mod p \Rightarrow c_i = 0$
$\therefore f(x) = c_0 + c_p x^p + \cdots + c_{ip}x^{ip} \quad(ip \le n)$
$= c_0 + c_p x^p + \cdots + c_{ip}(x^p)^i$
$\therefore \exists g(x) = c_0 + c_p x + \cdots c_{ip}x^i \in F[x] \text{ s.t. } f(x) = g(x^p)$
$\Leftarrow$
Let
$\alpha \in \bar F, f(\alpha) = 0$
then
$f(\alpha) = g(\alpha^p) = 0$
$\therefore h(y) \in \bar F[y], g(y) = (y - \alpha^p)h(y)$
$\therefore f(x) = g(x^p) = (x^p - \alpha^p)h(x^p)$
$x^p - \alpha^p = (x - \alpha)^p (\mod p)$
$\therefore f(x) = (x - \alpha)^p h(x^p)$
$\therefore f(x)$:non-Separable Polynomial
QED.

Proposition

素数標数の既約非分離多項式

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 $\Leftrightarrow f'(x) = 0 (\in F[x]) \quad(\because \text{Prop.既約分離多項式の微分})$
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
QED.