# 03.1.分離多項式

## Proposition

### 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

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

## Proposition

### 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 $\Leftrightarrow f'(x) = 0$

$\Rightarrow$
$\exists \alpha \in \bar F, f(x) = (x - \alpha)^2 g(x)$
$f'(x) = 2(x - \alpha)g(x) + (x - \alpha)^2 g'(x)$
$\therefore f(\alpha) = f'(\alpha) = 0$

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
$\text{deg}p(x) \le \text{deg}f'(x) \lt \text{deg}f(x)$
$\therefore\text{deg}p(x) \lt \text{deg}f(x)$
$\therefore \exists q(x), f(x) = p(x)q(x) \quad (\text{deg}q(x) \gt 0)$
contradiction to $f(x)$:Irreducible

$\Leftarrow$
$\forall \alpha \in \bar F, f(\alpha) = 0$
$f'(\alpha) = 0 \because f'(x) = 0$
$\therefore \exists g(x) \in \bar F[x] \text{ s.t. } f(x) = (x - \alpha)^2 g(x)$
$\therefore f(x)$:non-Separable Polynomial

QED.