Separable polynomials have distinct roots - Imported Knowlpedia Content

Let $K$ be a field and $f(x)\in K[x]$ be nonzero. In an algebraic closure $\overline K$ , the following are equivalent:

$f$ has no repeated roots in $\overline K$ , i.e. every root $\alpha\in\overline K$ occurs with multiplicity $1$ .
$\gcd(f,f')=1$ in $K[x]$ , where $f'$ is the formal derivative.
In the splitting field $L$ of $f$ over $K$ , the polynomial $f$ factors as a product of distinct linear factors.

When these conditions hold, $f$ is called separable. In particular, an algebraic element $\alpha$ is separable over $K$ precisely when its minimal polynomial (over $K$ ) has distinct roots in $\overline K$ .

Remarks

A useful characteristic- $p$ test: if $\mathrm{char}(K)=p>0$ , then $f'=0$ iff $f(x)=g(x^p)$ for some $g\in K[x]$ ; in that case $f$ cannot be separable unless $\deg(f)=1$ .

Examples

Characteristic $0$ : always distinct for irreducibles. Over $\mathbb{Q}$ , $f(x)=x^3-2$ has derivative $f'(x)=3x^2$ , and $\gcd(f,f')=1$ , so its three complex roots are distinct in its splitting field.

A purely inseparable example. Over $K=\mathbb{F}_p(t)$ , the polynomial $f(x)=x^p-t$ satisfies $f'(x)=px^{p-1}=0$ . In $\overline K$ , it has a single root $\alpha=t^{1/p}$ with multiplicity $p$ , so $f$ is not separable and $\alpha$ is not a separable element over $K$ .

A separable polynomial in characteristic $p$ . Over $K=\mathbb{F}_p$ , $f(x)=x^p-x$ has derivative $f'(x)=-1\neq 0$ , hence $\gcd(f,f')=1$ . It splits as $\prod_{a\in \mathbb{F}_p}(x-a)$ with distinct roots (indeed it cuts out the prime field).