Separable polynomials have distinct roots

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

1. $f$ has no repeated roots in $\overline K$, i.e. every root $\alpha\in\overline K$ occurs with multiplicity $1$.
2. $\gcd(f,f')=1$ in $K[x]$, where $f'$ is the formal derivative.
3. 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 $\overline K$.

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

1. 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 .

2. 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 $K$.

3. 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).