Proof. sketch [1704A]

    1. Omitted.

    2. Let $d(x) = \gcd(f(x), f'(x))$.

      Claim: $\alpha \in \overline{F}$ is a common root of $f(x)$ and $f'(x)$ if and only if $\alpha$ is a root of $d(x)$. Then, just follow part 1.

      Let’s prove the claim.

      $\impliedby$: It is clear that $d(\alpha) = 0 \implies f(\alpha) = f'(\alpha) = 0$.

      $\implies$: Since $F[x]$ is a PID, there exists $a(x), b(x) \in F[x]$ such that $a(x)f(x) + b(x)f'(x) = d(x)$. Then $d(\alpha) = a(\alpha)f(\alpha) + b(\alpha)f'(\alpha) = 0$.

    3. A direct proposition of part 2.

    4. By part 3.

    5. $\operatorname{ch}(F) = 0 \implies f'(x) \neq 0$, then use part 3. For every $K / F$ and $\alpha \in K$, by part 4, $m_{\alpha, F}(x)$ is separable, so $\alpha$ is separable over $F$, hence $K / F$ is separable.