If \(f \in K[x]\) is irreducible of degree \(m > 0\) and \(\mathrm{char}(K)\) does not divide \(m\), then \(f\) is separable.
Solution
Write \(f(x) = \sum_{k=0}^m a_k x^k\) with \(a_m \neq 0\), and let \(\overline K\) be an algebraic closure of \(K\).
1. \(\deg f' = m-1\); in particular \(f'\ne0\).
The formal derivative is \[f'(x) = \sum_{k=1}^m k a_k x^{k-1},\] with coefficient \((m \cdot 1_K) a_m\) of \(x^{m-1}\). Since \(\operatorname{char}(K) \nmid m\), \(m \cdot 1_K \neq 0\), and since \(K\) is a field and \(a_m\ne0\), this coefficient is nonzero.
2. \(\gcd(f, f') = 1\) in \(K[x]\).
The monic gcd \(d\) divides \(f\). Because \(f\) is irreducible, \(d\) is \(1\) or \(a_m^{-1} f\). The second case would give \(f\mid f'\), which is impossible because \(f'\ne0\) and \(\deg f' = m - 1 < m\) by step 1.
3. Q.E.D.
A root \(\alpha\in\overline K\) of \(f\) is a multiple root if and only if \(f'(\alpha)=0\). By step 2 there are \(u,v\in K[x]\) with \(uf+vf'=1\); evaluating at a common root \(\alpha\) of \(f\) and \(f'\) would give \(0=1\). Hence \(f\) has no multiple root in \(\overline K\), so \(f\) is separable.