A field \(k\) is perfect if either \(k\) has characteristic \(0\), or \(k\) has characteristic \(p>0\) and the Frobenius map \(k\to k\), \(a\mapsto a^p\), is surjective, that is, \(k^p = k\).
For a field \(k\), the following are equivalent:
An irreducible \(f\in k[x]\) has a repeated root if and only if \(\gcd(f,f')\neq 1\), which for irreducible \(f\) happens if and only if \(f'=0\). In characteristic \(0\) this never happens for nonconstant \(f\), so (1), (2), and (3) all hold.
Let \(\characteristic k=p>0\). If \(k^p=k\) and \(f'=0\), then \(f(x)=\sum_i a_ix^{pi}\) with \(a_i=b_i^p\) for some \(b_i\in k\), so \(f=\bigl(\sum_i b_ix^i\bigr)^p\) is not irreducible; hence (1) implies (2). If \(a\in k\setminus k^p\), then \(x^p-a\) is irreducible over \(k\) and equals \((x-\alpha)^p\) over a splitting field, where \(\alpha^p=a\), so it is not separable; hence (2) implies (1). Finally, (2) implies (3) because the minimal polynomial of each element of \(F\) is irreducible over \(k\), and (3) implies (2) because a root \(\alpha\) of an irreducible \(f\) generates the finite extension \(k(\alpha)/k\), in which \(\alpha\) is separable exactly when \(f\) is separable.