If \(\mathrm{char} K = p \neq 0\) and \([F: K]\) is finite and not divisible by \(p\), then \(F\) is separable over \(K\).
Solution
Suppose, toward a contradiction, that \(F/K\) is not separable.
1. There exists \(u\in F\) whose minimal polynomial over \(K\) is inseparable.
By definition, an algebraic extension is separable if every element is separable. Since \(F/K\) is finite, it is algebraic. Thus failure of separability provides such an element \(u\).
2. If an irreducible polynomial \(m(x)\in K[x]\) is inseparable incharacteristic \(p\), then \(p\) divides \(\deg m\).
An irreducible polynomial is inseparable exactly when its formal derivative is zero. Write \[ m(x)=\sum_j a_jx^j. \] The equation \(m'(x)=0\) means that \(j a_j=0\) for every \(j\). In characteristic \(p\), whenever \(a_j\ne0\) this forces \(p\mid j\). Hence all exponents occurring in \(m\) are multiples of \(p\), so \[ m(x)=g(x^p) \] for some \(g\in K[x]\). In particular \(p\mid\deg m\).
3. The degree \([K(u):K]\) is divisible by \(p\).
Let \(m_u\) be the minimal polynomial of \(u\) over \(K\). By 1 it is inseparable, so 2 gives \[ p\mid\deg m_u=[K(u):K]. \]
4. This contradicts the hypothesis that \(p\nmid[F:K]\).
The tower law for \[ K\subseteq K(u)\subseteq F \] gives \[ [F:K]=[F:K(u)][K(u):K]. \] By 3 the second factor is divisible by \(p\), so \(p\mid[F:K]\), contrary to hypothesis.
5. Therefore \(F/K\) is separable.
The assumption of inseparability led to the contradiction in 4.