Let \(F\) be a field of characteristic \(p>0\) such that every element of \(F\) has a \(p\)th root in \(F\). Show that every irreducible polynomial in \(F[x]\) is separable.
Solution
Let \(f\in F[x]\) be irreducible. Suppose, for contradiction, that \(f\) is inseparable. Since \(f\) is irreducible, inseparability implies \[ \gcd(f,f')=f. \] But \(\deg f'<\deg f\), so this forces \[ f'=0. \]
In characteristic \(p\), the derivative vanishes exactly when every exponent occurring in \(f\) is divisible by \(p\). Hence \[ f(x)=\sum_i a_i x^{pi} \] for coefficients \(a_i\in F\).
By hypothesis, each \(a_i\) is a \(p\)th power, say \[ a_i=b_i^p \] with \(b_i\in F\). Therefore, using the Frobenius identity in characteristic \(p\), \[ f(x)=\sum_i b_i^p x^{pi} =\left(\sum_i b_i x^i\right)^p. \] Since \(f\) is nonconstant, the polynomial inside the parentheses is nonconstant. Thus \(f\) is a nontrivial \(p\)th power in \(F[x]\), contradicting irreducibility.
Hence every irreducible polynomial over \(F\) is separable.