Let \(F\) be a field of characteristic \(p>0\). Suppose every irreducible polynomial in \(F[x]\) is separable. Show that every \(a\in F\) is a \(p\)-th power in \(F\).
Solution
Let \(a\in F\). If \(a=0\), take \(\beta=0\). Assume \(a\ne0\) and consider \[ f(x)=x^p-a. \] In an algebraic closure choose its unique root \(\beta\), so \[ \beta^p=a \qquad\text{and}\qquad f(x)=(x-\beta)^p. \]
The derivative is \(f'(x)=0\), so \(f\) is inseparable. By hypothesis no irreducible polynomial over \(F\) is inseparable; therefore \(f\) cannot be irreducible. Choose a monic irreducible proper factor \[ g(x)\mid f(x), \qquad 1\le \ell:=\deg g<p. \] Over the algebraic closure, every root of \(g\) is the unique root \(\beta\) of \(f\), hence \[ g(x)=(x-\beta)^\ell. \] Since \(g\in F[x]\), its constant coefficient gives \[ (-\beta)^\ell\in F, \] so \(\beta^\ell\in F\).
Because \(1\le\ell<p\) and \(p\) is prime, \[ \gcd(\ell,p)=1. \] Choose integers \(u,v\) with \[ u\ell+vp=1. \] Since \(\beta\ne0\), \[ \beta =\beta^{u\ell+vp} =(\beta^\ell)^u(\beta^p)^v \in F. \] Therefore \[ a=\beta^p \] with \(\beta\in F\). Thus Frobenius \(x\mapsto x^p\) is surjective on \(F\).