Here is a problem from Allan Clark's Abstract Algebra book:
Let $F$ be a field, $p > 0$ the characteristic of $F$ and $E$ a finite algebraic extension of $F$. Denote by $E^{(p)}$ the minimal subfield of $E$ containing $F$ such that $E^{(p)}$ contains every element of the form $\alpha^p$ for $\alpha \in E$. Prove that $E$ is separable if and only if $E = E^{(p)}$.
Firstly, I'm confused as to the meaning of "finite algebraic extension". Is it simply supposed to mean that $E$ is finite and algebraic? Because that would mean the "algebraic" part is superfluous, as all finite extensions are algebraic. Additionally, $E$ is then immediately perfect, so the condition $E^{(p)} = E$ holds trivially.
I've been able to deduce, here, that the field $E \setminus F$ is perfect in the case that $E^{(p)} = E$. This is because every element of $E \setminus F$ is of the form $\alpha^p$ (otherwise $E^{(p})$ would not be minimal) and so the mapping $\varphi(x) = x^p$ is an automorphism on $E \setminus F$. It does not follow, however, that $E$ itself is perfect, as some element of $F$ need not be of this form (we require $E^{(p)}$ to contain $F$ in its entirety, so the minimality argument does not apply).
I don't see how to deduce anything from this observation.