This is from Lang's Algebra (page 251)
Proposition 6.11 Let $E/F$ be a normal field extension. Let $E^G$ be the fixed field of $\operatorname{Aut}(E/F)$. Then, $E^G$ is purely inseparable over $F$ and $E$ is separable over $E^G$.
And below is a corollary of this theorem:
Corollary 6.12. Let $F$ be a field with characteristic $p\neq 0$ such that $F^p=F$. Then, every algebraic extension $E$ of $F$ is separable and $E^p=E$.
How this is a corollary of the above theorem?
Lang states that "Every algebraic extension is contained in a normal extension, so Proposition 6.11 can be applied to get this", but how?
Let $E$ be an algebraic extension of $F$. Then, there is a field extension $L$ of $E$ such that $L/F$ is normal.
Let $\phi\colon F\to F:a\mapsto a^p$.
Then, by hypothesis, $\phi$ is a field automorphism of $F$.
Then, $\phi$ can be extended to a field monomorphism $\sigma \colon \bar F \to \bar F$, but since $\phi$ is not fixing $F$, I don't get what this has to do with Proposition 6.11.
These are what all I know. How do I proceed to prove the corollary?