$F[X]$ is of the form $a_0+a_1X+...+a_nX^n$ where $a_0,a_1,...a_n\in F\text{ with characteristic p>0}.$ Express $f\in F[X]\ as\ g(X^{p^m})$, where the nonnegative integer m is a large as possible and $f$ is irreducible. Show that g is irreducible and separable.
This is a problem in section 3.4 of Basic Algebra by Robert Ash.
The question I had about is how to write $f(X)$ in form of $g(X^{p^m})$. I have tried this using the property $(a+b)^p=a^p+b^p$, since $F$ has characteristic p and the Frobenius Automorphism indicate if $\alpha \in F, \alpha = \beta ^p,$ for some $\beta \in F$. $g(X^{p^m})$ can be therefore written as $(b_0+b_1X+...+b_nX^n)^{p^m}.$ This is writing $g(X^{p^m})$ in in form of $f(X)$, for the other way around, is it just taking the $p^m$ root of the above expression: $$f(X)=\sqrt[p^m]{g(X^{p^m})}$$
The fact that $g$ is irreducible follows from the fact that $f$ is; i.e. if $g = h_1 h_2$ then $f = h_1(x^{p^m})h_2(x^{p^m})$.
Once we show existence, this proves that $g$ is separable too: an irreducible polynomial is separable iff $g' \neq 0$, and in characteristic $p$ the fact that $g' = 0$ in turn means that $g = \widetilde{g}(x^p)$ for some $\widetilde g$. We cannot have the latter, otherwise it would contradict the maximality of $m$.
Now, consider
$S = \{m \in \mathbb N_0 : f = g(x^{p^m}), \text{ for some $g$}\}.$
The set $S$ is non empty, because $0 \in S$, and it is bounded above by degree considerations. Hence there exists $m = \max S$ with an associated $g$, which is irreducible and separable by the previous remarks.