In Field and Galois theory of Patrick Morandi there's a proof about this proposition:
Let $f(x)\in F(x)$ be an irreducible polynomial. If $char(F)=p>0$, then $f(x)=g(x^{p^m})$ for some $m\geq 0$ and some $g(x)\in F(x)$ that is irreducible and separable over $F$.
My question is about the proof used:
Suppose that $char(F) = p$, and let $f(x)\in F[x]$. Let $m$ be maximal such that $f(x)\in F[x^{p^m}]$. Such an $m$ exists, since $f\in F[x^{p^0}]$ and $f$ lies in $F[x^{p^r}]$ for only finitely many $r$ because any nonconstant polynomial in $F[x^{p^r}]$ has degree at least $p^r$. Say $f(x)=g(x^{p^m})$. Then $g(x)\notin F[x^{p}]$ by maximality of $m$. Moreover, $g(x)$ is irreducible over $F$, since if $g(x) = h(x) \cdot k(x)$, then $f (x) = h(x^{p^m}) \cdot k(x^{p^m} )$ is reducible over $F$. Because of the statement, $g$ is separable over $F$.
This are my doubts:
- The affirmation he gives about the existence of $m$ isn't saying that $\deg(f)=p^k$ for certain integer $k$?
- If the first question is a no, why exists that $m$?
- Why because of the statement the polynomial $g$ is separable over $F$?
The proof can be read in the page $44$, but is the same as the posted here.