Show that $f(x)=g(x^{p^e})$ and each root has the same multiplicity

Question

Let $K$ a field of characteristic $p\neq 0$ and let $f(x) \in K[x]$ be an irreducible polynomial.

1. Show that $f(x)=g(x^{p^e}), e \geq 0$, where $g(x)$ is an irreducible separable polynomial of $K[x]$.
2. Show that each root of $f(x)$ in an algebraic closure $\overline{K}$ of $K$ has the same multiplicity $p^e$.

Could you give me some hints how to show the two sentences??

Answer 1

As you said, you are looking for hints instead of a full solution, so here is a possible plan:

1) Lemma: If $f\in K[x]$ is a polynomial, then $f$ has has no repeated roots in $\overline{K}$ if and only if $f$ and $f'$ has no common factor. Here, $f'$ is the derivative of $f$ (e.g. if $f(x)= x^n + a_{n-1}x^{n-1}+\cdots+a_0$, then $f'(x) = nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1$).

The product rule might be useful. Once you convince yourself of the lemma, then

Corollary: If $f$ is irreducible in the lemma above, then $f$ is separable if and only if $f'\neq 0$.

Now, use the corollary above to show that $f$ is not separable if and only if $f$ is a polynomial in $x^p$, i.e. $f(x)=f_1(x^p)$ for some $f_1$. Repeat the process on $f_1$.

2) Once you have 1), using the fact that $(a+b)^p=a^p+b^p$ in characteristic $p$, show that there exists $h\in \overline{K}[x]$ such that $h(x)^{p^e}=g(x^{p^e})=f(x)$.