I want to show the following theorem about characteristics:
Let $F$ be a field. Then $$\{f \in F[X] \mid f'=0\} = \Biggl\{\begin{array}{l} F, \qquad \qquad \qquad \qquad char(F)=0 \\ \{g(X^p) \mid g \in F[X]\}, \, \, char(F)=p>0 \end{array}$$
I've already managed to show the case of $char(F)=0$, but I don't know what to do, when the characteristic is $p$. I hope someone can help me there.
In all that follows, $F$ is a field of characteristic $p>0$ and $f,g\in F[X]$.
If $f(x) = g(X^p)$, then $f'(x) = g'(X^p)\cdot p X^{p - 1} = 0$.
Conversely, suppose that $f(X) = \sum_{i = 0}^n a_i X^i$ and $f'(X) = 0$. By definition, $$ f'(X) := \sum_{i = 1}^n a_i i X^{i - 1}, $$ and $f'(X) = 0$ if and only if $a_i i = 0$ for all $i$. If $p$ does not divide $i$, then $a_i$ must be $0$: $a_i i = 0$ holds over a field of characteristic $p$, so one of $a_i$ or $i$ must be $0$, but if $p$ doesn't divide $i$, then $i$ is invertible in $F$, so $a_i = (a_i i)\cdot i^{-1} = 0\cdot i^{-1} = 0$. Hence, $$ f(X) = \sum_{0\leq i\leq n,\,p\mid i} a_{i} X^{i}. $$ In other words, we may write $$ f(X) = \sum_{j = 0}^m a_{pj} X^{pj} = \sum_{j = 0}^m a_{pj} (X^p)^j $$ for some $m$. Set $g(X) = \sum_{j = 0}^m a_{pj} X^j$ to obtain the result.