$k$ be perfect field , char $p>0$ , $u=f(X)/g(X) \in k(X) ; f(X),g(X) \in k[X]$ relatively prime , $k(X)/k(u)$ separable ; to show $u \notin k(X)^p$

Question

Let $k$ be a perfect field of characteristic $p>0$ ( perfect means $k^p:=\{a^p : a \in k\}=k$ ) . Let $E=k(X)$ be the rational function field over $k$ , let $u=f(X)/g(X) \in E=k(X)$ such that $f(X),g(X) \in k[X]$ are relatively prime . Let $F=k(u)$ .

If $E/F $ is separable then how to show that $u \notin E^p=\{a^p : a\in E\}$ ?

I know that the minimal polynomial of $X \in E$ over $F$ is $p(T)=f(T)-ug(T) \in F[T]$ and $X$ is separable over $F$ so $f'(X) \ne u g'(X)$ ; but then I don't know whether this is useful or not . Please help . Thanks in advance

Answer 1

$E^p=k(X^p)$ and hence $u\in E^p$ implies $u= m(X^p)/n(X^p)$ but then we have $ K(u^{1/p}) $ is inseparable over $K(u)$ and hence this is a contradiction.