TITLE: If a monic $f\in\overline{K}[x]$ has a power $f^n\in K[x]$, where the characteristic of $K$ doesn't divide $n$, then must $f\in K[x]$?
QUESTION [5 upvotes]: Suppose you have a monic polynomial $f(x) \in \overline{K}[x]$, and some integer $n>1$, where  $\mathrm{char}(K)\nmid n$, and  $\big (f(x)\big )^n\in K[x]$. Does it imply $f(x) \in K[x]$?
The question seems very elementary, and my intuition immediately said: "yes", this is true!
However, after few attempts to provide an elementary proof, I got nothing.  I ended up heading to Galois/Kummer Theory, but nothing came out so far. I'm really wondering if such an elementary question has no straightforward answer( with very easy way to prove it).

REPLY [2 votes]: Let $R \subseteq S$ be a ring extension and $n \in R^*$. Then if some monic polynomial $f \in S[x]$ satisfies $f^n \in R[x]$, then $f \in R[x]$.

Proof: Write $f = x^d + a_{d-1} x^{d-1} + \dotsc + a_1 x + a_0$ with $a_i \in S$. We show by induction on $1 \leq i \leq d$ that $a_{d-i} \in R$. We begin by observing that $f^n = x^{nd} + na_{d-1} x^{nd-1}+\text{ lower terms}$. Hence $na_{d-1} \in R$, i.e. $a_{d-1} \in R$. Now assume that we already know $a_{d-1},\dotsc,a_{d-i+1} \in R$. The coefficient of $x^{nd-i}$ in $f^n$ is $n a_{d-i} + \text{some polyonomial in } a_{d-1},\dotsc,a_{d-i+1}$: The first summand comes from choosing the monomial $a_{d-i} x^{d-i}$ in one $f$, and $x^d$ in the other $n-1$ copies of $f$. In the other summands we have to choose higher monomials. Since a polynomial in $a_{d-1},\dotsc,a_{d-i+1}$ lies in $R$ by induction hypothesis, we conclude $n a_{d-i} \in R$, i.e. $a_{d-i} \in R$. $\square$