Field extension of a field of Characteristic $p$

Question

Let $K$ be extension of $F$ with characteristic of $F$ being $p$ and $f(x)$ a monic polynomial in $K[x]$ with $[f(x)]^m$ belongs to $F[x]$ such that $p$ does not divides $m$. Then show that $f(x)$ belongs to $F[x]$.

Answer 1

Write $P$ for the prime field of $F$ (and $K$). Let $f(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+\ldots+a_{n-1}x+a_n$ with $a_1,a_2,\ldots,a_n\in K$ and $a_n\neq 0$. If $g(x)=\big(f(x)\big)^m\in F[x]$ with $p\nmid m$, then I will show that $a_1,a_2,\ldots,a_n\in F$ by induction.

Start with $a_1$. The coefficient of $x^{mn-1}$ in $g(n)$ is $ma_1$. As $ma_1\in F$ and $\gcd(m,p)=1$, $m$ is an invertible element of $P$ (whence of $F$). Therefore, $m^{-1}\in P\subseteq F$ and $$a_1=m^{-1}(ma_1)\in F.$$

Suppose that $a_1,a_2,\ldots,a_{k-1}\in F$ and $k\leq n$. We want to show that $a_{k}\in F$ as well. Observe that the coefficient of $x^{mn-k}$ in $g(x)$ equals $$T_k(a_1,a_2,\ldots,a_{k-1})+ma_k$$ for some polynomial $T_k[x_1,x_2,\ldots,x_{k-1}]\in P[x_1,x_2,\ldots,x_{k-1}]$. That is, $$T_k(x_1,x_2,\ldots,x_{k-1})=\sum\binom{m}{r_0,r_1,r_2,\ldots,r_{k-1}}x_1^{r_1}x_2^{r_2}\cdots x_{k-1}^{r_{k-1}}$$ where the sum ranges over all tuple $(r_0,r_1,r_2,\ldots,r_{k-1})$ of non-negative integers such that $$r_0+r_1+r_2+\ldots+r_{k-1}=m$$ and $$nr_0+(n-1)r_1+(n-2)r_2+\ldots+(n-k+1)r_{k-1}=mn-k.$$ Clearly, $T_k(a_1,a_2,\ldots,a_{k-1})\in F$. Therefore, $ma_k\in F$ as well. The same argument as the previous paragraph shows that $a_k\in F$.

By induction, $a_1,a_2,\ldots,a_n\in F$. That is, $f(x)\in F[x]$.

If $f$ is not monic with leading coefficient $a_0\in K$ but $g(x)=\big(f(x)\big)^m\in F[x]$, then the assertion that $f(x)\in F[x]$ does not necessarily hold. (Take for example, $f(x)=\sqrt{-1}\ x$ with $F=\operatorname{GF}(3)$, $K=\operatorname{GF}(9)$, and $m=2$.) However, $a_0$ is in a finite extension of $F$, and $$a_0^{m-1}a_k\in F$$ for every $k=1,2,\ldots,n$.