Question: Let $K=F(u)$ be a separable extension of $F$ with $u^m\in F$ for some positive integer $m$. Show that if the characteristic of $F$ is $p$ and $m=p^tr$, then $u^r\in F$. (I think we also need to assume $p$ does not divide $r$).
Thoughts: Let $m_{u^r}(x)\in F[x]$ be the minimal polynomial of $u^r$ over $F$. We know $u^r$ is a root of $x^{p^t}-u^m$, so the minimal polynomial divides this. Note that $u^m=u^{p^tr}=u^{rp^t}=(u^r)^{p^t}$. Since char$F=p$, $x^{p^t}-u^m=x^{p^t}-(u^r)^{p^t}=(x-u^r)^{p^t}\in K[x]$. Now this is where I get stuck. Am I done? Can I just claim here that since $K$ is a separable extension of $F$ that the minimal polynomial must then be $x-u^r$, thus $u^r\in F[x]$? Any help would be greatly appreciated! Thank you.
The question has some hints here: An exponent of an element of a simple separable extension is contained in the base field., but I was hoping to maybe do it in the way above.
We know, the element $u^r$ must have a minimal polynomial in $F[x]$. Because the extension $F(u)$ is seperable, the extension $F(u^r)$ also must be seperable (because it is inbetween $F(u)$ and $F$). From that you know, that $m_{u^r}$ cannot have multiple zeros. Every polynomial that divides $x^{p^t} - u^m$ has either degree $1$, or multiple zeros. So, your conclusion that it already must be $x-u^r$ is correct.