An exponent of an element of a simple separable extension is contained in the base field.

Question

Let $L=F(\alpha)$ be a separable extension of F, with $\alpha^k \in F$. If char$(F)=p$ and $k=p^{t}n$ and $p\not |\ n$ then $\alpha^n \in F$

Attempt at a solution:

Taking the polynomial $p(x)=x^{p^t}-\alpha^k$ shows that $\alpha^n$ is a root of $p(x)$. Also the minimal polynomial of $\alpha^n$ over $F$ divides $p(x)$. Since char$(F)=p$, the derivative $p'(x)=0$ implying that $p(x)$ is inseparable and every root of $p(x)$ is a multiple root.

I'm stumped after this.

Answer 1

Hints:

1) The minimal polynomial of $\alpha^n$ over $F$ is separable.

2) You can factor $f$ (or $p$, whatever you decide to call it) over $L$.