Let $F$ be a field of characteristic $p > 0$, and let $K$ be a purely inseparable extension of F with $[K : F] = p^n$. Prove that $a^{p^n}\in F$

Question

Let $F$ be a field of characteristic $p > 0$, and let $K$ be a purely inseparable extension of $F$ with $[K : F] = p^n$. Prove that $a^{p^n}\in F$ for all $a\in K$.

Let $a\in K$ be, $a$ is the root of the polynomial $x^{p^n}-a^{p^n}=(x-a)^{p^n}$, then min$(a,F)$ divides $(x-a)^{p^n}$ and I know deg(min$(a,F)$)$=p^n$, but I do not know what else to say or if that's the right way, could anyone help me please? Thank you very much.

Answer 1

A purely inseparable extension is one where each element is purely inseparable over the base field. This means that if $a\in K$, then its minimum polynomial over $F$ is $X^{p^r}-b$ for some $r$. As $|K:F|=p^n$ then $r\le n$. Then $a^{p^r}=b\in F$ and so $a^{p^n}=b^{p^{n-r}}\in F$.