$n$th root of $\alpha$

Question

Let $n$ be a positive integer and $K$ be a field of positive characteristic coprime to $n$. Show that for any $a \in K^\times$, $[K(a^{1/n}):K]$ is a divisor of $n$. We are not assuming $\zeta_n\in K$.

I know a special case of this to be true like if $K=\mathbb{F}_p$, but how to show this in general, if $K$ is not finite?

Answer 1

With $a^{1/n}$ I mean a root of $x^n-a\in K[x]$ minimizing $[K(a^{1/n}):K]$. Since $$[K(a^{1/{pn}}):K]=[K((a^{1/n})^{1/p}):K(a^{1/n})][K(a^{1/{n}}):K]$$ It will suffice to check the case $F(b^{1/p})/F$ with $p$ prime and $b \in F$.

The $F(\zeta_p)$ minimal polynomial of $b^{1/p}$ divides $x^p-b\in F(\zeta_p)[x]$, it is $$f(x)=\prod_{j=1}^d (x-\zeta_p^{c_j} b^{1/p}), \qquad d=[F(\zeta_p,b^{1/p}):F(\zeta_p)]\ \le p$$ $$f(0)=\zeta_p^t b^{d/p}\in F(\zeta_p)$$

Thus, either $d=p$ or $b^{1/p}\in F(\zeta_p),d=1$.

• If $d=p$ then $x^p-b\in F[x]$ is irreducible.

• Otherwise for $b^{1/p}\in F(\zeta_p),d=1$ then $$N_{F(\zeta_p)/F}(b^{1/p})=\zeta_p^s b^{r/p}\in F,\qquad p\nmid r=[F(\zeta_p):F]$$ Whence $\zeta_p^{s r^{-1}} b^{1/p}\in F$ and $F$ contains a root of $x^p-b$. Since we chose a root of $x^p-b$ minimizing $[F(b^{1/p}):F]$ it means that $[F(b^{1/p}):F]=1$.

All this proves that $$[F(b^{1/p}):F]\ |\ p \qquad \implies\qquad [K(a^{1/n}):K]\ | \ n$$