# Thread: Fields, odd prime

1. ## Fields, odd prime

Suppose $F \subseteq K$ are fields, and that $[K:F]=p$, where $p$ is an odd prime. Let $\alpha \in K-F$. Prove that $K=F(\alpha^n)$ for any $n$, $1 \leq n .

This one seems simple, I am thinking to prove first that the basis for $F(\alpha)$ is 1, $\alpha$, $\ldots$, $\alpha^{p-1}$ and then this would mean that $F(\alpha)=F(\alpha^m)$ for any $n$, $1 \leq n . Is this a good way to go or is there an easier way I am not seeing now?

2. Originally Posted by xboxlive89128
Suppose $F \subseteq K$ are fields, and that $[K:F]=p$, where $p$ is an odd prime. Let $\alpha \in K-F$. Prove that $K=F(\alpha^n)$ for any $n$, $1 \leq n .

This one seems simple, I am thinking to prove first that the basis for $F(\alpha)$ is 1, $\alpha$, $\ldots$, $\alpha^{p-1}$ and then this would mean that $F(\alpha)=F(\alpha^m)$ for any $n$, $1 \leq n . Is this a good way to go or is there an easier way I am not seeing now?
Hint: $p=[K:F]=[K:F(\alpha^n)][F(\alpha^n): F].$