Thread: Fields, odd prime

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

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

Originally Posted by xboxlive89128

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

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

Hint: $\displaystyle p=[K:F]=[K:F(\alpha^n)][F(\alpha^n): F].$