Fields, odd prime

**xboxlive89128** · May 13th 2009, 05:55 PM

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 $n$ , $1 \leq n <p$ .

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 $n$ , $1 \leq n <p$ . Is this a good way to go or is there an easier way I am not seeing now?

**NonCommAlg** · May 13th 2009, 06:10 PM

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 $n$ , $1 \leq n <p$ .

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 $n$ , $1 \leq n <p$ . Is this a good way to go or is there an easier way I am not seeing now?

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

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