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Thread: Fields, odd prime

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by xboxlive89128 View Post
    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: $\displaystyle p=[K:F]=[K:F(\alpha^n)][F(\alpha^n): F].$
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