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Let $p$ and $q$ be two distinct prime integers. For a field $F$, assume that $\deg(\alpha, F)=p$. Is it necessarily true that $\deg(\alpha^q, F)=p$? Is there any counterexample?

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  • $\begingroup$ @Bhaskar Vashishth, do you mean a counterexmaple? I think that it is not a counterexampe. $\endgroup$
    – user279492
    Oct 12, 2015 at 11:10
  • $\begingroup$ where is this problem from, which book? $\endgroup$
    – Departed
    Oct 12, 2015 at 11:41
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    $\begingroup$ It is not a problem from book. $\endgroup$
    – user279492
    Oct 12, 2015 at 11:42

2 Answers 2

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It's not necessarily true, try $p=2$, $q=3$, $F=\mathbb{Q}$ and $\alpha=e^{2\pi i/3}$.

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Somewhat generalizing the answer of Jarek Kuben, let $p,q$ be primes with $q-1=pr$ for some integer $r$, let $\alpha=e^{2\pi i/q}$ be a nonreal $q$th root of unity, let $K={\bf Q}(\alpha)$. Then $K$ is of degree $pr$ over the rationals, and is cyclic, so it has a subfield $E$ of degree $r$ over the rationals. Then $\alpha$ is of degree $p$ over $E$, but $\alpha^q=1$ is of degree 1 over $E$.

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