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How to show that $G=\mathrm{Gal}[\Bbb Q(\xi_{p^n})/\Bbb Q(\xi_{p^m})]$ cyclic? Here $p \not = 2$ is a prime number and $n>m>0$ are natural numbers. $\xi_{p^n}$ and $\xi_{p^m}$ are cyclotomic roots.

I know that the order of $ G $ is $|G|=p^{n-m}$

How I found $a \in G$ such that $a^{p^{n-m}}=1$ and how generally I found such $a$ in other cyclic cases?

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  • $\begingroup$ why G is subgroup of $Gal(\Bbb Q(\xi_{p^n})/\Bbb Q)$? and I would like to know how to deal with the statement when $p=2$ and $m>1$ $\endgroup$ – Daniel Vainshtein Jun 21 '18 at 15:54
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    $\begingroup$ If $\sigma \in G$, then $\sigma$ is an automorphism of $\Bbb Q(\zeta_{p^n})$ fixing $\Bbb Q(\zeta_{p^m})$. In particular, it also fix $\Bbb Q$. Therefore, $\sigma \in \mathrm{Gal}(\Bbb Q(\zeta_{p^n}) / \Bbb Q)$ (the latter being a Galois extension). $\endgroup$ – Watson Jun 21 '18 at 15:56
  • $\begingroup$ @Watson yes indeed! $\endgroup$ – Daniel Vainshtein Jun 21 '18 at 16:26
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The group $G$ is a subgroup of $\mathrm{Gal}(\Bbb Q(\zeta_{p^n}) / \Bbb Q)$, which is isomorphic to $(\Bbb Z/p^n \Bbb Z)^{\times}$ (see this related question). This is a cyclic group when $p>2$ is prime. Since any subgroup of a cyclic group is cyclic, we are done.

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  • $\begingroup$ For $p=2$, the question has been addressed here. $\endgroup$ – Watson Jun 22 '18 at 8:00

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