Prove that for positive integers $d$ and $n$, the containment $\mathbb{Q}(\zeta_d) \subseteq \mathbb{Q}(\zeta_n)$ holds iff $d|n$. I have the reverse direction. Can someone give me a hint on how to prove the forward direction?
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Hint: Show that the only roots of unity in $\mathbf Q(\zeta_n)$ are the $\pm \zeta_n^k$'s. (Can you see how it follows from that?) In order to do this, use the fact that a finite subgroup of the multiplicative group of a field is cyclic. Could $\mathbf Q(\zeta_n)$ contain an $m$-th root of unity for $m>n$ if $n$ is even (resp $m>2n$ if $n$ is odd)?
Bruno Joyal
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Hint: $\mathbb{Q}[\zeta_d] \subseteq \mathbb{Q}[\zeta_n] \iff \zeta_d \in \mathbb{Q}[\zeta_n]$.
Kaj Hansen
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Dear Kaj: how is this but a reformulation of the statement? I don't see how this excludes the possibility that $\zeta_d \in \mathbf Q(\zeta_n)$ while $d \nmid n$. – Bruno Joyal May 23 '14 at 02:40
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Oftentimes the best hints are those that restate the problem in order to change one's perspective without actually giving anything away. From this angle, the natural question turns to whether it is possible to have a dth root of unity in this larger cyclotomic field that is not of the form $\zeta_n^k$. – Kaj Hansen May 23 '14 at 02:45
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Your response is spot on though (+1) – Kaj Hansen May 23 '14 at 02:47