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If $n$ divides $m$, prove that $\mathbb{Q}(\zeta_{n}) \subset \mathbb{Q}(\zeta_{m})$.

If $n$ divides $m$, so $m = nk$ and $\varphi(m) = \varphi(n)\varphi(k)(k,n)/\varphi((k,n))$, then $\varphi(n)$ divide $\varphi(m)$ and thus, $[\mathbb{Q}(\zeta_{n}):\mathbb{Q}]$ divides $[\mathbb{Q}(\zeta_{m}):\mathbb{Q}]$. But, this doesn't ensure that $\mathbb{Q}(\zeta_{n}) \subset \mathbb{Q}(\zeta_{m})$. How do I do this?

KCd
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Lucas
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  • Where did you see this notation for $\Bbb{Q}_n?$ – Chickenmancer Jun 29 '18 at 22:24
  • @Chickenmancer I got rid of that strange notation everywhere in the title and the question. – KCd Jun 29 '18 at 23:00
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    You had written from $m = nk$ that $\varphi(m) = \varphi(n)\varphi(k)$. That equation is correct when $(n,k) = 1$ but it is wrong in general if $(n,k) > 1$: just try $n = k \geq 2$. – KCd Jun 29 '18 at 23:01
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    @Chickenmancer, from Morandi's book $``$Field and Galois Theory$"$ – Lucas Jun 29 '18 at 23:49

1 Answers1

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I don't think you even need to talk about field extensions answer this. If $\omega$ is a primitive $n$th root of unity and $n$ divides $m$, then it'll be an $m$th root of unity too since $$\omega^m = \omega^{nk} =(\omega^n)^k = 1^k = 1$$ so $\omega$ must live in the $m$th cyclotomic field, which contains all the $m$th roots of unity.

KCd
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Mike Pierce
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    You don't need to talk about primitive roots of unity: $\omega^m = \omega^{nk} =(\omega^n)^k = 1^k = 1$ holds for every $n$-th root of unity $\omega$. – lhf Jun 29 '18 at 21:57