Hi everyone,
How can I prove that Q(Zeta36) = Q(Zeta12,Zeta18) does not contain ANY primitive 5th roots of unity?
[here, ZetaN = cos2pi/N + isin2pi/N ]
Many many thanks in advance. x
Okay I fixed it.
It can be shown that $\displaystyle \mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m) = \mathbb{Q}(\zeta_d)$ where $\displaystyle d=\gcd(n,m)$. Do you know this result? I can prove it for you.
If $\displaystyle \zeta_5\in \mathbb{Q}(\zeta_{36})$ then $\displaystyle \mathbb{Q}(\zeta_5)\subseteq \mathbb{Q}(\zeta_{36})$ and so $\displaystyle \mathbb{Q}(\zeta_5) \cap \mathbb{Q}(\zeta_{36}) = \mathbb{Q}(\zeta_5)$.
But this is a problem because $\displaystyle \gcd(36,5) = 1$.
By the way you can generalize this to prove that $\displaystyle \mathbb{Q}(\zeta_n)\subseteq \mathbb{Q}(\zeta_m)$ if and only if $\displaystyle n|m$.
One minor point though - you have done the question for Zeta_5 = cos2pi/5+isin2pi/5.
But the original question was for Zeta = ANY PRIMITIVE 5th root of unity.
So my question is does Q(Zeta) = Q(Zeta_5) ? Because if so then your argument still holds. :-)
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Let $\displaystyle \omega$ be a 5-th root of unity then the other primitive roots of unity are: $\displaystyle \omega,\omega^2,\omega^3, \omega^4$. Therefore, $\displaystyle \zeta_5 \in \{ \omega,\omega^2,\omega^3,\omega^4\}$. This means if $\displaystyle \omega \in \mathbb{Q}_{36}$ then $\displaystyle \{\omega,\omega^2,\omega^3,\omega^4\} \subset \mathbb{Q}_{36}$ and so $\displaystyle \zeta_5 \in \mathbb{Q}_{36}$. Now apply the argument above.
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