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Is $\sqrt[3]{2}$ contained in $\mathbb{Q}(\zeta_n)$ for some $n$, where $\zeta_n=e^{2\pi i/n}$?

I think the answer is no, but I can't give a full proof. Assume the contrary, we then have $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\zeta_n)$, can I say $\mathbb{Q}(\zeta_n)/ \mathbb{Q}$ is a cyclic extension but $\mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q}$ is not, so this is a contradiction?

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2 Answers 2

up vote 4 down vote accepted

Hint 1: Every subgroup of an abelian group is normal.

Hint 2: If $L/K$ is Galois, normal subgroups $H \subseteq \text{Gal}(L/K)$ correspond to normal extensions $L^H/K$.

Hint 3: Is $\Bbb{Q}(\sqrt[3]{2})$ a normal extension?

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$\mathbb{Q}(\sqrt[3]{2}) /\mathbb{Q}$ is not normal for $x^3-2$ did not factor into linear factors in $\mathbb{Q}(\sqrt[3]{2}) $, so $\mathbb{Q}(\sqrt[3]{2})$ is not contained in $\mathbb{Q}(\zeta_n) $. Am I right? –  yify Aug 5 '13 at 17:05
    
@yify Yes you are :D –  user38268 Aug 5 '13 at 17:07

Not all the extensions $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ are cyclic (e.g. $n=8$ or any $n$ with two odd prime factors).

They are all abelian, though. So all their subextensions are ________

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all their subextension are abelian as well, but $\mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q}$ is not even galois, so we are done? –  yify Aug 5 '13 at 16:59
    
Correct, @yify! –  Jyrki Lahtonen Aug 5 '13 at 19:25

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