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Question is to prove that $\mathbb{Q}(\sqrt[3]{2})$ is not in any cyclotomic extension of $\mathbb{Q}$.

As i was not so sure how to proceed i did the following thing(which may possibly be not so relevant to the Question but hope it would give me some hint to proceed). As $\mathbb{Q}(\sqrt[3]{2})$ is not Galois over $\mathbb{Q}$, I tried Computing Splitting Field and corresponding galois Group and ended up in concluding that Galois Group is isomorphic to $S_3$. As $S_3$ is not abelian Group there could possibly no Sub field $K$ of Cyclotomic field with $Gal(\frac{K}{\mathbb{Q}})\cong S_3$.

I am helpless after this.

Please let me know am i going in a correct path?? Any suggestion/hint would be appreciated :) Thank You


marked as duplicate by user26857, Andrey Rekalo, Davide Giraudo, Amzoti, user1337 Aug 8 '13 at 15:59

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    $\begingroup$ I think you essentially have the answer. $\endgroup$ – Gerry Myerson Aug 6 '13 at 10:26
  • $\begingroup$ @GerryMyerson Sir, I am not very sure about it. Please let me know if there are any gaps, even if they are minute. $\endgroup$ – user87543 Aug 6 '13 at 10:29
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    $\begingroup$ This question was recently handled here. You have made the essential observation. Well done! All the subfields of all the fields $\mathbb{Q}(\zeta_n)$ are fixed fields of normal subgroups (because the Galois groups are abelian). But $\mathbb{Q}(\root3\of2)/\mathbb{Q}$ is not normal. $\endgroup$ – Jyrki Lahtonen Aug 6 '13 at 11:00
  • $\begingroup$ @JyrkiLahtonen Thanks for another nice observation :) $\endgroup$ – user87543 Aug 6 '13 at 12:33

You have already solved it. I would phrase it as follows: Normal subextensions of abelian extensions are also abelian. Therefore, $\mathbb{Q}(\sqrt[3]{2})$ is not contained in any abelian extension, since then also its normal closure $\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ would be contained, but this has Galois group $S_3$.