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I need to show that $\sqrt[7]{2}\not\in\mathbb{Q}(\sqrt[7]{3})$ but I have no idea how to do it. I've tried to reason with the extension degrees but I haven't got any satisfactory result.

Could someone help me?

Thanks in advance.

QGM
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    Hint: It goes like this post, or this post, or this one. You can also compute the field discriminants. Even better, use some concepts from ANT, see the comment below! – Dietrich Burde Dec 10 '20 at 11:19
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    IMO the sleekest proof for this uses concepts from algebraic number theory. The prime ideal $(2)$ is unramified in $K=\Bbb{Q}(\root7\of3)$. Therefore $\root7\of2\notin K$. I also like Dietrich's suggestion to use discriminants. – Jyrki Lahtonen Dec 10 '20 at 11:19
  • Thank you @Watson for locating a suitable duplicate. GRJ, please comment on the theoretical tools available to you. Discriminants and ramification are often not covered in a first course on field theory. Yet, with higher order roots (here seventh) they give the simplest answers. Not ruling out the possibility of a proof using less technology. – Jyrki Lahtonen Dec 10 '20 at 13:10
  • We also have this answer by orangeskid that may help you. – Jyrki Lahtonen Dec 10 '20 at 13:16

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