# Prove that $\sqrt[7]{2}\not\in\mathbb{Q}(\sqrt[7]{3})$ [duplicate]

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?

• 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! Dec 10, 2020 at 11:19
• 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. Dec 10, 2020 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. Dec 10, 2020 at 13:10
• We also have this answer by orangeskid that may help you. Dec 10, 2020 at 13:16