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Let $K/\Bbb Q$ be a finite extension and $p$ be a prime such that $p\nmid |K:\Bbb Q|$ and $\zeta_p \notin K$. Can we say that the $p$-th cylotomic polynomial $\Phi_p(X)$ is irreducible over $K$?

The above question appears from this question of mine. I want to solve it based on the fact that there can only be finitely many roots of unity in $K$. Thus there exists a prime $p$ such that $p \nmid |K:\Bbb Q|$ and $\zeta_p \notin K$. I want to show that $K(\sqrt[p]{2})/K$ is not normal. Note that $X^p-2$ irreducible over $K$, since $p \nmid |K:\Bbb Q|$. It suffices to show that $\zeta_p \notin K(\sqrt[p]{2})$. If $\zeta_p \in K(\sqrt[p]{2})$ and if I can show that $\Phi_p(X)$ is irreducible over $K$ I get a contradiction since $(p-1)\nmid p$. I want some help. Thanks.

user371231
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    No. $\mathbb Q(\zeta_p)$ is Galois over $\mathbb Q$ with Galois group isomorphic to $\mathbb Z/(p-1)\mathbb Z$, so it contains plenty of proper prime to $p$ subextensions. – Mathmo123 Aug 09 '21 at 16:07
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    Indeed, the first case to look at is $p=5$, with $\zeta=\zeta_5$ and $\xi=\zeta+\zeta^{-1}$. Then Irr$(\xi, \Bbb Q[X]) = X^2+X-1$, so that $\xi=\frac{-1\pm\sqrt5}2$. Then over $\Bbb Q(\sqrt5,)=\Bbb Q(\xi)$, we have $\Phi_5(X)=(X^2-\xi X+1)(X^2+(1+\xi)X+1)$. – Lubin Aug 09 '21 at 19:48

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