Irreducibility of cyclotomic polynomials over number fields Let $K$ be a number field, i.e., a finite extension of $\mathbb{Q}$. For a positive integer $n$, let $\Phi_n(X)$ denote the $n$-th cyclotomic polynomial. 
Is it possible to say that there exist at most finitely many $n$ such that $\Phi_n(X)$ is reducible over $K$, i.e., all but finitely many $\Phi_n(X)$'s are irreducible over $K$? If not, then can we at least say that there are infinitely many $n$ such that $\Phi_n(X)$ is irreducible over $K$?
 A: We can say that there will always be infinitely many cyclotomic polynomials that remain irreducible.
This is seen as follows. The polynomial $\Phi_n(X)$ is irreducible over $K$ if and only if $[K(\zeta_n):K]=\phi(n)=\deg \Phi_n(X)$. Now assume that $p$ is a rational prime that is unramified in $K$. Undoubtedly you know that there are only finitely many ramified primes. I claim that if $n=p^k$, $k$ a positive integer, then this holds. This is because $p$ is totally ramified in the extension
$\Bbb{Q}(\zeta_n)/\Bbb{Q}$. As $p$ is unramified in $K/\Bbb{Q}$ we must, by multiplicativity of $e$, have $e(\mathfrak{P}\mid \mathfrak{p})=\phi(p^k)$ for all prime ideals $\mathfrak{p}$ above $p$ in $K$ and $\mathfrak{P}$ in $K(\zeta_n)$.
Because the ramification index is always bounded from above by the degree of the field extension, we must have $[K(\zeta_n):K]\ge \phi(p^k)$, and the claim follows.
So we can say that $\Phi_p(X)$ remains irreducible for all but finitely many primes (the unramified ones and possibly some others). 
However, we cannot conclude that $\Phi_n(x)$ would necessarily remain irreducible for all but finitely many natural numbers $n$. If one $\Phi_n(X)$ factors over $K$, then all $\Phi_{m}(X)$ such that $n\mid m$ factor over $K$ also. This is because  $[K(\zeta_m):K(\zeta_n)]$ is always at most $\phi(m)/\phi(n)$, when $n\mid m$. Consequently if $[K(\zeta_n):K]<\phi(n)$ then also $[K(\zeta_m):K]<\phi(m)$ whenever $n\mid m$.
A: What about the following argument?
We will show that for all but finitely many primes $p$, $\Phi_p(X)$ is irreducible over $K$. As above here also we work with a prime $p$ which does not ramify in $K$. Now the following three statements are equivalent.
1) The polynomial $\Phi_p(X)$ is irreducible over $K$.
2) $[K(\zeta_p):K]=p-1$.
3) $K \cap \mathbb Q(\zeta_p)=\mathbb Q$.
Now consider the following two towers of field extensions: $\mathbb Q \subseteq K \cap \mathbb Q(\zeta_p) \subseteq K$ and $\mathbb Q \subseteq K \cap \mathbb Q(\zeta_p) \subseteq \mathbb Q(\zeta_p)$. In the first case, $p$ is a prime which does not ramify. In the second case $p$ is the only prime that ramifies. Hence, if  $K \cap \mathbb Q(\zeta_p)\neq \mathbb Q$, by considering a prime $q$ which ramifies in $K \cap \mathbb Q(\zeta_p)$, we will arrive at a contradiction.
