Statement: Given a number field $K$, there exists infinitely many primes $p$ such that the field extension $K(\zeta_{p})$ has degree $(p-1)$ over $K$. Here, $\zeta_{p}$ is the $p^{th}$ root of unity.
I am wondering if the statement holds.
If $K$ is a number field, it contains finitely many roots of unity, say $n_{1}^{th}, n_{2}^{th}, \ldots, n_{k}^{th}$ roots of unity. I was thinking that if we choose a prime $p$ such that $p \nmid n_{j}$ for any $j = 1, 2, \ldots, k$, we will have $[K(\zeta_{p}):K ] = p-1$. But after a while I realized that $K = \mathbb{Q} (\zeta_{p} + \zeta_{p}^{-1})$ does not contain $\zeta_{p}$ but $[K(\zeta_{p}) : K] = 2$ only.
Just choosing $p$ so that $\zeta_{p} \not\in K$ is not enough. We may have to choose $p$ such that the set $\{1, \zeta_{p}, \zeta_{p}^{2}, \ldots, \zeta_{p}^{p-1}\}$ is linearly independent over $K$. In other words, we want to choose $p$ such that $$\sum_{j=1}^{p-1} a_{j}\zeta_{p}^{j} \in K$$ for some $a_{1}, a_{2}, \ldots, a_{p-1} \in K$ means $a_{1} = a_{2} = \ldots = a_{p-1} = 0$.
If I am not wrong, this is equivalent to asking whether given a number field $K$,one could choose infinitely many primes $p$ such that $K$ and $\mathbb{Q}(\zeta_{p})$ are linearly disjoint. Since I have not been able to prove/disprove my statement, I am wondering if anyone could offer some insight/help.