If $m=p^rn,(p,n)=1$, for any $K\subset Q[m]$, $p$ does not ramify in $K$, then $K\subset Q[n]$?

Question

Let $m=p^rn,(p,n)=1$. For any $K\subset Q[m]$ where $K$ is a number field contained in cyclotomic field of order $m$, suppose $p$ does not ramify in $K$,

$\textbf{Q:}$ Is $K\subset Q[n]$? I do not see any obvious reason this has to be the case.

Answer 1

The answer is yes. One way to see this is to notice that if $K$ were not contained in $\mathbb{Q}[n]$, then the compositum $K\mathbb{Q}[n]$ would be strictly larger than $\mathbb{Q}[n]$, unramified at $p$, and still contained in $\mathbb{Q}[m]$. But $\mathbb{Q}[m] = \mathbb{Q}[n] \cdot \mathbb{Q}[p^r]$, and these latter two fields are independent over $\mathbb{Q}$ and thus the ramification degree of $p$ in $\mathbb{Q}[m]$ is exactly $[\mathbb{Q}[p^r]:\mathbb{Q}]$ as $p$ is totally ramified here. This means that there can't be an unramified-at-$p$ subextension of $\mathbb{Q}[m]$ any larger than $\mathbb{Q}[n]$.