Let $\mu \in \mathbb{Q}(\zeta_n)$ lie above the rational prime $p\equiv 1 \text{ mod }n$, and let the prime ideal $\mathscr{P}\subset \mathbb{Z}[\zeta_n]$ have exponent $a$ in $(\mu)$.
Why is it then true that $\mathscr{P}$ has ramification index $\frac{n}{(a,n)}$ in $\mathbb{Q}(\zeta_n, \mu^{1/n})/\mathbb{Q}(\zeta_n)$?
I would appreciate a proof as all references I can find seem to have proofs of thing that reference back to umpteen different earlier propositions, which is hard to read.
Here’s my take on the situation: it seems to me to be a purely local question at $p$, so that it should be sufficient to translate everything to extensions of $\Bbb Q_p$.
Let’s do this translation, using the additional hypothesis that $n$ and $p$ are relatively prime: put $\mathfrak o=\Bbb Z_p[\zeta_n]$, ring of integers of an unramified extension $k$ of $\Bbb Q_p$. Use the standard additive valuation $v$ on $\Bbb Q_p$, so that $v(p)=1$, and extend to $k$. You have set $v(\mu)=a$, say $\mu=p^au$ where $u$ is a unit of $\mathfrak o$. Then $\mu^{1/n}=p^{a/n}u^{1/n}$. Now $\mathfrak o[u^{1/n}]$ is unramified over $\mathfrak o$: it just amounts to taking the $n$-th root of an element of $\mathfrak o/p\mathfrak o$, a finite field of characteristic $p$. But adjoining $p^{a/n}$ is cleasrly a totally ramified extension of degree $n/\!\gcd(a,n)$.