I've heard that this fact is "known" or "can be shown" but I am having trouble finding a proof of it myself.
Take a Galois extension $E/K$ of number fields and let $q\in E$ lie over a prime $p\in K$. Let $L$ be an intermediate field with $q\cap L = I$. If $D_q$ is the decomposition group of $q$, show that $L\subset E^{D_q}$ iff $e_I=f_I=1$; i.e., the ramification and inertia degrees of $I$ over $p$ in the extension $L/K$ are $1$.
It makes sense to me that this should be true because $E^{D_q}$ gets rid of the values $e_q$ and $f_q$. But I can't see how to transfer this information to $L$, and moreover, the fact that $L/K$ isn't necessarily Galois means that I don't have an intuitive sense of how to deal with $e_I$ and $f_I$ in the first place.
Those things become particularly clear in the $p$-adic completions $$E_q= Frac(\varprojlim O_E/q^n)=Frac(\varprojlim (O_E-q)^{-1} O_E/(\pi_q)^n)$$
(the localization $(O_E-q)^{-1} O_E$ at $q$ is a DVR whose maximal ideal is principal)
For complete discretely valued fields, because elements of $L_I$ are of the form $\sum_{m= -M}^\infty c_m \pi_I^m$ with $c_m$ in representatives of the residue field we always have $$f_Ie_I= [L_I:K_p], \qquad f_I = [O_{L_I}/(\pi_I) : O_{K_p}/(\pi_p)],\qquad \pi_p \in \pi_I^{e_I} O_{L_I}$$
$E_q/L_I/K_p $ where $E_q/K_p$ is Galois,
The natural map $E\to E_q$ makes $D_q \cong Gal(E_q/K_p)$
Then $L \subset E^{D_q}$ means $L_I \subset E_q^{Gal(E_q/K_p)}$ which means $L_I= K_p$ so that $$f_Ie_I=1$$