We know that in a normal extension of number fields $L/K$, for any prime $P$ of $K$ and any primes $Q\_1,Q\_2$ of $L$ lying over $P$, the ramification indices and inertial degrees are the same,
$$e(Q\_1|P)=e(Q\_2|P),\quad f(Q\_1|P)=f(Q\_2|P),$$
and that this is not necessarily the case if $L/K$ is not normal. Is it never the case when $L/K$ is not normal? That is, must there be at least one $P$ and $Q\_1,Q\_2$ for which $e(Q\_1|P)\neq e(Q\_2|P)$ or $f(Q\_1|P)\neq f(Q\_2|P)$?



Converse to basic result on prime decomposition
Forget the ramification indices: they're nearly always 1. Let's focus on the residue field degrees. Let $E$ be the Galois closure of $L$ over $K$. For a prime $P$ in $K$, write $f\_P(E/K)$ for the common residue field degree of all primes in $E$ over $P$. Since $E$ is determined (up to isom. as an extension of $K$) by $L$, one can imagine there might be a *formula* for $f\_P(E/K)$ in terms of the $f(Q\_i|P)$'s as $Q\_i$ runs over the primes in $K$ that lie over $P$. There is such a formula if $P$ is unramified in $E$: $f\_P(E/K) = \text{lcm } f(Q\_i|P)$. (This is basically because a composite of finite fields over a particular finite field has degree equal to the lcm of the degrees of the extensions.)


Now imagine that for all $P$ the numbers $f(Q\_i|P)$ are equal. Then their lcm is that common value, so when $P$ is unramified in $E$ we have $f\_P(E/K) = f(Q\_i|P)$ for all $Q\_i$ over $P$ in $L$. Then $f\_{Q\_i}(E/L) = 1$ for all $Q\_i$, so all but finitely many primes in $L$ are split completely in $E$. By Chebotarev's density theorem that implies $E = L$, so $L/K$ is Galois. (Strictly speaking we are using a result that is logically much weaker than Chebotarev: that the primes which split completely determine a Galois extension is due to Bauer from 1916 or so, before Chebotarev proved his general theorem.)


Since Chebotarev works with a density 0 set of primes taken out, instead of assuming for all $P$ that the numbers $f(Q\_i|P)$ (when $Q\_i|P$) are equal you only need to assume that for all but finitely many $P$ or even for all but a density 0 set of $P$. 


