Converse to basic result on prime decomposition

Question

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 $f(Q_1|P)\neq f(Q_2|P)$?

Answer 1

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 $P$.

Answer 2

The answer is yes. The inertial degrees of the primes lying over $P$ are controlled by the action of the Frobenius for $P$ on $G/H$ where $G$ is the Galois group of the normal closure of $L$ over $K$, and $H$ be the Galois group of the normal closure over $L$. The reason for this is that the primes above $p$ in the normal closure form the $G$-set $G/F$, where $F$ is the cyclic group generated by the Frobenius of $P$. The primes over $H$ correspond to $H$-orbits on this set, with the size of the orbit being proportional to the inertial degree. However $H$-orbits on $G/F$ and $F$-orbits on $G/H$ are in canonical bijection, preserving the ratios between sizes (think about $H\times F$-orbits on $G$ as an intermediate stage), and the latter are easier to think about as you vary primes.

Since every element of $G$ appears as Frobenius of something by Chebotarev, you are asking whether you could have finite groups $G$ and $H$, with $G$ acting faithfully on $G/H$, and have all $g\in G$ acting by orbits of all the same size. This is plainly impossible; look at the action of any non-zero element of $H$.