# Ramification index and inertia degree same for all the primes, then is the extension Galois

Let $L/K$ be a Galois extension of number fields. We know that if $Q,Q'$ are two primes of $L$ lying over a prime $P$ of $K$, then

1. e$(Q|P)$=e$(Q'|P)$ (the ramification indices are same )
2. f$(Q|P)$=f$(Q'|P)$ (the inertia degrees are same )

My question is this: Is the converse true ?

If $L/K$ is an extension which has the following property:

Let $P$ be any prime of $K$. If $Q$ and $Q'$ are primes of $L$ lying over $P$ then

1. e$(Q|P)$=e$(Q'|P)$ (the ramification indices are same )
2. f$(Q|P)$=f$(Q'|P)$ (the inertia degrees are same )

Can we say that $L/K$ is Galois extension ?

• Yes - this must follow from the arguments in Lenstra's and Stevenhagen's article on the Chebotarev density theorem. Since density arguments are involved, you may even drop condition 1 since it concerns only finitely many primes. got to run now . . . – franz lemmermeyer Oct 4 '16 at 8:31
• Related: mathoverflow.net/questions/34180 – Watson Jul 1 '18 at 17:00
• Possible duplicate of If primes split nicely, is it a Galois extension? – Watson Nov 12 '18 at 13:28