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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 ?

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    $\begingroup$ 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 . . . $\endgroup$ – franz lemmermeyer Oct 4 '16 at 8:31
  • $\begingroup$ Related: mathoverflow.net/questions/34180 $\endgroup$ – Watson Jul 1 '18 at 17:00
  • 1
    $\begingroup$ Possible duplicate of If primes split nicely, is it a Galois extension? $\endgroup$ – Watson Nov 12 '18 at 13:28

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