TITLE: If primes split nicely, is it a Galois extension?
QUESTION [5 upvotes]: Let $L/K$ be a finite extension of number fields.  When $L/K$ is Galois, the Galois group $\Gamma = \textrm{Gal}(L/K)$ acts transitively on the primes $\mathfrak P$ of $L$ lying over any given prime $\mathfrak p$.  
For $\mathfrak P \mid \mathfrak p$, the ramification index $e(\mathfrak P/\mathfrak p)$ and residue degree $f(\mathfrak P/\mathfrak p)$ depend only on the extension of local rings $(\mathcal O_K)_{\mathfrak p} \subseteq (\mathcal O_L)_{\mathfrak P}$.  As any $\sigma \in \Gamma$ induces an isomorphism of localized rings $(\mathcal O_L)_{\mathfrak P} \rightarrow (\mathcal O_L)_{\sigma \mathfrak P}$, we can conclude that the numbers $e(\mathfrak P/\mathfrak p)$ and $f(\mathfrak P/\mathfrak p)$ are constant for all $\mathfrak P \mid \mathfrak p$.
Is the converse true?  That is, if for each $\mathfrak p$, the numbers $e(\mathfrak P/\mathfrak p)$ and $f(\mathfrak P/\mathfrak p)$ are constant for all $\mathfrak P \mid \mathfrak p$, can we conclude that $L/K$ is Galois?

REPLY [4 votes]: As I learned in a question of mine on MathOverflow, the answer is yes, $L/K$ must be Galois.
It's basically due to Chebotarev's theorem.