If $L/K$ is a Galois extension, then any prime $\mathfrak{p}$ of $K$ splits into a product ${\mathfrak P}\_1^e\cdots {\mathfrak P}\_g^e$ of primes in $L$, and the exponents on the primes are equal since the Galois group acts transitively on the primes dividing $\mathfrak{p}$. 


Question: Is the converse true? Namely, if $L/K$ is an extension such that every prime of $K$ has a single associated exponent in this manner, then is $L/K$ necessarily Galois?


I believe the answer is yes but I was wondering in particular if there was an easy way to see this. 



Proof of a Simple Converse in Algebraic Number Theory
If you throw in the residue degrees as well, then you get Zev's question. Otherwise, the converse is not true. As an example, consider a quadratic field that has an unramified everywhere $A\_5$ Galois extension (see e.g. [this MO thread](https://mathoverflow.net/questions/44801/a-5-extension-of-number-fields-unramified-everywhere)). Since $A\_5$ is simple, any proper intermediate extension will not be Galois over the quadratic. But the ramification indices will all be 1, since they are all 1 in the big extension.


