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.

notanything like $B$. Don't $\mathfrak P$ and $\mathfrak B$ look completely different? :) – KConrad May 1 '12 at 2:07