TITLE: Places ramifying in an extension of number fields
QUESTION [5 upvotes]: I came across the following statement in a number theory paper: 
Let $L/K$ denote an arbitrary Galois extension of number fields with Galois
group $G$. Let $S$ be a finite non-empty set of places of L containing all archimedean
places of L (if any) and all those which ramify in the extension $L/K$.
What does it mean for a non-Archimedean place of $L$ to ramify in the extension $L/K$?  
Also can Archimedean places of $K$ ramify in $L$?
Many thanks for your help.

REPLY [2 votes]: $\DeclareMathOperator{\spec}{Spec}$
Here is another perspective. Instead of thinking about an injection of number fields $K\hookrightarrow L$, we should think about the map $O_K\to O_L$ of their rings of integers, and even better the induced morphism $f:\spec(O_L) \to \spec(O_K)$. Of course, a point in $\spec(O_L)$ is just a prime ideal $\mathfrak P\subset O_L$, and $f(\mathfrak P)=\mathfrak P\cap O_K$. So for instance we could say that for $\mathfrak p\subset O_K$, we have
$$
  \{\text{primes $\mathfrak P\subset O_L$ lying over $\mathfrak p$}\} = f^{-1}(\mathfrak p) \text{.}
$$
There is a scheme-theoretic definition of $f^{-1}(\mathfrak p)$, namely it is the fiber product 
$$
  f^{-1}(\mathfrak p)=\spec(O_L)\times_{\spec(O_K)} \spec(O_K/\mathfrak p) = \spec(O_L/\mathfrak p) \text{.}
$$
The extension $L/K$ is unramified at $\mathfrak p$ precisely when $f^{-1}(\mathfrak p)$ is a disjoint union of spectra of fields, i.e. when $O_L/\mathfrak p$ has no nilpotents. Since 
$$
  O_L/\mathfrak p\simeq \prod_{\mathfrak P\mid \mathfrak p} O_L/\mathfrak P^{e(\mathfrak P/\mathfrak p)}
$$
this comes down to $e(\mathfrak P/\mathfrak p)=1$ for all $\mathfrak P\mid \mathfrak p$. 
(What is going on here is that $L/K$ is unramified at $\mathfrak p$ if and only if $\spec(O_L) \to \spec(O_K)$ is etale at $\mathfrak p$.)