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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.

Mark Rodriguez
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2 Answers2

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If $\mathfrak P$ is a prime of $L$, and $\mathfrak p$ is the prime of $K$ that lies under it, then there is a maximal integer $e$ such that $\mathfrak B^e$ divides $\mathfrak p$. (Alternative, but equivalent, formulation: if we factor $\mathfrak p$ in $\mathcal O_L$, then it is a product of primes $\mathfrak B$ of $L$ to various powers, and the ramified primes are the ones where this power is $> 1$.)

We say that an archimedean prime is ramified if it is of the form $\mathbb C/\mathbb R$ (rather than $\mathbb C/\mathbb C$ or $\mathbb R/\mathbb R$).

Matt E
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  • I suppose one might add that treating $\mathbb C/\mathbb R$ as ramified, rather than inert, has the virtue of compatibility in various formulaic and other principles, rather than having an immediate rationalization in terms of the kind of ramification that literally occurs for finite places. – paul garrett Dec 29 '13 at 22:11
  • Dear Paul, Yes, that's a good remark. Cheers, – Matt E Dec 29 '13 at 22:51
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$\DeclareMathOperator{\spec}{Spec}$ 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 $\mathfrak p$.)

Daniel Miller
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  • Dear @Daniel Miller, what is the meaning of etale at $\mathfrak{p}$ – Math123 Mar 20 '14 at 07:24
  • @Sushma: there is a general notion of an etale morphism of schemes (see Wikipedia, for example) which is defined pointwise. Strictly speaking, I mean that $\operatorname{Spec}(O_L) \to \operatorname{Spec}(O_K)$ is etale at all points lying above $\mathfrak p$. – Daniel Miller Mar 20 '14 at 13:03