Ramification of primes in number fields in terms of valuations

Question

Let $L/K$ be a number field extension. Is there a definition of ramification of primes(both infinite and finite) in terms of the valuations induced?

This answer gives a definition for infinite primes:

Now fix an infinite place $v$ on $K$, let $L$ be a finite field extension of $K$, and let $w$ be an extension of $v$ to $L$. The extension is said to ramify at $w$ iff $\#\{\tau\in Gal(L,K)\mid w\circ\tau=w\}>1$. But in reality this all simplifies to what Keenan said. The only possibilities for $\tau$ satisfying $w\circ\tau=w$ are the identity map and complex conjugation.

The natural way to extend this to finite places does not quite seem to work(unless I am making a mistake):

Let $L = \Bbb Q[i], K = \Bbb Q$ and the prime $p = (3)$. This is inert in the extension and is fixed under the automorphism(complex conjugation) of $L/K$. According to the above definition, this would seem to imply that $(3)$ is ramified in $L/K$.

It is not very hard to find a definition for the case of finite places but I am having a little trouble finding one that works for both infinite and finite places. I feel like a small modification should make everything work in a unified way for both infinite and finite places.

Answer 1

I'm not sure if there is a "nice" completely parallel definition for ramification in both the infinite and finite cases. The finite places are very different to the infinite ones - in particular, for each finite place, we have a valuation ring with a uniformiser and a residue field. However, I'll try to explain some of the ideas in this answer.

---

Viewpoint 1: Valuations

Let $L/K$ be an extension of number fields, and let $w$ be a (normalised) place of $L$ above a place $v$ of $K$. Choose $x\in K$ such that $|x|_v \ne 1$.

We say that $w\mid v$ is unramified if $$|x|_w = |x|_v.$$

In the finite case, since any $x\in K$ can be written as $u\pi_K^m$ for some $u$ with $|u|_v = 1$ and $\pi_K$ a uniformiser, this is equivalent to saying $$v(\pi_K) = w(\pi_K).$$

In the infinite case, recall that if $w$ is complex, then $|x|_w = |x|_{\mathbb C}^2$, so $w\mid v$ is ramified if and only if $v$ is real and $w$ is complex.

---

Viewpoint 2: The Inertia Group

Since your definition implicitly assumes that $L/K$ is a Galois extension, I will do the same.

Let $G = \mathrm{Gal}(L/K)$ be the Galois group of $L/K$, and for each place $w$ (infinite or finite) of $L$ lying over a place $v$ of $K$, let $G_w\le G$ be the Galois group of $L_w/K_v$.

In fact, we can describe $G_w$ explicitly: $$G_w = \{\tau \in G:w\circ \tau = w\}$$ When $w$ is a finite prime, the group $G_w$ is the decomopostion group of $w\mid v$. Note that in the infinite case, this is the set you described in your definition of ramification. $L_w$ and $K_v$ are either $\mathbb R$ or $\mathbb C$, and the extension will be ramified if and only if $K_v = \mathbb R$ and $L_w=\mathbb C$. In this case, there isn't much more we can say.

The finite case is much more interesting. Let $k_v$ and $k_w$ be the residue fields associated to $v$ and $w$. The action of any $\tau\in G_w$ descends to an action of $k_w/k_v$ giving a (surjective) group homomorphism $$G_w\to \mathrm{Gal}(k_w/k_v)$$

Let $I_w$ be the kernel of this map. $I_w$ is called the inertia group of $w\mid v$. We say that $w\mid v $ is unramified if and only if $$\#I_w = 1.$$

In the infinite case, we can define $I_w$ to just be $G_w$, and again the definitions are the same.