Computing the degree of extension of local fields from the corresponding global fields

Question

Let $p$ be a prime and K be a finite Galois extension of $\mathbb{Q}$ (the set of rational numbers) of degree $p^r, r>0$. Let $v$ be a prime of $K$ lying above $p$. What can one say about the degree of $K_v\, / \, \mathbb{Q}_P$ where $K_v$ and $\mathbb{Q}_P$ denote the completions of $K$ and $\mathbb{Q}$ with respect to the primes $v$ and $p$ respectively ?

Answer 1

Let $G$ be the Galois group of $K$ over $\mathbb{Q}$. Let $\mathfrak p$ be the prime of $\mathcal O_K$ corresponding to $v$. The choice of $v$ (or $\mathfrak p$) gives an absolute value $| \cdot |$ on $K$ which restricts to the $p$-adic absolute value on $\mathbb Q$.

If $\sigma \in G$, then $|\sigma( x )| = |x |$ for all $x \in K$, if and only if $\sigma(\mathfrak p) = \mathfrak p$ (in general, $\sigma(\mathfrak p)$ will be a prime of $\mathcal O_K$ lying over $p$). The set of such $\sigma$ forms a subgroup of $G$, called the decomposition group of $\mathfrak p$:

$$G_{\mathfrak p} = \{ \sigma \in G : \sigma(\mathfrak p) = \mathfrak p\}$$

Proposition: If $\sigma \in G_{\mathfrak p}$, then $\sigma: K \rightarrow K$ extends uniquely to a $\mathbb Q_p$-automorphism of $K_v$. Every such automorphism comes from a unique $\sigma$.

Proof: $K$ is dense in its completion $K_v$, and $\sigma$ is an isometry ($|\sigma(x)| = |x|$) and is in particular uniformly continuous. Hence $\sigma$ extends to a unique continuous function $K_v \rightarrow K_v$ (in fact, $\sigma$ remains an isometry via the unique extension of $| \cdot |$ to $K_v$). Being continuous and fixing $\mathbb Q$ pointwise, it also fixes the closure of $\mathbb Q$ in $K_v$, which is $\mathbb Q_p$. Also by continuity, the extension of $\sigma$ still preserves addition and multiplication. Hence the extension of $\sigma$ lies in $\operatorname{Gal}(K_v/\mathbb Q_p)$.

If $\tau \in \operatorname{Gal}(K_v/\mathbb Q_p)$, then since $K$ is Galois over $\mathbb Q$, $\tau|_K$ is an automorphism of $K$. Now $\tau$, being an isometry, stabilizes the unique prime $\mathfrak p_v$ of $\mathcal O_{K_v}$. Hence it stabilizes $\mathfrak p_v \cap \mathcal O_K = \mathfrak p$. Thus $\tau|_K \in G_{\mathfrak p}$. $\blacksquare$

The proposition shows that the Galois group of $K_v/\mathbb Q_p$ can be identified naturally with the decomposition group $G_{\mathfrak p}$. Thus the degree of $K_v/\mathbb{Q}_p$ is $|G_{\mathfrak p}|$, which is $[K : \mathbb Q]$ divided by the number of primes of $\mathcal O_K$ lying over $p$.