Let $k$ be a global field, $p$ be a rational prime and let $S$ be a set of primes of $k$ with density $\delta(S) = 1$. Let $\mathfrak{p} \in S$ be a prime and denote by $k_\mathfrak{p}$ the completion of $k$ by $\mathfrak{p}$. For a number field $k$ denote by $\bar k$ the separable closure of $k$. Let's denote by $k_S(p)$ the maximal $p$-extension that is unramified outside $S$ and the Galois group by $G_S(k)(p) = Gal(k_S(p) | k)$. For the case of $S = \{\text{all primes}\}$, this notation simplifies to $G_k(p)$.

I'm interested in some basic facts about the behaviour of local Galois group $G_{k_\mathfrak{p}} = Gal(\overline{k_\mathfrak{p}} | k_\mathfrak{p})$ and the maximal pro-$p$-factor group $G_{k_\mathfrak{p}}(p) = Gal(\overline{k_\mathfrak{p}}(p) | k_\mathfrak{p})$.

  1. It is stated that there is a canonical surjective map

$$G_{k_\mathfrak{p}}(p) \rightarrow G_\mathfrak{p}(k_S(p) | k)$$ onto the decomposition group of $\mathfrak{p}$ in $k_S(p) | k$. Where does this map come from? I know that there is an isomorphism

$$G_\mathfrak{p}(K | k) \cong Gal(K_\mathfrak{p} | k_\mathfrak{p})$$

for every finite Galois extension $K$ of $k$.

Thanks a lot for you help,



1 Answer 1


No, there's a map, but it's not canonical, it depends on a choice of field embeddings over k. For each prime v, choose an embedding of k^sep into k_v^sep, and consequently an extension w of v to k^sep and an identification of G_v:=absolute Galois group of k_v with the decomposition group of w in G_k:=absolute Galois group of k. If G_S:=Gal(k_S/k), where k_S is the maximal extension unramified outside S, then G_S is a quotient of G_k, and we can define the composite map G_v --> G_k --> G_S. Same construction when considering (pro)-p-extensions. Note that although the map just constructed is not canonical, the induced cohomological maps do not depend on any choice.


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