Let $K$ be a local field, $k$ be its residue field, $G_K, G_k$ be the absolute Galois groups of $K, k$ and $I_K$ be the inertia group of $K$. In several books and papers, I found the following exact sequence:
$$ 1 \longrightarrow I_K \longrightarrow \ G_K \longrightarrow G_k \longrightarrow 1. $$
Now I would like to know why because my motivation is it to understand the definition of the Weil group $W_K$. If I recall correctly, this is defined by the exact sequence
$$ 1 \longrightarrow I_K \longrightarrow W_K \longrightarrow \mathbb{Z} \longrightarrow 0,$$
but I did not understand every detail of it.
First, let me assemble what I understood (or not understood entirely):
- I heard that $G_k$ is isomorphic to $\hat{\mathbb{Z}} = \varprojlim\limits_n \mathbb{Z}/n\mathbb{Z}$. I think this follows from the fact that $k$ is a finite field and the absolute Galois group is a profinite group but I was not yet able to put these facts together.
- Anyway, I think that $\mathbb{Z}$ is a subgroup of $\hat{\mathbb{Z}}$ but I do not know how the embedding would look like.
- One defines $W_K$ to be the inverse image of $\mathbb{Z}$ under the map $\phi: G_K \to \hat{\mathbb{Z}} $.
- I know that $I_K = \{ \sigma \in G_K : \sigma(x) \equiv x \mod{\mathcal{P}_K} \: \forall x \in \mathcal{O}_K \} $ where $\mathcal{P}_K = \{ x \in K : |x|<1 \}$ denotes the unique maximal ideal over $K$ and $\mathcal{O}_K$ denotes the ring of integers over $K$. As the first sequence is exact, this must mean that $I_K$ is the kernel of some map $G_K \to G_k$ resp. $G_K \to \hat{\mathbb{Z}}$. But I do now know how this map would look like.
- How does the Frobenius automorphism comes into play? The Frobenius automorphism is the map $\operatorname{Frob}_k \in G_k$ with $\operatorname{Frob}_k : x \mapsto x^q $ where $q$ is the size of the finite field $k$.
Could you please explain this to me? Thank you!
Let $K$ be your non-archimedean local field, $\mathcal O_K$ be the ring of integers, $\mathfrak p$ be the maximal ideal of the ring, and $k := \mathcal O_K / \mathfrak p$ be the residue field.
If $k$ is a finite field and $|k| = q$, then we assemble $\overline k$ this way: by the classification of finite fields, for every $n$ there is a unique field of order $q^n$ up to isomorphism, and the field of order $q^m$ is a subfield of the field of order $q^n$ iff $m \mid n$, so we take the "union" (really a direct limit) of the fields of order $q^n$, and that is $\overline k$.
And as to the Galois groups, if we denote by $\Bbb F_{q^n}$ the field of order $q^n$, then $\operatorname{Gal}(\Bbb F_{q^n} / \Bbb F_q) \cong \Bbb Z/n\Bbb Z$, with the canonical generator being the automorphism $x \mapsto x^q$. The Galois group of a direct limit is the inverse limit of the Galois groups, i.e.: $$\operatorname{Gal}(\overline{\Bbb F_q} / \Bbb F_q) \cong \operatorname{Gal}(\varinjlim \Bbb F_{q^n} / \Bbb F_q) \cong \varprojlim \operatorname{Gal}(\Bbb F_{q^n} / \Bbb F_q) \cong \varprojlim \Bbb Z/n\Bbb Z = \widehat{\Bbb Z}$$ And the topological generator is still the automorphism $x \mapsto x^q$. "Topological generator" means that this automorphism generates the dense cyclic subgroup $\Bbb Z$ of $\widehat{\Bbb Z}$.
So this clears the $\Bbb Z$ part of your query.
Now we discuss the map $G_K \to G_k$
The following three things are equivalent:
Moreover, from the classification of finite fields, $\Bbb F_{q^n}$ can be described as $\Bbb F_q(\mu_{q^n-1})$, where $\mu_{q^n-1}$ is the set of the $(q^n-1)$st roots of unity.
From the classification of unramified extensions, for each $n$ there is a unique unramified extension $L/K$ with $[L:K] = n$, and $L = K(\mu_{q^n-1})$. Again we have $\operatorname{Gal}(L/K) \cong \Bbb Z/n\Bbb Z$, the generator being the automorphism that sends a primitive $(q^n-1)$th root of unity to its $q$-th power (no, it doesn't send everything else to their $q$-th powers). We call this map Frobenius.
Using the same machinery, we can establish that $\operatorname{Gal}(K^{ur}/K) \cong \widehat{\Bbb Z}$.
Note that $K^{ur}$, being a union of normal extensions, is itself a normal extension. Therefore, we have a map $G_K \to \operatorname{Gal}(K^{ur}/K) \cong \operatorname{Gal}(\overline k/k) = G_k$ as required.
A caveat is that the topology on the Weil group is not the subspace topology, but finer than the subspace topology. We require that (the image of) the inertia group be open.