Why is a general element in $G(L/K)$ a power of the Frobenius automorphism when restricted to $L \cap \tilde{K}$?

Question

I'm going through Neukirch's Algebraic Number theory and I'm stuck on why we can say that a general element in $G(L/K)$ is a power of the Frobenius automorphism when restricted to $L \cap \tilde{K}$ in the proof of proposition 4.4.

Prop. : Given a finite Galois extension $L/K$, then the mapping $$\{\sigma \in G(\tilde{L}//K) | d_{K} (\sigma)=n \in \mathbb{N}^{+} \} \rightarrow G(L/K)$$ is surjective.

In the proof it is just stated that given $\sigma \in G(L/K)$ when restricted to $L \cap \tilde{K}$ it is a power of the Frobenius automorphism restricted to $L \cap \tilde{K}$. And I can't think of a reason why it should be true.

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Answer 1

We are in the case of local fields with finite residue field.

An unramified extension has Galois group isomorphic to the residue field extension. Finite extensions of finite fields are cyclic and generated by the Frobenius, therefore everything in the Galois group is a power of (the lift of) the Frobenius.

Answer 2

This is Chapter IV of Neukirch, where he works with abstract Galois theory. Here is an argument that any $\sigma \in G(L|K)$ is a power of the Frobenius in that setting, rather than the special case where $G$ is the Galois group of local fields.

In abstract Galois theory $G_K$, $G_L$ are closed subgroups of a profinite group $G$, and $G(L|K) = G_K/G_L$. Also there is a surjective map $d: G \rightarrow \widehat{\mathbb{Z}}$, $f_K = (\widehat{\mathbb{Z}} : d(G_K))$, $d_K = (1/f_K)d$, the Frobenius $\varphi_K$ is defined via $d_K(\varphi_K) = 1$, and he writes $\varphi_{L|K}$ for $\varphi_K \bmod G_L$. Unramified means $(d(G_K):d(G_L)) = (G_K : G_L)$.

It is enough to show that if $L$ is unramified over $K$ then $G(L|K) = ( \varphi_{L|K} )$, that is, $G(L|K)$ is the cyclic group generated by $\varphi_{L|K}$. For this, suppose $|G_K/G_L| = n$, then \begin{equation}n = (G_K : G_L) = (d(G_K) : d(G_L)) = (d_K(G_K) : d_K(G_L))\end{equation} Since $d_K(G_K) = \widehat{\mathbb{Z}}$, $d_K(G_L) = n\widehat{\mathbb{Z}}$ (using facts about $\widehat{\mathbb{Z}}$ presented in Example 4 on page 272). The $\{\varphi_{L|K}^j \}_{j=0}^{n-1}$ are distinct mod $G_L$, because $d_K(\varphi_{L|K}^j) = j$ and so $\varphi_{L|K}^j$ cannot be in $G_L$ since $j = d_K(\varphi_{L|K}^j) \notin d_K(G_L) = n\widehat{\mathbb{Z}}$ for $j < n$. So the $n$ powers $\{\varphi_{L|K}^j\}_{j=0}^{n-1}$ account for all of $G_K/G_L$.