Let $K$ be a local field, $k$ its residue field and $G_K$, $G_k$ be the absolute Galois groups of $K$ and $k$, respectively.

A Frobenius element is an element $\operatorname{Frob}_K \in G_K$ such that its image under the restriction homomorphism $G_K \to G_k$ is the Frobenius automorphism $x \mapsto x^q$ where $q$ denotes the cardinality of $k$.

Question: Why are all Frobenius elements conjugated to each other?

I have seen this statement quite often already, but I have not figured out why this is true.

I tried it to understand on the level of a finite Galois extension $F/K$, so a Frobenius element there would be an element $\operatorname{Frob}_{F/K} \in \operatorname{Gal}(F/K)$ such that under the restriction homomorphism $\operatorname{Gal}(F/K) \to \operatorname{Gal}(f/k)$ it gets mapped to $x \mapsto x^q$ as before ($f$ is the residue field of $F$). I also know that the residue field degree is exactly the cardinality of $\operatorname{Gal}(f/k)$. But I was not able to see any relation there.

Could you please help me here? Thank you!

  • 2
    $\begingroup$ I believe the notion of conjugacy of Frobenius elements only has relevance in the global case, where you have many primes lying over a given prime. I don't think there is anything to say about conjugacy in the local case. $\endgroup$ – D_S Oct 19 '18 at 3:49

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