This might be a easy question, but I couldn't get the point.

Let $F$ be a p-adic field, $\bar{F}$ a separable algebraic closure of $F$. Set $\Omega_F=Gal(\bar{F}/F)$. Use $F_{\infty}\subset \bar{F}$ to denote the maximal unramified extension of $F$. Use $E_{\infty} \supset F_{\infty} $ to denote the maximal tamely ramified extension of $F$.

Let $\mathcal{I}=Gal(\bar{F}/F_{\infty})$, $\mathcal{P}=Gal(\bar{F}/E_{\infty})$. In the book 28.3.2 says that there is a isomorphism

$$t_0:\mathcal{I}/\mathcal{P}\cong \prod_{l\ne p}\mathbb{Z}_l$$ where $l$ ranges over all prime numbers other than $p$, and $\mathbb{Z}_l$ denotes the ring of $l$-adic integers.

The statement I'm confused is that in the book above proposition 28.3.3, the group $Gal(F_{\infty}/F)$ acts on $Gal(E_{\infty}/F_{\infty})$ by conjugation, and moreover

$$t_0(\Phi_F\sigma\Phi_F^{-1})=q^{-1}t_0(\sigma)$$ for any $\sigma\in \mathcal{I}/\mathcal{P}$, where $\Phi_F$ is the geometric Frobenius in $Gal(F_{\infty}/F)$, and $q$ is the number of elements in the residue field of $F$.

I couldn't see how $Gal(F_{\infty}/F)$ acts on $Gal(E_{\infty}/F_{\infty})$ by conjugation,and thus don't know the meaning of $\Phi_F\sigma\Phi_F^{-1}$.

Thanks a lot for any help, and sorry if the question is not appropriate here.


Your real question is : When we have a short exact sequence of (multiplicatively written) groups $1\to C\to E\to G\to1$ in which $C$ is commutative, how does $G$ act on $C$ "by conjugation" ?

The answer is that you make $g\in G$ act on $C$ by $c\mapsto\tilde g c\tilde g^{-1}$, where $\tilde g\in E$ is any lift of $g$. It doesn't matter which lift you take, because any two lifts differ by an element of $C$, and the action of $C$ on itself by conjugation is trivial by the hypothesis that $C$ is commutative.

(By abuse of notation, one may write $gcg^{-1}$ for $\tilde g c\tilde g^{-1}$ because the thing only depends on $g$, not on the lift $\tilde g$).


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