$K$ is a local field. $L/K$ is Galois with $K_{ur}\subset L$, where $K_{ur}$ is the maximal unramified extension of $K$. Noting $\Phi\in Gal(L/K)$ a lift of the Frobenius $\phi\in Gal(K_{ur}/K)$ and $F$ the fixed field of $\Phi$, I just read as being obvious:
"As $\langle\Phi\rangle$ is dense in $Gal(L/F)$,..."
I have no idea why, even if i know that $\langle \phi\rangle$ is dense in $Gal(K_{ur}/K)$. (My explanation: this subgroup has the same projections as the Galois group viewed as projective limit.)
Can someone explain me why this density? Thank you.
I think I got it !
In fact it is a general property that could be asked as an exercise after the Galois theorem for infinite extensions.
Property : If $L/K$ is Galois and $G\subset Gal(L/K)$ then $G$ is dense in the Galois group of its fixed field $Fix(G)$.
Let $\overline G$ the closure of $G$ in $Gal(L/K)$ (with Krull topology). The theorem affirm that, as closed subgroup, $\overline G=Gal(L/Fix(\overline G))$. But, $G\subset\overline G\Rightarrow Fix(\overline G)\subset Fix(G)$ and $G\in Gal(L/Fix(G))$ which is closed, as Galois group of a subfield, and so $\overline G\subset Gal(L/Fix(G))$. $\overline G$ fix $Fix(G)$ henceforth $Fix(\overline G)=Fix(G)$ (to be fixed is a closed condition).
$\overline G=Gal(L/Fix(G))$ or equivalently $G$ is dense.