Let $I(L/K)$ be the <inertia group> and let $\varphi$ denote <Frobenius> on the residue-field extension. The <Relative Weil group> is
$$
W(L/K)=\{\sigma\in\operatorname{Gal}(L/K):\bar\sigma=\varphi^n\text{ for some }n\in\mathbb Z\}.
$$
Its <Weil-group topology> makes $I(L/K)$ an open <profinite group> with its usual topology and gives $W(L/K)/I(L/K)$ the discrete topology. Thus every inertia coset is an open copy of $I(L/K)$.

Take $L=K^{\mathrm{nr}}$, the <maximal unramified extension> of $K$. Then
$$
\operatorname{Gal}(L/K)\cong\widehat{\mathbb Z},
\qquad
W(L/K)\cong\mathbb Z.
$$
The subgroup $\{0\}\subset\mathbb Z$ is open in the discrete Weil-group topology, but it is not open in the <profinite topology> inherited from $\widehat{\mathbb Z}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 136 / 1 / a / Solution

{c}
{title2=$W(L/K)$}

For a Galois extension $L/K$ of non-Archimedean local fields, the relative Weil group consists of the automorphisms whose residue action is an integral power of <Frobenius>. It fits into
$$
1\longrightarrow I(L/K)\longrightarrow W(L/K)\longrightarrow\mathbb Z\longrightarrow1
$$
when the residue extension is the full algebraic extension.

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 Incoming links: Relative Weil group