Let $K$ be a number field or function field over $\mathbb{F}_q$ of transcendence degree 1. Let $\mathbb{A}$ be the finite adele ring or full adele ring, respectively. Let $\hat{\mathcal{O}} = \prod_p \mathcal{O}_p$, where $\mathcal{O}_p \subset K_p$ is the ring of integers inside the completion at $p$.
If $\vartheta : \mathbb{A}^* \to \text{Gal}(K^{ab}/K)$ is the idelic artin map, then I know that $K^*$ must be inside the kernel. But what about $\hat{\mathcal{O}}$ ? Must that be inside the kernel of $\vartheta$ as well? It would seem to me that the answer is yes, as $\vartheta$ must be compatible with the ideal Artin map. That is, $J$ is the fractional ideal group and $i : \mathbb{A}^* \to J$ is the map $(a_p)_p \mapsto \prod_p p^{v_p(a_p)}$ and $\theta : J \to \text{Gal}(K^{ab}/K)$ is the ideal artin map, then I must have $\vartheta = \theta \circ i$, right? And the image of $\hat{\mathcal{O}}$ under $i$ is trivial.
Definitely not! In order to be compatible with local class field theory, for each finite place $v$, the Artin map restricted to $\mathcal O_{K_v}^\times\subset \hat{\mathcal O}^\times$ must be the local Artin map. This maps onto the Inertia group $I_{K_v}\subset \mathrm{Gal}(K_v^{\text{ab}}/K_v)$.
In fact, the Artin map gives an isomorphism $$K^\times \backslash\mathbb A^\times_K/C_K\to \mathrm{Gal}(K^{\text{ab}}/K),$$ where if $(K_\infty^\times)^0$ is the identity connected component of the archimedean part $K_\infty^\times$, then $C_K$ is the closure of its image in $K^\times\backslash\mathbb A_K^\times$.