Let $k$ be a finite extension of $\mathbb{Q}_p$ very often we use the isomorphism that $Gal(\overline{k}/k)^{ab} \simeq \hat{(k^{\times})}$ given by local class field theory. My question would be do we have (and if yes how can I prove it) $ \hat{O_k^{\times}} \times \hat{\mathbb{Z}} \simeq \hat{(k^{\times})}$ ? and is the image of the inertia subgroup of $Gal(\overline{k}/k)$ in $Gal(\overline{k}/k)^{ab}$ isomorphic to $ \hat{O_k^{\times}}$ ? I imagine that the proof of the first point can come from the fact that $\hat{k^{\times}} \simeq \widehat{O_k^{\times} \times \mathbb{Z}}$ using the decomposition with an uniformizer. But then can I split the product ? Do we have $\widehat{O_k^{\times}} = O^{\times}_k$ ?

$\begingroup$ You can split the sequence by choosing an uniformizer. $\endgroup$– XarlesJul 6, 2018 at 8:02

$\begingroup$ I don't understand what you mean sorry. $\endgroup$– Pierre21Jul 6, 2018 at 10:23

$\begingroup$ The group $O_k^\times$ is already profinite, so its profinite completion is itself. $\endgroup$– KConradJul 6, 2018 at 10:59

$\begingroup$ Of wich system is it the inverse limit ? $\endgroup$– Pierre21Jul 6, 2018 at 11:20

$\begingroup$ $O_k^\times$ is the limit of the system of its quotients by $1 + P_k^n$, where $P_k$ is the prime ideal of $O_k$. $\endgroup$– LSpiceJul 6, 2018 at 14:17
1 Answer
Answering your questions: It is true that $$\widehat{(k^{\times})} \simeq \widehat{O_k^{\times} \times \mathbb{Z}} \simeq \widehat{O_k^{\times}} \times \widehat{\mathbb{Z}} \simeq O_k^{\times} \times \widehat{\mathbb{Z}}$$ as profinite groups, since
 $k^{\times}\simeq O_k^{\times} \times \mathbb{Z}$ as groups by choosing an uniformizer $\pi$ of $k$: any nonzero element $a$ of $k$ can be written as $a=\pi^v b$, where $b\in O_k^{\times}$ and $v\in \mathbb{Z}$ the valuation.
 Profinite completion commutes with products.
 $O_k^{\times}$ is already profinite.
Secondly, it is also true that the image of the inertia group by the (local Artin) isomorphism is $O_k^{\times}$. See for example Poonen notes.

$\begingroup$ Maybe two remarks (more useful to me than to other people, I guess) : 1) to avoid any confusion, here $\widehat{ \cdot }$ means the "topological" profinite completion, i.e. inverse limit over the closed normal subgroups of finite index (we can replace "closed" by "open"). 2) $O_k^{\times}$ is profinite, typically because inverse limit commutes with group of units, and $O_k$ is a profinite ring. $\endgroup$– WatsonJul 6, 2018 at 16:24