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Let $K$ be a local field and let $\widehat{K^*}$ and $\widehat{\mathbb{Z}}$ denote the profinite completions of $K$ and $\mathbb{Z}$. As the title suggests I'm having difficulties understanding the isomorphism ${\mathcal{O}^*_K} \times \widehat{\mathbb{Z}} \cong \widehat{K^*}$ that is given by $(u, z) \mapsto u \pi^z$ where $\pi$ is a uniformizer of ${\mathcal{O}^*_K}$. Concerning this, I have three questions:

How can we view $\pi^z$ as an element of $\widehat{K^*}$?

What does this map look like levelwise when derived as the map coming from the universal property of direct limits?

Why is $\widehat{\mathcal{O}_K^*} = \mathcal{O}_K^*?$

coconuthead
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    Note that $\mathcal O^\times_K$ is finite, and it follows from the definition that the profinite completion of a finite group is the group itself. Secondly, profinite completion commutes with finite direct products. – bsbb4 Feb 14 '24 at 17:36
  • Are you sure that $\mathcal{O}_K^*$ is finite? Also I know that completion commutes with finite products but still I fail to answer the questions that I listed above. – coconuthead Feb 15 '24 at 17:06
  • After looking at this again I have now understood that this isomorphism is just simply induced by the valuation of (on every level of the inverse limits). – coconuthead Mar 11 '24 at 23:47

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