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. $\endgroup$
    – Xarles
    Jul 6, 2018 at 8:02
  • $\begingroup$ I don't understand what you mean sorry. $\endgroup$
    – Pierre21
    Jul 6, 2018 at 10:23
  • $\begingroup$ The group $O_k^\times$ is already profinite, so its profinite completion is itself. $\endgroup$
    – KConrad
    Jul 6, 2018 at 10:59
  • $\begingroup$ Of wich system is it the inverse limit ? $\endgroup$
    – Pierre21
    Jul 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 $O_k$. $\endgroup$
    – LSpice
    Jul 6, 2018 at 14:17

1 Answer 1


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 non-zero 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. $\endgroup$
    – Watson
    Jul 6, 2018 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.