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I have known for a finite extension $E\supseteq Q_p$, the ring of integers $O_E$ is a profinite group as an additive group. And $O_E^\times$ is also profinite as a multiplicative group.

Questions: For the p-adic complex field $C_p=\widehat{\overline{Q_p}}$, is $O_{C_p}$ a profinite group or compact? And what about ${O_{C_p}}^\times$ ?

Thanks in advance!

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Hint: The maximal ideal of $O_{\mathbb C_p}$ is $m=\lbrace x \in O_{\mathbb C_p}: \lvert x \rvert_p < 1 \rbrace$. Is $m$ open? How many elements does the quotient $O_{\mathbb C_p} /m$ have? If you answer both questions, you should see a covering of $O_{\mathbb C_p}$ by (actually: a partition of $O_{\mathbb C_p}$ into) infinitely many, mutually disjoint, open sets.

Try an analogous idea for $O_{\mathbb C_p}^\times$ using the multiplicative subgroup $1+m$.

Added: Here's an entirely different argument. One knows that the $p$-adic value on $\mathbb C_p$, normed to $\lvert p \rvert_p=p^{-1}$, has non-discrete value group $\lvert \mathbb C_p^* \rvert_p=p^\mathbb Q$. Now choose a bounded (!) sequence of rationals $(r_n)_n$, all $r_n \le 0$, such that $r_n \neq r_m$ for all $n \neq m$. For each $n$, pick $x_n \in O_{\mathbb C_p}$ with $\lvert x_n \rvert_p = p^{r_n}$. Then show that the sequence $(x_n)_n$ has no convergent subsequence, using that for a sequence $(y_k)_k$ in an ultrametric field to converge to something of absolute value $\lvert y \rvert \neq 0$, one needs $\lvert y_k \rvert = \lvert y\rvert$ for high enough $k$.

Slight generalisations of these two arguments will show that for a field $K$ with non-archimedean valuation $\lvert \cdot\rvert$ to be locally compact, it is necessary that

  • the residue field is finite, and
  • the valuation is discrete (meaning, its value group $\lvert K^* \rvert$ is).

Both conditions are not satisfied for $\mathbb C_p$, and it's a good exercise to come up with many other fields which fail either, or both. -- An obvious third condition, which is met by $\mathbb C_p$, is that

  • the field is complete.

Conversely, if a field with a non-archimedean valuation satisfies all three conditions above, it is locally compact. This is another good exercise; and a last good exercise is to show that such a field is necessarily a finite extension of some $\mathbb Q_p$ or $\mathbb F_p((T))$, so that with all exercises combined, we have proven the non-archimedean part of the theorem mentioned in Lubin's answer. (I am quite sure I have seen this as a combined exercise somewhere in Schikhof's Ultrametric Calculus).

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  • $\begingroup$ Thank you, I think I can solve my problems with your hints now. $\endgroup$ – Sssss Jul 2 at 4:42
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The answer of @TorstenSchoeneberg is the right one, because it argues from first principles, without calling in any advanced theorem. I take the opposite tack below:

If $\mathcal O_{\Bbb C_p}$ were compact, then $\Bbb C_p$ would be locally compact. But it’s a “well-known” theorem that the only locally compact fields of characteristic zero are $\Bbb C$, $\Bbb R$, and the fields $\Bbb Q_p$ and their finite extensions.

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  • $\begingroup$ Yes, you are right, so $O_{C_p}$ is not compact. I remember the well-known theorem is proved in 《Fourier Analysis on Number Fields》by Ramakrishnan or 《Basic Number Theory》by Weil. But for $O_{C_p}^\times$, is there another method to prove the non-compactness? Thanks for your answers! $\endgroup$ – Sssss Jul 2 at 5:19
  • $\begingroup$ Good point to put it in context of course. I've added that to my answer. $\endgroup$ – Torsten Schoeneberg Jul 2 at 16:46

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