Is $O_{C_p}$ and ${O_{C_p}}^\times$ a profinite group or compact?

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 $${O_{C_p}}^\times$$ ?

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).

• Thank you, I think I can solve my problems with your hints now. – Sssss Jul 2 at 4:42

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.

• 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! – Sssss Jul 2 at 5:19
• Good point to put it in context of course. I've added that to my answer. – Torsten Schoeneberg Jul 2 at 16:46