# Ring of integers of $\mathbb{C}_p$

Sorry, maybe this is a really stupid question. Let $$\mathbb{C}_p$$ be the completion of the algebraic closure $$\overline{\mathbb{Q}_p}$$ of the field of $$p$$-adic numbers. We know that there exists a way to extend the $$p$$-adic valuation to $$\mathbb{C}_p$$, and in this way it is possible to define the ring of integers $$\mathcal{O}_{\mathbb{C}_p}$$ given by elements of valuation greater or equal than $$0$$. It is also possible to define the maximal ideal of this valuation ring given by elements of valuation strictly greater than $$0$$. Now, is this ideal finitely generated?

• Hint: elements of the maximal ideal can have valuations arbitrarily close to $0$. – Wojowu Dec 28 '18 at 17:12
• Let $K^{unr} = \mathbf{Q}_p(\zeta_{p^\infty-1})$ the maximal unramified extension of $\mathbf{Q}_p$ (so that $(p)$ is the unique maximal ideal of both rings of integers). Since $O_{K^{unr}}/(p)$ is algebraically closed then for any finite extension $L/K$ of degree $n$ you'll have $|\pi_L|_L= |p|_L^{1/n}$. Thus only finite extensions of $K$ are local fields and any infinite extension of $K$ will be dense in $\mathbb{C}_p$. – reuns Dec 28 '18 at 18:47
• @reuns, I do not believe that any infinite extension of $K=K^{unr}$ is dense in $\Bbb C_p$. What about the maximal tame extension of $K$? – Lubin Dec 29 '18 at 22:21
• @Lubin Thank you. You are saying that $p \nmid n = [K^{unr}(a):K^{unr}]$ implies that $|a|= p^{l/n}$ so $|a| \ne |p^{1/p}|$ and $|p^{1/p}-a| \ge |p^{1/p}|$ ? Whence with $L$ the maximal tame extension of $K^{unr}$ then $\inf_{a \in L} |p^{1/p} - a| = |p^{1/p}|$ and my last sentence was a big mistake and $\mathbb{C}_p$ is very different to $\mathbb{C}$ in this regard. – reuns Dec 30 '18 at 17:11
• Right you are. I’m pretty sure that even if your valuation group was all of $\Bbb Q$, that would not be enough to guarantee that the completion was all of $\Bbb C$. Wild ramification is truly a wild phenomenon! – Lubin Dec 30 '18 at 18:15

No it isn't - indeed, suppose it were, say $$\mathfrak{m}_{\mathbb{C}_p} = (x_1,\ldots,x_n)$$. Then $$v(x) \geq \min v(x_i) > 0$$ for any $$x\in\mathfrak{m}_{\mathbb{C}_p}$$. However, this is a contradiction as there exist elements of $$\mathfrak{m}_{\mathbb{C}_p}$$ with valuation strictly less than $$\min v(x_i)$$ (for example, one of the $$\sqrt{x_i}$$).