$\mathcal{O}_{\mathbb{C}_p}/ p \mathcal{O}_{\mathbb{C}_p}$ infinite

Question

Let $\mathcal{O}_{\mathbb{C}_p}$ the ring of integers of complex $p$-adic numbers $\mathbb{C}_p$ which are defined as completion of alg closure $\overline{\mathbb{Q}_p}$ with respect $\vert \cdot \vert_p$ (extended from $\mathbb{Q}_p$ to $\overline{\mathbb{Q}_p}$. It is known that $\mathcal{O}_{\mathbb{C}_p}$ is not a Noetherian ring (e.g. A. M. Robert, A Course in p-adic Analysis, page 135).

How this fact can be explicitely used to show that $\mathcal{O}_{\mathbb{C}_p}/ p \mathcal{O}_{\mathbb{C}_p}$ is infinite?

Answer 1

I believe it is not a good idea to show $\mathcal{O}_{\mathbb C_p}/p \mathcal{O}_{\mathbb C_p}$ is infinite from the fact that $\mathcal{O}_{\mathbb C_p}$ is not noetherian and then some makeshift ring theoretic proposition or argument. Rather, I would claim that any good proof which shows that $\mathcal{O}_{\mathbb C_p}$ is not noetherian already shows immediately -- namely, by exhibiting the ideal structure of the ring -- that the quotient $\mathcal{O}_{\mathbb C_p}/p \mathcal{O}_{\mathbb C_p}$ is not noetherian as well.

We have to use something more about the specific rings and ideals in question anyway, because obviously there are non-noetherian rings $A$ with ideals $I$ such that $A/I$ is finite; in fact, there are non-noetherian local domains $R$ of characteristic $0$, with $p$ contained in their maximal ideal and $R/p$ finite, so we would need to prove a severely fine-tuned extra proposition here, on top of proving non-noetherianness of $\mathcal{O}_{\mathbb C_p}$; that seems like total overkill.

Because I also believe that whatever proves that $\mathcal{O}_{\mathbb C_p}$ is not noetherian already proves much more than that, and in particular infinitude of $\mathcal{O}_{\mathbb C_p}/p\mathcal{O}_{\mathbb C_p}$. This "whatever", from which everything follows, is what one should actually understand about the ring in question, and I believe it is nothing else than the unique extension of the $p$-adic absolute value $\lvert \cdot \rvert_p$ (normed to, say, $\lvert p \rvert_p = p^{-1}$) from $\mathbb Q$ to $\mathbb Q_p$ to an algebraic closure $\overline{ \mathbb Q_p}$ to its completion $\mathbb C_p$.

Crucially, on $\overline{\mathbb Q_p}$ as well as on $\mathbb C_p$, its value group consists of ($0$ and) all rational powers of $p$. This, and the fact that it is an absolute (nonarchimedean) value, tell us everything about the ideal structure of the ring by translating it into some undergrad exercise in open or closed intervals. Indeed, it follows now that the ring

$$\mathcal{O}_{\mathbb C_p} = \{x \in \mathbb C_p: \lvert x \rvert_p \le 1 \}$$

has unit group

$$\mathcal{O}^\times_{\mathbb C_p} = \{x \in \mathbb C_p: \lvert x \rvert_p = 1 \}.$$

With that, it further follows that $\mathcal{O}_{\mathbb C_p}$ has two kinds of non-zero ideals: On the one hand,

$$I^c_q := \{x \in \mathbb C_p: \lvert x \rvert_p \color{red}{\le} p^{-q} \}$$

for non-negative rationals $q$; these are the principal ideals, and any $x \in \mathbb C_p$ with $\lvert x \rvert_p = p^{-q}$ is a generator. On the other hand

$$I^o_r := \{x \in \mathbb C_p: \lvert x \rvert_p \color{red}{<} p^{-r} \}$$

for any non-negative real number $r$; these ideals are not finitely generated; a set of generators is given by the underlying set of any sequence $x_n$ in $\mathbb C_p$ such that the sequence of the values $\lvert x_n\rvert_p$ is increasing and has limit $p^{-r}$.

The sheer existence of non-finitely generated ideals shows that ${O}_{\mathbb C_p}$ is not noetherian; of course any monotonic decreasing non-stabilising sequence of real or rational numbers now also translates to non-stabilising increasing ideal chains.

Actually, we also get $I^c_{q_1} \cdot I^c_{q_2} = I^c_{q_1+q_2}$ and $I^o_{r_1} \cdot I^o_{r_2} = I^o_{r_1+r_2}$ and $I^c_q \cdot I^o_r = I^o_{q+r}$ almost for free, as well as e.g.

$$\bigcup_{q_n} I^c_{q_n} = \begin{cases} I^c_m \text{ if } m = \min\{q_n\}\text{ exists} \\ I^o_{\inf(q_n)} \text{ otherwise } \end{cases} \qquad \text{ etc. etc.}$$

In particular, this implies that ${O}_{\mathbb C_p}$ is a local ring with maximal ideal

$$\mathfrak{m}_{\mathcal{O}_{\mathbb C_p}} = \{x \in \mathbb C_p: \lvert x \rvert_p < 1 \} = I^o_0$$

(and immediately the maximal ideal is idempotent, $\mathfrak{m}_{\mathcal{O}_{\mathbb C_p}} = \mathfrak{m}_{\mathcal{O}_{\mathbb C_p}}^n$ for all $n$), while of course $p \mathcal{O}_{\mathbb C_p} = I^c_1$, and it is obvious, not only that $\mathcal{O}_{\mathbb C_p}/p \mathcal{O}_{\mathbb C_p}$ is infinite, but that its ideals are in one-to-one correspondence to the $I^c_q$ and $I^o_r$ where $q$ and $r$ are any rational or real numbers between $0$ and $1$, in particular there are tons of infinitely ascending ideal chains in there, in particular that quotient is as far from being noetherian as the whole ring, in particular it is very infinite.

Check out related questions Ring of integers of $\mathbb{C}_p$ and Why is $O_{\mathbb{C}_p}/p$ not perfect?.

Answer 2

Torsten’s answer is excellent, and depends on the observation that there is infinite ramification. But there is also infinite residue-field extension degree. In particular, the field $\mathcal K=\mathcal O_{\Bbb C_p}/\mathcal M$ (where $\mathcal M$ is the maximal ideal of the integer ring here) contains representatives of every element of $\overline{\Bbb F_p}$, the algebraic closure of the prime field $\Bbb F_p$. Thus $\mathcal K$ is infinite, and this is is all the more so for $\mathcal O_{\Bbb C_p}/p\mathcal O_{\Bbb C_p}$.

Perhaps I should point out that noetherianness has little to do with the case: if $\mathcal U$ is the maximal unramified extension of $\Bbb Q_p$, with ring of integers $\mathcal O_{\mathcal U}$, this last is still a discrete valuation ring, with prime element $p$, and thus noetherian. But $\mathcal O_{\mathcal U}/p\mathcal O_{\mathcal U}$ is infinite.

Answer 3

Note that an algebraic closure of $\mathbb{Q}_p$ is contained in $\mathbb{C}_p$. In particular each $f(x) \in \mathbb{Z}_p[x]$ has a root in $\mathbb{C}_p$. But then it is clear that if $\bar{f} \in \mathbb{F}_p[x]$ then any lift $f$ to $\mathbb{Z}_p$ has a root in $\mathbb{C}_p$ - in particular $\bar{f}$ has a root in $\mathcal{O}_{\mathbb{C}_p}/p \mathcal{O}_{\mathbb{C}_p}$.

But this implies that $\mathcal{O}_{\mathbb{C}_p}/p \mathcal{O}_{\mathbb{C}_p}$ contains an algebraic closure of $\mathbb{F}_p$, which is infinite.