I’m not quite sure why you’re being downvoted. These are reasonable questions in my opinion.
So, you’ve singled out that you have a valuation ring $\mathcal{O}$ (for future reference let’s call $L:=\mathrm{Frac}(\mathcal{O})$) and you want to know
- is it true that $\mathcal{O}$ has only one non-zero prime,
- what is the Krull dimension of $\mathcal{O}$,
- whether the map $\mathrm{Spec}(L)\to\mathrm{Spec}(\mathcal{O})$ is an open embedding.
To answer your first question, let us recall the following fact.
Fact: Let $\mathcal{O}$ be a valuation ring. Then, the set of ideals of $\mathcal{O}$ is a linearly ordered set.
In particular, we see that the spectrum $\mathrm{Spec}(\mathcal{O})$ looks like a ‘path’, like this (assuming it’s finite dimensional)
$$(0)—\bullet-\cdots\bullet-\mathfrak{m}$$
(incidentally, spectra of valuation rings are exactly used as ‘paths’ in birational geometry of schemes/formal schemes). Moreover, it’s clear that the length of this path is the same thing as the Krull dimension of $\mathcal{O}$. Thus, $\mathcal{O}$ has a unique non-zero prime ideal (which is the maximal ideal) if and only if it is of dimension $1$.
Is $\mathcal{O}_{\overline{K}}$ of dimension $1$? The answer is yes since $\mathrm{Spec}(\mathcal{O}_{\overline{K}})\to\mathrm{Spec}(\mathcal{O}_K)$ is surjective and integral (see Tag 0CEG). Thus, your first (and second) questions are answered.
To answer your third it is useful to put things in a bigger context. Namely, we would like to know when the map $\mathrm{Spec}(L)\to\mathrm{Spec}(\mathcal{O})$ is an open embedding or, equivalently, when $\mathrm{Spec}(L)$ is an open subset of $\mathrm{Spec}(\mathcal{O})$.
To this end, recall that any time you have a valuation ring $\mathcal{O}$ you get a valuation
$$v:L^\times\to \Gamma$$
where $\Gamma$ is some totally ordered abelian group (in fact, one can take $\Gamma=L^\times/\mathcal{O}^\times$ in the usual way). Recall that we say that $\Gamma$ is of height 1 if there exists an embedding of totally ordered abelian groups $\Gamma\hookrightarrow (\mathbb{R},^+)$.
One is then able to put a valuation topology on $L$ by declaring that a basis of open sets is given by (let’s assume that $v$ is non-trivial!) $\{a\in L:v(a)\leqslant \gamma\}$ for some $\gamma$ in $\Gamma$.
We then have the following definition.
Definition A valuation ring $\mathcal{O}$ is called microbial if its valuation topology agrees with the valuation topology of some
height $1$ valuation $v’$ on $\mathcal{O}$.
In particular, note that being microbial doesn’t mean your valuation is height $1$, it only means that your ‘valuation topology is rank $1$’.
NB: Such valuation rings are very important in Huber’s theory of adic spaces, and Fujiwara—Kato’s theory of the birational geometry of formal schemes. You can look there for many natural examples of microbial valuation rings which aren’t height 1. You can also see [Morel, Example I.1.5.5] for a concrete example.
We then have the following nice proposition.
Proposition: Let $\mathcal{O}$ be a valuation ring. Then, the following are equivalent:
- $\mathcal{O}$ is microbial,
- there exists some non-zero element $\varpi\in \mathcal{O}$ such that $\mathcal{O}$ is $\varpi$-adically separated,
- there exists some non-zero element $\varpi\in\mathcal{O}$ such that $L=\mathcal{O}[\frac{1}{\varpi}]$,
- the map $\mathrm{Spec}(L)\to\mathrm{Spec}(\mathcal{O})$ is an open embedding.
Proof: See this and [Morel, Theorem I.1.5.4]. $\blacksquare$
In particular, I can look at your example of $\mathcal{O}_{\overline{K}}$ and immediately see that $\mathrm{Spec}(\overline{K})\to \mathrm{Spec}(\mathcal{O}_{\overline{K}})$ is an open embedding since either:
- the valuation ring $\mathcal{O}_{\overline{K}}$ is height 1, and so of course microbial,
- it is $p$-adically separated (exercise!).
NB: Note that $\mathcal{O}_{\overline{K}}$ is not $p$-adically complete, so it’s good that in the definition of microbial we only required separatedness!
References:
[Morel] Morel, S., 2019. Adic Spaces. Lecture Notes. https://web.math.princeton.edu/~smorel/adic_notes.pdf.
Best Answer
Any Dedekind domain which is not a P.I.D. will be a counter-example.
Indeed, it can be shown that any noetherian gcd domain is a U.F.D., and a Dedekind domain which is a U.F.D. is actually principal.
Now Stark-Heegner's theorem gives the list of the $9$ imaginary quadratic fields with a principal ring of integers. $\mathbf Q(\sqrt{-5})$ is not in the list; so its ring of integers, which is $$\mathbf Z[\sqrt{-5}]\quad\text{since }-5\not\equiv 1\bmod 4,$$is a counter-example: it is integrally closed, but not a gcd domain.