What is the Krull dimension of the ring of holomorphic functions on a complex manifold?

Question

Consider a connected holomorphic manifold $X$ and its ring of holomorphic functions $\mathcal O(X).$
My general question is simply: in which cases is the Krull dimension $\dim \mathcal O(X)$ known?

Of course if $X$ is compact $\mathcal O(X)=\mathbb C$ and that dimension is $0$.
There are also quite a lot of non-compact manifolds with $\mathcal O(Z)=\mathbb C$:
For example if $X$ is connected of dimension $\geq 2$ and $Y\subset X$ is an analytic subset of codimension at least $2$ ( or a small compact ball) , you will still have $\mathcal O(X\setminus Y)=\mathbb C$ .

But apart from these trivial examples I can't compute a single Krull dimension $dim \mathcal O(X)$ for, say, Stein manifolds of positive dimension.

Just in order to ask something definite, let me pose the ridiculous-sounding question:

Does there exist a connected holomorphic manifold $X$ with $0\lt \dim \mathcal O(X)\lt \infty$ ?

Answer 1

It follows from the proof in Sasane's paper that Krull dimension of a (connected) complex manifold $M$ is infinite iff $M$ admits a nonconstant holomorphic function $F: M\to {\mathbb C}$. Namely, using Sard's theorem find a sequence of points $a_k \in F(M)$ which are regular values of $F$ and so that $(a_k)$ converges to a point in $({\mathbb C}\cup \infty) \setminus F(M)$. Then, pick regular points $b_k\in V_k:=F^{-1}(a_k)$ of $F$ and define multiplicity of zero for a holomorphic function $h: M\to {\mathbb C}$ with respect to the germ of $V_k$ at $b_k$. (I.e., multiplicity of $h$ is determined by the largest $m$ so that $h=(F-a_k)^m g$ on the level of germs at $b_k$.) Now, the same proof as in Sasane's paper goes through, where you will be using functions $f_n\circ F$ instead of Sasane's functions $f_n$. The point is that Sasane's argument is essentially local at zeroes of the functions $f_n$. Actually, what Sasane proves is a lemma about a commutative ring $R$ with a sequence of valuations $m_k$ for which there exists a sequence of elements $f_i\in R$ so that $m_k(f_i)$ grows slower than $m_k(f_{i+1})$ for every $i$ as $k\to \infty$ (more precisely, in his case, the growth rate of $m_k(f_i)$ is $k^{i+1}$). Under this assumption, Krull dimension of $R$ is infinite.

Edit: I finally wrote a detailed proof here.

Edit. I wrote a proof that the Krull dimension of $H(M)$ (when it is positive) has cardinality at least continuum. The new proof uses surreal numbers instead of ultralimits. For the sake of completeness I am keeping the older proof as well.

Answer 2

My previous (non/)answer is now out of date because of Misha's revisions. I have written up an independent exposition of the "greater Kapovich theorem" that if a connected $\mathbb{C}$-manifold $M$ has a nonconstant holomorphic function then the cardinal Krull dimension of $\operatorname{Hol}(M)$ is at least $\mathfrak{c} = 2^{\aleph_0}$.

I have also shown -- by a straightforward reduction to the one-dimensional case -- that if $M$ is a Stein manifold then its cardinal Krull dimension is at least that of $\operatorname{Hol}(\mathbb{C})$ and thus -- by a result of Henriksen -- at least $2^{\aleph_1}$. Whether this stronger bound should occur under the much weaker hypothesis that there is a nonconstant holomorphic function I have no idea.

Added: The note has been published here.

Answer 3

Are you also looking for holomorphic manifolds with $\dim \mathcal O=\infty$?

In that case, in the paper by Sasane On the Krull Dimension of Rings of Transfer Functions [Acta Applicandae Mathematicae Volume 103, Number 2 (2008), 161-168] it is shown that the Krull dimension of $\mathcal{O}(\Omega)$ is infinite for any nonempty open subset $\Omega$ of $\mathbb{C}$ (see Corollary 2.3).

In particular the ring of entire functions $\mathcal{O}(\mathbb{C})$ has infinite Krull dimension.

Answer 4

I think the Krull dimension of $\mathcal O(X)$ is infinite if $\mathcal O(X)\neq\mathbb C$. Just take any non-constant holomorphic function $f$ in $\mathcal O(X)$. This has open image in $\mathbb C$ which we can assume to be unbounded (using for example the Riemann mapping theorem). Then pick a sequence of points $x_i$ in $X$ so that $f(x_i)$ converges to infinity. There is an infinite chain of prime ideals, the $n$th given by the functions which vanish on $x_{2^ni}$ for an infinite number of $i$.