TITLE: What is the Krull dimension of the ring of holomorphic functions on a complex manifold?
QUESTION [41 upvotes]: 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 $0\lt \dim \mathcal O(X)\lt \infty$ ?

REPLY [15 votes]: My previous (non/)answer is now out of date because of Misha's revisions.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.