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

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 $0\lt \dim \mathcal O(X)\lt \infty$ ?

• Interesting. I had no idea this problem could be hard. – Olivier Apr 19 '12 at 15:17
• When $X$ is the complex plane, see mathoverflow.net/questions/33945/… and in particular the answers of Andreas Blass and Kevin Ventullo, which imply that the dimension is infinite. Perhaps the argument generalizes? – Laurent Moret-Bailly Apr 19 '12 at 15:44
• Cher Laurent, thanks a lot for your link : strangely enough I had tried to check if a similar question had been asked here on MO, but I was unsuccessful with key words like ring of holomorphic functions and Krull dimension. It seems, from browsing (very superficially !) through Henriksen's article mentioned in the MO page you refer to, that there are in $\mathcal O(\mathbb C)$ chains of prime ideals of staggering cardinality. I'll have a more detailed look, since the article seems to be relatively easy to read. – Georges Elencwajg Apr 19 '12 at 17:20

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.

• Dear Misha, this is a very impressive answer and I like it a lot since it answers my "ridiculous-sounding question" in the negative, just as I hoped: no ring of holomorphic functions on a manifold has finite positive Krull dimension. I would like to strongly encourage you to write up the sketch above (which seems quite convincing) as an article containing a self-contained complete proof. Indeeed, your result is very counterintuitive and interesting. It would be a most welcome addition to the apparently very scarce literature on the subject. – Georges Elencwajg Apr 21 '12 at 10:08
• A follow-up question: Can the Krull dimension of such a ring be countable, or does it have to be uncountable? – Neil Epstein Apr 23 '12 at 10:15

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.

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.

• Dear Francesco: yes I am definitely looking for examples with $dim \mathcal O(X)=\infty$, since I believe that this is essentially always the case! So thanks a lot for your answer. Unfortunately I don't think I have access to the journal you mention (I must confess I didn't even know it existed !) . Could you perhaps expand your answer or give a more accessible link to (some version or variant of) the paper? – Georges Elencwajg Apr 19 '12 at 15:51
• Dear Georges, unfortunately I could not find a more accessible link to the paper. Since the proof seems quite technical, expanding the answer is not easy. My e-mail address is polizzi@mat.unical.it. Please feel free to contact me if you want that I send you a pdf-copy of the article. – Francesco Polizzi Apr 19 '12 at 16:03
• Caro Francesco, thanks a lot for this kind proposition: you've got mail! – Georges Elencwajg Apr 19 '12 at 17:14
• Thanks a lot Francesco: I really appreciate your friendliness. Sasane mentions the "weak Krull dimension" obtained by only considering chains of finitely generated prime ideals. For $\mathcal O(\mathbb C)$ that weak dimension is, satisfyingly, equal to $1$. I had never heard of weak Krull dimension, but perhaps it is a reasonable concept for rings of holomorphic functions. – Georges Elencwajg Apr 19 '12 at 18:14
• Dear Gunnar, thanks for your comments. I can only say that $\mathcal O(X)$ is a Bézout ring if $X$ is a domain in $\mathbb C$, or more generally a non-compact Riemann surface: this means that every finitely generated ideal of $\mathcal O(X)$ is principal. And that this is false in higher dimensions. – Georges Elencwajg Apr 19 '12 at 23:43

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 $i$.

• This is not an ideal (not closed under addition). But using $f$, you can probably adapt Kevin Ventullo's argument in mathoverflow.net/questions/33945/… to conclude. – Laurent Moret-Bailly Apr 20 '12 at 6:19
• Dear Brett, unfortunately I think your proof does not work because the $n$th set, say $S_n$, you describe is not even an ideal, let alone a prime ideal: indeed it is not stable under addition. Let me give an explicit example where $S_1$ is not stable under addition.Take $X=\mathbb C , f(z)=z$ and $x_i=i$. We have $g(z)=sin(\frac {\pi}{4}z)\in S_1$ because $g(x_{2i})=g(2i)=0$ for all $i=2j$. Also $1-g\in S_1$ since $(1-g)(2i)=1-sin(\frac {\pi}{4}(2i))=0$ for all $i=4j+1$. But of course $g+(1-g)=1\notin S_1$. – Georges Elencwajg Apr 20 '12 at 7:00
• Ah, I see Laurent posted the same objection while I was writing mine and pausing for breakfast . I suppose he gets up earlier or eats less or more probably thinks and writes faster than I do :-) – Georges Elencwajg Apr 20 '12 at 7:11
• Yes- that was silly of me! Thanks for pointing out the mistake! – Brett Parker Apr 23 '12 at 6:05