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

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

2012-04-19T14:51:34Z

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 by Francesco Polizzi for What is the Krull dimension of the ring of holomorphic functions on a complex manifold ?

2012-04-19T15:29:39Z

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 by Brett Parker for What is the Krull dimension of the ring of holomorphic functions on a complex manifold ?

2012-04-20T04:36:54Z

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 $i$.

Answer by Misha for What is the Krull dimension of the ring of holomorphic functions on a complex manifold ?

2012-04-20T20:41:56Z

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.