Krull dimension of the ring of holomorphic functions on an almost complex manifold

Can the Krull dimension of the ring of holomorphic functions on an almost complex manifold be non-zero and finite?

Can the $$\mathbb{C}$$-algebra of holomorphic functions on an almost complex manifold be isomorphic to $$\mathbb{C}[[z]]$$?

The almost complex structure cannot be integrable. The simplest smooth manifold to consider is probably the unit ball of real dimension 4.

• You might want to have a look at the MO question mathoverflow.net/questions/94537/…. I would think that the answers to your questions are 'no', since, if there is a nonconstant holomorphic function $f$ on the almost-complex manifold $M$, then the set of all $h\circ f$ (where $h$ is any holomorphic function $h$ on $\mathbb{C}$) forms a subring, and this subring already will already have infinite Krull dimension, won't it? – Robert Bryant Nov 12 '20 at 21:11
• Krull dimension is not necessarily increasing for inclusions of rings – Andrea Ferretti Nov 13 '20 at 8:05
• If we consider the subalgebra $\mathbb{C}[x, xy, xy^2, \ldots]\subset \mathbb{C}[x, y]$ I think it has infinite Krull dimension. – Nguyen Nov 13 '20 at 8:21
• Though Dr. Bryant's remark would seem to rule out the polynomial ring in one variable. – Nguyen Nov 13 '20 at 8:22
• @Nguyen: It can't be $\mathbb{C}[z]$, since, if $z$ is a non-constant holomorphic function on $M$, then $\mathrm{e}^z$ will be a non-constant holomorphic function on $M$ that does not belong to $\mathbb{C}[z]$. – Robert Bryant Nov 13 '20 at 10:35