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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.

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  • $\begingroup$ 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? $\endgroup$ – Robert Bryant Nov 12 '20 at 21:11
  • $\begingroup$ Krull dimension is not necessarily increasing for inclusions of rings $\endgroup$ – Andrea Ferretti Nov 13 '20 at 8:05
  • $\begingroup$ If we consider the subalgebra $\mathbb{C}[x, xy, xy^2, \ldots]\subset \mathbb{C}[x, y]$ I think it has infinite Krull dimension. $\endgroup$ – Nguyen Nov 13 '20 at 8:21
  • $\begingroup$ Though Dr. Bryant's remark would seem to rule out the polynomial ring in one variable. $\endgroup$ – Nguyen Nov 13 '20 at 8:22
  • $\begingroup$ @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 $\mathbb{C}[z]$. $\endgroup$ – Robert Bryant Nov 13 '20 at 10:35

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