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How is the conformal structure of regions of the complex plane determined by the integral domain of holomorphic functions defined on those regions?


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I understand all the words, but I still don't understand the question: what you do mean by "determining" the conformal structure? Also some context would help. – Pete L. Clark May 20 '11 at 3:46
Ah, here's a guess: are you asking whether any Riemann surface $X$ with field of meromorphic functions isomorphic to the field of meromorphic functions on $\mathbb{C}$ is actually isomorphic to $\mathbb{C}$? (That would be a good question...) – Pete L. Clark May 20 '11 at 3:52
I mean determined in the sense that given two regions of the complex plane such that there is an isomorphism between the fields of meromorphic functions defined on each region, then these regions will be conformally equivalent. – user7485 May 20 '11 at 3:59
.: okay, that's a perfectly good thing to mean. It doesn't seem to be what you said though, so you should probably edit your question for clarity. (And if this is what you mean, why is your question tagged riemann-surface?) – Pete L. Clark May 20 '11 at 4:42
Thanks Pete, I've edited the question- it turns out that it's the integral domain of holomorphic functions that determines the conformal structure of the regions (though in general, plane regions), and the field of meromorphic functions determines the conformal structure of Riemann surfaces. – user7485 May 20 '11 at 9:07

The conformal structure on the plane domain is completely determined by the ring of holomorphic functions. More precisely if two such rings are isomorphic, and isomorphism is identity on the constants, the regions are conformally equivalent. If the rings are only isomorphic as abstract rings, then the regions are either conformally equivalent or anti-conformally equivalent.

This is due to Bers (BAMS 54, 1948) for plane domains and to Rudin (BAMS 61, 1955) for open Riemann surfaces.

The idea is to consider the space of maximal ideals of the ring. They are in 1-1 correspondence with the points of the region.

Remark. BAMS is free onine.

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