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Is every connected non-compact Riemann surface biholomorphically equivalent to an affine algebraic curve in some ${\mathbb C}^n$? I suspect that surfaces of infinite genus probably are not but could not find a reference.

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In fact the disk is a simple counterexample. The point is that affine algebraic curves are the same thing as compact Riemann surfaces minus a finite set of points. In particular, they have no constant bounded holomorphic functions, where as the disk has plenty. – Donu Arapura May 11 '13 at 21:00
It should have finite genus and finitely many ends - in other words, it should have a compactification that's also a Riemann surface. Given an affine curve, one can take the projective closure and then normalize it, which always produces a manifold compactification. Any copmact Riemann surface is a projective algebraic curve, and removing one or more points gets you an affine algebraic curve. – Will Sawin May 11 '13 at 21:02
(can't edit) constant -> nonconstant. – Donu Arapura May 11 '13 at 21:07
Donu and Will, thanks a lot! MichaelE – MichaelE May 11 '13 at 21:12

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