TITLE: Can any countably generated k-algebra occur as the ring of global sections of some variety?
QUESTION [8 upvotes]: In the answer to  this  question we saw that  there exists  a nonsingular quasi-projective threefold over a field with non-finitely generated global sections.
I was talking about this previous question today and the following question came up - given any countably generated noetherian k-algebra R which is an integral domain and whose field of fractions has finite transcendence degree over k, where k is a field does there exist some quasi-projective variety X (by variety I mean an integral separated scheme of finite type over k) such that the ring of global sections of X is R?
It is possible one needs more hypotheses to make this work - if this is false I think it would be interesting to know the class of algebras which can occur.

REPLY [10 votes]: No. Take k[t] and invert countably many relatively prime polynomials. This obeys all of your adjectives (localization preserves noetherianness, the others are obvious.)
However, a ring of global sections must be a subring of some finitely generated k-algebra. (I pointed this out in the last discussion.) Hence, its unit group must be a subgroup of the group of units of a finitely generated k-algebra. If A is any finitely generated k-algebra, then Units(A)/Units(k) is a finitely generated abelian group, and thus can't contain the countably generated group of the above example.
I can't find a reference for the fact about Units(A)/Units(k) at the moment, Tevelev describes this as well known in

Compactifications of subvarieties of tori, American Journal of Mathematics 129 no. 4 (2007) pp. 1087–1104, doi:10.1353/ajm.2007.0029, arXiv:math/0412329

REPLY [2 votes]: No, but for somewhat trivial reasons.  Let R be the polynomial ring in countably many variables, with no relations.  This is a countably generated k-algebra, and it can't be the ring of functions on a quasi-projective.  Any quasi-projective has function field of finite transcendence degree over the base field, because they are birational to hypersurfaces in P^n.  Now, if you add the hypothesis that R is countably-generated, reduced, noetherian k-algebra, it might be true, though both reduced and noetherian are necessary hypotheses.