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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.

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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

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    $\begingroup$ The fact that Units(A)/Units(k) is finitely generated is easy, at least if A is integral: One may first assume that A is normal by replacing it by its normalisation. If X is a normal compactification of Spec(A) then any unit is determined by its divisor on X upto a unit in k and the divisors supported on the complement of Spec(A) clearly form a finitely generated abelian group. $\endgroup$
    – naf
    Commented Apr 16, 2010 at 9:02
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    $\begingroup$ Contd. Finite generation of Units(A)/Units(k) is not true if A is not reduced (consider the ring k[x]/x^2); the general reduced case can be deduced from the integral case by localisation. (One must also assume that k is the "constant field", otherwise A could simply be a finite extension of k.) $\endgroup$
    – naf
    Commented Apr 16, 2010 at 9:24
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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.

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    $\begingroup$ Good catch - I've added noetherian and reduced as hypotheses. $\endgroup$
    – Greg Stevenson
    Commented Oct 26, 2009 at 11:53

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