# Non-finitely generated ring of regular functions

It is remarked in Shafarevich's Basic Algebraic Geometry 1 that Rees and Nagata constructed examples of quasiprojective varieties such that the ring of regular functions is not finitely generated, but I cannot find the source he is referring to. Can anyone give such examples here? Does that mean we can't really say anything about the ring of regular functions of a quasi-projective variety?

It's a theorem that a quasi-projective variety is affine if and only if it is Stein (we're working over C, say) and its ring of functions is finitely generated. So find a Stein manifold that isn't affine, and that will do it.

And, after a bit of looking, it appears that Vakil may have rediscovered the Rees and Nagata example, here.

• I don't think that a Stein manifold that isn't affine will do the trick. For instance, if we take Serre's example ( P^1 bundle over an elliptic curve obtained as the projectivization of the unique non-trivial extension 0-> O -> V -> O -> 0 minus the section determined by O -> V) is Stein, and every regular function on it is constant since the section has zero self-intersection. – Jorge Vitório Pereira Oct 23 '09 at 11:49
• I was certain that I'd read that Stein + f.g. => Affine for varieties, but that seems like a counterexample. I must be missing a hypothesis. Asking a question to try to find out exactly what is true. Must be some nontriviality hypothesis for the ring of regular functions, I'm guessing (maybe separates points?) – Charles Siegel Oct 23 '09 at 12:52
• Ok, so according to Tony Pantev on my related question, what we need is a quasi-affine variety that is Stein. Then we have affine if and only if finitely generated. So it's a matter of looking for non-affine Stein manifolds which are quasi-affine. – Charles Siegel Oct 23 '09 at 21:26

"Does that mean we can't really say anything about the ring of regular functions of a quasi-projective variety?"

Since every variety contains an open affine, the ring of regular functions is always a subring of a finitely generated ring. (I assume that you consider varieties to be integral.) This is a nontrivial restriction. Also, the ring of regular functions will be noetherian, since any infinite ascending chain of ideals would give an infinite descending chain of subschemes. Wrong, see below.

• Since the question has been bumped to the front page, it is worth pointing out that, contrary to what you say, the ring of regular functions on a noetherian scheme (even quasiprojective variety) need not be noetherian. A simple counterexample is described here (namely the union of two planes in $\mathbb{P}^3$ meeting in a line, minus some other line in one of the planes). – Gro-Tsen Jun 19 '17 at 20:30
• (So I'll leave you the honor of explaining what is wrong with the argument you gave, because I'm sure there's something interesting to be learned there.) – Gro-Tsen Jun 19 '17 at 20:33
• The Nullstellensatz doesn't work. There might be ideals without zeros. – Friedrich Knop Jun 20 '17 at 6:40
• The part about being a subring of a f.g. ring is still valid, isn't it? This is all very confusing, I wish there were an introductory textbook addressing the question in some details, with various examples and counterexamples. – Gro-Tsen Jun 20 '17 at 14:47