# 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. Oct 23, 2009 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?) Oct 23, 2009 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. Oct 23, 2009 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). Jun 19, 2017 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.) Jun 19, 2017 at 20:33
• The Nullstellensatz doesn't work. There might be ideals without zeros. Jun 20, 2017 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. Jun 20, 2017 at 14:47