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 quasiprojective variety?
2 Answers
It's a theorem that a quasiprojective 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.

1$\begingroup$ 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 nontrivial 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 selfintersection. $\endgroup$ Oct 23, 2009 at 11:49

$\begingroup$ 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?) $\endgroup$ Oct 23, 2009 at 12:52

1$\begingroup$ Ok, so according to Tony Pantev on my related question, what we need is a quasiaffine variety that is Stein. Then we have affine if and only if finitely generated. So it's a matter of looking for nonaffine Stein manifolds which are quasiaffine. $\endgroup$ Oct 23, 2009 at 21:26
"Does that mean we can't really say anything about the ring of regular functions of a quasiprojective 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.

$\begingroup$ 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). $\endgroup$– GroTsenJun 19, 2017 at 20:30

$\begingroup$ (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.) $\endgroup$– GroTsenJun 19, 2017 at 20:33

2$\begingroup$ The Nullstellensatz doesn't work. There might be ideals without zeros. $\endgroup$ Jun 20, 2017 at 6:40

1$\begingroup$ 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. $\endgroup$– GroTsenJun 20, 2017 at 14:47