most recent 30 from http://mathoverflow.net 2013-05-20T18:23:48Z http://mathoverflow.net/feeds/question/2071 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2071/non-finitely-generated-ring-of-regular-functions Ho Chung Siu 2009-10-23T11:23:08Z 2009-10-23T12:16:37Z

<p>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?</p>

http://mathoverflow.net/questions/2071/non-finitely-generated-ring-of-regular-functions/2073#2073 Charles Siegel 2009-10-23T11:33:35Z 2009-10-23T11:33:35Z

<p>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.</p> <p>And, after a bit of looking, it appears that Vakil may have rediscovered the Rees and Nagata example, <a href="http://math.stanford.edu/~vakil/files/nonfg.pdf" rel="nofollow">here</a>.</p>

http://mathoverflow.net/questions/2071/non-finitely-generated-ring-of-regular-functions/2078#2078 David Speyer 2009-10-23T12:16:37Z 2009-10-23T12:16:37Z

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