most recent 30 from http://mathoverflow.net 2013-06-19T07:03:14Z http://mathoverflow.net/feeds/question/2083 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2083/stein-manifolds-and-affine-varieties Charles Siegel 2009-10-23T12:54:09Z 2009-11-12T16:23:49Z

<p>When is a Stein manifold a complex affine variety? I had thought that there was a theorem saying that a variety which is Stein and has finitely generated ring of regular functions implies affine, but in the comments to my answer <a href="http://mathoverflow.net/questions/2071/non-finitely-generated-ring-of-regular-functions" rel="nofollow">here</a>, Serre's counterexample was brought up. I'm guessing that the answer is that the ring of regular functions must be nontrivial somehow, like it must separate points, but I'm curious about what the exact condition is.</p>

http://mathoverflow.net/questions/2083/stein-manifolds-and-affine-varieties/2087#2087 Tony Pantev 2009-10-23T13:43:59Z 2009-11-12T16:23:49Z

<p>Charlie, it is funny answering this way but here it is. </p> <p>The criterion you are thinking about is a criterion that is relative to an embedding. It says that if $X$ is a quasi-affine complex normal variety, whose associated analytic space $X^{an}$ is Stein, then $X$ is affine if (and only if) the algebra $\Gamma(X,\mathcal{O}_{X})$ is finitely generated. This is a theorem of Neeman. </p> <p>You can reformulate the requirement of $X$ being quasi-affine as a separation of points property: for any point $x \in X$ consider the subset $S_{x} \subset X$ defined as the set of all points $y \in X$ such that all regular functions on $X$ have equal values at $x$ and $y$. Then by an old theorem of Goodman and Hartshorne $X$ is quasi-affine if $S_{x}$ is finite for all $x$. So you can say that $X$ is affine if it satisfies: 1) $X^{an}$ is Stein; 2) $S_{x}$ is finite for all $x \in X$; 3) $\Gamma(X,\mathcal{O}_{X})$ is finitely generated.</p>