most recent 30 from http://mathoverflow.net 2013-05-22T19:43:47Z http://mathoverflow.net/feeds/question/86471 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86471/can-every-finite-poset-be-realized-as-divisors-of-an-algebraic-curve Will Sawin 2012-01-23T18:57:31Z 2012-01-26T04:02:06Z

<p>Let $D_1$, ... , $D_n$ be a finite set of divisor classes on a nonsingular projective irreducible algebraic curve. We say that $D_1\geq D_n$ if the line bundle defined by $D_1-D_n$ has a section. This obviously satisfies the axioms of a partial order.</p> <p>Suppose ${x_1,....,x_n}$ is a finite partially ordered set. Does there exist a (projective, nonsingular) algebraic curve of sufficiently high genus, and a set of divisors on it, that are isomorphic as a partially ordered set to ${x_1,...,x_n}$?</p>

http://mathoverflow.net/questions/86471/can-every-finite-poset-be-realized-as-divisors-of-an-algebraic-curve/86480#86480 MP 2012-01-23T20:17:49Z 2012-01-24T20:21:03Z

<p>Choosing a general set of $n$ points on a curve of genus at least $n$, you can assume that the divisor they define has a unique global section. Let $P$ denote the poset generated by all the possible sums with coefficients $0,1$ of these $n$ points. This poset is the same as the poset of subsets of an $n$-element set.</p> <p>Since every finite poset seems to be a subposet of the poset of subsets of a finite set [EDIT: this is true-see the comments below], just embed your poset in a "power set poset" and remove the unwanted divisors, to deduce that what you want is true.</p>