TITLE: Can every finite poset be realized as divisors of an algebraic curve?
QUESTION [13 upvotes]: 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.
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 $\{x_1,...,x_n\}$?

REPLY [14 votes]: 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.
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.