Can every finite poset be realized as divisors of an algebraic curve?

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 $\{x_1,...,x_n\}$?

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Remark: this is only an order on divisor classes (assuming the curve to be projective and irreducible). – Laurent Moret-Bailly Jan 25 '12 at 8:06
ah, good point. I'll fix the notation. – Will Sawin Jan 25 '12 at 19:43

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.

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To replace "seems to be", note that you can just map each thing to the set of things it's bigger than or equal to. – Will Sawin Jan 23 '12 at 20:23
Note, furthermore, that the result to which Will alludes is a special case of the Yoneda lemma (applied to the poset viewed as a category, where $a \le b$ means that there is a single arrow $a \to b$). – Qiaochu Yuan Jan 24 '12 at 16:59