Let $R$ be a Dedekind domain with fraction field $K$. 


Say that a Dedekind domain $R$ has the **Riemann-Roch property** if: for every nonzero prime ideal $\mathfrak{p}$ of $R$, there exists an element $f \in (\bigcap\_{\mathfrak{q} \neq \mathfrak{p}} R\_{\mathfrak{q}}) \setminus R$, i.e., an element of $K$ which is integral at every prime ideal $\mathfrak{q} \neq \mathfrak{p}$ and is not integral at $\mathfrak{p}$. 



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> Do all Dedekind domains have the Riemann-Roch property?
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Motivation: for any subset $\Sigma \subset \operatorname{MaxSpec}(R)$, put $R\_{\Sigma} := \bigcap\_{\mathfrak{p} \in \Sigma} R\_{\mathfrak{p}}$. Then the maximal ideals of $R\_{\Sigma}$ correspond bijectively to the maximal ideals $\mathfrak{p}$ of $R$ such that $\mathfrak{p} R\_{\Sigma} \subsetneq R\_{\Sigma}$. Thus $\operatorname{MaxSpec}(R\_{\Sigma})$ may be viewed as containing $\Sigma$. $R$ has the Riemann-Roch property iff for all $\Sigma$, 
$\operatorname{MaxSpec}(R\_{\Sigma}) = \Sigma$. Equivalently, the mapping $\Sigma \mapsto R\_{\Sigma}$ is an injection. 


Remarks: $R$ has the Riemann-Roch property if its class group is torsion: then for every $\mathfrak{p} \in \operatorname{MaxSpec}(R)$ there exists $n \in \mathbb{Z}^+$ and $x \in R$ such that $\mathfrak{p}^n = (x)$, so take $f = \frac{1}{x}$. Also the coordinate ring $k[C]$ of a nonsingular, integral affine curve $C$ over a field $k$ has the Riemann-Roch property...by the Riemann-Roch theorem. Unfortunately this already exhausts the most familiar examples of Dedekind domains!



Do all Dedekind domains have the "Riemann-Roch property"?
Yes. Given a maximal ideal $P$ there exists $x \in K \backslash R\_P$. Let $S$ be the finite set of maximal ideals $Q$ so that $x \notin R\_{Q}$. For each $Q \in S$ such that $Q \neq P$ let $y\_Q \in Q\backslash P$. The element $f$ given by multiplying $x$ by large positive powers of all the $y\_Q$ has the desired property.


