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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 $\mathfrak{p}$.

Do all Dedekind domains have the Riemann-Roch property?

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 the Riemann-Roch theorem. Unfortunately this already exhausts the most familiar examples of Dedekind domains!

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up vote 6 down vote accepted

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.

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Thanks. It turned out to be an easy question, but I am glad to have the answer. – Pete L. Clark Apr 12 '10 at 8:07

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