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Suppose I have two domains, $A\subset B$, where $A$ is Dedekind and $\operatorname{Frac}(A)=\operatorname{Frac}(B)$. I also know that $B$ is both integrally closed and has height $1$. Is $B$ necessarily Dedekind? If not, I'd love to see a counterexample.

(Note: I'm adding the homework tag since this is motivated by (and would finish off) a homework problem about global function fields, although I now know another way, with slightly stronger hypotheses, to get the necessary result with Riemann-Roch.)

Brett Frankel
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  • When you say "height one", do you mean "of Krull dimension one" (every non-zero prime is maximal)? – Nils Matthes Dec 11 '12 at 20:21
  • @NilsMatthes Yes. The only remaining thing to show is the Noetherian condition. – Brett Frankel Dec 11 '12 at 20:22
  • Dear @Andrew, There are lots of instances where both $A$ and $B$ are Dedekind and $B$ is not finite over $A$. For example, $A=\mathbf{Z}$ and $B=\mathbf{Z}_{(p)}$ for some prime $p$. – Keenan Kidwell Dec 11 '12 at 20:33
  • Dear @KeenanKidwell, hum... nice example! These are both integrally closed as well, if I'm not mistaken. – Andrew Dec 11 '12 at 20:40
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    Dear @Andrew, In fact, if $B$ is finite over $A$ then it is integral over $A$, and since $B\subseteq\mathrm{Frac}(A)$, $A=B$. – Keenan Kidwell Dec 11 '12 at 20:41
  • If $A$ has torsion class group, then every non-trivial absolute value on $\mathrm{Frac}(A)$ which is bounded on $A$ comes from a maximal ideal of $A$ in the usual way, i.e., is $\mathfrak{m}$-adic for some $\mathfrak{m}$. If there were a Dedekind domain with non-torsion class group whose fraction field admitted a non-discrete (non-trivial) non-Archimedean absolute value, bounded on $A$, then the valuation ring of this absolute value would serve as an example where $B$ is not Noetherian. I have no idea if such a thing can exist though. – Keenan Kidwell Dec 11 '12 at 20:48
  • @Keenan You might find of interest the following results: for a domain D: every overring of D is Noetherian iff D is 1-dimensional Noetherian; every overring of D is integrally closed iff D is a Prufer domain. – Bill Dubuque Dec 12 '12 at 00:15

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In the multiplicative theory of ideals it is well known that the overrings of Dedekind domains are also Dedekind. See, for instance, Larsen and McCarthy, Multiplicative Theory of Ideals, Theorem 6.21 or these notes, Proposition 22.2.

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