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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.)

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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
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
up vote 4 down vote accepted

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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The overrings which satisfy some condition, I guess. – Mariano Suárez-Alvarez Dec 11 '12 at 23:00
An overring of a domain $R$ is a ring intermediate between $R$ and its fraction field $K$. This is the only condition. – user26857 Dec 11 '12 at 23:04
Ah! I generally see that term to mean the relation converse to subring. – Mariano Suárez-Alvarez Dec 11 '12 at 23:07
@Mariano That definition of overring is used frequently by rings theorists studying factorization / divisibility theory and related topics, e.g. see these various characterizations of Prufer domains. – Bill Dubuque Dec 11 '12 at 23:58

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