# Ring containing a Dedekind ring

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

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