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Can someone give me, with proof, an example of a Noetherian ring which has Krull dimension one but is not a Dedekind domain?

I think it would also be instructive to see other "near misses."

user26857
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Tony
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    A domain is Dedekind if and only if it is integrally closed, Noetherian, and has Krull dimension $1$. So if you want to stick to domains, just take one which is not integrally closed but is all the rest. Wouldn't $\mathbb{Z}[\sqrt{-3}]$ work, or more generally, $\mathbb{Z}[\sqrt{d}]$ with $d\neq 1$, squarefree, and $d\equiv 1 \pmod{4}$? – Arturo Magidin Jan 03 '11 at 01:35
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    The affine ring of an irreducible singular affine curve is such an example. – Makoto Kato Jun 05 '12 at 02:43

4 Answers4

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The ring $\mathbb{Z}[2i]$ is an example. It satisfies all the properties of a Dedekind domain except that it is not integrally closed. (To see that it satisfies these properties, note that it is integral over $\mathbb{Z}$, so the Krull dimension is one, as integral extensions preserve dimension. It is also clearly noetherian.)

Akhil Mathew
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  • As for the proof that integral extensions preserve dimension, the following is my proof. Is there an easier way to see it? Let's suppose that B is integral over A. By Lying-Over, dim B \geq dim A. Conversely, it is a standard argument that if p \subset q is a strict containment in B, then the containment is still strict after intersecting. – Tony Jan 04 '11 at 14:46
  • @Tony: That's the standard argument, I believe; I'm afraid I know of no others. – Akhil Mathew Jan 04 '11 at 16:22
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The ring $R_1 = \mathbb{Z} \times \mathbb{Z}$ gives a counterexample to your claim (it is not a domain). However $\operatorname{Spec}(R_1)$ is a Dedekind scheme, so this is a somewhat cheap counterexample. The counterexample $R_2 = \mathbb{Z}[t]/(t^2)$ is more serious.

Pete L. Clark
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If by "other near misses" you mean other rings that satisfy two of the three conditions for being a Dedekind domain, another such ring is k[x, y], which is Noetherian (by the Hilbert basis theorem), integrally closed (it is a UFD, and UFDs are integrally closed) but not every prime ideal is maximal (for example (x)).

Vitaly Lorman
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Consider $f(x,y)$ = $x^3-y^2$ $\in$ $\mathbb{C}[x,y]$. Then the quotient ring $\mathbb{C}[x,y]/(f)$ is a Noetherian ring with Krull dimension 1 but is not a Dedekind Domain. In general for an algebraically closed field $\bar k$, coordinate ring of the affine plane curve $Z_{f}(\bar k)$ is integrally closed in its field of fractions if and only if the curve $Z_{f}(\bar k)$ is nonsingular.

user737722
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