5
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What is the simplest example of an integrally closed Noetherian domain in which not every height one prime is locally principal? The latter condition is equivalent to the ring being locally factorial.

So the question is to find an integrally closed Noetherian domain which is not locally factorial (hence not factorial) and a non-locally principal ideal of height one in it.

But when it comes to nonfactorial rings, I can't come up with anything better than $\mathbb{Z}[\sqrt{-5}]$; however, this ring is locally factorial.

Note that locally principal ideal means an ideal whose localizations at all primes are principal ideals in the corresponding localized ring (another way of saying this is that such an ideal is just a locally free module of rank one).

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  • $\begingroup$ I suggest you the following example: $\mathbb C[X,Y,Z]/(XY-Z^2)\simeq\mathbb C[U^2,UV,V^2]$. $\endgroup$ – user26857 Dec 24 '16 at 16:57
  • $\begingroup$ You need to consider domains with larger dimension. If your domain has Krull dimension $1$, then, with your other assumptions, it is a Dedekind domain, and Dedekind domains are locally DVRs, so every ideal is locally principal. $\endgroup$ – André 3000 Dec 24 '16 at 20:05
  • $\begingroup$ Okay, I think I can see why this ring is integrally closed, Noetherian, and non-factorial, but what is an example of a height one non-locally principal ideal? $\endgroup$ – Cary Dec 25 '16 at 20:34
  • $\begingroup$ @Cary Have you identified a prime ideal of height one in that ring? $\endgroup$ – user26857 Dec 25 '16 at 22:38
  • $\begingroup$ I'm afraid I haven't... $\endgroup$ – Cary Dec 27 '16 at 8:06

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