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As the title says, I am looking for a noetherian local ring $R$ of dimension 0 which is reduced (and thus Cohen-Macaulay) but not Gorenstein.

Due to Bruns, Herzog $-$ Cohen-Macaulay Rings Theorem 3.2.10 every noetherian local ring which is not Gorenstein fails to be Cohen-Macaulay or fails to be of type 1. Since every reduced ring of dimension $\leq 1$ is Cohen-Macaulay (see Stacks-Reference), we are thus looking for a noetherian, reduced local ring of dimension 0 that fails to be of type 1.

What constitutes a simple example of such a ring?

I am grateful for any kind of help or input! Cheers!

windsheaf
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2 Answers2

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If you mean Krull dimension $0$, then I guess there is no example.

A reduced ring with Krull dimension $0$ is von Neumann regular, and a local VNR ring is a field.

The reducedness condition really kills things. $F_2[x,y]/(x,y)^2$ satisfies everything you said except it is not reduced.

rschwieb
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  • Yes, I mean Krull dimension zero. I was suspecting that there is no example, but I wasn't sure about that! Do you have a reference for your statement "Local +Reduced + Krull-Dim 0 $\implies$ Field"? – windsheaf Jun 15 '20 at 15:06
  • Here is a reference that shows it is VNR. That a local VNR is a field is trivial, right? – rschwieb Jun 15 '20 at 15:12
  • I have found a direct reference: https://stacks.math.columbia.edu/tag/00EU – windsheaf Jun 15 '20 at 15:15
  • @windsheaf Yup that works. – rschwieb Jun 15 '20 at 15:18
  • So now we have $\operatorname{Hom}_{R}(R,R) \cong R$ and thus $R$ is of type 1, correct? – windsheaf Jun 15 '20 at 15:20
  • @windsheaf I don’t know what type 1 means. $Hom_R(R,R)\cong R$ for every commutative ring with identity, right? – rschwieb Jun 15 '20 at 16:28
  • @windsheaf Trying to read between the lines a little bit, it looks to me like you're looking for simple examples of Cohen-Macaulay (CM) local rings which are not Gorenstein, equivalently, as you say, that do not have type $1$. In particular, you are looking to rings of small dimension to help you meet the Cohen-Macaulay condition. While it is true that reduced rings of dimension $\le 1$ are CM, this is really only relevant in dimension $1$, as every ring of dimension $0$ is CM. The example given by rschiwieb is such an example; if $k$ is a field, the ring $R=k[x,y]/(x,y)^2$ has type $2$. – metalspringpro Jun 15 '20 at 22:19
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    @metalspringpro is that claim about $0$ dimensional rings missing some qualification? Noetherian maybe? It appears there are examples of $0$ dimensional nonCM rings – rschwieb Jun 15 '20 at 23:30
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    @rschwieb Yes I do want Noetherian. To me Noetherian-ness is part of what it means to be local; I usually am good about saying "Noetherian local" to be clear to those taking a broader definition, but in this instance I forgot. But you are correct, though I think Cohen-Macaulayness is somewhat ill-defined for non-Noetherian rings; the usual characterizations of this condition for Noetherian rings are no longer equivalent, so one has several different possible definitions for what CM could mean, and, depending on the context, each of these may be appropriate. – metalspringpro Jun 15 '20 at 23:43
  • @rschwieb Since I was quite cavalier with my assumptions in my comment, I'm just going to move it to an answer since I can then edit it appropriately. – metalspringpro Jun 16 '20 at 00:00
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Trying to read between the lines a little bit, it looks to me like you're looking for simple examples of Cohen-Macaulay local rings which are not Gorenstein, equivalently, as you say, that do not have type 1. In particular, you are looking to rings of small dimension to help you meet the Cohen-Macaulay condition.

While it is true that reduced Noetherian local rings of dimension at most $1$ are Cohen-Macaulay, this is really only relevant in dimension 1, as every Notherian local ring of dimension $0$ is Cohen-Macaulay, and, as others have already mentioned, a Noetherian local ring of dimension $0$ is Artinian, and an Artinian reduced local ring is a field.

The example given by rschiwieb in their answer is exactly the sort of example you seek; if k is a field, the ring $R=k[x,y]/(x,y)^2$ has type $2$.

This example can be extended to a larger family of examples:

Let $k$ be a field and let $R=k[x_1,\dots,x_n]/(x_1,\dots,x_n)^m$ where $n,m \ge 2$. Then $R$ is a Noetherian local ring of dimension $0$ (thus is Cohen-Macaulay) and the type $r(R)$ of $R$ is not $1$. In fact, one can compute explicitly that $r(R)=\mu_R[(x_1,\dots,x_n)^{m-1}]={m+n-2\choose n-1}$.

metalspringpro
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