# Example of a certain type of Cohen-Macaulay ring

Let $$k$$ be a perfect field. I am looking for a $$k$$-algebra $$R$$ with the following properties.

• $$R$$ is of finite type over $$k$$ and is a domain;

• for all $${\mathfrak p}\in{\rm Spec}(R)$$, the local ring $$R_{\mathfrak p}$$ is Cohen-Macaulay and $${\rm emb.\, dim.}(R_{\mathfrak p})-{\rm dim}(R_{\mathfrak p})\leq1$$;

• for at least one $${\mathfrak p}\in{\rm Spec}(R)$$, the local ring $$R_{\mathfrak p}$$ is not Gorenstein.

I would also be happy with the weaker condition "not a complete intersection" in place of "not Gorenstein".

A related question: is there an example of reduced local noetherian ring $$T$$, which is Cohen-Macaulay, has $${\rm emb.\, dim.}(T)-{\rm dim}(T)=1$$, but is not Gorenstein (resp. not a complete intersection)?

Any help would be appreciated.

• I think your second condition forces your ring to be a local complete intersection May 9 at 16:43
• @Mohan. Thank you for looking into this. My question is in fact just that: is there such a ring, which is not a complete intersection? and if not, why? I would be happy to see the argument. May 9 at 16:46

There is no such example. A Cohen-Macaulay local ring with $$\operatorname{embdim}(R)-\dim(R) \le 1$$ is a hypersurface, which is in particular a complete intersection. Indeed, we may pass to the completion to suppose $$R$$ is complete. Then by Cohen's structure theorem, $$R \cong S/I$$ where $$(S,\mathfrak{n})$$ is a regular local ring and $$I \subseteq \mathfrak{n}^2$$. In this situation, $$\dim S=\operatorname{embdim}(R)$$, and as $$R$$ is Cohen-Macaulay, the Auslander-Buchsbaum formula implies $$\operatorname{pd}_S(R) \le 1$$. If it is $$0$$, then $$R$$ is regular and we're done. If it is $$1$$, then from the exact sequence $$0 \to I \to S \to R \to 0$$ we see that $$I$$ is a free $$S$$ module. As $$I$$ is an ideal, it must be principal, so $$R \cong S/(f)$$ for some $$f$$.