Example of a certain type of Cohen-Macaulay ring

Question

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.

Answer 1

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 $f$.