4
$\begingroup$

Let $(R, \mathfrak m)$ be a local Cohen-Macaulay ring. If $\dim R=\mu (\mathfrak m)-1$ , then is it true that $R \cong S/(f)$ for some regular local ring $S$ and some (non-invertible) regular element $f \in S$ (note that since $S$ is a domain, any non-zero element in the maximal ideal of $S$ is regular) ? If this is not true in general, how about if we also assume $R$ is Gorenstein ?

$\endgroup$
0

1 Answer 1

2
$\begingroup$

A local ring $(R,\mathfrak{m})$ with $\operatorname{depth} R \ge \mu_R(\mathfrak{m})-1$ is often termed an abstract hypersurface (note this includes the case $\operatorname{depth} R=\mu(\mathfrak{m})$ i.e. when $R$ is regular). Sometimes this condition is defined instead to be that the completion is a hypersurface in the absolute sense, i.e., that $\hat{R} \cong S/(f)$ for some regular local ring $S$ and $f \in S$. That these are equivalent will be shown below. In some sense, this second definition is more natural since one defines an abstract complete intersection in the analogous manner.

We observe the following:

Suppose $R$ is an abstract hypersurface and suppose $R$ is the homomorphic image of a regular local ring. Then $R$ is a hypersurface in the absolute sense, i.e., $R \cong S/(f)$ for some regular local ring $S$ and $f \in S$.

Proof: We may write $R \cong S/I$ where $(S,\mathfrak{n})$ is a regular local ring and $I \subseteq \mathfrak{n}^2$. By assumption, $\operatorname{depth} S-\operatorname{depth} R=\mu_R(\mathfrak{m})-\operatorname{depth} R \le 1$, and thus, by Auslander-Buchsbaum, $\operatorname{pd}_S R \le 1$. If $\operatorname{pd}_S R=0$, then $R \cong S$ and we are done. Otherwise, $\operatorname{pd}_S R=1$. But we have an exact sequence $0 \to I \to S \to R \to 0$. As $\operatorname{pd}_S R=1$, this forces $I$ to be a free $S$-module, and thus $I$ is principal.

In particular, the above theorem shows that the completion of any abstract hypersurface is always a hypersurface in the absolute sense, since $\mu_R(\mathfrak{m})$ and $\operatorname{depth} R$ are preserved by completion, and thus in turns shows that abstract hypersurfaces enjoy many of the same homological properties enjoyed by hypersurfaces in the absolute sense; in particular they are always Gorenstein.

To answer your question directly, no; an example was given by Heitmann and Jorgensen of an abstract hypersurface which is not a hypersurface in the absolute sense. See the (excellently titled) paper ``Are complete intersections complete intersections?" https://arxiv.org/pdf/1109.4921.pdf.

$\endgroup$
2
  • $\begingroup$ Could you add a proof as to why $\hat R \cong S/(f)$ for a regular local ring $S$ ? $\endgroup$
    – uno
    Dec 26, 2019 at 1:05
  • $\begingroup$ Yes, I added one. $\endgroup$ Dec 26, 2019 at 9:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .