A quotient of a Cohen-Macaulay ring

Question

Let $S$ be a Cohen-Macaulay local ring, $I$ an ideal of $S$ and $R=S/I$. If we know that $I$ is generated by $\dim S-\dim R$ elements could we infer that $R$ is Cohen-Macaulay?

Thanks in advance!

Answer 1

Since $S$ is a Cohen-Macaulay local ring we have $\operatorname{grade} I=\dim S-\dim S/I$ (see Bruns and Herzog, Theorem 2.1.2 (b)). This shows that $I$ is generated by $\operatorname{grade} I$ elements. From a well known result (see, e.g. Kaplansky, Theorem 125), we can find a system of generators for $I$ forming an $S$-sequence. Now it's clear that $R=S/I$ is also Cohen-Macaulay.

Answer 2

I think so:
$μ(I)=\dim (S)-\dim(R) = \mathrm{grade} (I,S)$ (Bruns-Herzog, 2.1.2). So $I$ is generated by a sequence (Bruns-Herzog 1.6.19). Now use Bruns-Herzog, 2.1.3.