I know the fact that a local ring $(R,m)$ with $\dim(R)=d$ is Cohen-Macaulay (C-M) if and only if any one of the following holds:
1) $\operatorname{grade}(m)=\operatorname{height}(m)$
2) $\operatorname{grade}(p)=\operatorname{height}(p)$ for all prime ideals $p$ of $R$
3) $\operatorname{grade}(I)=\operatorname{height}(I)$ for all ideals $I$ of $R$
How could we deduce from the above that:
A ring $R$ is C-M if and only if any of the above conditions holds for $R$? (Of course, the first condition for any all maximal ideal ideals $m$ of $R$.)
I also know that $R$ is C-M if and only if $R_m$ is so, for any maximal ideal $m$ of $R$. Thanks a lot.
Let $m$ be a maximal ideal of $R$.
If $R_m$ is CM, then $\operatorname{depth}R_m=\dim R_m$, and therefore $\operatorname{grade}m=\operatorname{ht}m$.
Conversely, assume that $\operatorname{grade}m=\operatorname{ht}m$. Then $\operatorname{depth}R_m=\dim R_m$, that is, $R_m$ is CM.
(I've used that $\operatorname{grade}m=\operatorname{grade}mR_m$ and $\operatorname{ht}m=\dim R_m$.)