\ \! (I, A)={\rm grade} \ \! (I_{\mathfrak m}, A_{\mathfrak m})$, see [@BrunsHerzog Prop. 1.5.15 (e)]: this takes care of the three terms involved in equation (a). For (b), one uses three times the general formula $\dim A=\dim A_{\mathfrak m}$, for which we refer to the proof of Prop. 5.1 in [@Go-To] (note that the argument found there uses the assumption that $R/{\mathfrak m}$ is a field).

A group action $G\times M \to M$ is Cohen-Macaulay in the sense of Def. \[def:cm\] if and only if $H^*_G(M; \bR)$ is Cohen-Macaulay as a module over $H^*(BG; \bR)$.

One applies Proposition \[commalg\] by taking $R=H^*(BG)$, $S=H^*(BG; \bR)$, $A=H^*_G(M)$, and $B=H^*(BG;\bR)$, in each case with the obvious module structure; also, take $\varphi :  H^*(BG)\to H^*(BG; \bR)$ the inclusion map.

\[pair\] [ The same argument shows that if $N$ is a $G$-invariant subspace of $M$, then the $G$-equivariant cohomologies $H^*_G(M, N)$ and $H^*_G(M, N; \bR)$ as modules over $H^*(BG)$, respectively $H^*(BG; \bR)$, have the same depth, the same Krull dimension, and are thus simultaneously Cohen-Macaulay.]{}

An immediate observation is that any equivariantly formal action is Cohen-Macaulay. The converse implication is in general not true, as one can easily see in concrete situations (for example, 