Let $(R,\frak m)$ denote a commutative Noetherian local ring and $M$ a finitely generated $R$-module. We say $R$ is Cohen-Macaulay, provided $\operatorname{dim}R=\operatorname{depth}R$. Similarly, $M$ is a Cohen-Macaulay $R$-module, if $\operatorname{dim}_RM=\operatorname{depth}_RM$. Also in the case that $\operatorname{depth}_RM=\operatorname{dim}_RM=\operatorname{dim}R$, we say that $M$ is maximal Cohen-Macaulay.
It is easy to see that if $M$ is a Cohen-Macaulay $R$-module, it is maximal Cohen-Macaulay as a $R/\operatorname{Ann}_RM$-module, and so it is Cohen-Macaulay. I proved some conditions under which $R/\operatorname{Ann}_RM$ is a Cohen-Macaulay ring. For example, if $R$ is a Cohen-Macaulay ring and $\operatorname{Ann}_RM$ is generated by an $R$-regular sequence, then $R/\operatorname{Ann}_RM$ is a Cohen-Macaulay ring.
An $R$-module $M$ is indecomposable, if $M=A\oplus B$, where $A$, $B$ are submodules of $M$, we conclude that $A=0$ or $B=0$.
My question is: When $R/\operatorname{Ann}_RM$ is a Cohen-Macaulay ring in the case that $M$ is an indecomposable $R$-module?