Example 4

Let $M$ be a finitely generated $A$-module, and let $a$ be an ideal of $A$ such that $aM = 0$. In order that the $A$-module $M$ be Macaulay, it is necessary and sufficient that it be Macaulay as an $(A/a)$-module. In fact, put $B = A/a$; let $n$ be a maximal ideal of $B$ and $m$ its inverse image in $A$; then $\dim_{A_m}(M_m) = \dim_{B_n}(M_n)$ and $\mathrm{prof}_{A_m}(M_m) = \mathrm{prof}_{B_n}(M_n)$ ($§ 1$, No. 3, cor. of prop. 4).