Proposition 8

Let $\rho : A \to B$ be a homomorphism of Noetherian rings and let $N$ be a $B$-module which is a finitely generated $A$-module. In order that the $A$-module $N$ be Macaulay, it is necessary and sufficient that the $B$-module $N$ be Macaulay and that, for every pair $(n, n')$ of maximal ideals of $\mathrm{Supp}_B(N)$ such that $\rho^{-1}(n) = \rho^{-1}(n')$, one has $\dim_{B_n}(N_n) = \dim_{B'_n}(N'_n)$.

The $A$-module $B/\mathrm{Ann}_B(N)$ is isomorphic to a submodule of the finitely generated $A$-module $\mathrm{End}_A(N)$, and is therefore of finite type. Replacing $A$ by $A/\mathrm{Ann}_A(N)$ and $B$ by $B/\mathrm{Ann}_B(N)$, we are reduced to the case where $\rho$ is injective and makes $B$ into a finite $A$-algebra, and where one has $\mathrm{Supp}_A(N) = \mathrm{Spec}(A)$ and $\mathrm{Supp}_B(N) = \mathrm{Spec}(B)$. The mapping $f : \mathrm{Spec}(B) \to \mathrm{Spec}(A)$ deduced from $\rho$ is then surjective, and a prime ideal $q$ of $B$ is maximal if and only if $f(q)$ is a maximal ideal of $A$ (V, § 2, No. 1, th. 1 and prop. 1).

Let $m$ be a maximal ideal of $A$. By the above remark, the prime ideals of the ring $B_m$ containing $mB_m$ are the ideals of the form $qB_m$, where $q \in \mathrm{Spec}(B)$ is an ideal of $B$ (necessarily maximal) such that $f(q) = m$. One has

(§ 1, No. 3, prop. 4 and No. 5, prop. 8), and

(VIII, § 2, No. 3, th. 1 and § 1, No. 4, prop. 9). Since one has $\mathrm{prof}_{B_q}(N_q) \leq \dim_{B_q}(N_q)$ for every $q \in f^{-1}(m)$, the proposition follows from these equalities.