Proposition 5

Let $A$ be a ring, $J$ an ideal of finite type of $A$ and $M$ an $A$-module. Let $\Omega$ denote the set of maximal ideals of $A$ belonging to $\operatorname{Supp}(M)$ and containing $J$. Then

Let $x = (x_i)_{i \in I}$ be a finite generating family of $J$. Let $p$ be a prime ideal of $A$; the ideal $J_p$ is generated by the image $x_p$ of the family $(x_i)$ in $A_p$. For every $p \geq 0$, the $A_p$-module $(H^p(x, M))_p$ is isomorphic to $H^p(x_p, M_p)$ (Lemma 1); in view of Theorem 1, we therefore have $\operatorname{prof}_A(J; M) \leq \inf_{p \in \operatorname{Spec}(A)} \operatorname{prof}_{A_p}(J_p; M_p) \leq \inf_{m \in \Omega} \operatorname{prof}_{A_m}(J_m; M_m)$.

Let $p$ be an integer strictly less than $\operatorname{prof}_{A_m}(JA_m; M_m)$ for every $m \in \Omega$. Then $H^p(x_m, M_m) = 0$ for every maximal ideal $m$ of $A$: this results from Theorem 1 if $m \in \Omega$, from the fact that $M_m = 0$ if $m \not\in \operatorname{Supp}(M)$, and from the fact that the ideal $JA_m$, which annihilates $H^p(x_m, M_m)$ (A, X, p. 148, Cor. 2), is equal to $A_m$ if $m \not\in V(J)$. Hence $(H^p(x, M))_m = 0$ for every maximal ideal $m$ of $A$, which implies $H^p(x, M) = 0$ (II, § 3, No. 3, Cor. 2 of Theorem 1). The proposition then results from Theorem 1.