Theorem 3

Let $A$ be a noetherian local ring, $q$ an ideal of $A$ distinct from $A$, and $M$ a finitely generated $A$-module such that $M/qM$ is of finite length. Then the integer $d_q(M)$ is the dimension of the $A$-module $M$ ($§ 1$, No. 4, Def. 8).

One may suppose $M \neq 0$. Let us prove the inequality $d_q(M) \leq \dim_A(M)$. By Cor. 2 to Prop. 9 of $§ 3$, No. 5, there exist $x_1, ..., x_r \in q$, with $r = \dim_A(M)$ and $\operatorname{long}(M / \sum_{i=1}^r x_i M) < +\infty$; let us put $x = \sum_{i=1}^r x_i A$. By Prop. 4 of No. 3, one has $d_x(M) \leq r$; one has $x \subset q$, whence $H^{(1)}_{M,q} \leq H^{(1)}_{M,x}$ and therefore (Lemma 2 of No. 1)

Let us now prove, by induction on $\dim_A(M)$, the inequality $\dim_A(M) \leq d_q(M)$, which is obvious when $\dim_A(M) = 0$.

Suppose that $\dim_A(M) > 0$, and that $\dim_A(N) \leq d_q(N)$ for every finitely generated $A$-module $N$ such that $\dim_A(N) < \dim_A(M)$. If $0 = M_0 \subset M_1 \subset ... \subset M_n = M$ is a composition series of $M$, one has $\dim_A(M) = \sup(\dim_A(M_i/M_{i-1}))$ ($§ 1$, No. 4, prop. 9) and $d_q(M) = \sup(d_q(M_i/M_{i-1}))$ (No. 3, prop. 5). By IV, $§ 1$, No. 4, th. 1, one may therefore suppose that $M$ is of the form $A/p$, where $p$ is a prime ideal of $A$, and one has $p \neq m_A$ since $\dim_A(M) > 0$. Let $x \in m_A - p$; the homothety $x_M$ of $M = A/p$ is injective, and one has the exact sequence

By $§ 3$, No. 2, prop. 3, one has $\dim_A(M/xM) = \dim_A(M) - 1$; by prop. 5 of No. 3, and the preceding exact sequence, one has $d_q(M/xM) \leq d_q(M) - 1$. By the induction hypothesis, one therefore has

which completes the proof.