Theorem 2

Let $A$ be a noetherian ring, $J$ an ideal of $A$ and $M$ a finitely generated $A$-module.

a) Suppose that $\operatorname{prof}_A(J; M)$ is finite. Then every $M$-regular sequence of elements of $J$ can be completed to an $M$-regular sequence of length $\operatorname{prof}_A(J; M)$ of elements of $J$.

b) The depth of $M$ relative to $J$ is the least upper bound of the lengths of $M$-regular sequences formed of elements of $J$.

c) For $\operatorname{prof}_A(J; M)$ to be finite, it is necessary and sufficient that the support of $M$ meet $V(J)$, or again that one have $JM \neq M$.

Let $(x_1, \ldots, x_r)$ be an $M$-regular sequence of elements of $J$. One has $r \leqslant \operatorname{prof}_A(J; M)$ (Prop. 7); suppose that the inequality is strict. Let $N$ denote the $A$-module $M/(x_1M + \ldots + x_rM)$. One has $\operatorname{prof}_A(J; N) > 0$ (loc. cit.), so that there exists an element $x$ of $J$ such that the homothety $x_N$ is injective (No. 1, Remark 2), that is to say, such that the sequence $(x_1, \ldots, x_r, x)$ is $M$-regular. It follows by induction that for every integer $s$ such that $r \leqslant s \leqslant \operatorname{prof}_A(J; M)$ the sequence $(x_1, \ldots, x_r)$ can be completed to an $M$-regular sequence of length $s$, which entails assertions a) and b). Assertion c) follows from Remark 1 of No. 1 and Corollary 1 of Theorem 1 of No. 3.