Theorem 1

Let $A$ be a noetherian ring, $q$ an ideal of $A$ contained in the radical of $A$ and $M$ a finitely generated $A$-module such that $M/qM$ is of finite length. Let $x_1, ..., x_m$ be a finite sequence of elements of $q$. Put $x = Ax_1 + \cdots + Ax_m \subset q$.

a) One has $\dim_A(M/xM) \geq \dim_A(M) - m$.

b) If $\dim_A(M/xM) = \dim_A(M) - m$, then $e_q(M/xM) \geq e_q(M)$.

c) If $m < \dim_A(M)$, and if for $i = 1, ..., m$, the element $x_i$ of $A$ is superficial for $M/(x_1M + \cdots + x_{i-1}M)$ relative to $q$, then one has

d) If $m = \dim_A(M)$, and if, for $i = 1, ..., m$, the element $x_i$ of $A$ is superficial for $M/(x_1M + \cdots + x_{i-1}M)$ relative to $q$, then one has

Parts $a), b),$ and c) follow, for $m = 1$, from Prop. 8, and the general case is deduced by induction. Suppose the hypotheses of d) are satisfied and set $x' = Ax_1 + \cdots + Ax_{m-1}$ and $M' = M/x'M$, so that $M/xM$ is identified with $M'/x_mM'$. Then, by c), one has $\dim_A(M') = 1$ and $e_q(M) = e_q(M')$. By Prop. 8, $M/xM$ is of finite length and one has $e_q(M') = e_{x_mA}(M') \leq \operatorname{long}(M/xM)$. But, since $x_m^nM' = x^nM'$ for every $n$, one has $e_{x_mA}(M') = e_x(M')$. On the other hand, one has $e_x(M') \geq e_x(M)$: this follows from b), where $m$ is replaced by $m - 1$, $x$ by $x'$, and $q$ by $x$. Consequently, one has

Since $x$ is contained in $q$, this implies $e_x(M) = e_q(M)$ (No. 1, Remark 1), and completes the proof.