Theorem 1

Let $A$ be a noetherian local ring, $M$ a nonzero finite $A$-module and $S$ a subset of the maximal ideal $m_A$ of $A$.

a) If $M/SM$ is of finite length, one has $\mathrm{Card}(S) \geqslant \dim_A(M)$; if $S$ is secant for $M$, one has $\mathrm{Card}(S) \leqslant \dim_A(M)$.

b) Every secant subset for $M$ is contained in a maximal secant subset for $M$.

c) The following properties are equivalent: (i) $S$ is a maximal secant subset for $M$; (ii) $S$ is a secant subset for $M$ and $\mathrm{Card}(S) = \dim_A(M)$; (iii) $M/SM$ is of finite length and $\mathrm{Card}(S) = \dim_A(M)$; (iv) $S$ is a secant subset for $M$ and $M/SM$ is of finite length.

Since one has $S \subset m_A$, Nakayama's lemma shows that one has $M/SM \neq \{0\}$, whence $\dim_A(M/SM) \geqslant 0$ with equality if and only if $M/SM$ is of finite length. Assertion a) then results from formulas (8) and (9), as well as the equivalence of properties (ii), (iii) and (iv).

Assertion b) results from the fact that the cardinal of every subset of $m_A$ secant for $M$ is bounded above by the integer $\dim_A(M)$.

By a), every secant subset for $M$, of cardinal equal to $\dim_A(M)$, is maximal. It remains to prove that, if $S$ is secant for $M$ and if $\mathrm{Card}(S) < \dim_A(M)$, then $S$ is not maximal. Let $a$ be the annihilator of $M$, and $B$ the noetherian local ring $A/(a + SA)$. By cor. 2 of prop. 2 of No. 1, there exists an element $x$ of $m_A$ such that $\dim(B/xB) = \dim(B) - 1$ whence $x \notin S$. By remark 2, the subset $S \cup \{x\}$ of $m_A$ is secant for $A/a$, hence for $M$ by remark 1.