Proposition 9

Let $A$ be a noetherian ring, $M$ a finitely generated $A$-module, $a$ a subset of $A$ stable under addition and multiplication, $q_1, ..., q_m$ prime ideals of $A$ not containing $a$, and $r \geq 1$ an integer such that

There exists a sequence $(x_1, ..., x_r)$ of elements of $a$, belonging to none of the ideals $q_1, ..., q_m$, and such that the sequence $(x_1/1, ..., x_r/1)$ of elements of $A_p$ is secant for the $A_p$-module $M_p$, for every prime ideal $p \in V(a) \cap \operatorname{Supp}(M)$.

We reason by induction on $r$. Suppose first that $r = 1$, and let $\Phi$ be the (finite) set of minimal elements of $\operatorname{Supp}(M)$. We argue by contradiction and suppose that there exists no element $x_1$ of $a$ satisfying the conditions of the statement. Let $x \in a$, not belonging to $q_1 \cup ... \cup q_m$. By hypothesis there exists an element $p$ of $V(a) \cap \operatorname{Supp}(M)$ such that the image $x/1$ of $x$ in $A_p$ is not secant for $M_p$. Let $\Psi$ be the set of ideals $q \in \Phi$ contained in $p$; then the minimal elements of $\operatorname{Supp}(M_p)$ are the prime ideals $qA_p$ of $A_p$, where $q$ runs through $\Psi$. By Prop. 3 of No. 2, there therefore exists an element $q$ of $\Psi$ such that $x/1 \in qA_p$, whence $x \in q$. In other words, we have $a \subset q_1 \cup ... \cup q_m \cup \bigcup_{q \in \Phi} q$. Since $a \not\subset q_j$ for $1 \leq j \leq m$, Prop. 2 of II, § 1, No. 1 proves the existence of an element $q$ of $\Phi$ containing $a$, whence

Since $V(a)$ contains an irreducible component of $\operatorname{Supp}(M)$, this contradicts the hypothesis

Suppose now that $r \geq 2$. By the induction hypothesis, one can find a sequence $(x_1, ..., x_{r-1})$ of elements of $a$, not belonging to $q_1 \cup ... \cup q_m$, and such that, for every $p \in V(a) \cap \operatorname{Supp}(M)$, the sequence $(x_1/1, ..., x_{r-1}/1)$ of elements of $A_p$ is secant for $M_p$. Put $N = M / \sum_{i=1}^{r-1} x_i M$. It is enough to construct an element $x_r$ of $a$ not belonging to $q_1 \cup ... \cup q_m$ and such that, for every $p \in V(a) \cap \operatorname{Supp}(M)$, the element $x_r/1$ of $A_p$ is secant for $N_p$. By the first part of the proof, it is enough to establish the two relations

Now one has

whence $X' = \operatorname{Supp}(N)$.