Corollary 1

Let $A$ be a ring, $M$ a finitely generated $A$-module and $a$ an element of $A$. For $a$ to belong to every prime ideal of the support of $M$, it is necessary and sufficient that the homothety of $M$ with ratio $a$ be nilpotent.

It follows from Proposition 17 that the intersection of the prime ideals belonging to $\operatorname{Supp}(M)$ is the radical of the annihilator $a$ of $M$ (no. 3, Proposition 11 (ii)). To say that $a$ belongs to this radical is equivalent to saying that there exist a power $a^k \in a$ and hence that $a^k M = 0$.