Proposition 7

Let $M$ be a finitely generated module over a Noetherian ring $A$. The following properties are equivalent: (a) $M$ is of finite length. (b) Every ideal $p \in \operatorname{Ass}(M)$ is a maximal ideal of $A$. (c) Every ideal $p \in \operatorname{Supp}(M)$ is a maximal ideal of $A$.

Let $(M_i)_{0 \leq i \leq n}$ be a composition series of $M$ such that, for $0 \leq i \leq n - 1$, $M_i/M_{i+1}$ is isomorphic to $A/p_i$, where $p_i$ is prime ($§ 1$, no. 4, Theorem 1). If $M$ is of finite length, so is each of the $A$-modules $A/p_i$, which implies that each of the rings $A/p_i$ is Artinian; but as $A/p_i$ is an integral domain, it is therefore a field, in other words $p_i$ is maximal; we conclude that (a) implies (b) ($§ 1$, no. 4, Theorem 2). Condition (b) implies (c) by $§ 1$, no. 3, Proposition 7. Finally, if all the ideals of $\operatorname{Supp}(M)$ are maximal, so are the $p_i$ ($§ 1$, no. 4, Theorem 2), hence the $A/p_i$ are simple $A$-modules and $M$ is of finite length, which completes the proof.