Remark 2

Let $A$ be a noetherian ring, $M$ an $A$-module, $e : M \to I$ an injective envelope of $M$. The set Ass(M) is equal to Ass(I): in fact, the inclusion $Ass(M) \subset Ass(I)$ is obvious. On the other hand, if $p$ is an element of $Ass(I)$, $I$ contains a submodule $N$ isomorphic to $A/p$; since the $A$-module $e^{-1}(N)$ is nonzero, one has $Ass(e^{-1}(N)) = \{p\}$ (IV, § 1, No. 1, Prop. 1). This proves that $p$ is associated with $e^{-1}(N)$, hence with $M$, whence our assertion.

Let us retain the notation of the theorem, and suppose further that $Ass(M)$ is finite. For every $q \in Ass(M)$, let $Q_q$ denote the intersection with $M$ of $\bigoplus_{p \in Ass(M) - \{q\}} I(p)^{(a_p)}$. Then $(Q_q)_{q \in Ass(M)}$ is a reduced primary decomposition of 0 in M (IV, § 2, No. 3, Def. 3). For indeed $\cap Q_q = 0$; since $M/Q_q$ is identified with a nonzero a-submodule of $I(q)^{(a_q)}$, we have $Ass(M/Q_q) = \{q\}$, and it is enough to apply Prop. 4 of loc. cit.