Proposition 8

Let $M$ be a module of finite length over a Noetherian ring $A$.

(i) There only exists a single primary decomposition of $\{0\}$ with respect to $M$ indexed by $\operatorname{Ass}(M)$ (necessarily reduced) — let $\{0\} = \bigcap_{p \in \operatorname{Ass}(M)} Q(p)$ be this decomposition, where $Q(p)$ is $p$-primary with respect to $M$.

(ii) There exists an integer $n_0$ such that, for all $n \geq n_0$ and all $p \in \operatorname{Ass}(M)$, $Q(p) = p^n M$.

(iii) For all $p \in \operatorname{Ass}(M)$, the canonical mapping of $M$ to $M_p$ is surjective and its kernel is $Q(p)$.

(iv) The canonical injection of $M$ into $\bigoplus_{p \in \operatorname{Ass}(M)} (M/Q(p))$ is bijective.

As every element $p \in \operatorname{Ass}(M)$ is minimal in $\operatorname{Ass}(M)$ (Proposition 7), assertion (i) follows from no. 3, Proposition 5. As $M$ is finitely generated, there exists $n_0$ such that $p^n M \subset Q(p)$ for all $p \in \operatorname{Ass}(M)$ and all $n \geq n_0$ (no. 1, Remark); but as $p$ is a maximal ideal, $p^n M$ is $p$-primary with respect to $M$ (no. 1, Examples 2 and 3) and, as $\bigcap_{p \in \operatorname{Ass}(M)} p^n M = \{0\}$, it follows from (i) that necessarily $p^n M = Q(p)$ for all $p \in \operatorname{Ass}(M)$; whence (ii). As the $p^n$, for $p \in \operatorname{Ass}(M)$, are relatively prime in pairs (Chapter 11, § 1, no. 2, Proposition 3), the canonical mapping $M \to \bigoplus_{p \in \operatorname{Ass}(M)} (M/p^n M)$ is surjective (Chapter 11, § 1, no. 2, Proposition 6), whence (iv). Then $\operatorname{Ass}(Q(p)) = \operatorname{Ass}(M) - \{p\}$ and $\operatorname{Ass}(M/Q(p)) = \{p\}$ (no. 3, Proposition 4); as the elements of $\operatorname{Ass}(M)$ are maximal ideals, $p$ is the only element of $\operatorname{Ass}(M)$ which does not meet $A - p$; $Q(p)$ is therefore the kernel of the canonical mapping $j : M \to M_p$ ($§ 1$, no. 2, Proposition 6). If $s \in A - p$, the homothety of $M/Q(p)$ with ratio $s$ is injective by virtue of the relation $\operatorname{Ass}(M/Q(p)) = \{p\}$ (no. 1, Proposition 1); since $M/Q(p)$ is Artinian, this homothety is bijective (Algebra, Chapter VIII, § 1, no. 2, Lemma 3). The canonical mapping $M \to M/Q(p)$ is then written $f \circ j$, where $f : M_p \to M/Q(p)$ is an $A$-homomorphism (Chapter II, § 2, no. 2, Proposition 3); as $\operatorname{Ker}(j) = \operatorname{Ker}(f \circ j) = Q(p)$, $f$ is injective; we conclude that $j$ is surjective and $f$ bijective.