Proposition 4

Let $M$ *be a module over a Noetherian ring, N a submodule of M, $N = \bigcap_{i \in I} Q_i$ a primary decomposition of N in M and, for all $i \in I$, let $\{p_i\} = \operatorname{Ass}(M/Q_i)$. For this decomposition to be reduced, it is necessary and sufficient that the $p_i$ be distinct and belong to $\operatorname{Ass}(M/N)$; then

(2)

(3)

If the condition of the statement is fulfilled, $N = \bigcap_{j \neq i} Q_j$ cannot hold, for we would deduce that $\operatorname{Ass}(M/N) \subset \bigcup_{j \neq i} \{p_j\}$ (§ 1, no. 1, Corollary 2 to Proposition 3) contrary to the hypothesis; the primary decomposition $(Q_i)_{i \in I}$ of $N$ is then certainly reduced. Conversely, $\operatorname{Ass}(M/N) \subset \bigcup_{i \in I} \{p_i\}$ always holds (§ 1, no. 1, Corollary 2 to Proposition 3); on the other hand, for all $i \in I$, let us write $P_i = \bigcap_j Q_j$; then $P_i \cap Q_i = N$ and $P_i \neq N$ if $(Q_i)_{i \in I}$ is reduced, hence $P_i/N$ is non-zero and is isomorphic to the submodule $(P_i + Q_i)/Q_i$ of $M/Q_i$, whence $\{p_i\} = \operatorname{Ass}(P_i/N)$ (§ 1, no. 1, Proposition 3 and Corollary 1 to Proposition 2); as $P_i/N \subset M/N, p_i \in \operatorname{Ass}(M/N)$, which completes the proof of the necessity of the condition in the statement and formula (2). Finally, as $N = \bigcap_{j \neq i} (Q_j \cap Q_i), \operatorname{Ass}(Q_i/N) \subset \bigcup_{j \neq i} \operatorname{Ass}(Q_i/(Q_j \cap Q_i))$ (§ 1, no. 1, Corollary 2 to Proposition 3); but $Q_i/(Q_j \cap Q_i)$ is isomorphic to the submodule $(Q_i + Q_j)/Q_j$ of $M/Q_j$, hence $\operatorname{Ass}(Q_i/(Q_j \cap Q_i)) \subset \{p_j\}$ and