Proposition 3

Let A be a Noetherian ring, S a multiplicative subset of A, p a prime ideal of A, M an A-module, N a submodule of M and $i = i_A^S$ the canonical mapping of M to $S^{-1}M$.

(i) Suppose that $p \cap S \neq \varnothing$. If N is p-primary with respect to M, then $S^{-1}N = S^{-1}M$.

(ii) Suppose that $p \cap S = \varnothing$. For N to be p-primary with respect to M, it is necessary and sufficient that N be of the form $i_1(N')$, where $N'$ is a sub-$S^{-1}A$-module of $S^{-1}M$ which is $(S^{-1}p)$-primary with respect to $S^{-1}M$; then $N' = S^{-1}N$.

(i) If $p \cap S \neq \varnothing$ and N is p-primary with respect to M, then

(§ 1, no. 2, Corollary to Proposition 5) and hence $S^{-1}(M/N) = 0$ (§ 1, no. 1, Corollary 1 to Proposition 2), whence $S^{-1}M/S^{-1}N = 0$.

(ii) Suppose that $p \cap S = \varnothing$. If N is p-primary with respect to M, then $\operatorname{Ass}_{S^{-1}A}(S^{-1}(M/N)) = \{S^{-1}p\}$ (§ 1, no. 2, Corollary to Proposition 5) and hence the submodule $N' = S^{-1}N$ of $S^{-1}M$ is $(S^{-1}p)$-primary; moreover, if $s \in S$ and $m \in M$ are such that $sm \in N$, then $m \in N$, for the homothety with ratio s in M/N is injective, whence $N = i_1(N')$ (Chapter II, § 2, no. 4, Proposition 10). Conversely, let $N'$ be a submodule of $S^{-1}M$ which is $(S^{-1}p)$-primary with respect to $S^{-1}M$; let us write $N = i_1(N')$; then $N' = S^{-1}N$ (Chapter II, § 2, no. 4, Proposition 10) and $\operatorname{Ass}_{S^{-1}A}(S^{-1}(M/N)) = \operatorname{Ass}_{S^{-1}A}((S^{-1}M)/N') = \{S^{-1}p\}$. As the canonical mapping $M/N \to S(M/N)$ is injective, no prime ideal of A associated with M/N meets S (§ 1, no. 2, Proposition 6); it follows that $\operatorname{Ass}(M/N) = \{p\}$ (§ 1, no. 2, Corollary to Proposition 5), so that N is p-primary with respect to M.