Proposition 5

Let $A$ be a ring, $S$ a multiplicative subset of $A$, $\Phi$ the set of prime ideals of $A$ which do not meet $S$ and $M$ an A-module. Then: (i) The mapping $p \mapsto S^{-1}p$ is a bijection of $\operatorname{Ass}_A(M) \cap \Phi$ onto a subset of $\operatorname{Ass}_{S^{-1}A}(S^{-1}M)$. (ii) If $p \in \Phi$ is a finitely generated ideal and $S^{-1}p \in \operatorname{Ass}_{S^{-1}A}(S^{-1}M)$, then $p \in \operatorname{Ass}_A(M)$.

Recall (Chapter II, § 2, no. 5, Proposition 11) that the mapping $p \mapsto S^{-1}p$ is a bijection of $\Phi$ onto the set of prime ideals of $S^{-1}A$. If $p \in \operatorname{Ass}_A(M) \cap \Phi$, $p$ is the annihilator of a monogenous submodule $N$ of $M$; then $S^{-1}p$ is the annihilator of the monogenous submodule $S^{-1}N$ of $S^{-1}M$ (Chapter II, § 2, no. 4, formula (9)) and hence $S^{-1}\mathfrak{p} \in \mathrm{Ass}_{S^{-1}A}(S^{-1}M)$. Conversely, suppose that $\mathfrak{p} \in \Phi$ is finitely generated and such that $S^{-1}\mathfrak{p}$ is associated with $S^{-1}M$; then there exists $x \in M$ and $t \in S$ such that $S^{-1}\mathfrak{p}$ is the annihilator of $x/t$. Let $(a_i)_{1 \leq i \leq n}$ be a system of generators of $\mathfrak{p}$; then $(a_i/1)(x/t) = 0$ and hence there exists $s_i \in S$ such that $s_i a_i x = 0 \ (1 \leq i \leq n)$. Let us write $s = s_1 s_2 \ldots s_n$; for all $a \in \mathfrak{p}$, $s a x = 0$, whence $p \subset \mathrm{Ann}(s x)$; on the other hand, if $b \in A$ satisfies $b s x = 0$, then $b/1 \in S^{-1}\mathfrak{p}$ by definition, whence $b \in \mathfrak{p}$. Then $\mathfrak{p} = \mathrm{Ann}(s x)$ and $\mathfrak{p} \in \mathrm{Ass}_A(M)$.