Proposition 7

Let A be a ring and M an A-module.

(i) Every prime ideal $\mathfrak{p}$ of A containing an element $\in \operatorname{Ass}(M)$ belongs to $\operatorname{Supp}(M)$.

(ii) Conversely, if A is Noetherian, every ideal $\mathfrak{p} \in \operatorname{Supp}(M)$ contains an element $\in \operatorname{Ass}(M)$.

If $\mathfrak{p}$ contains an element $q$ of $\operatorname{Ass}(M)$, then $q \cap (A - \mathfrak{p}) = \varnothing$ and hence, if we write $S = A - \mathfrak{p}$, $S^{-1}\mathfrak{p}$ is a prime ideal associated with $S^{-1}M = M$, (no. 2, Proposition 5) and a fortiori $M, \# 0$, hence $\mathfrak{p} \in \operatorname{Supp}(M)$. Conversely, if A is Noetherian, so is A, (Chapter II, § 2, no. 4, Corollary 2 to Proposition 10). If $M_{\mathfrak{p}} \neq 0$, then $\operatorname{Ass}_{A_{\mathfrak{p}}}(M_{\mathfrak{p}}) \neq 0$ (no. 1, Corollary 1 to Proposition 2) and hence there exists $q \in \operatorname{Ass}_A(M)$ such that $q \cap (A - \mathfrak{p}) = \varnothing$ (no. 2, Corollary to Proposition 5).