TITLE: Associated Prime Ideals in a Noetherian Ring; Exercise 6.4 in Matsumura
QUESTION [11 upvotes]: Let $I$ and $J$ be ideals of a Noetherian ring $A$. If $JA_P\subseteq IA_P$ for every $P\in \operatorname{Ass}_A(A/I)$, then $J\subseteq I$.

I'm reading Matsumura's Commutative Ring Theory book on my own. This is exercise $6.4$ in that book, and I fail to prove it. Could you help me with this? Thanks in advance.

REPLY [5 votes]: Let $I+J/I\neq 0$. Then $\operatorname{Ass}({I+J}/I)\neq\emptyset$ and it is a subset of $\operatorname{Ass}(R/I)$.
Therefore, there exist a prime ideal $\mathfrak p$ contained in $\operatorname{Ass}({I+J}/I)$ and so ${\mathfrak p}R_{\mathfrak p}\in\operatorname{Ass}({I+J}/I)_{\mathfrak p}$. But $({I+J}/I)_{\mathfrak p}=0$ and this is a contradiction. So ${I+J}/I=0$, that is, $J\subseteq I$.