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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.

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up vote 4 down vote accepted

The problem reduces easily to the following:

Let $R$ be a noetherian ring and $\mathfrak a$ an ideal of $R$ such that $\mathfrak aR_{\mathfrak p}=0$ for all $\mathfrak p\in\operatorname{Ass}R$. Prove that $\mathfrak a=0$.

Since $R$ is noetherian there exists a (reduced) primary decomposition $0=\mathfrak q_1\cap\cdots\cap\mathfrak q_t$. It follows that $\operatorname{Ass}R=\{\mathfrak p_1.\dots,\mathfrak p_t\}$, where $\mathfrak p_i=\sqrt{\mathfrak q_i}$. Now take $a\in\mathfrak a$, $a\neq 0$. There exists $\mathfrak q_i$ such that $a\notin\mathfrak q_i$. Then the ideal $(\mathfrak q_i:a)$ is $\mathfrak p_i$-primary, in particular $\sqrt{(\mathfrak q_i:a)}=\mathfrak p_i$. Now localize this equality at $\mathfrak p_i$ and obtain $\sqrt{(\mathfrak q_iR_{\mathfrak p_i}:aR_{\mathfrak p_i})}=\mathfrak p_iR_{\mathfrak p_i}$. But $aR_{\mathfrak p_i}=0$ and thus $(\mathfrak q_iR_{\mathfrak p_i}:aR_{\mathfrak p_i})=R_{\mathfrak p_i},$ a contradiction.

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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, $J\subseteq I$.

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