Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 3 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.

share|improve this answer

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, $J\subseteq I$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.