Let $R$ be a commutative Noetherian ring such that all associated primes are minimal primes. Let $r\in R$ be a zero-divisor such that $r/1=0$ in $R_P$ for every minimal prime $P$ of $R$. Then, is it true that $r=0$?
I know this is true if $R$ is reduced (even without assuming $r$ is a zero-divisor): If $R$ is reduced, then the intersection of all minimal primes is $0$ which gives the injectivity of the first natural map in the following, whereas the injectivity of the second map comes from the natural embedding $R/P\to Q(R/P)\cong R_P/PR_p$ $$R\to \prod R/P \to \prod R_P/PR_P,$$ where the product is taken over minimal primes. Thus if $r/1=0$ in $R_P$ for all minimal prime $P$, then $r=0$ in $R$.
I am not sure about the situation in my original question.
Please help.
Best Answer
It is true that $r=0$ in this situation. If $r \not= 0$ we must have $\mathrm{ann}(r) \not= R$. Since $R$ is noetherian there is an annihilator maximal among annihilators that contains $\mathrm{ann}(r)$. It is a well known (and easy) result that the maximality of this annihilator implies it is prime. Let $P$ be such a prime ideal then $P$ is an associated prime ideal (since it is an annihilator). Since $\mathrm{ann}(r) \subset P$ we have $r/1 \not= 0$ in $R_P$. But it is given that no such $P$ exists, hence $\mathrm{ann}(r)=R$ which shows that $r=0$.