# A zero-divisor whose every localization is zero in a Noetherian ring whose every associated prime is minimal

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.

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 $$r=0$$.
• Long story short: a minimal prime over $\mathrm{ann}(r)$ is associated, hence, by hypothesis, minimal. Commented Sep 1, 2021 at 12:19