depth $R_{\mathfrak p}=0$ implies $\mathfrak pR_{\mathfrak p}$ is associated to zero

Let $$R$$ be a commutative Noetherian ring with unity. I want to prove the fact that depth $$R_{\mathfrak p}=0$$ implies $$\mathfrak pR_{\mathfrak p}$$ is associated to zero.

Since depth $$R_{\mathfrak p}=0$$ we have that $$\mathfrak pR_{\mathfrak p}$$ consists of zero divisors. But to prove that it is a prime associated to $$0$$, I need to show that it is an annihilator ideal of some element in $$R_{\mathfrak p}$$.

How do I show this?

Thank you

• The set of zero-divisors is the union of associated primes. – user26857 May 9 at 15:02
• Yes. You can check Atiyah Prop. 4.7 for a reference – pyrogen May 10 at 9:12