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

82 Views Asked by At

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