TITLE: Prime ideal containing only zero divisors
QUESTION [5 upvotes]: Let $R$ be a Noetherian ring and $p\subset R$ be a prime ideal containing only zero divisors. Does it follow that $p$ is an associated prime of $R$? In other words is there some $x\in R$ such that $p=Ann(x)$?

REPLY [2 votes]: A simple (counter)example: $R=K[X,Y,Z]/(X^2,XY,XZ)$, $\mathfrak p=(x,y)$. 
Let me give some details: the ideal $I=(X^2,XY,XZ)$ can be written as $(X)\cap(X^2,Y,Z)$ (see here). This is a primary decomposition of $I$ and thus the associated primes of $I$ are $(X)$ and $(X,Y,Z)$. Now consider the prime ideal $(X,Y)$ which is strictly contained between the two associated primes.