Suppose that $R$ is a commutative ring and $I$ an irreducible ideal of $R$.
Let $J$ be the set of zero divisors of $\frac{R}{I}$, i.e., $J=\{ r\in R \; |\; rs\in I\; \text{for some s}\; \in R\setminus I\}$.
It is easy to show that $J$ is an ideal of $R$. But my question is that is $J$ a prime ideal of $R$?
First, note that $ 1 \notin J $. Next, suppose that $ x y \in J $, so $ x y s \in I $ for some $ s \notin I $. If $ y s \in I $, then $ y \in J $; else if $ y s \notin I $, then since $ x y s \in I $, $ x \in J $. Either way, $ x \in J $ or $ y \in J $, so $ J $ is prime.
It seems to me that the primality of $ J $ has nothing in particular to do with the irreducibility of $ I $ as such (or indeed even whether $ I $ is an ideal); $ I $ only needs to be irreducible to make $ J $ into an ideal in the first place.