In Atiyah/Macdonald's Introduction to Commutative Algebra, the following proposition is given (4.7):
Let $\mathfrak{a}$ be a decomposable ideal in a ring $A$, let $\mathfrak{a} = \bigcap_{i=1}^n\mathfrak{q}_i$ be a minimal primary decomposition, and let $r(\mathfrak{q}_i) = \mathfrak{p}_i$. Then
$$
\bigcup_{i=1}^n \mathfrak{p}_i = \{x \in A : (\mathfrak{a}:x) \neq \mathfrak{a}\}.
$$
In particular, if the zero ideal is decomposable, the set $D$ of zero-divisors of $A$ is the union of the prime ideals belonging to $0$.
The proof begins by noting that we can reduce to just proving the last sentence without loss of generality, because if $\mathfrak{a} = \bigcap_{i=1}^n\mathfrak{q}_i$, then in $A/\mathfrak{a}$, $0 = \bigcap_{i=1}^n\bar{\mathfrak{q}}_i$, and each $\bar{\mathfrak{q}}_i$ is primary, so $0$ is decomposable in the quotient. There are still some details missing here–to really be able to prove the first part given the last sentence, we also have to know how the primes belonging to $0$ in the quotient relate to the $\mathfrak{p}_i$ belonging to $\mathfrak{a}$ in the original ring. I want to say the primes belonging to $0$ in the quotient are just $\bar{\mathfrak{p}}_i$ but I'm not sure; seems like this involves some annoying checks about how radicals play with quotients. Am I on the right track?
Best Answer
Remember that for ideals $a \subset b$ of a ring $A$, then the prime ideals of $A/a$ containing $b/a$ are precisely the quotients $\bar{p}$ for the prime ideals $p$ of $A$ containing $b$.
Also, check that $rad(b/a) \subset A/a$ is just the quotient of $rad(b) \subset A$.