Showing the set of zero-divisors is a union of prime ideals
Let $\frak{m}$ be a maximal element in $\Sigma$. We want to show it is prime, i.e. that if $x\notin\frak{m}$ and $y\notin\frak{m}$, then $xy\notin\frak{m}$.
If $x\notin\frak{m}$ and $y\notin\frak{m}$, then ${\frak{m}}+(x)$ and ${\frak{m}}+(y)$ are both ideals of $A$ that strictly contain $\frak{m}$, and therefore each must contain non-zero-divisors ($\frak{m}$ is maximal among ideals consisting only of zero-divisors, so any ideal strictly containing $\frak{m}$ cannot consist only of zero-divisors). Thus the ideal $({\frak{m}}+(x))({\frak{m}}+(y))\subseteq{\frak{m}}+(xy)$ contains non-zero-divisors (because there is at least one non-zero-divisor in each of ${\frak{m}}+(x)$ and ${\frak{m}}+(y)$, and the product of two non-zero-divisors is a non-zero-divisor). But the fact that ${\frak{m}}+(xy)$ contains non-zero-divisors implies that ${\frak{m}}+(xy)$ strictly contains $\frak{m}$, hence $xy\notin\frak{m}$. Thus $\frak{m}$ is prime.
Hint $ $ The non-zero-divisors form a saturated monoid $\rm\:M\:$ (i.e. $\rm\:ab\in M\!\iff\! a\in M\:$ & $\rm\:b\in M$) so its complement is a union of prime ideals (this can be proved either by localization, or by a direct elementary proof using that an ideal maximal wrt the exclusion of a monoid is prime). For a nice exposition see the first few pages of Kaplansky: Commutative Rings.
Note $ $ For generalizations, see Lam and Reyes: Oka and Ako Ideal Families in Commutative Rings mentioned by Zev, and the paper reviewed below.
MR 95i:13023 13G05 06F20
Anderson, D. D.; Zafrullah, Muhammad
On a theorem of Kaplansky.
Boll. Un. Mat. Ital. A (7) 8 (1994), no. 3, 397--402.
The complement of a saturated multiplicatively closed set is a union of prime ideals (Bourbaki). $\ $ The authors apply this fact to a property (*) of non-zero elements in an integral domain such that the elements satisfying (*) form a saturated multiplicatively closed set. Suitable choices of (*) yield characterizations of UFDs [I. Kaplansky, Commutative rings, Univ. Chicago Press, Chicago, IL, 1974; MR 49 #10674], GCD domains, valuation and Prüfer domains. Lattice ordered groups and Riesz groups are characterized similarly. [Reviewed by C. P. L. Rhodes]
Zbl 816.13001
Analyzing the proof of Kaplansky's theorem, that an integral domain is a unique factorization domain if and only if every nonzero prime ideal contains a nonzero principal prime ideal, the authors state - leaving the proof to the reader:
Let D be an integral domain. Let (*) be a property of elements in D. Assume that the set S of nonzero elements in D with (*) is a nonempty saturated multiplicatively closed set. Then every nonzero element in D has (*) if and only if every nonzero prime ideal contains a nonzero element with (*). This observation is then applied to various situations, to characterize
- integral domains that are UFD's,
- integral domains that are valuation domains,
- integral domains that are Pruefer domains,
- directed partially ordered groups that are lattice-ordered,
- directed partially ordered groups that are Riesz groups.
[ K.Roggenkamp (Stuttgart)]
The following steps lead to a solution:
(1) If $a\in\Sigma$ is not a prime ideal, we will show that $a$ is not a maximal element of $\Sigma$. Since $a$ is not a prime ideal, there exists $x,y\not\in a$ such that $xy\in a$. The ideal $(a:x)=\{z\in A:xz\in a\}$ can be considered.
(2) Prove that $a\subseteq (a:x)$ and that this inclusion is proper.
(3) Prove that $(a:x)$ consists entirely of zero-divisors. (Hint: If not, then there exists $z\in (a:x)$ such that $z$ is not a zero divisor. Deduce that $(a:z)\in \Sigma$.)
(4) Show that $a$ is not maximal. (Hint: The inclusion $a\subseteq (a:z)$ is proper. Why?)