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

  1. integral domains that are UFD's,
  2. integral domains that are valuation domains,
  3. integral domains that are Pruefer domains,
  4. directed partially ordered groups that are lattice-ordered,
  5. 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?)