I'm working on an exercise from Atiyah and MacDonald's Commutative Algebra, and have hit a bump on Exercise 14 of Chapter 1.

In a ring $A$, let $\Sigma$ be the set of all ideals in which every element is a zero-divisor. Show that set $\Sigma$ has maximal elements and that every maximal element of $\Sigma$ is a prime ideal. Hence the set of zero-divisors in $A$ is a union of prime ideals.

I see by an application of Zorn's Lemma that $\Sigma$ has maximal elements. I take $\mathfrak{m}$ to be maximal in $\Sigma$, with $xy\in\mathfrak{m}$. Since $xy$ is a zero divisor, $xyz=0$ for some $z\neq 0$. If $yz=0$, then $y$ is a zero divisor, otherwise $x$ is a zero divisor. So I guess I then want to show $x\in\Sigma$ or $y\in\Sigma$. If neither is, then $\mathfrak{m}$ is properly contained in both $(\mathfrak{m},x)$ and $(\mathfrak{m},y)$. However, I'm not sure how to show either of these ideals is again in $\Sigma$.

If my understanding is correct, elements of $(\mathfrak{m},x)$ are finite sums of the form $\sum_ia_im_i+bx$ for $a_i\in A$, $m_i\in\mathfrak{m}$ and $b\in A$. To show this sum is a zero divisor, my hunch is that if $c_im_i=0$ for $c_i\neq 0$ and $dx=0$ for $d\neq 0$, then $$ \left(\sum_ia_im_i+bx\right)(d\prod_ic_i)=0. $$ My concern is that perhaps $d\prod_ic_i=0$, so the above wouldn't show that $(\mathfrak{m},x)$ consists of only zero divisors. How can I get around this? Or is there perhaps a better approach? Thank you for your help.

  • 1
    $\begingroup$ Note that the "Hence" in the problem statement requires not just that $\Sigma$ has maximal elements, but that every ideal in $\Sigma$ is contained in a maximal ideal in $\Sigma$. (Your application of Zorn proves this stronger thing.) $\endgroup$ – 6005 Mar 28 '17 at 17:51
  • $\begingroup$ Possibly related: math.stackexchange.com/questions/518 $\endgroup$ – Watson Nov 28 '18 at 20:30

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.

  • 2
    $\begingroup$ Thank you Zev, this was nice and clear. $\endgroup$ – yunone Jun 10 '11 at 4:08
  • 1
    $\begingroup$ @yunone: No problem! This is a classic method of proving that ideals that are maximal with respect to having some property are prime. In fact there are general theorems about what properties such an argument can be made about, see here for example. $\endgroup$ – Zev Chonoles Jun 10 '11 at 4:12
  • 1
    $\begingroup$ How one can conclude that the set of zero divisors is an union of prime ideals? $\endgroup$ – Epsilondelta Nov 26 '14 at 15:42
  • 1
    $\begingroup$ @ZevChonoles : How does one conclude that the set of zero divisors is a union of prime ideals? $\endgroup$ – user228168 May 14 '15 at 14:15
  • 1
    $\begingroup$ @SaunDev: Read the original question by "yunone", and then read my answer. $\endgroup$ – Zev Chonoles May 14 '15 at 15:15

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 very nice exposition see the first few pages of Kaplansky: Commutative Rings.

Note $\ $ For generalizations, in addition to the Lam and Reyes paper Oka and Ako Ideal Families in Commutative Rings mentioned by Zev, see also 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?)

  • 1
    $\begingroup$ Very nice, I'd not seen this approach before. $\endgroup$ – Zev Chonoles Jun 10 '11 at 4:08
  • 1
    $\begingroup$ Thanks Amitesh for the walkthrough. So for (2), $y\in (a:x)-a$, so the inclusion is proper. For (3), taking $b\in (a:z)$, then $bzc=0$ for $c\neq 0$. But $zc\neq 0$ since $z$ is not a zero divisor, so $b$ is a zero-divisor, so $(a:z)\in\Sigma$. Finally for (4), $zx\in a$ from (2) so $x\in (a:z)-a$, and the inclusion is proper. So either way $a$ is contained in either $(a:x)$ or $(a:z)$, contradicting maximality. So $a$ must be prime. $\endgroup$ – yunone Jun 10 '11 at 4:20
  • $\begingroup$ How one can conclude that the set of zero divisors is an union of prime ideals? $\endgroup$ – Epsilondelta Nov 26 '14 at 15:43

An elements of $(\mathfrak{m},x)$ can be described as $m_1+a_1 x$ where $m_1 \in \mathfrak{m}$ and $a_1 \in A$, a little simpler than yours. And I think it is better to assume that $x \notin \mathfrak{m}, y \notin \mathfrak{m} $ then deduce a contradiction, since $(\mathfrak{m},x)$ alone may properly contain $\mathfrak{m}$, but both $(\mathfrak{m},x) \supsetneq \mathfrak{m}$, $(\mathfrak{m},y) \supsetneq \mathfrak{m}$ makes a contradiction.

So assume that $xy \in \mathfrak{m}$ with $(\mathfrak{m},x) \supsetneq \mathfrak{m}$, $(\mathfrak{m},y) \supsetneq \mathfrak{m}$, then there are $m_1+a_1 x \in (\mathfrak{m},x)$ and $m_2+a_2 y \in (\mathfrak{m},y)$ which are not 0-divisors. Then $(m_1+a_1 x)(m_2+a_2 y) \in \mathfrak{m}$ since $xy$ is in $\mathfrak{m}$. But this is a contradiction, since a product of non 0-divisors is not a 0-divisor.

  • $\begingroup$ Good answer! I suppose it is the same as Zev's answer; instead of considering a product of ideals, you have considered a product of elements. However, sometimes this can be more conceptual. $\endgroup$ – Amitesh Datta Jun 10 '11 at 4:44

Zev's answer is excellent. However, here are some exercises that provide practice with the technique Zev introduced (all rings are commutative and have a multiplicative identity):

Exercise 1: Let $A$ be a ring with at least one non-principal ideal. Let $\Sigma$ be the set of all ideals of $A$ that are not principal. Prove that $\Sigma$ has maximal elements and that maximal elements of $\Sigma$ are prime. (Hint: study my answer carefully.)

Exercise 2: Let $A$ be a ring with at least one non-finitely generated ideal. Let $\Sigma$ be the set of all ideals of $A$ that are not finitely generated. Prove that $\Sigma$ has maximal elements and that maximal elements of $\Sigma$ are prime. (Hint: let $I$ be a maximal element of $\Sigma$. If $I$ is not prime, then there exists $x,y\not\in I$ such that $xy\in I$. Use the fact that $I+(x)$ is finitely generated and that $(I:x)$ is finitely generated. Why are these ideals finitely generated?)

Exercise 3: Let $S$ be a multiplicatively closed subset of $A$ such that $0\not\in S$. (Definition: $S$ is closed under multiplication and $1\in S$.) Let $\Sigma$ be the set of all ideals $I$ such that $I\cap S = \emptyset$. Note that $\Sigma\neq \emptyset$ since $0\in \Sigma$. Prove that $\Sigma$ has maximal elements and that a maximal element $I$ of $\Sigma$ is a prime ideal. (Hint: use the technique Zev introduced.)

  • $\begingroup$ Hi! These exercises are very nice, and... maybe you'd like to tell us where are them from? $\endgroup$ – user263834 Jun 12 '17 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.