Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is in follow-up to this question.

Let $R$ be a commutative ring with identity and consider the set $Z \subset R$ of zero divisors. If the ideal $I\subset Z$ is maximal with respect to the constraint, need it be prime?

Since Zorn's Lemma applies equally well to proving that either minimal prime ideals exist or that maximal ideals contained in $Z$ exist, this would provide an alternative proof of that question if true.

A naive approach to proving this is to assume $ab \in I$ with $a, b \notin I$. Since $ab \in I$, it follows that $ab$ is a zero divisor, so $abx = 0$ for some $x \neq 0$, and therefore either $a$ is a zero divisor annihilated by $bx \neq 0$ or $b$ is a zero divisor annihilated by $x$. In either case, the obvious thing to consider would be $aR+I$ or $bR+I$, but there's no obvious reason why either of these ideals should consist of zero divisors.

According to rschwieb, the answer is yes for reduced Noetherian rings.

share|improve this question

2 Answers 2

up vote 4 down vote accepted

The proof I recall uses the ideal quotient $(\mathfrak{a}:\mathfrak{b})$, where if $\mathfrak{a}$ and $\mathfrak{b}$ are ideals of a commutative unital ring $A$, then $$ (\mathfrak{a}:\mathfrak{b})=\{x\in A:x\mathfrak{b}\subseteq\mathfrak{a}\}. $$ If $\mathfrak{b}=(x)$ is principal, I write $(\mathfrak{a}:x)$ in place of $(\mathfrak{a}:(x))$.

Let $\mathfrak{a}$ be a maximal element in $\Sigma$, the set of all ideals consisting of zero-divisors, ordered by inclusion. I claim $\mathfrak{a}$ is prime in $A$.

Suppose $x,y\notin\mathfrak{a}$, but $xy\in\mathfrak{a}$. Then $y\in(\mathfrak{a}:x)$, so $\mathfrak{a}\subsetneq(\mathfrak{a}:x)$. By maximality of $\mathfrak{a}$, $(\mathfrak{a}:x)\notin\Sigma$, so there exists $z\in(\mathfrak{a}:x)$ which is not a zero divisor.

Consider $(\mathfrak{a}:z)$. If $w\in(\mathfrak{a}:z)$, $zw\in\mathfrak{a}$. Thus there exists $v\neq 0$ such that $vzw=(vw)z=0$. Since $z$ is not a zero divisor, $vw=0$, so $w$ is a zero divisor. Thus $(\mathfrak{a}:z)\in\Sigma$. But $x\in(\mathfrak{a}:z)$, so $\mathfrak{a}\subsetneq (\mathfrak{a}:z)$, a contradiction to maximality of $\mathfrak{a}$. So $\mathfrak{a}$ is prime. Hopefully that works.

share|improve this answer
makes sense! Thanks! –  Dustan Levenstein Jan 5 at 7:17

Hint with the stuff you've done so far:

$$I\lneqq aR+I\,,\,bR+I\;\;,\;\;\text{but}\;\;(aR+I)(bR+I)\le abR+aI+bI+I^2\le I$$

Take now non-zero divisors $\;x\in aR+I\;,\;\;y\in bR+I\implies xy\in I\;$ , which cannot be (why?)

share|improve this answer
That's quite simple, thanks! –  Dustan Levenstein Jan 5 at 7:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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