Is an ideal which is maximal with respect to the property that it consists of zero divisors necessarily prime?

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.

Solution 1:

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.

Solution 2:

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?)