TITLE: Proper ideal $I \implies \exists $ prime ideals $P_i$ such that $P_1 \cdots P_n \subset I$.
QUESTION [5 upvotes]: Let the below ideals be in a commutative Noetherian ring $R$.
Corollary 22.  (3)  There are prime ideals $P_1, \dots, P_n$ (not necc. distinct) $\supset I$ such that $P_1\cdots P_n \subset I$.
(Out of D&F)

Prove (3) of Corollary 22 directly by considering the coll. $\mathcal{S}$ of ideals that do not contain a finite product of prime ideals.  [If $I$ is a maximal element in $\mathcal{S}$, show that since $I$ is not prime there are ideals $J, K$ properly containing $I$ (hence not in $\mathcal{S}$) with $JK \subset I$.]

I know:

$I$ is not prime $\implies \exists$ ideals $J,K$ such that $JK \subset I$ yet $J \not\subset I$ and $K \not\subset I$.
$I$ is not prime $\implies$ in particular not maximal $\implies$ $I$ properly contained in some maximal ideal $J$.
From examining proof to Proposition 20 the proof of this would go something like if $\mathcal{S}$ were not empty, then since $R$ is Noetherian, all chains in $\mathcal{S}$ are upper bounded and so $\mathcal{S}$ contains a maximal element $I$.

I can't piece it together from these facts alone, what am I missing?

REPLY [2 votes]: by maximality of $I$ in $S$, in the ring $R/I$ every ideal $K \subset R/I$ is contained in each of a finite set of prime ideals $K_i$ such that some product formed from them (possibly with repetitions) is contained in $K$. we may call the $K_i$ a primal set for $K$ and the product contained within $K$ as a primal product for $K$
since $I$ is not prime the quotient $R/I$ has zero-divisors. let us say $PQ=0$ for the corresponding principal ideals.
let $P_i,Q_i$ be primal sets for $P$ and $Q$ respectively. since any $T'$ (the pre-image of a prime ideal $T \subset R/I$) is prime in $R$ we have that $P'_i,Q'_j$ are primal sets for $P',Q'$. 
since $P'Q' \subset I$, $I \subset P'_i$ and $I \subset  Q'_j$ the $P'_i$ and $Q'_j$ together form a primal set for  $I$ (with the corresponding primal product being the product of the primal products for $P'$ and $Q'$  separately
this contradicts the assumption $I \subset S$ showing that $S$ is empty