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?