I am concerned with the following proposition:
Let $R$ be a commutative ring with unity, and $I\subsetneqq R$ be a proper ideal. Then there exist prime ideals $P_1,\dots,P_d$ containing $I$ so that $I=P_1 P_2 \cdots P_d$.
The notes that I took down argues as follows. Let $\mathcal S$ be the collection of all proper ideals $I$ for which the above is FALSE. Then there is a maximal element $I\in \mathcal S$. We claim that $I$ is prime. Suppose not. Then there are $a,b\notin I$ with $ab\in I$. Let $J_1:=\langle a \rangle+I$, $J_2:=\langle b \rangle+I$. Then $J_1,J_2$ both strictly contain $I$, and are not in $\mathcal S$ by maximality. Thus $J_1,J_2$ satisfies the factorisation property. Take $P_1,\dots,P_d$ all containing $J_1$ so that $J_1= P_1\cdots P_d$, and take $Q_1,\dots,Q_s$ all containing $J_2$ so that $J_2= Q_1\cdots Q_s$. Then note that \begin{align*} P_1\cdots P_d\cdot Q_1\cdots Q_s &= J_1 J_2=(\langle a \rangle+I)(\langle b \rangle+I)\\ &=\left\{\sum_{i,j}c_i a d_j b+c_i a s_j+d_j b r_i+ r_i s_j\right\}\\ &\subseteq I,\text{ since }ab\in I \text{ and $I$ is an ideal}. \end{align*} Since $I\subsetneqq J_1$, $I$ is a proper ideal of $R$. Since $I$ admits such factorisation as shown, we see that $I\notin \mathcal S$, which is a contradiction. Thus $I$ is prime.
Lastly, since $I$ is prime, take $P_1=I$ so that $I=P_1$ is a trivial factorisation. Hence $I\notin \mathcal S$, which is a contradiction. Thus $\mathcal S=\varnothing$.
The only place where I got stuck is the last line in the computation of $P_1\cdots P_d\cdot Q_1\cdots Q_s$, since it only proves one direction. What about the other direction? Or is such proposition actually false?