Theorem $:$
Let $R$ be a Noetherian domain such that it is integrally closed and every non-zero prime ideal of $R$ is maximal. Then every non-zero proper ideal of $R$ can be written as a product of finitely many prime ideals, determined uniquely, upto order.
Proof $:$
Let $S$ denote the collection of all non-zero proper ideals of $R$ which cannot be factored into product of finitely many prime ideals. if possible let $S \neq \varnothing.$ Let $A \in S$ be such that $A \supseteq p_1p_2 \cdots p_m$ with $m$ minimal. Clearly $m \neq 1.$ $\exists$ a prime ideal $p$ such that $p_1p_2 \cdots p_m \subseteq p.$ $\implies$ some $p_i$ say $p_1 \subseteq p.$ But since every non-zero prime ideal of $R$ is maximal so we have $p_1 = p.$ But then $p_2p_3 \cdots p_m \subseteq Ap^{-1} \subseteq pp^{-1} = R.$ Therefore $\text {Ap}^{-1} \notin \text S.$ So $Ap^{-1} = q_1 q_2 \cdots q_r \implies A = pq_1q_2 \cdots q_r,$ a contradiction. Hence $S = \varnothing.$
In the above proof I dont understand why $Ap^{-1} \notin S.$ Would anybody please help me in understanding this?
Thank you very much.