Let $R$ be a ring and $I$ is an invertible ideal and $P$ a prime ideal of $R$ such that $P\subset I$. My question is $I/P$ an invertible ideal of $R$?
My attempt: From $P\subset I$ and $I$ is invertible we infer that $P=IP$. If $P$ is invertible, then $I=R$. So we assume that $P$ is not invertible.
Now, as $I$ is invertible so $IJ=dR$ for some regular element $d$ and an ideal $J$ of $R$. Here I know that $P\subset I$. But I am not sure whether $P\subset J$ or not. Also, I wonder $d$ is outside of $P$ or not. (Basically, I am looking at the equation $(I/P)(J/P)=(dR/P)$).
We know the definition; $I$ is an invertible ideal of $R$ if and only if it is finitely generated and locally principal generated by a regular element. Let $M$ be a maximal ideal containing $I$. Let $I_M=xR_M$ where $x$ is a regular element of $R$. It is well known that $(I/P)_{(M/P)}=I_M/P_M$. We deduce that $(I/P)_{M/P}=xR_M/P_M$. The only thing remains to check that $x\not\in P$. On the contrary assume that $x\in P$, then $P_M=I_M$. So $P=P_M\cap R=I_M\cap R\supseteq I$, a contradiction to our hypothesis that $P\subset I$.