The above proposition says:
Let $R$ be a Noetherian ring and let $I$ and $P$ be ideals of $R$ where $P$ is a prime ideal. If $IP=I$, then $I=0$.
I feel that this is false. After passing to localization at $P$ and using Nakayama, we can easily get $S^{-1}I=0$ ($S=R-P$), from which we can conclude that $sI=0$ for some $s\in S$. How can we conclude that $I=0$?
Alternatively, we can use Corollary 2.5 of Atiyah and Macdonald to directly conclude that $xI=0$, where $x=1+p$ with $p\in P$.
The proof given there says since $I$ is a subset of $S^{-1}I$, $I=0$. This is false unless $R$ is a domain, isn't it?
Yes, the highlighted proposition is false as shown by the following counterexample:
Given a field $k$, take $R=k\times k$ and $I=P=k\times \{0\}$. Then $IP=I$ and yet $I\neq0$.