Can we prove the converse of Krull’s principle ideal theorem in the case when $R$ is integral over $\mathbb{Z}$?
To be more precise, can we prove the following statement: In the above case, when $P$ is a prime ideal with height 1 in $R$, then $P$ contains a non zero-divisor?
@Mohan has shown in the comment that the statement above doesn’t hold for some case.
So I want to change the condition for ring $R$ to be finitely generated free $\mathbb{Z}$-module. Does the statement holdin this case?