$\DeclareMathOperator\ht{ht}$All rings are commutative Noetherian with identity.  

Exercise 9.8 of Matsumura's book *Commutative ring theory*: Let $A$ be a ring and $A\subset B$ an integral extension. If $P$ is a prime ideal of $B $ with $\mathfrak p = P\cap A$ then $\ht P \leq \ht \mathfrak p$.  

There are many examples that equality occurs; i.e. $\ht P = \ht \mathfrak p$. But the issue is different for the case of examples for which $\ht P \lt \ht \mathfrak p$. Can you give such examples, please?  

Thank you.



Examples of integral ring extensions that $\operatorname{ht}P \lt \operatorname{ht}P\cap A$
Take $A=\mathbb{Z}$ and $B=\mathbb{Z}[X]/(X^2+X,2X)$ $=$ $A[x]$, where $x$ denotes the residue class of $X$. Clearly, $x$ is integral over $A$. Every prime ideal of $B$ either contains $x$ or both $x+1$ and $2$. Hence $P$ = $(2,x+1)$ is a minimal prime ideal, so of height $0$. (Note that $P$ is also a maximal ideal of $B$). Since $2$ $\in$ $P$, it follows that $P\cap A$ $=$ $2\mathbb{Z}$, which is of height $1$.


