# $B \subseteq A$ be an integral extension of integral domains where $B$ is normal; if $P \in Spec A$ and $Q=P\cap B$, then $\dim A_P=\dim B_Q$?

Let $B \subseteq A$ be an integral extension of integral domains where $B$ is normal (i.e. integrally closed in its own fraction field ). Let $P$ be a prime ideal of $A$ and $Q=P \cap B$ . Then how to show that $\dim A_P=\dim B_Q$ ?

I know that the extension $B \subseteq A$ satisfies both Going-Up and Going-down property (along with in comparability ). But I am not sure how to derive the dimension equality from that. Please help.

The dimension of $B_Q$ is the largest length of a chains of the primes in $B$ of the form $Q_1 \subset \cdots \subset Q_n$ such that $Q_n = Q$. Same for $A_P$ and $P$.
If you begin with a chain $Q_1 \subset \cdots \subset Q_n = Q$ in $B$, incomparability (a property of integral extensions) gives you a chain $Q_1 \cap A \subset \cdots Q_n \cap A = P$ in $A$ with the same length, with the last term equal to $P$. This shows that $\operatorname{Dim} B_Q \leq \operatorname{Dim} A_P$.
If you begin with a chain $P_1 \subset \cdots \subset P_n = P$ in $A$, and you let $Q_n =Q$, then going-down tells you that there exists a chain $Q_1 \subset \cdots \subset Q_n = Q$ of primes in $B$ with $Q_i$ lying over $P_i$. In particular, there is a chain of $B$, ending at $Q$, with the same length. So $\operatorname{Dim} A_P \leq \operatorname{Dim} B_Q$.