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.