(Krull) dimension of dense open subset of finite type algebra over a domain

Question

Let $D\to A$ be a finite type algebra with $D$ a domain. Suppose $V\subset \operatorname{Spec}A$ is open and dense. Is it true that $\dim V=\dim A$?

I know that if $X\to \operatorname{Spec}\Bbbk$ is an integral scheme of finite type over a field then for any non-empty open $U\subset X$ we have $\dim U=\dim X$. Is this possible to globalize to domains?

Answer 1

No. For instance, let $D=A=\mathbb{Z}_p$ (or any other DVR). Then the open set $V\subset\operatorname{Spec} A$ where $p$ does not vanish is open and dense, but $V=\operatorname{Spec}\mathbb{Q}_p$ so $\dim V=0\neq 1=\dim A$.