Let $R$ be a domain and $\tilde R$ its integral closure in its fraction field: $R\subset \tilde R\subset Frac(R)$.
Is it true that a prime ideal $ \tilde {\mathfrak p} \subset \tilde R$ and its trace $\mathfrak p= \tilde {\mathfrak p}\cap R\subset R$ are related by the equality of heights $$ ht(\tilde {\mathfrak p})=ht(\mathfrak p)\quad (?)$$ This is true for example if $R$ is finitely generated over a field, since then we have the relation $$\operatorname {dim }(R)=\operatorname {dim }(R/\mathfrak p)+ht(\mathfrak p)$$ and the similar relation $$\operatorname {dim} (\tilde {R})=\operatorname {dim }(\tilde R/\tilde {\mathfrak p})+ht(\tilde {\mathfrak p})$$ Since dimension is conserved in integral ring extensions we have $$\operatorname {dim }(\tilde R)=\operatorname {dim }(R)\quad, \quad \operatorname {dim }(\tilde R/\tilde {\mathfrak p})=\operatorname {dim }(R/\mathfrak p)$$ from which the questioned equality $ ht(\tilde {\mathfrak p})=ht(\mathfrak p)$ follows.
But is the equality $(?)$ true in general, i.e. without the hypothesis of finite generation over a field?
[The motivation for my question comes in part from this answer and the comments it provoked]