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]