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]

1$\begingroup$ I just wanted to comment that the question you linked has an affirmative answer for any Noetherian domain. This is Lemma 4.7 in ; On finite generation of $R$subalgebras of $R[X]$ , Amartya K. Dutta; Nobuharu Onoda; Journal of Algebra 320 (2008) 57 80. google.co.in/url?sa=t&source=web&rct=j&url=http://… $\endgroup$ – user111492 May 20 '18 at 18:09
The answer for a general commutative Noetherian domain is "no". The following is adapted from a paper I wrote with Jay Shapiro a few years back.
Consider Nagata's book Local Rings, Appendix, Example 2 (see also Stacks Project, tag 02JE). I'm changing notation and specializing a bit (by setting his parameter $m$ to $0$), but it's basically like this. He first constructs $S$, a normal Noetherian domain with exactly two maximal ideals ${\frak m}_1, {\frak m}_2$, such that height$({\frak m}_i)=i$ for $i=1,2$, and a field $k$ along with an injective ring map $k \hookrightarrow S$ such that the composite maps $k \rightarrow S/{\frak m}_1$, $k \rightarrow S/{\frak m}_2$ are isomorphisms. (I think this is already impossible for excellent $k$algebras.) Then let $J := {\frak m}_1 \cap {\frak m}_2$ and $R := k+J$. Then $S$ is modulefinite over $R$ because of how pullbacks of rings work, and it is elementary that their fraction fields coincide, so $S=\tilde R$, and by the EakinNagata theorem $R$ is Noetherian.
But then ${\frak m}_1 \cap R = J$, which has height 2 because it is the unique maximal ideal of the twodimensional local ring $R$, even though ${\frak m}_1$ has height 1.
On the other hand, if $R$ is universally catenary, then the answer is "yes": height is preserved when contracting primes from $\tilde R$ to $R$. This follows from two results of Ratliff. First, use [Notes on three integral dependence theorems, J. Algebra, 1980, Corollary 2.5], which says that all we need to check is that for any $f\in \widetilde{R[X]}$, height is preserved when contracting primes from $R[X,f]$ to $R[X]$. But since $R[X]$ is universally catenary, Ratliff's theorem on (given as Theorem 15.6 in Matsumura's Commutative Ring Theory) says that the Dimension Formula (the one that involves transcendence degrees) holds between $R[X]$ and $R[X,f]$ (since $R[X,f]$ is finitely generated as an $R$algebra). Since both of the transcendence degrees involved are then 0, it follows that contraction of primes between these two rings preserves height.

2$\begingroup$ Thank you for solving my problem Neil. I find it puzzling that a naïve question about a basic notion, which could have been asked and solved 60 years ago, is not evoked in a single book on commutative algebra or algebraic geometry that I know of. $\endgroup$ – Georges Elencwajg Feb 9 '18 at 17:00

2$\begingroup$ @GeorgesElencwajg Yes. And it's probably a question Ratliff in particular has known the answer to for 40 years, if he ever thought about it. But most general algebraists probably just assume it's false, whereas people who are happy to assume everything is excellent (e.g. f.g. over a field) just assume it's true in good situations, and everyone is satisfied. But my coauthor and I really needed the answer, in particular in the case of height 1, in our initial study on perinormality. So we found it. $\endgroup$ – Neil Epstein Feb 9 '18 at 19:01