# Is height preserved in a normalization?

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]

• 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://… – user111492 May 20 '18 at 18:09

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 module-finite 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 Eakin-Nagata 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 two-dimensional 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.