This question asks for the necessity of a noetherian hypothesis in a certain relation between properties of rings concerning chains of prime ideals. We use the following definitions.


1. A ring $R$ is called **universally catenary** if every $R$-algebra of finite type is catenary. (Note that $R$ need not be noetherian.)
2. A ring $R$ is called a **going-between ring** if for every *integral* ring extension $R\subseteq S$, every saturated chain of primes in $S$ contracts to a saturated chain of primes in $R$.


From results by Ratliff we get the following.


**Theorem:** Every noetherian universally catenary ring is a going-between ring.


The proof is rather complicated, as it relies on several non-trivial results concerning relations between different chain conditions. In particular, it is not clear to me whether one can get through without noetherianness.



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Is every universally catenary ring a going-between ring?
OK, let $R \subset S$ be an integral ring extension with $R$ universally catenary. Let $\mathfrak q \subset \mathfrak q'$ be primes in $S$ such that there is no prime strictly in between them. We have to show that the same is true for the corresponding primes $\mathfrak p \subset \mathfrak p'$ of $R$.


Reduction to the finite case. Suppose to the contrary that $\mathfrak p''$ is strictly in between. Write $S$ as the filtered colimit $S = \text{colim} S\_i$ of its finite sub $R$-algebras. Denote $E\_i$ the set of primes of $S\_i$ which lie over $\mathfrak p''$ and are strictly in between $\mathfrak q \cap S\_i$ and $\mathfrak q' \cap S\_i$. If we know the result in the finite case then $E\_i$ is nonempty. The inclusions between the rings $S\_i$ define transition maps between the sets $E\_i$ (this is a standard fact about finite ring maps). Now $E = \text{lim} E\_i$ is nonempty as a cofiltered limit of finite nonempty sets. An element of $E$ is the same thing as a prime of $S$ strictly between $\mathfrak q$ and $\mathfrak q'$.


Finite case. Choose a surjection $P = R[x\_1, \ldots, x\_n] \to S$. The primes $\mathfrak q \subset \mathfrak q'$ correspond to primes $\mathfrak r \subset \mathfrak r'$ of $P$ with no prime strictly in between. Note that $\mathfrak r$ defines a closed point of the fibre of $\text{Spec}(P) \to \text{Spec}(R)$ over $\mathfrak p$. Hence there is a length $n$ saturated chain of primes starting with $\mathfrak p P$ and ending with $\mathfrak r$. Thus we have a saturated chain of length $n + 1$ between $\mathfrak p P$ and $\mathfrak r'$.


But if we have $\mathfrak p''$ as above, then we can make a chain longer than this (and arrive at a contradiction). Namely, we can first consider $\mathfrak p P \subset \mathfrak p'' P \subset \mathfrak p' P$ and then make a chain between $\mathfrak p' P$ and $\mathfrak r'$ in the fibre as before.


