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.

Is it known whether or not we can omit the noetherian hypothesis in the above result?


1 Answer 1


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.

  • 1
    $\begingroup$ Would you start stating what you're going to prove? $\endgroup$
    – YCor
    Apr 28, 2019 at 18:40
  • $\begingroup$ Dear @darx, thanks a lot. I still have to check some of the details, but this looks good! $\endgroup$ Apr 29, 2019 at 7:53
  • $\begingroup$ @darx: Ok - checked the details, and everything is fine. Thanks again. (Well, it is not clear to me that $E$ equals the set of primes strictly between $\mathfrak{q}$ and $\mathfrak{q}'$, but as we only need that $E$ is contained in this set, it does not matter.) $\endgroup$ May 1, 2019 at 9:28

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