Catenarity of Nagata Rings in Pullback Construction: A Characterization Approach

Abstract

In this paper, we investigate the catenarity of Nagata rings arising from pullback constructions of type $(T,I,D)$, where I is the intersection of finitely many maximal ideals of T. Assuming that D is a Prüfer domain, T is universally catenarian, and the residue field extensions satisfy suitable transcendence degree conditions, we establish necessary and sufficient conditions for the Nagata rings $R\left(n\right)$ to be catenarian for all positive integers n. Our characterization extends previous results obtained for pullbacks defined by a single maximal ideal and provides a unified treatment of a broader class of pullback constructions. As applications, we exhibit examples highlighting the distinction between the catenarity of Nagata rings and that of the corresponding Serre conjecture rings, and we provide an affirmative answer to an open question from the literature.

1. Introduction

The study of catenarity occupies a central position in dimension theory. Since the pioneering work of Nagata [1], considerable effort has been devoted to understanding the behavior of catenarity under polynomial extensions and related constructions. Among these, Nagata rings have proved to be particularly effective tools, as they often preserve significant dimension-theoretic properties while providing a more tractable description of the prime spectrum. Important contributions in this direction include [1,2,3,4,5,6,7], where various aspects of the spectrum and dimension theory of Nagata rings were investigated.

A major step toward understanding catenarity in Nagata rings associated with pullback constructions was achieved in [5]. There, Ben Nasr, Jarboui, and Zeyada introduced two conditions, denoted ${(\ast )}_n$ and ${\left(\mathfrak{J}\right)}_n$ (cf. [5], Definition 3), which govern the behavior of heights of prime ideals in $R\left[n\right]$ and their relationship with the corresponding primes of $T\left[n\right]$. They established that these conditions are intimately connected with the catenarity of the Nagata ring $R\left(n\right)$. In particular, the catenarity of $R\left(n\right)$ was shown to imply a precise additivity formula for heights in $R\left[n\right]$, yielding condition ${(\ast )}_n$ (see [5], Lemma 1). Under additional hypotheses involving prime lifting and transcendence degree conditions, condition ${\left(\mathfrak{J}\right)}_n$ was also obtained, leading to a characterization of the catenarity of $R\left(n\right)$ in terms of the catenarity of $D\left(n\right)$ together with structural properties of the underlying pullback construction (cf. [5], Theorem 2 and Corollary 7).

Despite these advances, the scope of the existing theory remains restricted. In particular, conditions ${(\ast )}_n$ and ${\left(\mathfrak{J}\right)}_n$ have not yet been fully exploited, and their consequences for broader classes of pullback constructions remain largely unexplored. Moreover, the catenarity of Nagata rings associated with general $(T,I,D)$-constructions, in the sense of Cahen [6,7], has not been systematically investigated.

The primary objective of this paper is to fill this gap. More precisely, we study the catenarity of Nagata rings $R\left(n\right)$ arising from a $(T,I,D)$-construction (or a pullback diagram of type $\left({\square }_{\cap }\right)$) in which I is the intersection of finitely many maximal ideals of T, including the classical case where I is a single maximal ideal (see Theorems 1 and 2). Our results extend the framework developed in [5], provide new criteria for the catenarity of $R\left(n\right)$, and considerably enlarge the class of pullback constructions for which the catenarity problem can be completely analyzed. As a further application, Theorem 1 provides an affirmative answer to [8] (Question 1).

At the same time, strong S-domains and stably strong S-domains, introduced by Kaplansky and further developed by Malik and Mott, Kabbaj, Cahen, Ayache, Jarboui and others [9,10,11,12,13,14,15,16,17], provide a natural framework for studying chains of prime ideals in polynomial rings. Notably, Noetherian rings and Prüfer domains belong to this class, and many questions concerning catenarity have been formulated in this setting.

An intriguing phenomenon was discovered by Yengui in [18,19]. In particular, ref. [19] (the second example) exhibits a three-dimensional stably strong S-domain R such that the Nagata ring $R\left(n\right)$ is catenarian for every positive integer n, whereas the Serre conjecture ring $R⟨n⟩$ is not catenarian for any positive integer n. Although this example shows a striking discrepancy between Nagata rings and Serre conjecture rings, the underlying structural reason remained unclear.

The second purpose of this paper is to provide a general theoretical explanation of this phenomenon and to place it within a unified framework based on pullback constructions of type $\left({\square }_{\cap }\right)$ (see Theorem 3). These constructions have proved to be particularly effective in the study of catenarity and dimension theory [6,7,10,12,13,20,21].

We begin by establishing a transfer result for catenarity through Nagata extensions. We prove that if A is a strong S-domain and its Nagata ring $A\left(1\right)$ is catenarian, then A is catenarian (Lemma 2). This provides a useful tool for detecting catenarity from Nagata extensions.

Next, under suitable hypotheses involving Prüfer domains, universally catenarian overrings, and algebraic residual extensions, we obtain a complete characterization of the catenarity of Nagata rings arising from pullback diagrams of type $\left({\square }_{\cap }\right)$. In Theorem 2, we show that $R\left(1\right)$ is catenarian if and only if $R\left(n\right)$ is catenarian for every $n\ge 1$, and that this property can be described entirely in terms of the heights of maximal ideals in a distinguished subset $\mathsf{\Omega }\subseteq Max\left(T\right)$.

Our main result, Theorem 3, provides necessary and sufficient conditions for the phenomenon that all Nagata rings $R\left(n\right)$ are catenarian while all Serre conjecture rings $R⟨n⟩$ fail to be catenarian. We prove that this occurs precisely when $\mathsf{\Omega }$ contains both low and high maximal ideals and all high maximal ideals have the same height equal to ${ht}_R\left(I\right)$.

These results show that the phenomenon observed in [19] is not accidental, but rather reflects a general structural behavior governed by the configuration of maximal ideals in the pullback diagram. As an application, we construct for every integer $k\ge 2$ a k-dimensional stably strong S-domain R such that $R\left(n\right)$ is catenarian for all $n\ge 1$, whereas $R⟨n⟩$ is not catenarian for any $n\ge 1$ (Example 1). This significantly extends the known examples and shows that such behavior occurs in arbitrary dimensions.

Finally, we fix some notation. The prime spectrum (resp. maximal spectrum) of a ring A is denoted by $Spec\left(A\right)$ (resp. $Max\left(A\right)$). If $A\subseteq B$ are domains, then $tr.deg[B:A]$ denotes the transcendence degree of the quotient field of B over that of A. Let $\mathfrak{p}$ be a prime ideal of a ring A, and let n be a nonnegative integer (or $n=\infty$). We denote by $\mathfrak{p}\left[n\right]$ the extension of $\mathfrak{p}$ to $A\left[n\right]$, that is, the set of polynomials with coefficients in $\mathfrak{p}$. Moreover, $\mathfrak{p}\left(n\right)$ (resp., $\mathfrak{p}⟨n⟩$) denotes the localization of $\mathfrak{p}$ in $A\left(n\right)$ (resp., $A⟨n⟩$). We write $ht\left(\mathfrak{p}\right)$ for the height of $\mathfrak{p}$, and, following [22], define the valuative height of $\mathfrak{p}$, denoted by ${ht}_v\left(\mathfrak{p}\right)$, to be the valuative dimension of the localization $A_{\mathfrak{p}}$.

2. Preliminaries and Background

We begin by recalling several notions and constructions that will be used throughout the paper.

An integral domain A is called an S-domain if, for every height-one prime ideal p of A, the extended ideal $p\left[X\right]$ has height one in $A\left[X\right]$. A ring A is called a strong S-ring if the quotient $A/p$ is an S-domain for every prime ideal p of A. Equivalently, for every pair of consecutive prime ideals $p\subset q$ in A, the extended ideals $p\left[X\right]\subset q\left[X\right]$ remain consecutive in $A\left[X\right]$ (see [16,17]). Strong S-rings are stable under localization and passage to factor rings, but they are not, in general, preserved under polynomial extensions. A ring A is called stably strong S if $A\left[n\right]$ is a strong S-ring for every integer $n\ge 1$. Classical examples include Noetherian rings and Prüfer domains. These classes have been widely studied and their structural properties are well understood (see, for instance, [10,12,14,15,16,17,23,24,25,26,27,28]).

A domain A is said to be catenarian if, for every pair of prime ideals $p\subset q$, all saturated chains of prime ideals between p and q have the same finite length. Equivalently, the additivity of heights holds for every pair $p\subset q$ of consecutive prime ideals:

$$ht\left(q\right)=ht\left(p\right)+1<\infty .$$

A ring A is called universally catenarian if $A\left[n\right]$ is catenarian for any positive integer n (see, for instance, [10]).

Among the various extensions of polynomial rings, two constructions play a fundamental role in both algebraic and geometric contexts: the Serre conjecture ring and the Nagata ring.

Let U denote the multiplicative set of monic polynomials in $A\left[X\right]$. The Serre conjecture ring is defined by the localization

$$A⟨X⟩:=U^{-1}A\left[X\right],$$

which was introduced by Quillen in his proof of Serre’s conjecture on projective modules [29]. This construction has since become an important tool in the study of projective modules and homological properties of non-Noetherian rings (see, e.g., [30,31,32,33]). Iteratively, one defines

$$A⟨X_1,\dots ,X_n⟩:=A⟨X_1,\dots ,X_{n-1}⟩⟨X_n⟩.$$

Equivalently, $A⟨n⟩$ can be viewed as the localization of the polynomial ring $A\left[n\right]=A[X_1,\dots ,X_n]$ at the multiplicative set

$$U_n=\{f\in A\left[n\right]\mid \mathrm{LC}\left(f\right)=1\},$$

where $\mathrm{LC}\left(f\right)$ denotes the leading coefficient of f with respect to the lexicographic order $X_1<X_2<\cdots <X_n$. Although the order of variables may affect the resulting ring (see [34], Theorem 10), this issue will not play any role in this work, and we simply write $A⟨n⟩$.

In parallel, the Nagata ring is defined as follows. Let S be the set of polynomials in $A\left[X\right]$ whose coefficients generate the unit ideal of A. The localization

$$A\left(X\right):=S^{-1}A\left[X\right]$$

is called the Nagata ring of A in one indeterminate. Inductively, one sets

$$A\left(n\right):=\left(A(X_1,\dots ,X_{n-1})\right)\left(X_n\right).$$

This construction is well behaved with respect to disjoint sets of indeterminates, since

$$A(X\cup Y)=A\left(X\right)\left(Y\right)=A\left(Y\right)\left(X\right)$$

(see [35]). For convenience, we also set $A:=A\left[0\right]:=A\left(0\right):=A⟨0⟩$. It is known that $A\left(n\right)$ is the localization of $A\left[n\right]$ at the multiplicative set

$$S_n=\{f\in A\left[n\right]\mid c\left(f\right)=A\}=A\left[n\right]\setminus \bigcup _{M\in Max\left(A\right)}M\left[n\right],$$

and that we have the inclusions

$$A\left[n\right]\subseteq A⟨n⟩\subseteq A\left(n\right),$$

with $A\left(n\right)$ being a localization of $A⟨n⟩$.

Nagata rings have been extensively studied due to their ability to reflect and preserve important properties of the base ring. For instance, every maximal ideal of $A\left(X\right)$ is of the form $MA\left(X\right)$ for some maximal ideal M of A [1] (6.17), and an integrally closed domain A is Prüfer if and only if $A\left(X\right)$ is Prüfer [36] (Theorem 33.4). Thus, Nagata rings often transform local problems into more global and tractable ones, making them a fundamental tool in dimension theory and multiplicative ideal theory.

Nagata’s classical construction [1] already illustrates the subtle behavior of dimension in pullback settings. It produces a local domain that fails to satisfy the altitude formula by gluing maximal ideals of different heights inside a larger domain. This idea can be formalized using pullback constructions. Concretely, one considers a diagram

$$\begin{array}{ccc}R& ⟶& D\\ \downarrow & & \downarrow \\ T& ⟶& T/I=E\end{array}$$

where T is a domain, I is a nonzero ideal, $\phi :T\to E=T/I$ is the canonical projection, $D\subset E$ is a subdomain, and $R={\phi }^{-1}\left(D\right)$. We write $R=(T,I,D)$ and refer to such a diagram as a pullback of type $(\square )$ following [6,7]. When $I={\bigcap }_{M\in \mathsf{\Omega }}M$ with $\mathsf{\Omega }$ a nonempty finite subset of $Max\left(T\right)$, the diagram is said to be of type $\left({\square }_{\cap }\right)$. Nagata’s original construction fits naturally into this framework.

Throughout the paper, the data $T,D,I,$ and R will always refer to such a pullback diagram unless otherwise specified.

3. Catenarity of Nagata Rings Arising from Pullback Constructions: The Case $\left|\mathsf{\Omega }\right|=\mathbf{1}$

In this section, we investigate the catenarity of Nagata rings associated with pullback diagrams of type $\left({\square }_{\cap }\right)$ in the special case where $\left|\mathsf{\Omega }\right|=1$. Our main result provides a complete characterization of the catenarity of the Nagata ring $R\left(1\right)$ when the coefficient ring D is a field. As an application, it settles positively the open question posed in [8] (Question 1).

For a ring A and an ideal J of A, let

$$D_A\left(J\right)=\{P\in Spec\left(A\right)\mid J⊈P\}.$$

denote the distinguished open subset of $Spec\left(A\right)$ determined by J. If R is a pullback of type $(T,I,D)$, then the canonical map

$$Q⟼Q\cap R$$

induces an order-preserving homeomorphism

$$D_T\left(I\right)\cong D_R\left(I\right)$$

(see [6], Proposition 0). Consequently, it preserves inclusions, chains, and consecutive prime ideals. Moreover, polynomial extensions preserve pullback diagrams, namely, if $R=(T,I,D)$, then

$$R\left[n\right]=\left(T\right[n],I[n],D[n\left]\right).$$

We first recall the following lemma of Cahen [6] (Proposition 4).

Lemma 1.

For a diagram of type $(\square )$, if

$$P_0\subset \cdots \subset P_n$$

is a chain of prime ideals of R such that $P_n$ is minimal among the prime ideals of R containing both I and $P_{n-1}$, then this chain lifts to a chain of prime ideals of T.

These fundamental facts will be used repeatedly throughout the remainder of the paper.

Theorem 1.

Let R be a $(T,M,D)$-construction ring, where T is a Prüfer domain, M is a maximal ideal of T, and D is a subfield of the residue field $K:=T/M$. Then, the following statements are equivalent:

(1) 

The Nagata ring $R\left(1\right)$ is catenarian.

(2) 

The Serre conjecture ring $R⟨1⟩$ is catenarian.

(3) 

Either $\mathrm{tr}.\mathrm{deg}[K:D]=0$ or ${\mathrm{ht}}_{R\left[1\right]}\left(M\left[1\right]\right)={\mathrm{ht}}_T\left(M\right)+1$.

Proof. 

(1)⇒(2). Let $P^{\prime }\subset P$ be two consecutive prime ideals of $R\left[1\right]$ such that P survives in $R⟨1⟩$. Our task is to show that $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(P^{\prime }\right)+1$. Assume first that $P⊉M\left[1\right]$. By [6] (Proposition 0), the chain $P^{\prime }\subset P$ lifts to a saturated chain $Q^{\prime }\subset Q$ of prime ideals of $T\left[1\right]$. Since $T\left[1\right]$ is catenarian, we have $\mathrm{ht}\left(Q\right)=\mathrm{ht}\left(Q^{\prime }\right)+1.$

Moreover, [6] (Proposition 0) yields $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(Q\right)$ and $\mathrm{ht}\left(P^{\prime }\right)=\mathrm{ht}\left(Q^{\prime }\right).$ Hence, $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(P^{\prime }\right)+1.$ Next, assume that $M\left[1\right]\subseteq P^{\prime }$. Then $P^{\prime }/M\left[1\right]$ and $P/M\left[1\right]$ are consecutive prime ideals of $D\left[1\right]$. Furthermore, by [9] (Lemma 2), the prime ideal $P/M\left[1\right]$ survives in $D⟨1⟩$. Since $D⟨1⟩$ is catenarian, we obtain $\mathrm{ht}\left(P/M\left[1\right]\right)=\mathrm{ht}\left(P^{\prime }/M\left[1\right]\right)+1.$ Applying [21] (Lemma 2.5), we deduce that $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(P^{\prime }\right)+1.$

Finally, assume $P\supseteq M\left[1\right]$ and $P^{\prime }⊉M\left[1\right].$ Since D is a field and P survives in $R⟨1⟩$, necessarily, $P=M\left[1\right]$. Indeed, $M⟨1⟩$ is a maximal ideal of $R⟨1⟩$, and

$$M⟨1⟩\subseteq U^{-1}P,U=\{f\in R\left[1\right]\mid LC\left(f\right)=1\}.$$

Hence,

$$M⟨1⟩=U^{-1}P;$$

or equivalently, $U^{-1}M\left[1\right]=U^{-1}P$.

By the correspondence,

$$\{Q\in Spec\left(R\left[1\right]\right)\mid Q\cap U=⌀\}⟷Spec\left(R⟨1⟩\right),Q⟼U^{-1}Q.$$

we deduce that $P=M\left[1\right]$. It follows that P survives in $R\left(1\right)$. As $R\left(1\right)$ is catenarian, $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(P^{\prime }\right)+1$.

(2)⇒(3). Assume that $tr.deg[K:D]\ne 0$. We aim to show that ${\mathrm{ht}}_{R\left[1\right]}\left(M\left[1\right]\right)={\mathrm{ht}}_T\left(M\right)+1$. Let $a\in T$ be such that $\overline{a}=a+M$ is algebraically independent over D in $T/M$. Let $q^{\prime }\in Spec\left(T\right)$ satisfy $q^{\prime }\subset M$, and set $P^{\prime }=\left(q^{\prime }\left[1\right]+(X-a)T\left[1\right]\right)\cap R\left[1\right]$ and $P=M\left[1\right]$. It is straightforward to check that $P^{\prime }$ and P are prime ideals of $R\left[1\right]$, and that $P^{\prime }\subset P$ (see [8], Lemma 1). They are not necessarily consecutive, so let $P_1$ be maximal among the prime ideals of $R\left[1\right]$ such that $P^{\prime }\subseteq P_1\subset P$ and $P_1\neg \supseteq M\left[1\right]$. Then, $P_1\subset P$ are consecutive prime ideals of $R\left[1\right]$. By [6] (Proposition 4), the chain $P^{\prime }\subseteq P_1\subset P$ lifts to a chain in $T\left[1\right]$:

$$Q^{\prime }=q^{\prime }\left[1\right]+(X-a)T\left[1\right]\subseteq Q_1\subset Q.$$

Since $Q_1\subset Q$ are consecutive, and $M\subseteq Q\cap T$ with M maximal in T, we obtain $Q\cap T=M$. Moreover, $X-a\in Q$, hence $M\left[1\right]\subset Q$, and therefore $\mathrm{ht}\left(Q\right)={\mathrm{ht}}_{T\left[1\right]}\left(M\left[1\right]\right)+1.$ Since T is a strong S-domain (in fact any Prüfer domain is a strong S-domain according to [17] (Theorem 3.5)), ${\mathrm{ht}}_{T\left[1\right]}\left(M\left[1\right]\right)={\mathrm{ht}}_T\left(M\right)$, so $\mathrm{ht}\left(Q\right)={\mathrm{ht}}_T\left(M\right)+1.$

Because both $R⟨1⟩$ and $T\left[1\right]$ are catenarian, we have $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(P_1\right)+1$ and $\mathrm{ht}\left(Q\right)=\mathrm{ht}\left(Q_1\right)+1$. Using [6] (Proposition 0), we obtain $\mathrm{ht}\left(P_1\right)=\mathrm{ht}\left(Q_1\right)$, hence $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(Q\right)$. This yields

$${\mathrm{ht}}_{R\left[1\right]}\left(M\left[1\right]\right)={\mathrm{ht}}_T\left(M\right)+1.$$

(3)⇒(1). If $\mathrm{tr}.\mathrm{deg}[T/M:D]=0$, then R is universally catenarian by [10] (Corollary 2.3). In particular, the Nagata ring $R\left(1\right)$ is catenarian. One can also apply directly [5] (Corollary 8). Suppose now that ${\mathrm{ht}}_{R\left[1\right]}\left(M\left[1\right]\right)={\mathrm{ht}}_T\left(M\right)+1$. Let $P^{\prime }\subset P$ be two consecutive prime ideals of $R\left[1\right]$ such that P survives in $R\left(1\right)$. Since P survives in $R\left(1\right)$, there exists a maximal ideal $M^{\prime }$ of R such that $P\subseteq M^{\prime }\left[1\right]$. We shall show that $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(P^{\prime }\right)+1$. The case $P^{\prime }\supseteq M\left[1\right]$ cannot occur. Indeed, in this situation, one has $M\left[1\right]\subseteq P^{\prime }\subset P\subseteq M^{\prime }\left[1\right]$. Since M is a maximal ideal of R, it follows that $M^{\prime }=M$, which contradicts the fact that $P^{\prime }$ is properly contained in P. If $P⊉M\left[1\right]$, then the desired conclusion follows exactly as in the proof of (1)⇒(2). Therefore, it remains to consider the case where $P\supseteq M\left[1\right]$ and $P^{\prime }⊉M\left[1\right]$. Since $M\left[1\right]\subseteq P\subseteq M^{\prime }\left[1\right]$ and M is maximal in R, we necessarily have $M^{\prime }=M$. Consequently $M\left[1\right]\subseteq P\subseteq M\left[1\right]$, and hence $P=M\left[1\right]$. By [6] (Proposition 4), the chain $P^{\prime }\subset P$ lifts to a chain $Q^{\prime }\subset Q$ in $T\left[1\right]$, where Q is minimal among primes containing both $Q^{\prime }$ and $M\left[1\right]$. Hence, $Q^{\prime }\subset Q$ is a saturated chain, so $\mathrm{ht}\left(Q\right)=\mathrm{ht}\left(Q^{\prime }\right)+1$. Again by [6] (Proposition 0), $\mathrm{ht}\left(Q^{\prime }\right)=\mathrm{ht}\left(P^{\prime }\right)$, so it remains to prove $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(Q\right)$. Assume that $M\left[1\right]=Q$. Since T is a Prüfer domain, we have $Q^{\prime }=q^{\prime }\left[1\right]$ for some prime ideal $q^{\prime }$ of T. Consequently, $P^{\prime }=p^{\prime }\left[1\right]$ where $p^{\prime }=P^{\prime }\cap R$. We claim that $tr.deg[K:D]\ne 0$. Suppose, to the contrary, that $tr.deg[K:D]=0$. As $R_M:=(T_M,M{T}_M,\mathrm{qf}\left(D\right))=(T_M,M{T}_M,D)$, we get, from [24] (Proposition 2.5), ${\dim }_v\left(R_M\right)={\dim }_v\left(T_M\right)+tr.deg[K:D]={\dim }_v\left(T_M\right)$; or equivalently, ${\mathrm{ht}}_{v_R}\left(M\right)={\mathrm{ht}}_{v_T}\left(M\right)={\mathrm{ht}}_T\left(M\right)$. This implies ${\mathrm{ht}}_{R\left[1\right]}\left(M\left[1\right]\right)\le {\mathrm{ht}}_{v_R}\left(M\right)={\mathrm{ht}}_T\left(M\right)$, contradicting the assumption that ${\mathrm{ht}}_{R\left[1\right]}\left(M\left[1\right]\right)={\mathrm{ht}}_T\left(M\right)+1$. Thus, $tr.deg[K:D]\ne 0$. Choose $a\in T$ whose image in $K=T/M$ is transcendental over D. By [8] (Lemma 1), we obtain the strict chain $P^{\prime }=p^{\prime }\left[1\right]\subset (q^{\prime }\left[1\right]+(X-a)T\left[1\right])\cap R\left[1\right]\subset P=M\left[1\right]$, which contradicts the fact that $P^{\prime }$ and P are consecutive prime ideals. Therefore, $M\left[1\right]\subset Q$. Consequently,

$$\mathrm{ht}\left(Q\right)={\mathrm{ht}}_T\left(M\right)+1={\mathrm{ht}}_{R\left[1\right]}\left(M\left[1\right]\right)=\mathrm{ht}\left(P\right).$$

Therefore, $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(P^{\prime }\right)+1$. This completes the proof.    □

The preceding theorem completely settles the case where D is a field. Nevertheless, a satisfactory characterization remains elusive in the general setting. Indeed, when D is not a field, the behavior of the Nagata rings $R\left(n\right)$ appears to be substantially more intricate. We therefore conclude this section with the following question.

Question 1.

Let $R=(T,M,D)$, where M is a maximal ideal of T and D is not a field. Under what conditions is the Nagata ring $R\left(1\right)$ catenarian? More generally, characterize the catenarity of $R\left(n\right)$ for an arbitrary positive integer n.

4. Catenarity of Nagata Rings Arising from Pullback Constructions: The Case $\left|\mathsf{\Omega }\right|\ge \mathbf{2}$

The main purpose of this section is to characterize the catenarity of Nagata rings associated with pullback diagrams of type $\left({\square }_{\cap }\right)$ in the algebraic residue-field extension case. Assuming that D is a Prüfer domain, that T is universally catenarian, and that $tr.deg[T/M:D]=0$ for every maximal ideal $M\in \mathsf{\Omega }$, we prove that the catenarity of $R\left(n\right)$ for all positive integers n is equivalent to the catenarity of $R\left(1\right)$. We further express these equivalent conditions in terms of the heights of the high maximal ideals of $\mathsf{\Omega }$, yielding a complete and explicit criterion.

We begin by proving a general result that will play a key role in the sequel.

Lemma 2.

Let A be a strong S-domain. If the Nagata ring $A\left(1\right)$ is catenarian, then A is catenarian.

Proof. 

Let $p^{\prime }\subset p$ be two consecutive prime ideals of A. Since A is a strong S-domain, their extensions $p^{\prime }\left[1\right]\subset p\left[1\right]$ form a pair of consecutive prime ideals in $A\left[1\right]$. Moreover, the prime ideal $p\left[1\right]$ survives in the Nagata ring $A\left(1\right)$, because $p\left[1\right]\subseteq M\left[1\right]$ for some maximal ideal M of A. Since $A\left(1\right)$ is catenarian and $p^{\prime }\left[1\right]\subset p\left[1\right]$ are consecutive prime ideals, the additivity of heights yields

$$ht\left(p\left[1\right]\right)=ht\left(p^{\prime }\left[1\right]\right)+1.$$

Using the fact that heights are preserved under polynomial extension in strong S-domains (cf. [15], Theorem 2.4 and [17]), that is,

$$ht\left(q\right)=ht\left(q\right[1\left]\right)\mathrm{for}\mathrm{all}q\in Spec\left(A\right),$$

we immediately obtain

$$ht\left(p\right)=ht\left(p^{\prime }\right)+1.$$

Therefore, A is catenarian.    □

In the sequel, we assume that the pullback diagram is of type $\left({\square }_{\cap }\right)$, with $\left|\mathsf{\Omega }\right|>1$, where $\mathsf{\Omega }$ is the finite set of maximal ideals of T such that $I={\bigcap }_{M\in \mathsf{\Omega }}M$. In this context, for each $M\in \mathsf{\Omega }$, one has $M\cap R=I$. Indeed, since

$$T/I\cong \prod _{M\in \mathsf{\Omega }}T/M,$$

the ideal $M/I$ corresponds to a maximal ideal of $T/I$ and contracts to $\left(0\right)$ in $D\cong R/I$, showing that each M lies over I in a completely unramified manner.

Finally, recall that a maximal ideal is called low if its height is one, and high if its height is at least two.

Theorem 2.

Consider a pullback diagram of type $\left({\square }_{\cap }\right)$. Assume that D is a Prüfer domain, that T is universally catenarian, and that

$$tr.deg[T/M:D]=0$$

for every maximal ideal $M\in \mathsf{\Omega }$. Then, the following statements are equivalent:

(a) 

$R\left(n\right)$ is catenarian for every positive integer n;

(b) 

$R\left(1\right)$ is catenarian;

(c) 

Either all maximal ideals of Ω have the same height, or Ω contains both low and high maximal ideals, in which case every high maximal ideal has height ${ht}_R\left(I\right)$.

Proof. 

Under the hypotheses, R is a stably strong S-domain by [12] (Theorem 2.6). In particular, R is a strong S-domain.

• $\left(a\right)\Rightarrow \left(b\right)$. This is immediate.
• $\left(b\right)\Rightarrow \left(c\right)$. Assume that $R\left(1\right)$ is catenarian. Since R is a strong S-domain, Lemma 2 yields that R is catenarian. If every maximal ideal of $\mathsf{\Omega }$ is low, then condition $\left(c\right)$ is satisfied. Otherwise, $\mathsf{\Omega }$ contains at least one high maximal ideal. Since R is catenarian, all high maximal ideals of $\mathsf{\Omega }$ have the same height, namely, ${ht}_R\left(I\right)$ (see the third line of the proof of [37] (Proposition 2)). Hence, if all maximal ideals of $\mathsf{\Omega }$ are high, they necessarily have the same height; whereas if $\mathsf{\Omega }$ contains both low and high maximal ideals, then all its high maximal ideals have height ${ht}_R\left(I\right)$. Therefore, condition $\left(c\right)$ holds.
• $\left(c\right)\Rightarrow \left(a\right)$. Assume that condition $\left(c\right)$ holds.

If all maximal ideals of $\mathsf{\Omega }$ have the same height, then R is universally catenarian by [37] (Proposition 2). Consequently, $R\left(n\right)$ is catenarian for every positive integer n.

Assume now that $\mathsf{\Omega }$ contains both low and high maximal ideals and that every high maximal ideal has height ${ht}_R\left(I\right)$. We prove that $R\left(n\right)$ is catenarian.

Let $P^{\prime }\subset P$ be consecutive prime ideals of $R\left[n\right]$ such that P survives in $R\left(n\right)$.

If $P^{\prime }\supseteq I\left[n\right]$, then since $D\left(n\right)$ is catenarian, we get

$$ht(P/I\left[n\right])=ht(P^{\prime }/I\left[n\right])+1.$$

By [21] (Lemma 2.5), it follows that

$$ht\left(P\right)=ht\left(P^{\prime }\right)+1.$$

Assume next that $P⊉I\left[n\right]$. Then, the chain $P^{\prime }\subset P$ lifts to a saturated chain $Q^{\prime }\subset Q$ in $T\left[n\right]$. Since T is universally catenarian, the polynomial ring $T\left[n\right]$ is catenarian. Therefore,

$$ht\left(Q\right)=ht\left(Q^{\prime }\right)+1.$$

Since heights are preserved under the lifting correspondence, we obtain

$$ht\left(P\right)=ht\left(P^{\prime }\right)+1.$$

Finally, suppose that $P^{\prime }⊉I\left[n\right]$ and $P\supseteq I\left[n\right].$ Let $Q^{\prime }\subset Q$ be the corresponding lifted chain in $T\left[n\right]$.

Note that $R\left[n\right]/I\left[n\right]\cong D\left[n\right]$ is a domain; hence, $I\left[n\right]$ is a prime ideal of $R\left[n\right]$. Under the assumptions $tr.deg[T/M:D]=0$ and D is a Prüfer domain, every prime ideal of $T\left[n\right]$ lying over $I\left[n\right]$ is equal to $M\left[n\right]$ (cf. [6] (Proposition 0) and [5]). Therefore, these assumptions ensure that $Q=M\left[n\right]$ and $P=I\left[n\right]$. Without the condition $tr.deg[T/M:D]=0$, this characterization no longer follows, leaving the problem open (Question 2).

If M is low, then $ht\left(Q\right)=1$. Hence, $Q^{\prime }=\left(0\right)$ and therefore $P^{\prime }=\left(0\right)$. Consequently,

$$ht\left(P\right)=1=ht\left(P^{\prime }\right)+1.$$

Assume now that M is high. Since all high maximal ideals of $\mathsf{\Omega }$ have height ${ht}_R\left(I\right)$,

$$ht\left(M\right)={ht}_R\left(I\right).$$

Therefore,

$$ht\left(P\right)={ht}_{R\left[n\right]}\left(I\left[n\right]\right)={ht}_R\left(I\right)=ht\left(M\right)=ht\left(Q\right).$$

Since $Q^{\prime }\subset Q$ is a saturated chain in the catenarian ring $T\left[n\right]$,

$$ht\left(Q\right)=ht\left(Q^{\prime }\right)+1.$$

It follows that

$$ht\left(P\right)=ht\left(P^{\prime }\right)+1.$$

Therefore, $R\left(n\right)$ is catenarian for every positive integer n.    □

Theorem 2 yields a complete characterization of the catenarity of the Nagata rings $R\left(n\right)$ under the assumption that $tr.deg[T/M:D]=0$ for every maximal ideal $M\in \mathsf{\Omega }$. This hypothesis plays a crucial role in the proof, as it ensures a precise control of heights in polynomial and Nagata extensions. In contrast, when positive transcendence degree occurs, the techniques developed here no longer apply, and the catenarity of the rings $R\left(n\right)$ remains largely unexplored. This naturally leads to the following question.

Question 2.

Assume that there exists a maximal ideal $M\in \mathsf{\Omega }$ such that $tr.deg[T/M:D]>0$. Can one characterize the catenarity of $R\left(1\right)$, or more generally of $R\left(n\right)$ for every integer $n\ge 1$?

Theorem 3.

Consider a diagram of type $\left({\square }_{\cap }\right)$ and assume that D is a Prüfer domain and that T is universally catenarian. Suppose further that $tr.deg[T/M:D]=0$ for every maximal ideal $M\in \mathsf{\Omega }$. Then, the following conditions are equivalent:

(a) 

$R\left(1\right)$ is catenarian, but $R⟨1⟩$ is not.

(b) 

$R\left(n\right)$ is catenarian for every positive integer n, but $R⟨1⟩$ is not.

(c) 

$R\left(n\right)$ is catenarian for every positive integer n, but $R⟨n⟩$ is not.

(d) 

$R\left(1\right)$ is catenarian, but $R⟨n⟩$ is not for every positive integer n.

(e) 

D is not a field and the set Ω contains at least one low and one high maximal ideal. Moreover, all high maximal ideals in Ω have the same height, equal to ${ht}_R\left(I\right)$.

Proof. 

(a)⇒(b). Follows from Theorem 2.

• (b)⇒(c). As $R⟨1⟩$ is not catenarian, we can consider the extended primes in $R⟨n⟩$ using that R is a stably strong S-domain. Hence, $R⟨n⟩$ is not catenarian for every $n\ge 1$.
• (c)⇒(d). Trivial.
• (d)⇒(e). We begin by showing that D cannot be a field. Suppose, to the contrary, that D is a field. We shall prove that $R⟨1⟩$ is catenarian, contradicting assertion (d).

Let $P^{\prime }\subset P$ be a pair of consecutive prime ideals of $R\left[1\right]$ such that P survives in $R⟨1⟩$. We distinguish three cases.

Assume first that $P⊉I\left[1\right]$. By [6] (Proposition 0), the chain $P^{\prime }\subset P$ lifts to a saturated chain $Q^{\prime }\subset Q$ of prime ideals of $T\left[1\right]$. Since $T\left[1\right]$ is catenarian, it follows that $\mathrm{ht}\left(Q\right)=\mathrm{ht}\left(Q^{\prime }\right)+1$. Moreover, the same proposition yields $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(Q\right)$ and $\mathrm{ht}\left(P^{\prime }\right)=\mathrm{ht}\left(Q^{\prime }\right)$, whence $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(P^{\prime }\right)+1$.

Next, suppose that $I\left[1\right]\subseteq P^{\prime }$. Then, $P^{\prime }/I\left[1\right]$ and $P/I\left[1\right]$ are consecutive prime ideals of $D\left[1\right]$. Since P survives in $R⟨1⟩$, [9] (Lemma 2) ensures that $P/I\left[1\right]$ survives in $D⟨1⟩$. As $D⟨1⟩$ is catenarian, we obtain $\mathrm{ht}\left(P/I\left[1\right]\right)=\mathrm{ht}\left(P^{\prime }/I\left[1\right]\right)+1$. Therefore, by [21] (Lemma 2.5), $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(P^{\prime }\right)+1$.

Finally, assume that $P^{\prime }⊉I\left[1\right]$ and $P\supseteq I\left[1\right]$. Since D is a field and P survives in $R⟨1⟩$, we necessarily have $P=I\left[1\right]$. Indeed, $I⟨1⟩$ is a maximal ideal of $R⟨1⟩$, and $I⟨1⟩\subseteq U^{-1}P$, where $U=\{f\in R[1]\mid LC(f)=1\}$. Hence, $I⟨1⟩=U^{-1}P$. By the correspondence between prime ideals of $R\left[1\right]$ disjoint from U and prime ideals of $R⟨1⟩$, it follows that $P=I\left[1\right]$. In particular, P survives in $R\left(1\right)$. Since $R\left(1\right)$ is catenarian, we obtain once again $\mathrm{ht}\left(P\right)=\mathrm{ht}\left(P^{\prime }\right)+1$. Thus, every pair of consecutive prime ideals of $R\left[1\right]$ whose upper member survives in $R⟨1⟩$ satisfies the height condition above. Consequently, $R⟨1⟩$ is catenarian, contradicting assertion (d). Therefore, D is not a field.

If all the maximal ideals in $\mathsf{\Omega }$ are low, then by [37] (Proposition 2), R would be universally catenarian. In particular, $R⟨n⟩$ would be catenarian for all $n\ge 1$, which is a contradiction. If all of them are high, then since R is catenarian (Lemma 2), they would have the same height. Again, by virtue of [37] (Proposition 2), R would be universally catenarian, which is a contradiction. Thus, the set $\mathsf{\Omega }$ contains at least one low and one high maximal ideal. Moreover, as R is catenarian, all high maximal ideals in $\mathsf{\Omega }$ have the same height, equal to ${ht}_R\left(I\right)$.

(e)⇒(a). The fact that $R\left(1\right)$ is catenarian follows readily from Theorem 2. Next, we aim to show that $R⟨1⟩$ is not catenarian. Assume by way of contradiction that $R⟨1⟩$ is catenarian. By assumption, there exists a low maximal ideal, say M, and at least one high maximal ideal in $\mathsf{\Omega }$. Thus,

$${\mathrm{ht}}_T\left(M\right)=1<{\mathrm{ht}}_R\left(I\right).$$

• Step 1. Construction of the elements x and y. Because $M+{\bigcap }_{M^{\prime }\in \mathsf{\Omega }\setminus \left\{M\right\}}{M}^{\prime }=T,$ we may choose $x\in {\bigcap }_{M^{\prime }\in \mathsf{\Omega }\setminus \left\{M\right\}}{M}^{\prime }$ and $y\in M$ such that $x+y=1$. Since D is not a field, $I\notin Max\left(R\right)$, so there exists a maximal ideal $\mathfrak{m}$ of R with $I\subset \mathfrak{m}$. Pick $z\in \mathfrak{m}\setminus I$. It is clear that $x/z\in T_M$ since $x\in T$ and $z\notin M$ (because $z\in R\setminus I$ and $M\cap R=I$).
• Step 2. Construction of the prime ideals $Q^{\prime }$ and Q. Consider the prime ideals $Q^{\prime }\subset Q$ in $T\left[1\right]$, where $Q^{\prime }=(zX-x)T\left[1\right]$ and $Q=M\left[1\right]+(zX-x)T\left[1\right]$. Obviously, $Q^{\prime }$ is an upper to $\left(0\right)$ and Q is an upper to M. It is worth noticing that $Q^{\prime }=\{f\in T\left[1\right]\mid f(x/z)=0\}$ and $Q=\{f\in T\left[1\right]\mid f(x/z)\in M{T}_M\}.$ Let $P^{\prime }=Q^{\prime }\cap R\left[1\right]$ and $P=Q\cap R\left[1\right].$ Then, $I\left[1\right]\neg \subseteq P^{\prime }$ and $I\left[1\right]\subseteq P$, and in fact, $P\cap R=I$. The primes $P^{\prime }\subset P$ need not be adjacent. Let $P_1^{\prime }$ be a maximal prime with $P^{\prime }\subseteq P_1^{\prime }\subset P,I\left[1\right]\neg \subseteq P_1^{\prime },$ and let $P_1$ be a minimal prime with $P_1^{\prime }\subset P_1\subseteq P,I\left[1\right]\subseteq P_1.$ Then, $P_1^{\prime }\subset P_1$ are consecutive, and the chain $P^{\prime }\subseteq P_1^{\prime }\subset P_1$ lifts to $Q^{\prime }\subseteq {\mathfrak{Q}}_1^{\prime }\subset {\mathfrak{Q}}_1$ in $T\left[1\right]$ (cf. [6], Proposition 0). As $zX-x\in {\mathfrak{Q}}_1$ and $I\subseteq {\mathfrak{Q}}_1$, then $x\notin {\mathfrak{Q}}_1$. To see this, assume, by way of contradiction, that $x\in {\mathfrak{Q}}_1$. Then, $zX=(zX-x)+x\in {\mathfrak{Q}}_1$. Thus, either $z\in {\mathfrak{Q}}_1$ or $X\in {\mathfrak{Q}}_1$. If $z\in {\mathfrak{Q}}_1$, then $z\in P_1$, hence $z\in P$, implying $z\in P\cap R=I$, a contradiction. If $X\in {\mathfrak{Q}}_1$, then $X\in P_1\subseteq P$, so $x/z\in M{T}_M$. Writing $x/z=\alpha /\beta$ with $\alpha \in M$, $\beta \in T\setminus M$, we obtain $x\beta =\alpha z\in M$, hence $x\in M$, also a contradiction. Since ${\mathfrak{Q}}_1\cap T$ contains I, it must contain a maximal ideal in $\mathsf{\Omega }$, and the unique such ideal not containing x is M. Therefore, ${\mathfrak{Q}}_1$ lies over M. More precisely, ${\mathfrak{Q}}_1$ is an upper to M because $zX-x\in {\mathfrak{Q}}_1\setminus M\left[1\right]$. It follows that $P_1$ lies over $I=M\cap R$.
• Step 3. The prime $P_1$ is an upper to I.

We now show that $P_1$ is an upper to I. To this end, consider the polynomial

$$\zeta \left(X\right)=(zX-x)(zX-y)=z^2{X}^2-z(x+y)X+xy=z^2{X}^2-zX+xy,$$

Note that $\zeta \left(X\right)\in R\left[1\right]$ because $xy\in \left({\bigcap }_{M^{\prime }\in \mathsf{\Omega }\setminus \left\{M\right\}}{M}^{\prime }\right)\cap M=I$ and $z\in R$. Moreover, $\zeta \left(X\right)\in {\mathfrak{Q}}_1$. Thus, $\zeta \left(X\right)\in {\mathfrak{Q}}_1\cap R\left[1\right]=P_1$. As $z^2\notin I$, then $\zeta \left(X\right)\notin I\left[1\right]$. Therefore, $I\left[1\right]\subset P_1$. Next, we show that $P_1$ survives in $R⟨1⟩$.

• Step 4. The prime $P_1$ survives in $R⟨1⟩$. Suppose, for contradiction, that $P_1$ does not survive in $R⟨1⟩$. Then, $P_1$ contains a monic polynomial

  $$\delta \left(X\right)=X^k+r_{k-1}{X}^{k-1}+\cdots +r_{1}X+r_0,$$

  where $r_0,r_1,\cdots ,r_{k-1}\in R$. Since $\delta \left(X\right)\in P_1$, we have $\delta (x/z)\in M{T}_M$, and therefore,

  $${\left({\displaystyle \frac{x}{z}}\right)}^k+r_{k-1}{\left({\displaystyle \frac{x}{z}}\right)}^{k-1}+\cdots +r_1\left({\displaystyle \frac{x}{z}}\right)+r_0\in M{T}_M.$$

  Equivalently,

  $$x^k+z{r}_{k-1}{x}^{k-1}+\cdots +z^{k-1}{r}_{1}x+r_0{z}^k=\alpha z^k,$$

  (1)

  for some $\alpha \in M{T}_M$. Write $\alpha =u/v$ with $u\in M$ and $v\in T\setminus M$. Multiplying (1) by v yields

  $$\left(x^k+z{r}_{k-1}{x}^{k-1}+\cdots +z^{k-1}{r}_{1}x+r_0{z}^k\right)=u{z}^k\in M.$$

Using $x=1-y$, we obtain

$$v\left(1+z{r}_{k-1}+\cdots +z^{k-1}{r}_1+r_0{z}^k\right)\in M.$$

Since $v\notin M$, it follows that

$$1+z{r}_{k-1}+\cdots +z^{k-1}{r}_1+r_0{z}^k\in M\cap R=I\subset \mathfrak{m},$$

whence $1\in \mathfrak{m}$, a contradiction. Thus, $P_1$ indeed survives in $R⟨1⟩$. As $P_1$ and ${\mathfrak{Q}}_1$ are uppers to I and M, respectively, we get $ht\left(P_1\right)={ht}_{R\left[1\right]}\left(I\left[1\right]\right)+1$ and $ht\left({\mathfrak{Q}}_1\right)={ht}_{T\left[1\right]}\left(M\left[1\right]\right)+1$. Because $R⟨1⟩$ is catenarian, we obtain

$$ht\left(P_1\right)=ht\left(P_1^{\prime }\right)+1=ht\left({\mathfrak{Q}}_1^{\prime }\right)+1\le ht\left({\mathfrak{Q}}_1\right).$$

Hence,

$${ht}_{R\left[1\right]}\left(I\left[1\right]\right)\le {ht}_{T\left[1\right]}\left(M\left[1\right]\right).$$

The reverse inequality is immediate (one can use [6] (Lemme 4)), and therefore,

$${ht}_{R\left[1\right]}\left(I\left[1\right]\right)={ht}_{T\left[1\right]}\left(M\left[1\right]\right).$$

This implies

$${ht}_R\left(I\right)={ht}_T\left(M\right).$$

This is the desired contradiction, completing the proof.    □

Theorem 3 provides a conceptual explanation for the phenomenon observed by Yengui in [19]. In particular, it reveals that the failure of catenarity of the Serre conjecture rings $R⟨n⟩$ is governed by the presence of low and high maximal ideals, while the corresponding Nagata rings $R\left(n\right)$ may remain catenarian. The next example illustrates this mechanism and produces such rings in every dimension.

Example 1.

For each integer $k\ge 2$, we exhibit a stably strong S-domain R such that $R\left(n\right)$ is catenarian for every positive integer n, whereas the Serre conjecture ring $R⟨n⟩$ fails to be catenarian.

Let T be the two-dimensional Prüfer domain constructed by Cahen (see [7], Exemple A). More precisely, T is a semilocal Prüfer domain with exactly two maximal ideals $M_1$ and $M_2$ such that $ht\left(M_1\right)=1$, $ht\left(M_2\right)=2$, and $T/M_1\simeq T/M_2\simeq K$.

Let D be a valuation domain with quotient field K and rank $k-2$. Such a valuation domain can be obtained by considering a valuation of K with value group ${\mathbb{Z}}_{\mathrm{lex}}^{k-2}$. The existence of valuation domains with prescribed value groups follows from the general theory of valuations (see, for instance, [38] (Chapter VI) or [36] (Chapter 19)).

Set

$$I=M_1\cap M_2\mathrm{and}R:=(T,I,D).$$

Then, $\dim \left(R\right)=k$ and ${\mathrm{ht}}_R\left(I\right)=2$ (see [36], Proposition 4). Moreover, by [12] (Theorem 2.6), R is a stably strong S-domain. On the other hand, Theorem 3 implies that $R\left(n\right)$ is catenarian for every positive integer n, while $R⟨n⟩$ is not catenarian.

5. Concluding Remarks

The main objective of this paper was to investigate the catenarity of Nagata rings arising from pullback constructions of type $(T,I,D)$, where I is the intersection of finitely many maximal ideals of T. Under the assumptions that D is a Prüfer domain, T is universally catenarian, and

$$tr.deg[T/M:D]=0$$

for every maximal ideal $M\in \mathsf{\Omega }$, we obtained a complete characterization of the catenarity of the Nagata rings $R\left(n\right)$ for all positive integers n. In particular, our results extend the known theory for pullbacks of the form $(T,M,D)$ and provide an affirmative answer to an open question from the literature.

Despite these advances, several natural questions remain open. A first direction for future research is represented by Question 1, which concerns the characterization of the catenarity of Nagata rings associated with the classical pullback construction $(T,M,D)$ when D is not a field.

A second challenging problem is raised in Question 2. The techniques developed in this paper rely heavily on the assumption that

$$tr.deg[T/M:D]=0$$

for every maximal ideal $M\in \mathsf{\Omega }$. When positive transcendence degree occurs, the behavior of heights in polynomial and Nagata extensions becomes considerably more delicate, and a satisfactory characterization of the catenarity of $R\left(n\right)$ is still lacking.

We hope that the methods introduced here will contribute to the resolution of Questions 1 and 2 and stimulate further research on the interplay between pullback constructions, Nagata rings, and catenarity.