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Annali di Matematica Pura ed

Applicata

ISSN 0373-3114

Volume 191

Number 4

Annali di Matematica (2012)

191:611-629

DOI 10.1007/s10231-011-0199-9

Jump of a domain and spectrally balanced

domains

Mohamed Khalifa & Ali Benhissi

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123

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Annali di Matematica (2012) 191:611–629

DOI 10.1007/s10231-011-0199-9

Jump of a domain and spectrally balanced domains

Mohamed Khalifa · Ali Benhissi

Received: 24 November 2010 / Accepted: 16 March 2011 / Published online: 6 April 2011

© Fondazione Annali di Matematica Pura ed Applicata and Springer-Verlag 2011

Abstract

invariants jR(a) = inf{heightP − heightQ}, where P and Q range over prime ideals of R

such that Q ⊂ aR ⊆ P, and j(R) = sup{jR(a)} (called the jump of R), where a range over

nonzero nonunit elements of R. We study the jump of polynomial ring and power series ring,

we give many results involving jump, and specially we give more interest to LFD-domain R

such that j(R) = 1. We prove that if R is a finite-dimensional divided domain, then R is a

Jaffard domain if and only if for all integer n, j(R[x1,...,xn]) = 1.

Let R bealocallyfinite-dimensional(LFD)integraldomain.Weinvestigatetwo

Keywords

domain · APVD · Polynomial ring · Power series ring

Jaffard domain · Krull dimension · Jump of a domain · Prüfer

Mathematics Subject Classification (2000)

13A15 · 13C15 · 13F05 · 13B25 · 13F25

0 Introduction

Allringsconsideredinthispaperarecommutativewithidentityelement,withoutzerodivisor

(i.e., integral domain), and the dimension of a ring means its Krull dimension. Let R be an

integral domain. Recall that R is said to be locally finite-dimensional (for short, LFD) if

each prime ideal of R has finite height. Following [11], we say that R satisfies the principal

ideal theorem (PIT) if each minimal prime ideal on a nonzero principal ideal of R has height

one. The most standard examples of integral domains satisfying PIT are arbitrary Noetherian

domains [25, Theorem 142], Krull domains, and one-dimensional domains. Since, in an

integral domain R satisfying PIT, each nonzero proper principal ideal aR of R is contained

between two adjacent prime ideals of R (i.e., (0) ⊂ aR ⊆ P for some height-one prime

ideal P of R), and in order to shed some light on positions of nonunits elements of a ring

in its prime spectrum , it is of interest to study domains R in which each nonzero nonunit

M. Khalifa (B ) · A. Benhissi

Department of Mathematics, Faculty of Sciences, 5000 Monastir, Tunisia

e-mail: kmhoalg@yahoo.fr

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612 M. Khalifa, A. Benhissi

element a ∈ R satisfies Q ⊂ aR ⊆ P for some prime ideals Q and P of R such that

ht(P)−ht(Q) = 1. So that, we start by introducing some background material that we need

in our study.

Let R be an LFD-domain, and a be a nonzero nonunit element of R. We define the basic

height of a in R and we note htb,R(a) = sup{ht(Q), Q ∈ ?R(a)} where ?R(a) is the set of

prime ideals Q of R such that Q ⊂ aR. Recall that the height of a in R is defined and noted

asfollows:htR(a) = inf{ht(P), P ∈ VR(a)}where VR(a)isthesetofprimeidealsof R con-

taining a [16, p 508]. We define the jump of a in R, and we note jR(a) = htR(a)−htb,R(a).

It is easy to see that always jR(a) ≥ 1. We show that there is an LFD domain in which there

is a nonzero nonunit element that has jump > 1 (Examples 3.8). This forces us to ask which

LFD-domains in which each nonzero nonunit element has jump 1 ? To clarify our purpose

in this paper, we make the following definition:

Definition Let R beanLFD-domain.Wesaythatanonzerononunita of R isaspectrallybal-

ancedelement(forshort,anSB-element)if jR(a) = 1.Wesaythat R isaspectrallybalanced

domain (for short, R is an SB-domain) if each nonzero nonunit of R is an SB-element.

We show that the set of nonzero nonunit elements that are not SB-elements of R is a

multiplicative system of R (Remark 1.4), so it will be convenient to extend the SB-property

to the zero and units of R as follows: jR(0) = 1 and jR(u) = 0 for each unit u of R. In order

to see how far an LFD-domain R from being an SB-domain, we define the jump of R, and

we note j(R) = sup{jR(a),a ∈ R}.

The purpose of this paper is to study SB-domains, domains which their polynomial rings

are SB-domains, domains which their formal power series rings are SB-domains, jump of a

domain, jump of a polynomial ring, and jump of a power series ring. Also, we study jump of

certain kinds of pullback rings and jump of polynomial ring over some kinds of pullback. In

order to shorten this introduction, we have chosen to recall relevant definitions and facts as

needed throughout the paper.

1 Basic facts and examples of spectrally balanced domains

Let R be an LFD-domain. We list without proof some elementary observations concerning

jump of an element in R.

(i)For each nonzero nonunit a of R, jR(a) = inf{ht(P) − ht(Q)|P, Q are prime ideals

of R such that Q ⊂ aR ⊆ P} (there is no loss of equality if we add ”P is a minimal

prime of a”) and jR(a) ≥ 1.

Let a be a nonzero nonunit of R. Then, a is an SB-element of R if and only if there

exist two prime ideals P and Q of R such that Q ⊂ aR ⊆ P and ht(P) − ht(Q) = 1.

Let a be a nonzero nonunit of R, u a unit of R, and n > 0 an integer. Then, jR(an) =

jR(a) = jR(ua).

If a is a prime element of R, then a is an SB-element of R. Also?∞

R is a unique factorization domain (UFD) if and only if R is an atomic (i.e., each

nonzero nonunit is a product of a finite number of irreducible elements (atoms) of R)

and each irreducible element π of R with jR(π) = 1, is prime [25, Theorem 5].

If R satisfies PIT, then R is an SB-domain (by (ii)).

(ii)

(iii)

(iv)

n=0anR = (0) if

and only if ht(aR)=1 [29, Corollary 1.4].

(v)

(vi)

Consequently, all 1-dimensional and Noetherian and Krull domains (so UFDs) are SB-

domains. We shall see later (Remark 1.2 and Remarks 1.10) that the converse of (vi) is not

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true. Following [31], we say that a domain R is Archimedean in case?∞

arbitrary completely integrally closed domains [12, Theorem 2.1]; and domains satisfying

PIT [11, Proposition 3.7].

n=0anR = (0)

for each nonunit a ∈ R. The most natural examples of Archimedean integral domains are

Proposition 1.1 Let R an LFD-domain. Then, R is an Archimedean SB-domain if and only

if each nonzero nonunit a of R belongs to at least a height-one prime ideal of R.

Proof (⇒) Let a be a nonzero nonunit of R and P ∈ ?R(a), then P ⊆?∞

R such that a ∈ Q and ht(Q) = 1.(⇐)R is an SB-domain by (ii), and is Archimedean by

[29, Corollary 1.4].

n=0anR, and so

htb,R(a) = 0. Since htR(a) = jR(a) − htb,R(a) = 1 − 0 = 1, there is a prime ideal Q of

? ?

Remark 1.2 Let R be an LFD-domain. The concept ”each nonzero nonunit a of R belongs

to at least a height-one prime ideal of R” has appeared in [3, Theorem 2.6]. Recall that an

LFD-domain R, satisfying PIT, is an Archimedean SB-domain. But the converse is not true:

an example is given by R := Int(Z) = { f ∈ Q[X]| f (Z) ⊆ Z} the ring of integer-valued

polynômials over Z. Indeed, by [17, Corollary 1.3], dim(R) = 2 < ∞, and so R is an

LFD-domain. In [3, Example 2.8], P. J. Cahen proved that R does not satisfy PIT and each

nonzero nonunit of R is contained in a height-one prime ideal, and so R is an Archimedean

SB-domain.

Theorem 1.3 Let R be an LFD-domain. For all a,b ∈ R, jR(ab) ≥ inf{jR(a), jR(b)}.

Proof The result is clear if a or b is a unit of R. We can assume that a,b are nonun-

its of R. Also we can assume that a,b ?= 0 because jR(0)=1. Since ?R(ab)=?R(a)

∩ ?R(b), htb,R(ab) ≤ inf{htb,R(a),htb,R(b)}. It is easy to see that VR(ab)=VR(a) ∪

VR(b). Thus, htR(ab)= inf{htR(a),htR(b)}. If htR(a) ≤ htR(b), then jR(ab)=htR(a) −

htb,R(ab) ≥ htR(a)−htb,R(a)= jR(a).Similarly,ifhtR(a) > htR(b),then jR(ab) ≥ jR(b).

? ?

Remark 1.4 Immediately, we deduce (from the last result) that the set of spectrally unbal-

anced elements of an LFD-domain R, denoted Sunb(R) := {a ∈ R|jR(a) ?= 1}, is a

multiplicative set of R. It is easy to see that R is an SB-domain if and only if R = RSunb(R).

Corollary 1.5 Let R be an LFD-domain. If R is Archimedean, then for all nonunits

a,b∈ R, jR(ab) = inf{jR(a), jR(b)}. In particular, Sunb(R)= R\?{P|P is a height-one

Proof For each nonzero nonunit a ∈ R, htb,R(a) = 0 and hence jR(a) = htR(a). A replay

of the last theorem’s proof completes the proof of the first part. By the statement (ii), given

at the beginning of the present section, a nonunita ∈ R is an SB-element (i.e., a / ∈ Sunb(R))

if and only if htR(a) = 1 if and only if a belongs (at least) to a height-one prime ideal of R.

Hence, Sunb(R) is saturated by [25, Theorem 2].

prime ideal of R} is a saturated multiplicative set of R.

? ?

Themostwellknownexamplesofatomicdomainsarearbitraryaccp-domains(i.e.,domain

which satisfies the ascending chain condition on principal ideals) [1].

Corollary 1.6 Let R beanLFD-domain.If R isanaccp-domain,then j(R) = sup{jR(π)|π

irreducible element of R}.

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Proof Let a be a nonzero nonunit element of R. Since R is atomic, a = π1...πmfor some

irreducibles π1,...,πmof R. By [12, Theorem 2.1], R is Archimedean. Hence, jR(a) =

jR(πi) for some i by Corollary 1.5.

? ?

Following [1], a (saturated) multiplicative subset S of a domain R is a splitting multipli-

cative set if for each x ∈ R, x = as for some a ∈ R and s ∈ S such that aR ∩tR = atR for

all t ∈ S. In [1], it was shown that if S is a splitting multiplicative set generated by primes of

R (i.e., generated by a subset of the set of prime elements of R), then RSis an atomic (resp.

accp, UFD) domain if and only if R is an atomic (resp. accp, UFD) domain.

Theorem 1.7 Let R be an LFD-domain and S a splitting multiplicative set generated by

primes. Then, j(RS) = j(R). In particular, RS is an SB-domain if and only if R is an

SB-domain.

Proof Firstwewishtoshowthat:ifa isanonzerononunitof R suchthataR∩sR = asR for

all s ∈ S, then jRS(a) = jR(a). Let P, Q be two prime ideals of R such that Q ⊂ aR ⊆ P

and P is minimal prime of a. Suppose that P ∩ S is nonempty. Thus, P contains (at least) a

prime element π of R. Then, πnh ∈ aR for some integer n > 0 and h ∈ R\P. Suppose that

a / ∈ πR and take b ∈ R such that πnh = ab. Then, b ∈ πR and so πn−1h = ac for some

c ∈ R. We repeat the last sketch until we arrive at h ∈ aR. So h ∈ P, a contradiction. Then,

a ∈ πR and so a ∈ aR ∩ πR = aπR, a contradiction because πR ?= R. Hence, P ∩ S is

empty. Since QRS ⊂ aRS ⊆ PRS, htR(P) − htR(Q) ≥ jRS(a). Then, jR(a) ≥ jRS(a).

For the reverse inequality, let Q ⊂ P be two prime ideals of R such that P ∩ S is empty

and QRS ⊂ aRS ⊆ PRS. Thus, Q = QRS∩ R ⊂ aRS∩ R = aR ⊆ PRS∩ R = P

(sketch: if b ∈ aRS∩ R, then bs ∈ aR for some s ∈ S. Thus, bs ∈ aR ∩ sR = asR. Then,

b ∈ aR). Now, we have htRS(PRS) − htRS(QRS) = htR(P) − htR(Q) ≥ jR(a). Hence,

jRS(a) = jR(a).

Let x be a nonzero nonunit of RSand a ∈ R, s ∈ S such that x =a

t ∈ S such that a = bt and bR ∩ cR = bcR for all c ∈ S. Since s and t are units in

RS, jRS(x) = jRS(a) = jRS(b) = jR(b) ≤ j(R). Then, j(RS) ≤ j(R). For the reverse

inequality, let z be a nonzero nonunit of R. Take y ∈ R and s ∈ S such that z = ys and

yR ∩ tR = ytR for all t ∈ S. If s is a unit in R, then jR(z) = jR(y) = jRS(y) ≤ j(RS). If

not,thenz ∈ πR forsomeprimeelementπ of R.Bythestatement(iv),givenatthebeginning

of the present section, and [1, Proposition 1.6], htR(πR) = 1. Hence, jR(z) = 1 ≤ j(RS).

s. Let b ∈ R and

? ?

By [1, Corollary 1.7], any multiplicative set generated by primes of an atomic domain, is

a splitting multiplicative set.

Corollary 1.8 Let R beanLFD-domainand S amultiplicativesetof R generatedbyprimes.

If R is an atomic domain, then j(RS) = j(R). In particular, RSis an SB-domain if and only

if R is an SB-domain.

We shall see later (Examples 3.8) that (1): Corollary 1.8 does not remain true for arbitrary

multiplicative set of an atomic LFD-domain R and (2): For every integer m ≥ 2, there is

an LFD-domain R such that m ≤ j(R) ≤ 2m (so, R is not an SB-domain) and contains an

element a such that jR(a) = m.

Following [19], a domain R is said to be divided if each prime ideal P of R satisfies

PRP= P (⇔ P is comparable with every ideal of R). An immediate consequence is that

the prime spectrum of a divided domain is totally ordered.

Proposition 1.9 Let R be a domain. Then, R is an SB-domain with linearly ordered prime

ideals if and only if R is finite-dimensional divided domain.

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Proof (⇒)R isfinite-dimensionalbecauseitisLFDandquasi-local.Let P beaprimeidealof

R anda anonzerononunitof R suchthata / ∈ P.Since jR(a) = 1,thereare Q1, Q2twoprime

ideals of R such that Q1⊂ aR ⊆ Q2and ht(Q2) = ht(Q1) + 1. As a ∈ Q2, then Q2? P.

So P ⊂ Q2. Since P and Q1are comparable, P ⊆ Q1because there is no prime ideal of

R properly between Q1and Q2. Then, P ⊆ aR. Hence, R is a divided domain. (⇐) Let a

be a nonunit nonzero of R. Denote m = dim(R) and (0) = P0⊂ P1⊂ ··· ⊂ Pmthe prime

spectrumof R.Letk bethesmallestintegeri between1andm suchthata ∈ Pi.Then,a ∈ Pk

and a / ∈ Pk−1. So Pk−1⊂ aR ⊆ Pk. Thus, jR(a) ≤ ht(Pk) − ht(Pk−1) = k − (k − 1) = 1.

Hence, a is an SB-element of R.

? ?

Remark 1.10 (1)

domain with finite rank is an SB-domain. Then, the converse of (vi) is not true (it suf-

fices to take a rank two valuation domain). We shall see later (Examples 3.8) that there

is an SB-domain that is not divided and does not satisfy PIT.

(2)Recall that a ring R is said to be catenary if for each pair P ⊂ Q of prime ideals of

R, all saturated chains of prime ideals of R between P and Q have a common finite

length. In [19, Example 2.9], D. E. Dobbs showed that for each integer m ≥ 2, there is

an m-dimensional quasi-local going down domain R (so with linearly prime ideals (by

[19, Theorem 2.5]) and hence is catenary) which is not divided (so is not an SB-domain

by the last proposition). Then, a domain can be catenary and have not the SB-property.

Ogoma [28] gives an example of a noncatenary Noetherian finite-dimensional domain

R. Hence, a domain can be an SB-domain and not be catenary.

It is well known that any valuation domain is divided. So any valuation

Following [5], an extension of commutative rings R ⊂ S is called a root extension if for

each s ∈ S there is an integer n > 0 such that sn∈ R. The integral closure of an integral

domain R will be denoted by R.

Proposition 1.11 Let R beanLFD-domain.If R ⊂ R isarootextension,then j(R) ≤ j(R).

Proof Let x be a nonzero nonunit of R and n > 0 an integer such that xn∈ R. Let P1, P2

be two prime ideals of R such that P1⊂ xnR ⊆ P2. For i = 1,2, Pi = {z ∈ R|zk∈ Pi

for some integer k ≥ 1} is a prime ideal of R lying over Pi and htR(Pi) = htR(Pi) by

[5, Theorem 2.1]. If z ∈ P1, then zk∈ P1for some integer k > 0. Since P1⊂ xnR, (z

R. Thus, z ∈ xR. Then, P1⊆ xR ⊆ P2. Moreover, x / ∈ P1because xn/ ∈ P1= P1∩ R.

Hence, P1⊂ xR ⊆ P2. It follows that jR(x) ≤ jR(xn) ≤ j(R). Then, j(R) ≤ j(R).

Immediately, it follows that:

xn)k∈

? ?

Corollary 1.12 Let R beadomainsuchthat R ⊂ R isarootextension.If R isanSB-domain,

then R is an SB-domain.

Proposition 1.13 Let R ⊂ T be an extension of LFD-domains. If Spec(R) = Spec(T) then

j(R) = j(T). In particular, R is an SB-domain if and only if T is an SB-domain.

Proof By [4, Theorem 3.10], R and T are quasi-local with the same maximal ideal. Denote

M this maximal ideal. Let 0 ?= a ∈ M. Then, jR(a) = jT(a) because for each two prime

ideals P, Q of R, P ⊂ aR ⊆ Q if and only if P ⊂ aT ⊆ Q.

Corollary 1.14 Let T be a quasi-local domain of the form K + M, where K is a field and

M is the maximal ideal of T. Let F be a proper subfield of K, and set R = F + M. Then,

j(R) = j(T). In particular, R is an SB-domain if and only if T is an SB-domain.

? ?

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Proof Combine Proposition 1.13 and [4, Corollary 3.11].

? ?

Theorem 1.15 Let T be a quasi-local LFD-domain with maximal ideal M and residue field

K, φ : T −→ K thenaturalsurjection,and R = φ−1(D),where D isanLFD-subringof K.

Then, R is an LFD-domain and j(R) = max{j(T), j(D)}. In particular, R is an SB-domain

if and only if T and D are both SB-domains.

Proof by [2, Lemma 2.1-(e)], it is easy to see that {P ∈ Spec(R)|P ⊆ M} = Spec(T)

and each element P of this last set satisfies htR(P) = htT(P) < ∞. By [2, Lemma 2.1],

M is a divided prime of R and R/M∼= D. Then, htR(P) = htR(M) + htD(P/M) < ∞

for each prime ideal P of R such that M ⊆ P. So R is LFD. Thus, the jump function on

R, j : R −→ N is well defined.

Claim 1 if 0 ?= a ∈ M, jR(a) = jT(a).

Let P, Q be two prime ideals of R such that P ⊂ aR ⊆ Q and Q is a minimal prime ofa.

Thus, Q ⊆ M (because Q iscomparablewith M),andso P, Q aretwoprimeidealsof T and

P ⊂ aT ⊆ Q. Thus, jT(a) ≤ htT(Q)−htT(P) = htR(Q)−htR(P). Then, jR(a) ≥ jT(a).

For the reverse inequality, let P, Q be two prime ideals of T such that P ⊂ aT ⊆ Q. So

P, Q are prime ideals of R and P ⊂ aR ⊆ Q (sketch: if x ∈ P, then x = at for some

t ∈ T. Since a / ∈ P, t ∈ P ⊂ R. Thus, x ∈ aR). Then, jR(a) ≤ htR(Q) − htR(P). Hence,

jR(a) = jT(a).

Claim 2 if a is a nonunit of R such that a / ∈ M, then jR(a) = jD(? a), where? a denotes the

Let P, Q betwoprimeidealsof R suchthat P ⊂ aR ⊆ Q,andhtR(Q)−htR(P) = jR(a).

Since M is divided in R, M ⊂ aR. Thus, M ⊆ P, because if not, P ⊂ M ⊂ aR ⊆ Q, then

htR(Q)−htR(P) > htR(Q)−htR(M) ≥ jR(a),acontradiction.Then,? P ⊂? aD ⊆? Q,where

R such that M ⊆ P, htR(P) = htR(M)+htD(P/M) because M is a divided prime ideal of

R. Then, jD(? a) ≤ htD(? Q)−htD(? P) = jR(a). For the reverse inequality, let P ⊂ Q be two

jD(? a).Thus, P ⊂ aR+M = aR ⊆ Q.So jR(a) ≤ htR(Q)−htR(P).Then, jR(a) = jD(? a).

Asanapplication,weapplythelasttheoremtodomainsoftheform D+M.Westrengthen

Corollary 1.14 as follows:

class of a modulo M.

? P := P/M and? Q := Q/M are two prime ideals of D. Note that for any prime ideal P of

prime ideals of R such that M ⊆ P, P/M ⊂ aD ⊆ Q/M and htD(Q/M) − htD(P/M) =

Hence, the result follows from Claim 1 and Claim 2.

Corollary 1.16 Let T be an LFD-quasi-local domain of the form K + M, where K is a field

and M is the maximal ideal of T. Let D be a proper LFD-subring of K, and set R = D+ M.

Then, j(R) = max{j(T), j(D)}. In particular, R is an SB-domain if and only if T and D

are both SB-domains.

Proof Letφ : T −→ K bethenaturalsurjection.Itiseasytoseethatφ−1(D) = D+M = R.

Then, Theorem 1.15 applies.

? ?

Example 1.17 Let D be an LFD-subring of a field K and set R = D + XK[[X]]. Since

K[[X]] is an SB-domain, j(R) = j(D) by Corollary 1.16. In particular, R is an SB-domain

ifandonlyif DisanSB-domain.Asaconsequence,if Disaone-dimensionalnon-Noetherian

subring of K, then D + XK[[X]] is a non-Noetherian SB-domain.

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2 Jump of polynomial ring and universally spectrally balanced domain

Throughout this section, R denotes a domain X,andx1,...,xn,... denote a countable

set of independent algebraically indeterminates over R. For each integer n > 0, Xn :=

{x1,...,xn} and we write R[Xn] rather than R[x1,...,xn] (the polynomial ring in n inde-

terminates). Following [25], a domain R is said to be an S(eidenberg)-domain if, for each

height 1 prime ideal P of R, P[X] has height 1 in R[X]. Note that, for any domain R, R[X]

is always an S-domain [21, Chapter VI-Proposition 6.3.13]. The first result show that each

nonconstant polynomial element of R[Xn], is an SB-element of R[Xn].

Theorem 2.1 Let R be an LFD-domain and n > 0 an integer. Then,

j(R[Xn]) = sup{jR[Xn](a)|a ∈ R}.

Proof It suffices to show that any nonunit nonconstant polynomial of R[Xn] is an SB-ele-

ment. Let f be a nonunit nonconstant element of R[Xn]. Let k be the smallest integer

i between 1 and n such that f ∈ R[Xi]. Since f ∈ R[Xk]\R[Xk−1] (if k = 1, then

R[Xk−1] means R), f has degree ≥ 1 in xk, and so f R[Xk] ∩ R[Xk−1] = (0). Then, by

Zorn lemma and [25, Theorem 1], there is a prime ideal P of R[Xk] such that f ∈ P and

P ∩ R[Xk−1] = (0). By [25, Theorem 36], htR[Xk](P) = 1. Since R[Xk],..., R[Xn−1] are

S-domains, htR[Xn](P[xk+1,...,xn]) = 1. Hence, jR[Xn]( f ) = 1.

It is natural to ask when R[X] is an SB-domain and it will be more better to ask when

R[Xn] is an SB-domain for all n > 0. We make the following definition:

Definition 2.2 A domain R is called a universally spectrally balanced (for short, USB-)

domain if, for each integer n > 0, the polynomial ring R[Xn] is an SB-domain.

The following result show that an LFD Prüfer domain is an USB-domain if and only it is

an SB-domain.

Proposition 2.3 Let R be an LFD Prüfer domain and n > 0 an integer. Then, j(R[Xn]) =

j(R).

? ?

Proof Let a be a nonzero nonunit of R (so also of R[Xn]) and let Q, P be two prime ide-

als of R such that Q ⊂ aR ⊆ P. Thus, Q[Xn] ⊂ aR(Xn] ⊆ P[Xn] and so jR[Xn](a) ≤

ht(P[Xn])−ht(Q[Xn]) = ht(P)−ht(Q)by[9,Corollary2.8].Then, jR[Xn](a) ≤ jR(a).For

thereverseinequality,let P1,P2betwoprimeidealsof R[Xn]suchthat P1⊂ aR[Xn] ⊆ P2

and P2isaminimalprimeofa.Set Pi= Pi∩R fori = 1,2.Sincea ∈ P2[Xn] ⊆ P2, P2=

P2[Xn]. Again by [9, Corollary 2.8], P1= P1[Xn]. Then, jR(a) ≤ ht(P2)−ht(P1) because

P1⊂ aR ⊆ P2. Hence, jR(a) = jR[Xn](a).

Corollary 2.4 Let R be an LFD Prüfer domain. The following assertions are equivalent:

(1) R is an USB-domain.

(2) R[X] is an SB-domain.

(3) R is an SB-domain.

? ?

Proof It follows immediately from the last result.

? ?

It follows also that a finite-rank valuation domain is an USB-domain. Now, we will look

into divided domain and search what happens ? Following [2], a domain R is said to be a

Jaffard domain if dim(R) = dimv(R) < ∞ (where dimvmeans the valuative dimension),

equivalently if dim(R) < ∞ and dim(R[Xn]) = n+dim(R) for all integer n > 0. The

following result shows that the two concepts “USB” and “Jaffard” are the same on a divided

domain.

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Theorem 2.5 Let R be a divided domain with finite dimension and n > 0 an integer. Then,

R[Xn] is an SB-domain if and only if dim(R[Xn]) = n + dim(R). In particular, R is an

USB-domain if and only if R is a Jaffard domain.

Proof Leta beanonunitnonzeroof R.Denotem =dim(R)and(0) = P0⊂ P1⊂ ··· ⊂ Pm

the prime spectrum of R. Let k be the smallest integer i between 1 and m such that a ∈ Pi.

Then, a ∈ Pk and a / ∈ Pk−1. So Pk−1 ⊂ aR ⊆ Pk. Since Pk−1[Xn] ⊂ aR[Xn] ⊆

Pk[Xn], jR[Xn](a) ≤ ht(Pk[Xn]) − ht(Pk−1[Xn]). Let Q,P be two prime ideals of R[Xn]

such that Q ⊂ aR[Xn] ⊆ P. Since Pkis the unique minimal prime of a, Pk⊆ P ∩ R. Thus,

Pk[Xn] ⊆ P. Denote? R = R/Pk−1and? Pk= Pk/Pk−1. Since? Pkis a height 1 prime ideal

ht(P)−ht(Q) ≥ ht(Pk[Xn])−ht(Pk−1[Xn]).Then, jR[Xn](a) = ht(Pk[Xn])−ht(Pk−1[Xn]).

By Theorem 1, j(R[Xn]) = max{ht(Pi[Xn])−ht(Pi−1[Xn])|1 ≤ i ≤ m}. Hence, it is easy

to see that R[Xn] is an SB-domain if and only if ht(Pi[Xn]) = 1 + ht(Pi−1[Xn]) for all

integer 1 ≤ i ≤ m if and only if ht(Pi[Xn]) = i for all integer 1 ≤ i ≤ m if and only if

ht(Pm[Xn]) = m if and only if dim(R[Xn]) = n + m (by [9, Corollary 2.9]).

The most simple examples of divided domains are one-dimensional quasi-local domains.

of? R and ? a (the class of a modulo Pk−1) ∈? Pk,?∞

s=0? as? R = (? 0) by [29, Corollary 1.4].

Thus, Pk−1=?∞

s=0asR.Then, Q ⊆?∞

s=0(aR[Xn])s= (?∞

s=0asR)[Xn] = Pk−1[Xn].So

? ?

Corollary 2.6 Let R be one-dimensional quasi-local domain. The following assertions are

equivalent:

(1) R is an USB-domain.

(2) R[X] is an SB-domain.

(3) R is an S-domain.

(4) R is a Jaffard domain.

Proof (1) ⇒ (2) Clear. (2) ⇒ (3) Let M be the unique nonzero prime ideal of R. Then,

ht(M[X]) = j(R[X]) (see the proof of Theorem 2.5). Hence, ht(M[X]) = 1. (3) ⇒ (4)

follows from [15, Corollary 6.3]. (4) ⇒ (1) follows from the last Theorem.

Following Hedstrom and Houston [23], a domain R is called a pseudo-valuation domain

(PVD) in case each prime ideal P of R is strongly prime, in the sense that xy ∈ P, x, y are

inthequotientfieldof R implies x ∈ P or y ∈ P.In[10],BadawiandHoustongaveagener-

alization of PVDs: a domain R is said to be an almost pseudo-valuation domain (APVD) in

caseeachprimeideal P of R isstronglyprimary,inthesense xy ∈ P, x, y areinthequotient

field of R implies x ∈ P or yn∈ P for some integer n > 0. Recall (additionally) that a

domain R isanAPVDifandonlyif R isquasi-localwithmaximalideal M suchthat(M : M)

is a valuation domain with M primary to the maximal ideal of (M : M) [10, Theorem 3.4].

Recall that valuation domain ⇒ PVD ⇒ APVD ⇒ divided domain [10, Proposition 3.2]. In

the following result, we evaluate the jump of R[Xn] in case R is a finite-dimensional APVD.

Corollary 2.7 Let R be an APVD with maximal ideal M and finite dimension. Let n > 0

be an integer and denote d the transcendence degree of (M : M)/√M over R/M. Then,

j(R[Xn]) = 1 + inf{d,n} = dim(R[Xn]) − dim(R) − n + 1. In particular, the following

assertions are equivalent:

? ?

(1) R is an USB-domain.

(2) R[X] is an SB-domain.

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(3) (M : M)/√M is algebraic over R/M.

(4) R is a Jaffard domain.

Proof Denotem = dim(R)and(0) = P0⊂ P1⊂ ··· ⊂ Pmtheprimespectrumof R.Since

R is a divided domain, j(R[Xn]) = max{ht(Pi[Xn]) − ht(Pi−1[Xn])|1 ≤ i ≤ m} (see the

proof of Theorem 2.5). Let i ∈ {1,...,m − 1}. Since Pi = PiRPi, htR[Xn](Pi[Xn]) =

htRPi[Xn](Pi[Xn]). By [10, Theorem 2.11] and [24, Proposition 7], RPiis a valuation

domain, and so ht(Pi[Xn]) = ht(Pi). Thus, ht(Pi[Xn]) − ht(Pi−1[Xn]) = 1 for each i ∈

{1,...,m−1}.Then, j(R[Xn]) = ht(Pm[Xn])−ht(Pm−1[Xn]) = ht(Pm[Xn])−(m−1) =

dim(R[Xn])−n−m+1 by [9, Corollary 2.9]. Since there is a unique prime ideal of the val-

uation domain (M : M) which contain M (in fact, this prime is√M and is the maximal ideal

of (M : M)), dim(R[Xn]) = dim(R) + inf{d,n} + n by [16, Sect. 3, Corollary 1] and [27,

Lemma 1.1]. Hence, j(R[Xn]) = 1 + inf{d,n}. It follows that dim(R[Xn]) = n + dim(R)

if and only if d = 0 if and only if j(R[Xn]) = 1.

The last result helps us to show that: R is an SB-domain ? R[X] is an SB-domain. For

each integer m > 0, there is an SB-domain R with dimension m such that R[X] is not an

SB-domain (and of course, R is not an USB-domain).

? ?

Example 2.8 Let F ⊂ K betwofieldssuchthat K isnotalgebraicover F.Letm beapositive

integer, it is well known that there exists valuation domain V with finite rank m of the form

K + N where N is the maximal ideal of V. Set R = F + N. Then, R is an m-dimensional

PVD (and so an APVD) by [23, Example 2.1]. Since R is divided, R is an SB-domain. Since

V/N∼= K is not algebraic over F∼= R/M, R is not an USB-domain and R[X] is not an

SB-domain by Corollary 2.7.

For a nonzero nonunit a of a domain R, MinR(a) denotes the set of minimal prime ideals

of a in R. It is easy to see that MinR[Xn](a) = {P[Xn]|P ∈ MinR(a)}. Now, we study and

determine necessary and sufficient conditions for certain polynomial ring over ”pull-back-

type” constructions to be an SB-domain. First, we need the following lemma:

Lemma 2.9 Let R ⊂ T be two domains with nonzero common ideal I. If P is a prime ideal

of T such that I ? P, then htT(P) = htR(P ∩ R).

Proof Let u : R/I −→ T/I be the embedding map and v : T −→ T/I the canonical

surjection. One sees easily that the map f : r ?−→ (r,r) is an isomorphism of rings from

R to the pullback D := R/I ×T/I T = {(r,t) ∈ R/I × T;u(r) = v(t)} of R/I and

T over T/I. Since f (I) = {0} × I, the map Q ?−→ f−1(Q) is an isomorphism of par-

tially ordered set (with respect to the set-theoretic inclusion) from Spec(D) − VD({0} × I)

to Spec(R) − VR(I). Let u?: D ?−→ T the restriction of the canonical projection. By

[20, Corollary 1.5 (2) and (3)], the map P ?−→ u?−1(P) is an isomorphism of partially

ordered set (with respect to the set-theoretic inclusion) from Spec(T)−VT(I) to Spec(D)−

VD({0} × I). Then, the map P ?−→ f−1(u?−1(P)) = P ∩ R is an isomorphism of partially

ordered set from Spec(T) − VT(I) to Spec(R) − VR(I). This completes the proof.

Theorem 2.10 Let T be a quasi-local finite-dimensional domain (which is not a field) with

maximalideal M andresiduefield K, φ : T −→ K thenaturalsurjection,and R = φ−1(D),

where D is a finite-dimensional subring of K. Set ? := {a ∈ T|M is minimal prime of a

in T}.

(1) If ? is nonempty, then:

? ?

j(R[X]) = sup{j(T[X]), j(D[X]),htR[X](M[X]) − htT[X](M[X]) + sup

a∈?

jT[X](a)}.

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(2) If ? is empty, then j(R[X]) = sup{j(T[X]), j(D[X])}.

(3) Let n > 0 be an integer and suppose that D[Xn] is an SB-domain.

(i) If ? is nonempty, then

j(R[Xn]) = sup{j(T[Xn]),htR[Xn](M[Xn]) − htT[Xn](M[Xn])

+ sup

a∈?

(ii) If ? is empty, then j(R[Xn]) = j(T[Xn]).

jT[Xn](a)}.

Proof R is finite-dimensional by [2, Lemma 2.1], and so is R[Xn] by [25, Theorem 38].

Thus, the jump function on R[Xn], j : R[Xn] −→ N is well defined.

Claim 1 If 0 ?= a ∈ M, then htb,R[Xn](a) = htb,T[Xn](a).

LetP beaprimeidealof R[Xn]suchthatP ⊂ aR[Xn].Since M[Xn]isanonzerocommon

idealof R[Xn]and T[Xn]andsinceP ⊂ M[Xn],thenthereisaprimeideal Q of T[Xn]such

that Q ∩ R[Xn] = P by [16, Proposition 4]. If f ∈ Q, then af ∈ M[Xn] ⊂ R[Xn] and so

af ∈ P.Thus, f ∈ P.Then,P = QisaprimeidealofT[Xn].Let?beaprimeidealofT[Xn]

such that ? ⊂ aT[Xn]. Since ? ⊂ M[Xn], ? is a prime ideal of R[Xn] and ? ⊂ aR[Xn]

(because ? = a?). So ?R[Xn](a) = ?T[Xn](a), and hence, htb,R[Xn](a) = htb,T[Xn](a) by

the last lemma.

Claim 2 if 0 ?= a ∈ M such that M / ∈ MinR(a), jR[Xn](a) = jT[Xn](a).

Let Q ∈ MinT[Xn](a) such that htT[Xn](a) = htT[Xn](Q). Then, Q = (Q ∩ T)[Xn] ⊂

M[Xn]. So Q is a prime ideal of R[Xn]. By the last lemma, htT[Xn](Q) = htR[Xn](Q). Thus,

htT[Xn](a) ≥ htR[Xn](a). Let ? ∈ MinR[Xn](a) such that htR[Xn](a) = htR[Xn](?). Thus,

? = (? ∩ R)[Xn] ⊂ M[Xn]. Then, by [2, Lemma 2.1-(e)], ? ∩ R is a prime ideal of T and,

so ? is a prime ideal of T[Xn]. Hence, htT[Xn](a) ≤ htR[Xn](a) by the last lemma. The result

follows from Claim 1.

Claim 3 if0 ?= a ∈ M suchthat M ∈ MinR(a), jR[Xn](a) = jT[Xn](a)+htR[Xn](M[Xn])−

htT[Xn](M[Xn]).

It is easy to show that for each 0 ?= a ∈ M, M ∈ MinR(a) if and only if MinR(a) = {M}

if and only if MinT(a) = {M} if and only if M ∈ MinT(a) [2, Lemma 2.1-(e)]. Then, M is

theuniqueprimeof R whichisminimalovera because M isadividedprimeidealof R.Also

M istheuniqueprimeofT whichisminimalovera.Thus,htR[Xn](a) = htR[Xn](M[Xn])and

htT[Xn](a) = htT[Xn](M[Xn]).Then,byClaim1, jR[Xn](a)−jT[Xn](a) = htR[Xn](M[Xn])−

htT[Xn](M[Xn]).

Claim 4 If P is a prime ideal of R[Xn] such that M ⊆ P, then ht(P) = ht(M[Xn]) +

ht(P/M[Xn]).

Set P := P∩R.Then,by[22,Theorem30.18]and[2,Lemma2.2],ht(P)=ht(P[Xn])+

ht(P/P[Xn])=ht(M[Xn]) + ht(P[Xn]/M[Xn]) + ht(P/P[Xn]) ≤ ht(M[Xn]) + ht(P/

M[Xn]). The reverse inequality is trivial.

Claim 5 If a is a nonunit of R such that a / ∈ M, then jR[X](a) = jD[X](? a) and jR[Xn](a) ≤

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jD[Xn](? a), where? a denote the class of a modulo M.

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Let P, Q betwoprimeidealsof R[Xn]suchthat M[Xn] ⊆ P and P/M[Xn] ⊂? aD[Xn] ⊆

Thus, jR[Xn](a) ≤ jD[Xn](? a) by Claim 4. For the reverse inequality, let P, Q be two prime

Q/M[X] and so jD[X](? a) ≤ ht(Q) − ht(P) (again by Claim 4). If not, then ht(P) ≤

ht(Q) − ht(M[X]) ≤ ht(Q) − ht(P). Then, jD(X](? a) = jR[X](a).

Remarks 2.11 With the notation of the last theorem:

(1)We can replace htR[X](M[X]) − htT[X](M[X]) by dim(R[X]) − dim(T[X]) −

dim(D[X])+1andhtR[Xn](M[Xn])−htT[Xn](M[Xn])bydim(R[Xn])−dim(T[Xn])−

dim(D[Xn])+n.Indeed:By[9,Corollary 2.9]and[2,Lemma2.2],htT[Xn](M[Xn]) =

dim(T[Xn]) − n and there is a maximal ideal N of R such that dim(R[Xn]) =

ht(N[Xn]) + n = htR[Xn](M[Xn]) + ht((N/M)[Xn]) + n ≤ htR[Xn](M[Xn]) +

dim(D[Xn]) ≤ dim(R[Xn]) (because D

htT[Xn](M[Xn]) = dim(R[Xn]) − dim(D[Xn]) − dim(T[Xn]) + n.

(2)We can choose T such that ? is empty. It is easy to show that if T is a quasi-local

Noetherian domain such that dim(T) > 1, then ? is empty by [25, Theorem 152].

We can choose T such that ? is nonempty. For examples, choose T to be a finite-rank

valuation domain or 1-dimensional quasi-local domain.

Q/M[Xn]. Thus, P ⊂ aR[Xn] + M[Xn] = aR[Xn] ⊆ Q. So jR[Xn](a) ≤ ht(Q) − ht(P).

ideals of R[X] such that P ⊂ aR[X] ⊆ Q. If M[X] ⊆ P, then P/M[X] ⊂ ? aD(X] ⊆

ht((P ∩ R)[X])+1 ≤ ht(M[X]) because P ∩ R ⊂ M. Hence, jD(X](? a) ≤ ht(Q/M[X]) =

Hence, the result follows from Claims 2, 3, and 5.

? ?

∼=

R/M). Then, htR[Xn](M[Xn]) −

By combining the last theorem and last remarks, we have:

Corollary 2.12 With the same hypothesis as in the last theorem:

(1) If ? is empty, then

(a) R[X] is an SB-domain if and only if T[X] and D[X] are both SB-domains.

(b) If D is an USB-domain, then R is an USB-domain if and only if T is an USB-

domain.

(2) If ? is nonempty, then

(a) R[X] is an SB-domain if and only if T[X] and D[X] are both SB-domains and

dim(R[X]) = dim(T[X])+dim(D[X]) − 1.

(b) If D isanUSB-domain,then R isanUSB-domainifandonlyifT isanUSB-domain

and for each integer n > 0, dim(R[Xn]) = dim(T[Xn]) + dim(D[Xn]) − n.

The following result shows that in some cases, we do not need to look into the set ? to

determine when R[X] is an SB-domain.

Corollary 2.13 With the same hypothesis as in the last theorem: If K is algebraic over D,

then

(1) j(R[X]) = max{j(T[X]), j(D[X])}. In particular, R[X] is an SB-domain if and only

if T[X] and D[X] are both SB-domains.

(2) If D is an USB-domain, then j(R[Xn]) = j(T[Xn]) for each integer n > 1, and in

particular, R is an USB-domain if and only if T is an USB-domain.

Proof Let n > 0 an integer. Since the transcendence degree of K over D is zero,

dim(R[Xn]) ≤ max{dim(T[Xn]), htT[Xn](M[Xn]) + dim(D[Xn])} = htT[Xn](M[Xn]) +

dim(D[Xn]) by [16, Théorème 2]. Thus, dim(R[Xn]) ≤ dim(T[Xn]) − n + dim(D[Xn]),

and so htR[Xn](M[Xn]) = htT[Xn](M[Xn]) by [16, Proposition 5] and Remarks 2.11. The

result follows from the last theorem.

? ?

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Corollary 2.14 With the same hypothesis as in the last theorem, set d := t.d(K/D) the

transcendence degree of K over D. If the set ? is nonempty and T is a Jaffard domain, then

(1) j(R[X]) = sup{j(T[X]), j(D[X]),inf{d,1} + sup

an SB-domain if and only if T[X] and D[X] are both SB-domains and K is algebraic

over D.

(2) If D[Xn]isanSB-domain,then j(R[Xn]) = sup{j(T[Xn]),inf{d,n}+sup

In particular, if D is an USB-domain, then R is an USB-domain if and only if T is an

USB-domain and K is algebraic over D.

a∈?

jT[X](a)}. In particular, R[X] is

a∈?

jT[Xn](a)}.

Proof Set d := t.d(K/D) the transcendence degree of K over D and let n > 0 be an inte-

ger. Since R and T have a common nonzero ideal M, dim(R[Xn]) = dim(T[Xn]) − n +

inf {d,n}+dim(D[Xn]) by [16, Théorème 2]. Then, htR[Xn](M[Xn])−htT[Xn](M[Xn]) =

inf{d,n}. Hence, the result follows from Theorem 2.10.

Corollary 2.15 With the same hypothesis as in the last theorem: Set d := t.d(K/D) the

transcendence degree of K over D. If T is a valuation domain, then

(1) j(R[X]) = sup{j(D[X]),1 + inf{d,1}}. In particular, R[X] is an SB-domain if and

only if D[X] is an SB-domain and K is algebraic over D.

(2) If D[Xn] is an SB-domain, then j(R[Xn]) = 1 + inf{d,n}. In particular, if D is an

USB-domain, then R is an USB-domain if and only if K is algebraic over D.

? ?

Proof Any finite-rank valuation domain is an USB-domain and a Jaffard domain. Hence, we

apply the last corollary.

? ?

Remark 2.16 A consequence of the last result, if R is a PVD with nonzero finite dimension

and maximal ideal M, then j(R[Xn]) = 1 + inf{d,n}, where d is the transcendence degree

of (M : M)/M over R/M. (In fact, this result can be derived from Corollary 2.7). Indeed:

it is well known that (M : M) is a valuation domain and Spec((M : M)) = Spec(R). Let

φ : (M : M) −→ (M : M)/M be the natural surjection. Then, R = φ−1(R/M) by [4,

Proposition 2.6]. Since R/M is a field, (R/M)[Xn] is an SB-domain. Hence, one can apply

Corollary 2.15.

Note that, if R is a finite-dimensional domain, then j(R) ≤ dim(R). This inequality may

be equality (in case one-dimensional domain) and may be strict (a divided domain with finite

dimension> 1(seeProposition1.9)).Wewillshowthatthisinequality,incaseofpolynomial

ring, is always strict. Seidenberg proved in [30, Theorem 2], that if a domain R has finite

dimension, then dim(R)+1 ≤ dim(R[X]) ≤ 2.dim(R)+1. In particular, dim(R[X]) < ∞

if and only if dim(R) < ∞. Accordingly, it is natural to ask if there is similar inequalities

and equivalence between j(R) and j(R[X])?.

Lemma 2.17 Let R be an LFD-domain and a is a nonzero nonunit of R. Then

(1) htR(a) ≤ htR[X](a) ≤ 2.htR(a).

(2) htb,R(a) ≤ htb,R[X](a) ≤ 2.htb,R(a) + 1.

(3) jR[X](a) ≤ jR(a) + htR(a).

Proof (1) follows from [25, Theorem 38] and the fact that MinR[X](a) = {P[X]|P ∈

MinR(a)}. If P is a prime ideal of R[X] such that P ⊂ aR, then P[X] ⊂ aR[X]. So the

first inequality of (2) is clear. For the second one, if Q is a prime ideal of R[X] such that

Q ⊂ aR[X], then P ⊂ aR, where P = Q ∩ R. Since ht(Q) ≤ 2.ht(P) + 1 (again by

[25, Theorem 38]), htb,R[X](a) ≤ 2.htb,R(a)+1. The last, (3) follows immediately from (1)

and (2).

? ?

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Proposition 2.18 Let R be a domain with nonzero finite dimension. Then,

j(R[X]) ≤ inf{j(R) + dim(R), dim(R[X]) − 1}.

Proof By Lemma 2.17, if a is a nonzero nonunit of R, then jR[X](a) ≤ jR(a) + htR(a) ≤

j(R) + dim(R). Then, j(R[X]) ≤ j(R) + dim(R). Suppose that j(R[X]) = dim(R[X])

and choose a nonzero nonunit a of R such that jR[X](a) = dim(R[X]). Thus, htR[X](a) =

dim(R[X]). Then, there is a maximal ideal M of R[X] such that M is a minimal prime of

a and ht(M) = dim(R[X]). Set M = M ∩ R. Since a ∈ M ⊆ M[X] ⊆ M, M = M[X].

Thus, ht(M[X]) = dim(R[X]), a contradiction by [9, Corollary 2.9]. Hence, j(R[X]) <

dim(R[X]).

Remark 2.19 The inequality (in the last result) may be strict (take a rank two valuation

domain) and may be equality (take a rank one valuation domain).

? ?

Theorem 2.20 Let R be an LFD-domain. If R is Archimedean, then

(1) j(R) ≤ j(R[X]) ≤ 2.j(R). In particular, j(R[X]) < ∞ if and only if j(R) < ∞.

(2) Assume, furthermore, that R is an S-domain, then the following assertions are equiva-

lent:

(a) R is an USB-domain.

(b) R[X] is an SB-domain.

(c) R is an SB-domain.

Proof Note that in an Archimedean LFD-domain, the jump of a nonzero nonunit is equal

to its height. By [14, Chapter 3, Exercise 17], R[X] is also Archimedean. Then, (1) follows

from Lemma 2.17. For (2), we suppose that R is an S-domain. (a) ⇒ (b) Clear. (b) ⇒ (c):

Leta beanonzerononunitof R (andsoof R[X]).Thus,a ∈ Q forsomeheight1primeideal

Q of R[X] by Proposition 1.2. Since Q ∩ R ?= (0), Q = (Q ∩ R)[X]. So ht(Q ∩ R) = 1.

Then, R is an SB-domain (again by Proposition 1.2). (c) ⇒ (a) Since R[Xn] is an S-domain

and Archimedean for each integer n > 0, it suffices to show that R[X] is an SB-domain. Let

a be a nonzero nonunit of R and let P be a height-one prime ideal of R such that a ∈ P.

Since R is an S-domain, ht(P[X]) = 1. Hence, R[X] is an SB-domain by Proposition 1.2. ? ?

Immediately, it follows that:

Corollary 2.21 If R is a one-dimensional S-domain, then R is an USB-domain.

Lemma 2.22 Let R be an LFD-domain.

(1) Leta beanonzerononunitof R.Ifa isanSB-elementof R,then?∞

(2) If R is catenary and R[X] is an SB-domain, then R is an SB-domain.

Proof (1) It suffices to show that?∞

ideals of R such that P ⊂ aR ⊆ Q and ht(Q) = 1 + ht(P). Thus, P ⊆

Since Q/P is a height-one prime ideal of R/P,?∞

By (1), (?∞

where P =?∞

adjacent in Spec(R) because ht((Q ∩ R)[X]) = 1 + ht(P[X]). Then, jR(a) = 1.

i=1aiR istheunique

largest prime ideal strictly contained in aR.

i=1aiR is a prime ideal of R because for each prime

i=1aiR). Since jR(a) = 1, there is P, Q two prime

i=1? ai? R = (? 0) where? R = R/P and ? a

ideal P of R, (P ⊂ aR) ⇒ (P ⊆?∞

is the class of a modulo P. Hence,?∞

containedinaR[X].Thereisaminimalprime Q ofa in R[X]suchthatht(Q) = 1+ht(P[X])

i=1aiR. Necessarily, Q = (Q ∩ R)[X], and so P ⊂ aR ⊆ Q ∩ R. Thus,

jR(a) ≤ ht(Q ∩ R) − ht(P) = ht(Q ∩ R)/P because R is catenary. But P ⊂ Q ∩ R are

?∞

i=1aiR.

i=1aiR = P. (2) Let a be a nonzero nonunit of R.

i=1(aiR[X]) is the unique largest prime ideal of R[X] strictly

i=1aiR)[X] =?∞

? ?

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Theorem 2.23 Let R be a domain such that R[X] is catenarian. Then,

(1) j(R[X]) < ∞ if and only if j(R) < ∞.

(2) If j(R) < ∞, then j(R) − 1 ≤ j(R[X]) ≤ 2.j(R).

(3) R[X] is an SB-domain if and only if R is an SB-domain.

Proof (1) and (2): Assume that j(R[X]) is finite and that there is a nonzero nonunit a

of R such that jR(a) > 1 + j(R[X]). Let Q,? be two prime ideals of R[X] such that

Q ⊂ aR[X] ⊆ ? and ht(?) − ht(Q) = jR[X](a). Set P0 = Q ∩ R and P1 = ? ∩ R.

Thus, ? = P1[X]. Then, ht(P1/P0) ≤ ht((P1/P0)[X]) = ht(P1[X]/P0[X]) = ht(P1[X])−

ht(P0[X]) ≤ ht(?)−ht(Q)+1 = jR[X](a)+1 ≤ j(R[X])+1 < jR(a) ≤ ht(P1)−ht(P0),

because P0 ⊂ aR ⊆ P1. Then, R is not catenarian, a contradiction. Hence, j(R) ≤ 1 +

j(R[X]) < ∞. Now, assume that j(R) is finite and that there is a nonzero nonunit a of R

such that jR[X](a) > 2.j(R). Let P1, P0be two prime ideals of R such that P0⊂ aR ⊆ P1

andht(P1)−ht(P0) = jR(a).So P0[X] ⊂ aR[X] ⊆ P1[X].Then,ht(P1[X])−ht(P0[X]) ≥

jR[X](a) > 2j(R) ≥ 2.jR(a) = 2(ht(P1) − ht(P0)) = 2.ht(P1/P0) ≥ ht(P1[X]/P0[X]), a

contradiction because R[X] is catenarian. Hence, j(R[X]) ≤ 2.j(R) < ∞. (3): Since R is

catenarian,(⇒)followsfromLemma2.22.(⇐):Leta beanonzerononunitof R and P0, P1

be two prime ideals of R such that P0⊂ aR ⊆ P1and ht(P1)−ht(P0) = 1. Thus, P0⊂ P1

are adjacent in Spec(R) and so is P0[X] ⊂ P1[X] in Spec(R[X]) by [15, Lemma 2.3]. Then,

ht(P1[X]) = 1+ht(P0[X]) and so jR[X](a) = 1 because P0[X] ⊂ aR[X] ⊆ P1[X]. Hence,

a is an SB-element of R[X] and the result follows from Theorem 2.1.

Following [15], a domain R is said to be universally catenarian if, for each integer n > 0,

the polynomial ring R[Xn] is catenarian.

Corollary 2.24 Let R be a universally catenarian domain and n > 0 an integer. Then,

j(R[Xn]) < ∞ if and only if j(R) < ∞. In this case, j(R) − n ≤ j(R[Xn]) ≤ 2nj(R).

Proof We deduce this result by induction on n and from the last Theorem.

? ?

? ?

It is well known that if R is 1-dimensional domain, then R is universally catenarian if and

only if R is an S-domain [15, Corollary 6.3]. Now, we strengthen Corollary 2.21.

Corollary 2.25 If R is universally catenarian, then R is an USB-domain if and only if R is

an SB-domain.

Proof The result follows immediately by induction on n and from the last theorem.

? ?

Remarks 2.26 (1) It is well known that there exists a finite-dimensional Noetherian domain

which is not catenarian [28]. We read Corollary 2.25 as follows: For any SB-domain R, if R

is universally catenarian, then R is an USB-domain, then Ogoma’s example shows that the

converse is false.

(2)Let R beadomain.Following[2,Definition1.4], R issaidtobelocallyJaffarddomain

if RPis a Jaffard domain for each prime ideal P of R. It is well known that R is a locally

Jaffard domain if and only if ht(P[Xn]) = ht(P) for all prime ideal P of R and all integer

n > 0 [18, Remarques-1-(2)]. It is easy to see that if R is an SB-domain which is a locally

Jaffard domain, then R is an USB-domain.

Indeed: Let a be a nonzero nonunit of R and P0, P1be two prime ideals of R such that

P0 ⊂ aR ⊆ P1and ht(P1) − ht(P0) = 1. Thus, P0[Xn] ⊂ aR[Xn] ⊆ P1[Xn], and so

jR[Xn](a) ≤ ht(P1[Xn]) − ht(P0[Xn]) = ht(P1) − ht(P0) = 1. Hence, jR[Xn](a) = 1.

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Now,westudytheSB-propertyonsomewell-knownlocalizationof R[X].For f ∈ R[X],

let c( f ) be the ideal of R generated by the coefficients of f . We denote by N = { f ∈

R[X];c( f ) = R} the set of polynomials over R with unit content. By [22, Proposition 33.1],

N is a multiplicative set of R[X]. Let R(X) = R[X]N be R[X] localized at N. For each

prime ideal P of R, P(X) := P[X]N. Recall that a prime ideal Q of R(X) is said to be

extended from R if Q = P(X) for some prime ideal P of R.

Proposition 2.27 Let R be an LFD-domain. For each a of R, jR(X)(a) = jR[X](a). In

particular, j(R(X)) ≥ j(R[X]).

Proof Let Q1, Q2be two prime ideals of R[X] such that Q1⊂ aR[X] ⊆ Q2and Q2∈

MinR[X](a).Then, Q2= P[X],where P := Q2∩R.So Q2∩N isemptyby[22,Proposition

33.1]. Thus, (Q1)N,(Q2)N= P(X) are two prime ideals of R(X) and (Q1)N⊂ aR(X) ⊆

P(X). Then, jR(X)(a) ≤ ht((Q2)N) − ht((Q1)N) = ht(Q2) − ht(Q1). Hence, jR(X)(a) ≤

jR[X](a). For the reverse inequality, let Q ⊂ P be two prime ideals of R[X] such that P ∩ N

is empty, QN ⊂ aR(X) ⊆ PN. If f ∈ Q, f h ∈ aR[X] for some h ∈ N. Looking at the

contents of these polynomials, since c(h) = R, c( f ) = c( f h) ⊆ aR. Thus, f ∈ aR[X].

Then, Q ⊂ aR[X] ⊆ P. So jR[X](a) ≤ ht(P) − ht(Q) = ht(PN) − ht(QN). Hence,

jR[X](a) = jR(X)(a). By Theorem 2.1 j(R[X]) = sup

b∈R

jR[X](b) ≤ j(R(X)).

? ?

It is easy to see that, if V is a finite-rank valuation domain, then V(X) is also a valuation

domain and rank(V(X)) = rank(V), and hence, V(X) is an SB-domain. More generally,

we have the following result:

Proposition 2.28 Let R be an LFD Prüfer domain. Then, R(X) is an SB-domain if and only

if for each finitely generated nonzero proper ideal F of R, there exists P, Q two prime ideals

of R such that Q ⊂ F ⊆ P and ht(P) − ht(Q) = 1.

Proof (⇒) let F = a0R + ··· + anR be a finitely generated nonzero proper ideal of R

and set f = a0+ ··· + anXn. Since (0) ?= F = c( f ) ?= R, f is a nonzero nonunit

of R(X). Since jR(X)( f ) = 1, there exists P1 ⊂ P2two prime ideals of R[X] such that

P2∩ N is empty, (P1)N ⊂ f R(X) ⊆ (P2)N and ht((P2)N) − ht((P1)N) = 1. Since R

is a Prüfer domain, it is well known that every prime ideal ideal of R(X) is extended from

R [22, Proposition 33.1] and [9, Corollary 2.8]. Then, Pi = Pi[X] for i = 1,2, where

Pi:= Pi∩ R. By [14, Chapter 13-Corollary 7.10], f R(X) = F.R(X). Then, P1⊂ F ⊆ P2

and ht(P2) − ht(P1) = ht(P2[X]) − ht(P1[X]) = 1.

(⇐) let f = a0+ ··· + anXnbe a nonzero nonunit of R[X] such that f / ∈ N. Thus,

c( f ) = a0R + ··· + anR is a finitely generated nonzero proper ideal of R. Let P1, P2two

prime ideals of R such that P1 ⊂ c( f ) ⊆ P2and ht(P2) − ht(P1) = 1. Then, P1(X) ⊂

c( f )R(X) = f R(X) ⊆ P2(X). Since ht(P2(X)) − ht(P1(X)) = 1, jR(X)( f ) = 1.

? ?

Corollary 2.29 If R is a 1-dimensional Prüfer domain, then R(X) is an SB-domain.

Corollary 2.30 Let R be an LFD Bezout domain. Then, R(X) is an SB-domain if and only

if R is an SB-domain.

Proof The result follows immediately from the last proposition because every finitely gen-

erated ideal of R is principal.

? ?

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3 Jump of power series ring

Throughout this section, R denotes an LFD-domain which is not a field, X;x1,...,xn,...

denote a countable set of algebraically independent indeterminates over R. For each inte-

ger n > 0, Xn := {x1,...,xn} and we write R[[Xn]] (the power series ring) rather than

R[[x1,...,xn]].

The following result shows that if j(R[[X]]) < ∞, then j(R[[X]]) ?= dim(R[[X]]).

Proposition 3.1 If dim(R[[X]]) < ∞, then j(R[[X]]) ≤ dim(R[[X]]) − 1.

Proof Let f =

jR[[X]]( f ) = 1 because f ∈ XR[[X]] and ht(XR[[X]]) = 1. Assume that f0 ?= 0. It

is well known that a maximal ideal of R[[X]] has the form M + XR[[X]], where M is a

maximal ideal of R [14, Chapter 1, Corollary 1.6],. It suffices to show that for any maximal

ideal M of R such that f0∈ M, M + XR[[X]] is not a minimal prime of f . If not, there

exists an integer n > 0, h,g ∈ R[[X]] such that h0 / ∈ M (where h0is the constant term

of h) and Xnh = fg. Thus, h = f φ for some φ ∈ R[[X]]. So h0∈ f0R ⊆ M, a contra-

diction. Then, any minimal prime of f , is not maximal. Hence, jR[[X]]( f ) ≤ htR[[X]]( f ) ≤

dim(R[[X]]) − 1.

For every sequence ( fn)n≥0of elements of R[[X]], we define and denote?∞

?∞

where fn=?∞

(?), and the (usual) formal series?∞

Lemma 3.2 Let f, f0, f1,.... be a countable set of elements of R[[X]]. Then

(1) For each integer k ≥ 0,?∞

Proof For each integer n ≥ 0, set fn =?∞

If t ≤ k, then (the coefficient of Xtin φ)=(the coefficient of Xtin f0+ ··· + ftXt) =

a0,t+a1,t−1+···+at,0=?

easy to see that c =?

?

n ≥ 0, f fn=?∞

An ideal I of R is called an ideal of strong finite type (for short, SFT-ideal) if there exists

a finitely generated ideal J ⊆ I and a positive integer k such that ak∈ J for each a ∈ I.

The ring R is called an SFT-ring if each ideal of R is an SFT-ideal. These concepts were

?∞

i=0fiXibe a nonzero nonunit element of R[[X]]. If f0 = 0, then

? ?

n=0fnXnas

follows:

n=0fnXn=

?∞

k=0(

?

i+j=kai,j)Xk∈ R[[X]]

(?)

i=0an,iXifor every integer n ≥ 0.

Let (an)n≥0be a sequence of elements of R. It is easy to see that?∞

of confusion).

n=0anXn, defined as

n=0anXn, are the same formal series (there is no danger

n=0fnXn= f0+ ··· + fkXk+ Xk+1?∞

i=0an,iXi. (1) Let k ≥ 0 be an integer and

n=0fn+k+1Xnand φ = f0+ ··· + fkXk+ Xk+1ψ. Let t ≥ 0 be an integer.

n=0fn+k+1Xn.

(2) f?∞

set ψ =?∞

n=0fnXn=?∞

n=0f fnXn.

i+j=tai,j. Now, assume that t ≥ k +1. Then, (the coefficient

i+j=t−k−1ai+k+1,j =?t−k−1

i+j=tai,j. (2) Set f =

of Xtin f?∞

?

of Xtin φ) = a0,t+a1,t−1+···+ak,t−k+c where c is the coefficient of Xt−k−1in ψ. It is

?k

i+j=tbi

i=0

i+j=t

i=0

ai+k+1,t−k−1−i =?t

l=k+1al,t−l.

l=0al,t−l =

Then, (the coefficient of Xtin φ)=

n=0fnXn) =?

?

l=0al,t−l +?t

l=k+1al,t−l =

?t

?∞

i=0biX. Let t ≥ 0 be an integer. Then, (the coefficient

?

u+i+v=tbuai,v.

l+s=jal,s =?

i+l+s=tbial,s. For each integer

?

u+v=ibuan,vXi. Hence, (the coefficient of Xtin?∞

n=0f fnXn) =

u+v=jbuai,v=?

? ?

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introduced by J. T. Arnold in [6]. The condition that R is an SFT-ring plays a key role in

computingtheKrulldimensionofthepowerseriesring R[[X]].RecallthatArnold[7]proved

that if R is a Prüfer domain, then dim(R[[X]]) < ∞ if and only if R is an SFT-ring (and in

this case, dim(R[[X]]) = 1 + dim(R)).

Theorem 3.3 If R is an SFT-Prüfer domain with finite dimension and f =?∞

Proof Let f =

jR[[X]]( f ) = 1 = jR(0). Assume that f0 ?= 0 and let P, Q be two prime ideals of R

such that P ⊂ f0R ⊆ Q. Thus, P ⊆?∞

Set h =

Lemma 3.2. Hence, P[[X]] ⊂ f R[[X]]. Since f ∈ Q + XR[[X]] and Q + XR[[X]]

is not a minimal prime of f because f0 ?= 0 (see the proof of the last proposition),

there exists a minimal prime P of f such that P ⊂ Q + XR[[X]]. Since R[[X]] is

catenarian [8, Theorem 4] and by [7, Corollary 3.6], jR[[X]]( f ) ≤ ht(P) − ht(P[[X]]) ≤

ht(Q + XR[[X]]) − 1 − ht(P[[X]]) = ht(Q[[X]]) − ht(P[[X]]) = ht(Q) − ht(P). Then,

jR[[X]]( f ) ≤ jR( f0). For the reverse inequality, let P1,P2be two prime ideals of R[[X]]

such that P1 ⊂ f R[[X]] ⊆ P2 and P2 is a minimal prime of f . Set Pi = Pi ∩ R

for i = 1,2. Since R is an SFT-ring, Pi[[X]] ⊆ Pi for i = 1,2 by [6, Theorem 1].

Thus, P1 ⊆ f0R. If P1 = f0R, then P1[[X]] ⊂ f R[[X]] ⊆ P1+ XR[[X]] and so

jR[[X]]( f ) ≤ ht(P1+ XR[[X]]) − ht(P1[[X]]) = 1 ≤ ht(P2) − ht(P1). Now, we can

suppose that P1⊂ f0R.

Case 1: P2 = P2[[X]]. Thus, P1 ⊂ f0R ⊆ P2. Then, jR( f0) ≤ ht(P2) − ht(P1) =

ht(P2) − ht(P1) (again by [7, Corollary 3.6]).

Case2: P2[[X]] ?= P2.Since P2+XR[[X]]isnotaminimalprimeof f, P2?= P2+XR[[X]].

Thus, P1⊆ P2[[X]] by [8, Lemma 2]. So P1= P1[[X]] (again by [7, Corollary 3.6]). Let ?

be a prime ideal of R[[X]] such that P2is a prime just below ?. Since R/P2is an SFT-Prüfer

domain, P2⊂ P := ? ∩ R are adjacent and P2⊂ P + XR[[X]] by [8, Lemma 1]. Since

f0∈ P, jR( f0) ≤ ht(P)−ht(P1) = 1+ht(P2)−ht(P1) = 1+ht(P2[[X]])−ht(P1[[X]]) =

ht(P2) − ht(P1). Hence, jR[[X]]( f ) = jR( f0).

Corollary 3.4 Let R be a Prüfer domain with finite dimension. The following statements are

equivalent:

(1) R[[X]] is an SB-domain.

(2) R is an SB-domain satisfying the SFT-property.

i=0fiXi∈

R[[X]], then jR[[X]]( f ) = jR( f0). In particular, j(R[[X]]) = j(R).

?∞

?∞

i=0fiXibe a nonzero nonunit element of R[[X]]. If f0 = 0, then

i=0fi

i=0aiXi∈ P[[X]], then for all i, ai = f hi for some hi ∈ R[[X]].

i=0hiXi∈ R[[X]]. Thus, g =

0R, and so P ⊆ f R[[X]] by [13, Lemma

?∞

2.1]. Let g =

?∞

i=0f hiXi= f?∞

i=0hiXi= f h by

? ?

Proof By [6, Theorem 1], if R[[X]] is an LFD-domain, then R satisfies the SFT-property.

The result follows immediately from the last theorem.

? ?

It is well known that a finite-rank valuation domain V is an SFT-ring if and only if it is

discrete.

Corollary 3.5 If V is finite-rank discrete valuation domain, then V[[X]] is an SB-domain.

Following [26], R is said to be a star-domain (for short, ∗-domain) if Min(R) (i.e., the set

of height-one prime ideals of R) is nonempty and for each element P of this set, R satisfies

the acc on P-principal ideals of R (i.e., ideals of the form aP, where a ∈ R). We proved

[26, Theorem 2.9] that if R is an APVD with nonzero finite dimension, then dim(R[[X]]) <

∞ if and only if R is a residually star-domain (i.e., for each nonmaximal prime ideal P of

R, R/P is a ∗-domain).

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Theorem 3.6 Let R beanAPVDwithfinitenonzerodimension.If R isaresidually∗-domain,

then R[[X]] is an SB-domain.

Proof Note that R[[X]] is finite-dimensional and catenarian by [26, Theorem 2.9 and The-

orem 3.4]. For each prime ideal Q of R, since Q[[X]] ⊂ Q + XR[[X]] are adjacent prime

ideals of R[[X]], ht(Q + XR[[X]]) = 1 + ht(Q) by [26, Corollary 2.17]. We replay the

same first part of Theorem 3.3’s proof, we obtain that jR[[X]]( f ) ≤ jR( f0) for each nonzero

nonunit f =?∞

i=0fiXi∈ R[[X]]. Hence, the proof is complete since any APVD is a

divided domain (and so an SB-domain by Proposition 1.9).

? ?

Corollary 3.7 Let R be a PVD with finite dimension. If R is an SFT-ring, then R[[X]] is an

SB-domain.

Proof Combine the last theorem with [26, Corollary 1.8].

? ?

In the following examples, we show that:

•

•

There is an SB-domain that is not divided and not satisfy PIT.

For every integer m ≥ 2, there is an LFD-domain T such that m ≤ j(T) ≤ 2m (so, T is

not an SB-domain) and contains an element a such that jT(a) = m.

Corollary 1.8 does not remain true for arbitrary multiplicative set of an atomic LFD-

domain R.

ThereexistsanSB-domain R suchthatthepowerseriesring R[[X]]isfinite-dimensional

domain (so LFD) but is not an SB-domain. This example proves also that R[[X]] is an

SB-domain ? R[[X,Y]] is an SB-domain (where R is an LFD-domain, and X,Y are

two independent algebraically indeterminates over R).

•

•

Examples 3.8 (1) Let R be 2-dimensional APVD with maximal ideal M such that R is a

residually ∗-domain [27, Example 1.5]. Set A = R[[X]]. Then, A is an SB-domain by

Theorem 3.6. The domain A is not divided because M[[X]] ? X A and X / ∈ M[[X]]. By

[26, Theorem 1.6 and Proposition 1.4] , M ?= M2, and for each a ∈ M\M2, M6⊆ aR.

So M is a minimal prime of a. Note that if T is an SFT-ring and a is a nonunit of T, then

MinT[[X]](a) = {P[[X]], P ∈ MinT(a)} by [6, Theorem 1]. Then, M[[X]] is a minimal

prime of a in A. Since ht(M[[X]]) = 2 (by [26, Corollary 2.17]), A does not satisfy PIT.

(2) Let R be a one-dimensional APVD which is a ∗-domain as in [27, Example 2.8], m > 0

be an integer and Tm= R[[Xm]] the formal power series ring in m indeterminates over

R. By [27, Theorem 2.4] , dim(Tm) = 2m < ∞, and so Tmis an LFD-domain and

j(Tm) ≤ 2m. Let M be the unique nonzero prime ideal of R and 0 ?= a ∈ M. Then,

M[[Xm]] is the unique minimal prime of a in Tm, and so htTm(a) = ht(M[[[Xm]]) = m

by[27,Corollary2.5].Since R isArchimedean, TmisalsoArchimedeanby[14,Chapter

3, Exercise 17]. Thus, htb,Tm(a) = 0 and jTm(a) = m.

(3) With the same notation as in (2) and fix m = 2: Set S = R\(0) (S is a multiplicative set

of the domain T2). Since 2 ≤ j(T2) ≤ 4, T2is not an SB-domain. By [27, Lemma 1.2],

R is an accp-domain, and so is T2by [14, Chapter 2- Proposition 1.7]. Then, T2is an

atomic domain. But (T2)Sis Noetherian (and so is an SB-domain) by [26, Theorem 2.2].

(4) With the same notation as in (2). Then, T1is an SB-domain (by Theorem 3.6) but

T1[[x2]] = T2is not an SB-domain.

Acknowledgment

The authors thank the referee for his/her helpful comments and suggestions.

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