Jump of a domain and spectrally balanced domains
ABSTRACT Let R be a locally finitedimensional (LFD) integral domain. We investigate two invariants $${j_R(a)={{\rm inf}}\{{\rm height}P{\rm height} Q\}}$$, where P and Q range over prime ideals of R such that $${Q\subset aR\subseteq P}$$, and $${j(R)={\rm sup}\{j_R(a)\}}$$ (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 LFDdomain R such that j(R) = 1. We prove that if R is a finitedimensional divided domain, then R is a Jaffard domain if and only if for all integer $${n,\,j(R[x_1,\ldots,x_n])=1}$$.
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Article: Factorization in integral domains
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ABSTRACT: In this paper, we study factorization in an integral domain R, that is, factoring elements of R into products of irreducible elements. We investigate several factorization properties in R which are weaker than unique factorization.Journal of Pure and Applied Algebra 12/2010; · 0.53 Impact Factor  SourceAvailable from: Muhammad Zafrullah
Article: Almost Bezout Domains .II
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ABSTRACT: An integral domain R is an almost Bezout domain (respectively, almost valuation domain) if for each pair a;b 2 Rnf0g, there is a positive integer n = n(a;b) such that (an;bn) is principal (respectively, an j bn or bn j an): We show that a …nite intersection of almost valuation domains with the same quotient …eld is an almost Bezout domain. This generalizes the result that a …nite intersection of valuation domains with the same quotient …eld is a Bezout domain. We use our work to give a new characterization of CohenKaplansky domains.Journal of Algebra  J ALGEBRA. 01/1994; 167(3):547556.  SourceAvailable from: Muhammad Zafrullah[Show abstract] [Hide abstract]
ABSTRACT: Les AA. continuent l’étude des propriétés des anneaux intègres, qui sont plus faibles que les propriétés de anneaux factoriels, commencé dans la première partie du papier [cf. J. Pure Appl. Algebra 69, No. 1, 119 (1990; Zbl 0727.13007)]. Dans cette deuxième partie sont étudiés les comportements des propriétés pour les extensions d’anneaux intègres: localisations, en particulier par un système multiplicatif engendré par des éléments primes, et pour les réunions filtrantes d’anneaux.Reviewer: N.Radu (Bucureşti)Journal of Algebra 01/1992; 152(1):7893. · 0.58 Impact Factor
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Annali di Matematica Pura ed
Applicata
ISSN 03733114
Volume 191
Number 4
Annali di Matematica (2012)
191:611629
DOI 10.1007/s1023101101999
Jump of a domain and spectrally balanced
domains
Mohamed Khalifa & Ali Benhissi
Page 2
123
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Annali di Matematica (2012) 191:611–629
DOI 10.1007/s1023101101999
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 SpringerVerlag 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 LFDdomain R
such that j(R) = 1. We prove that if R is a finitedimensional divided domain, then R is a
Jaffard domain if and only if for all integer n, j(R[x1,...,xn]) = 1.
Let R bealocallyfinitedimensional(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 finitedimensional (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 onedimensional 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 heightone 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
email: kmhoalg@yahoo.fr
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612M. 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 LFDdomain, 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
LFDdomains in which each nonzero nonunit element has jump 1 ? To clarify our purpose
in this paper, we make the following definition:
Definition Let R beanLFDdomain.Wesaythatanonzerononunita of R isaspectrallybal
ancedelement(forshort,anSBelement)if jR(a) = 1.Wesaythat R isaspectrallybalanced
domain (for short, R is an SBdomain) if each nonzero nonunit of R is an SBelement.
We show that the set of nonzero nonunit elements that are not SBelements of R is a
multiplicative system of R (Remark 1.4), so it will be convenient to extend the SBproperty
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 LFDdomain R from being an SBdomain, we define the jump of R, and
we note j(R) = sup{jR(a),a ∈ R}.
The purpose of this paper is to study SBdomains, domains which their polynomial rings
are SBdomains, domains which their formal power series rings are SBdomains, 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 LFDdomain. 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 SBelement 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 SBelement 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 SBdomain (by (ii)).
(ii)
(iii)
(iv)
n=0anR = (0) if
and only if ht(aR)=1 [29, Corollary 1.4].
(v)
(vi)
Consequently, all 1dimensional 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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Jump of a domain613
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 LFDdomain. Then, R is an Archimedean SBdomain if and only
if each nonzero nonunit a of R belongs to at least a heightone 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 SBdomain 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 LFDdomain. The concept ”each nonzero nonunit a of R belongs
to at least a heightone prime ideal of R” has appeared in [3, Theorem 2.6]. Recall that an
LFDdomain R, satisfying PIT, is an Archimedean SBdomain. But the converse is not true:
an example is given by R := Int(Z) = { f ∈ Q[X] f (Z) ⊆ Z} the ring of integervalued
polynômials over Z. Indeed, by [17, Corollary 1.3], dim(R) = 2 < ∞, and so R is an
LFDdomain. 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 heightone prime ideal, and so R is an Archimedean
SBdomain.
Theorem 1.3 Let R be an LFDdomain. 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 LFDdomain R, denoted Sunb(R) := {a ∈ RjR(a) ?= 1}, is a
multiplicative set of R. It is easy to see that R is an SBdomain if and only if R = RSunb(R).
Corollary 1.5 Let R be an LFDdomain. If R is Archimedean, then for all nonunits
a,b∈ R, jR(ab) = inf{jR(a), jR(b)}. In particular, Sunb(R)= R\?{PP is a heightone
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 SBelement (i.e., a / ∈ Sunb(R))
if and only if htR(a) = 1 if and only if a belongs (at least) to a heightone prime ideal of R.
Hence, Sunb(R) is saturated by [25, Theorem 2].
prime ideal of R} is a saturated multiplicative set of R.
? ?
Themostwellknownexamplesofatomicdomainsarearbitraryaccpdomains(i.e.,domain
which satisfies the ascending chain condition on principal ideals) [1].
Corollary 1.6 Let R beanLFDdomain.If R isanaccpdomain,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 LFDdomain and S a splitting multiplicative set generated by
primes. Then, j(RS) = j(R). In particular, RS is an SBdomain if and only if R is an
SBdomain.
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 beanLFDdomainand S amultiplicativesetof R generatedbyprimes.
If R is an atomic domain, then j(RS) = j(R). In particular, RSis an SBdomain if and only
if R is an SBdomain.
We shall see later (Examples 3.8) that (1): Corollary 1.8 does not remain true for arbitrary
multiplicative set of an atomic LFDdomain R and (2): For every integer m ≥ 2, there is
an LFDdomain R such that m ≤ j(R) ≤ 2m (so, R is not an SBdomain) 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 SBdomain with linearly ordered prime
ideals if and only if R is finitedimensional divided domain.
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Proof (⇒)R isfinitedimensionalbecauseitisLFDandquasilocal.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 SBelement of R.
? ?
Remark 1.10 (1)
domain with finite rank is an SBdomain. 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 SBdomain 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 mdimensional quasilocal 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 SBdomain
by the last proposition). Then, a domain can be catenary and have not the SBproperty.
Ogoma [28] gives an example of a noncatenary Noetherian finitedimensional domain
R. Hence, a domain can be an SBdomain 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 beanLFDdomain.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 ∈ Rzk∈ 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 isanSBdomain,
then R is an SBdomain.
Proposition 1.13 Let R ⊂ T be an extension of LFDdomains. If Spec(R) = Spec(T) then
j(R) = j(T). In particular, R is an SBdomain if and only if T is an SBdomain.
Proof By [4, Theorem 3.10], R and T are quasilocal 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 quasilocal 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 SBdomain if and only if T is an SBdomain.
? ?
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Proof Combine Proposition 1.13 and [4, Corollary 3.11].
? ?
Theorem 1.15 Let T be a quasilocal LFDdomain with maximal ideal M and residue field
K, φ : T −→ K thenaturalsurjection,and R = φ−1(D),where D isanLFDsubringof K.
Then, R is an LFDdomain and j(R) = max{j(T), j(D)}. In particular, R is an SBdomain
if and only if T and D are both SBdomains.
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 LFDquasilocal domain of the form K + M, where K is a field
and M is the maximal ideal of T. Let D be a proper LFDsubring of K, and set R = D+ M.
Then, j(R) = max{j(T), j(D)}. In particular, R is an SBdomain if and only if T and D
are both SBdomains.
Proof Letφ : T −→ K bethenaturalsurjection.Itiseasytoseethatφ−1(D) = D+M = R.
Then, Theorem 1.15 applies.
? ?
Example 1.17 Let D be an LFDsubring of a field K and set R = D + XK[[X]]. Since
K[[X]] is an SBdomain, j(R) = j(D) by Corollary 1.16. In particular, R is an SBdomain
ifandonlyif DisanSBdomain.Asaconsequence,if DisaonedimensionalnonNoetherian
subring of K, then D + XK[[X]] is a nonNoetherian SBdomain.
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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 Sdomain [21, Chapter VIProposition 6.3.13]. The first result show that each
nonconstant polynomial element of R[Xn], is an SBelement of R[Xn].
Theorem 2.1 Let R be an LFDdomain 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 SBele
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
Sdomains, htR[Xn](P[xk+1,...,xn]) = 1. Hence, jR[Xn]( f ) = 1.
It is natural to ask when R[X] is an SBdomain and it will be more better to ask when
R[Xn] is an SBdomain 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 SBdomain.
The following result show that an LFD Prüfer domain is an USBdomain if and only it is
an SBdomain.
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 USBdomain.
(2) R[X] is an SBdomain.
(3) R is an SBdomain.
? ?
Proof It follows immediately from the last result.
? ?
It follows also that a finiterank valuation domain is an USBdomain. 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 SBdomain if and only if dim(R[Xn]) = n + dim(R). In particular, R is an
USBdomain 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 SBdomain 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 onedimensional quasilocal 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 onedimensional quasilocal domain. The following assertions are
equivalent:
(1) R is an USBdomain.
(2) R[X] is an SBdomain.
(3) R is an Sdomain.
(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 pseudovaluation 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 pseudovaluation 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 isquasilocalwithmaximalideal 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 finitedimensional 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 USBdomain.
(2) R[X] is an SBdomain.
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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 SBdomain ? R[X] is an SBdomain. For
each integer m > 0, there is an SBdomain R with dimension m such that R[X] is not an
SBdomain (and of course, R is not an USBdomain).
? ?
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 mdimensional
PVD (and so an APVD) by [23, Example 2.1]. Since R is divided, R is an SBdomain. Since
V/N∼= K is not algebraic over F∼= R/M, R is not an USBdomain and R[X] is not an
SBdomain 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 ”pullback
type” constructions to be an SBdomain. 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 settheoretic 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 settheoretic 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 quasilocal finitedimensional domain (which is not a field) with
maximalideal M andresiduefield K, φ : T −→ K thenaturalsurjection,and R = φ−1(D),
where D is a finitedimensional subring of K. Set ? := {a ∈ TM 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 SBdomain.
(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 finitedimensional 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 quasilocal
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 finiterank
valuation domain or 1dimensional quasilocal 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 SBdomain if and only if T[X] and D[X] are both SBdomains.
(b) If D is an USBdomain, then R is an USBdomain if and only if T is an USB
domain.
(2) If ? is nonempty, then
(a) R[X] is an SBdomain if and only if T[X] and D[X] are both SBdomains and
dim(R[X]) = dim(T[X])+dim(D[X]) − 1.
(b) If D isanUSBdomain,then R isanUSBdomainifandonlyifT isanUSBdomain
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 SBdomain.
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 SBdomain if and only
if T[X] and D[X] are both SBdomains.
(2) If D is an USBdomain, then j(R[Xn]) = j(T[Xn]) for each integer n > 1, and in
particular, R is an USBdomain if and only if T is an USBdomain.
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 SBdomain if and only if T[X] and D[X] are both SBdomains and K is algebraic
over D.
(2) If D[Xn]isanSBdomain,then j(R[Xn]) = sup{j(T[Xn]),inf{d,n}+sup
In particular, if D is an USBdomain, then R is an USBdomain if and only if T is an
USBdomain 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 SBdomain if and
only if D[X] is an SBdomain and K is algebraic over D.
(2) If D[Xn] is an SBdomain, then j(R[Xn]) = 1 + inf{d,n}. In particular, if D is an
USBdomain, then R is an USBdomain if and only if K is algebraic over D.
? ?
Proof Any finiterank valuation domain is an USBdomain 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 SBdomain. Hence, one can apply
Corollary 2.15.
Note that, if R is a finitedimensional domain, then j(R) ≤ dim(R). This inequality may
be equality (in case onedimensional 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 LFDdomain 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 LFDdomain. 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 Sdomain, then the following assertions are equiva
lent:
(a) R is an USBdomain.
(b) R[X] is an SBdomain.
(c) R is an SBdomain.
Proof Note that in an Archimedean LFDdomain, 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 Sdomain. (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 SBdomain (again by Proposition 1.2). (c) ⇒ (a) Since R[Xn] is an Sdomain
and Archimedean for each integer n > 0, it suffices to show that R[X] is an SBdomain. Let
a be a nonzero nonunit of R and let P be a heightone prime ideal of R such that a ∈ P.
Since R is an Sdomain, ht(P[X]) = 1. Hence, R[X] is an SBdomain by Proposition 1.2. ? ?
Immediately, it follows that:
Corollary 2.21 If R is a onedimensional Sdomain, then R is an USBdomain.
Lemma 2.22 Let R be an LFDdomain.
(1) Leta beanonzerononunitof R.Ifa isanSBelementof R,then?∞
(2) If R is catenary and R[X] is an SBdomain, then R is an SBdomain.
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 heightone 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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