Prüfer-Like Conditions in Subring Retracts and Applications

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Prüfer-Like Conditions in Subring Retracts and Applications

Chahrazade Bakkari ^a , Najib Mahdou ^a & Hakima Mouanis ^b

a Department of Mathematics, Faculty of Science and Technology of Fez , University S. M.

Ben Abdellah Fez , Morocco

b Department of Mathematics, Faculty of Science of Rabat , University Mohammed V Rabat , Morocco

Published online: 09 Oct 2009.

To cite this article: Chahrazade Bakkari , Najib Mahdou & Hakima Mouanis (2009) Prüfer-Like Conditions in Subring Retracts and Applications, Communications in Algebra, 37:1, 47-55, DOI: 10.1080/00927870802182515

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PRÜFER-LIKE CONDITIONS IN SUBRING RETRACTS AND APPLICATIONS

Chahrazade Bakkari¹, Najib Mahdou¹, and Hakima Mouanis²

1Department of Mathematics, Faculty of Science and Technology of Fez, University S. M. Ben Abdellah Fez, Morocco

2Department of Mathematics, Faculty of Science of Rabat, University Mohammed V Rabat, Morocco

In this article, we consider ﬁve possible extensions of the Prüfer domain notion to the case of commutative rings with zero divisors. We investigate the transfer of these Prüfer-like properties between a commutative ring and its subring retract. Our results generate new families of examples of rings subject to a given Prüfer-like conditions.

Key Words: Arithmetical rings; Gaussian rings; Nagata rings; Prüfer rings, Semihereditary rings;

Subring retract, Trivial ring extensions; Weak global dimension of rings.

1. INTRODUCTION

Throughout this article all rings are commutative with identity element and all modules are unital.

Prüfer (1932) introduced a new class of integral domains, namely, those domains R in which all ﬁnitely generated ideals are invertible. Through the years, Prüfer domains acquired a great many equivalent characterizations, each of which can, and was, extended to rings with zero-divisors in a number of ways. More precisely, we consider the following Prüfer-like properties on a commutative ring (Bazzoni and Glaz, 2006, 2007):

(1) Ris semihereditary, i.e., every ﬁnitely generated ideal is projective;

(2) The weak global dimension ofRis at most one;

(3) Ris an arithmetical ring, i.e., every ﬁnitely generated ideal is locally principal;

(4) R is a Gaussian ring, i.e., C_Rfg=C_RfC_Rg for any polynomials f g with coefﬁcients inR, whereC_Rfis the ideal ofRgenerated by the coefﬁcients off called the content ideal off;

(5) R is a Prüfer ring, i.e., every ﬁnitely generated regular ideal is invertible (equivalently, every two-generated regular ideal is invertible).

Received October 8, 2007; Revised November 13, 2007. Communicated by I. Swanson.

Address correspondence to Najib Mahdou, Department of Mathematics, Faculty of Science and Technology of Fez, Université de Fès, B.P. 229, Fès, Morocco; E-mail: mahdou@hotmail.com

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48 BAKKARI ET AL.

In Glaz (2005b), it is proved that each one of the above conditions implies the following next one (i.e., 1⇒2⇒3⇒4⇒5), and examples are given to show that in general the implications cannot be reversed. Moreover, an investigation is carried out to see which conditions may be added to some of the preceding properties in order to reverse the implications.

Recall that in the domain context, the five classes of Prüfer-like rings collapse to the notion of Prüfer domain. From Bazzoni and Glaz (2007, Theorem 3.12), we note that a Prüfer ring R satisfies one of the five conditions if and only if the total ring of quotients TotR of R satisfies the same condition. See for instance Bazzoni and Glaz (2006, 2007), Butts and Smith (1967), Glaz (2005a,b), Griffin (1970), Jensen (1966), Lucas (2005), and Tsang (1965).

For two ringsA⊆B, we say thatAis a module retract (or a subring retract) of Bif there exists anA-module homomorphism B−→Asuch thatA=idA;is called amodule retraction map. If such a mapexists,BcontainsAas anA-module direct summand.

Considerable works have been concerned with the descent and ascent of a variety of ﬁniteness and related homological properties between a ring and its subring retract. See for instance Bergman (1971a,b), Glaz (1992), Jondrup (1974), Kaplansky (1969), and Mahdou and Mouanis (2004).

A special application of subring retract is the notion of trivial ring extension.

Let A be a ring, E anA-module and R=A∝E, the set of pairs a ewith a∈A and e∈E, under coordinatewise addition and under an adjusted multiplication deﬁned bya ea e=aa ae+aefor alla a∈A e e∈E. ThenRis called thetrivial ring extensionofAbyE. It is clear thatAis a module retract ofR, where the module retraction mapis deﬁned byx e=x.

Trivial ring extensions have been studied extensively; the work is summarized in Glaz (1989) and Huckaba (1988). These extensions have been useful for solving many open problems and conjectures in both commutative and noncommutative ring theory. See for instance Glaz (1989), Huckaba (1988), and Kabbaj and Mahdou (2004).

In this article we investigate the transfer of the Prüfer-like properties between a commutative ring and its subring retract. Our results generate new and original examples which enrich the current literature with new families of Prüfer-like rings with zero-divisors.

2. PRÜFER-LIKE PROPERTIES IN SUBRING RETRACT

In this section we investigate the transfer of Gaussian, Prüfer, and arithmetical properties between a ring and its subring retract.

We begin by studying the transfer of Gaussian property. Recall thatNilRis the set of nilpotent elements in a ringR.

Theorem 2.1. LetRbe a ring andAa subring retract of R.

1) IfRis a Gaussian ring, then so isA.

2) Assume thatA Mis a local ring andR =A∝A/Mbe the trivial ring extension ofAbyA/M. ThenRis a Gaussian ring if and only if so isA.

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Proof. 1) Assume that R is a Gaussian ring, and let fX=n

i=0a_iXⁱ, gX=

m

i=0b_iXⁱ be two polynomials of AX, where n and m are two positive integers. Our aim is to prove that C_AfC_Ag⊆C_Afg. Let be an element of C_AfC_Ag, we have ∈C_RfC_Rg so, since R is Gaussian, ∈C_Rfg, i.e., =n+m

k=0

i+j=ka_ib_jr_k where r_k is an element of R for any 0≤k≤nm. Then, ==n+m

k=0

i+j=ka_ib_jr_k where is the module retraction map, which prove that∈C_Afg. Thus,Ais a Gaussian ring.

2) Assume thatA Mis a local ring andR =A∝A/Mbe the trivial ring extension ofA byA/M. If R is Gaussian, then so is A by 1). Conversely, the fact thatRis Gaussian in caseAis Gaussian follows easily from the characterization of local Gaussian rings given by Tsang (1965): a local ringR with maximal idealM is Gaussian if and only if for any two elementsa binM the following two conditions hold: 1)a b²=a²orb²; 2) ifa b²=a²andab=0, then b²=0.

The necessity of the conditions imposed in Theorem 2.1 will be proved in Examples 2.4 and 2.7.

Secondly, we study the transfer of Prüfer property between a ring and its subring retract. It is clear that each total ring of quotients is a Prüfer ring. Recall that an R-moduleE is called a torsion-freeif for every regular element a∈R and e∈E such thatae=0, we havea=0 or e=0.

Theorem 2.2. LetRbe a ring andAa subring retract ofR.

1) Assume that the module retraction map R−→Averiﬁes Keris torsion-free.

IfRis a Prüfer ring, then so isA.

2) Assume thatA Mis a local total ring of quotients, whereM is its maximal ideal;

and assume that the module retraction map veriﬁesM ker=0 and ker⊆ NilR. ThenRis a total ring of quotients; in particular,Ris Prüfer.

Proof. 1) Assume that Ker is torsion-free, where R−→A is the module retraction map, and R is a Prüfer ring. Let I=_n

i=1a_iA be a ﬁnitely generated regular ideal of A and a be a regular element of I. We tent to prove that I is invertible. Letbbe an element ofRsuch thatba=0, we haveba=0 sob= 0 sincea is a regular element of A, i.e., b∈Ker. On the other hand, by setting b=a+v∈R, wherea∈A andv∈Ker (sinceKer is a direct summand of R), we obtain that 0=aa+va, and so aa=va=0, therefore a=0 and v=0 as a is regular in A and Ker is torsion-free; which proves that a is a regular element ofRand so the idealJ =_n

i=1a_iRis a ﬁnitely generated regular ideal ofR.

Hence, sinceRis Prüfer,J is invertible inRand so the polynomialfX=_n

i=1a_iXⁱ is Gaussian inR(sinceJ=C_Rf). Using the proof of Theorem 2.1(1), we ﬁnd that fXis Gaussian inA; hence, asI =C_Afis a regular ideal ofA, it is invertible in A(by Bazzoni and Glaz, 2006, Theorem 4.2(2)). Thus,Ais Prüfer.

2) Assume that A M is a local total ring of quotients, where M is its maximal ideal; and assume that the module retraction mapveriﬁes M ker=0 andker⊆NilR. SetV =Ker.

In order to show thatRis a total ring of quotients, we have to prove that each elementa+vof Ris invertible or zero-divisor element. Indeed:

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a) Ifa∈M, then a is a noninvertible element of A; that is, a is zero-divisor in A (sinceAis a total ring of quotients). Hence there existsbnonzero element ofM such thatab=0. Therefore,ba+v=0 asMV =0, which means thata+vis a zero-divisor element inR.

b) If a M, thenais invertible inA and so inR; hence,a+vis invertible inRas sum of an invertible element and a nilpotent one.

Thus,Ris a total ring of quotients.

In the following example we prove that the retraction is not sufﬁcient to transfer the Prüfer property.

Example 2.3. Let A M be a non-Prüfer local ring and E a nonzero A-module such thatME=0. LetR =A∝Ebe the trivial ring extension ofAbyE. Then:

1) R is a total ring of quotients (since R is local with maximal ideal M∝E and M∝E0 e=0 for eache∈E). In particular,Ris Prüfer.

2) Ais a non-Prüfer subring retract ofR.

In the next example we ensure the necessity of the conditions imposed in Theorems 2.1(2) and 2.2(2).

Example 2.4. LetV Mbe a rank-one discrete valuation domain such that 2∈M (for instance, V =2). Then R =V ∝V is not Prüfer. In particular, R is not Gaussian.

Proof. It sufﬁces to show that R is not Prüfer. Let I =R20+R21 be a ﬁnitely generated ideal ofR. It is clear thatIis regular (since20is regular). Since R is local, the 2-generated regular ideal I is invertible if and only if it is principal.

Then, again sinceRis local,Iis principal if and only if it is generated by one of the two generators and this is false, so the conclusion follows easily.

We study now the transfer of arithmetical property between a ring and its subring retract.

Theorem 2.5. Let Rbe a ring and A a subring retract ofR. If Ris an arithmetical ring, then so isA.

Proof. By Jensen (1966, Theorem 2), it sufﬁces to show that for any pair of ideals I andJ ofA such thatI⊆J andJ is ﬁnitely generated, there should exist an ideal H ofAfor whichI=HJ.

We have IR⊆JR and JR is a ﬁnitely generated ideal of R; so, as R is arithmetical, there exists an ideal L of R such that IR=LJR that is IR=LJ. Therefore,I=IR=LJ=LJ and soAis arithmetical.

In the following example we prove that, under the same conditions as in Theorem 2.1(2), we cannot transfer the arithmetical property fromAto R.

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Example 2.6. Let A Mbe a valuation domain which is not a ﬁeld, whereM is its maximal ideal. Set R=A∝A/M be the trivial ring extension of A byA/M. Then:

1) Ais an arithmetical subring retract of the local ringR;

2) Ris not arithmetical.

Proof. 1) is clear. Also, we claim that R is not arithmetical. Let I =Ra0+ R0 ebe a ﬁnitely generated ideal ofR, whereais any nonzero element ofM and eis any nonzero element ofA/M. SinceR is local,I is principal if and only if it is generated by one of the two generators and this is false, so the conclusion follows

easily.

The following example proves that the condition “Ais a subring retract ofR”

cannot be removed in the proof of Theorems 2.1(1) and 2.5.

Example 2.7. Let K be a ﬁeld, KX Y the polynomial ring where X and Y are two indeterminate elements, and let QKX be the quotient ﬁeld of KX. Then QKXYis a Prüfer domain containing the subringKX Ywhich is not a Prüfer domain.

3. APPLICATIONS

In this section we give two applications to the results obtained in Section 2.

The ﬁrst application is devoted to trivial ring extensionR =A∝E of a ringA by anA-moduleE. Recall thatAis a module retract ofR, where the module retraction mapis deﬁned byx e=xandKer=0∝E.

Proposition 3.1. LetAbe a ring,EanA-module andR =A∝Ebe the trivial ring extension ofAbyE. Then:

1) a) Assume thatE =Keris torsion-free. IfRis a Prüfer ring, then so isA;

b) Assume that A M is a local ring, where M is its maximal ideal such that ME=0. ThenRis a total ring of quotients. In particular,Ris a Prüfer ring;

2) a) IfRis Gaussian, then so isA;

b) Assume thatE =A/M, whereM is a maximal ideal ofA. ThenRis a Gaussian ring if and only if so is A;

3) IfRis arithmetical, then so isA;

4) wdimR >1.

Proof. Let’s remark ﬁrst that R =A∝E, where A M is a local ring and ME=0, is a total ring of quotients (sinceRis a local ring with maximal idealM∝ EandM∝E01=0_R). By Section 2, it remains to show thatwdimR >1.

Letf ∈E− 0andJ =R0 f =0∝Af. Consider the exact sequence of R-modules:

0−→Keru−→R−→^u J −→0

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where ua e=a e0 f=0 af. Hence, Keru=Annf∝E. We claim that J is not ﬂat. Deny. Then, by Rotman (1979, Theorem 3.55), J =J∩Keru= JKeru=0∝AfAnnf∝E=0∝Annff=0, a contradiction. Therefore,

J is not ﬂat andwdimR >1.

As shown below, Proposition 3.1 enriches the literature with new examples of non-Gaussian Prüfer rings.

Example 3.2. Let A M be a non-Prüfer local domain, where M is its maximal ideal, and letE be a nonzero A-module such that ME=0. Let R =A∝E be the trivial ring extension of AbyE. Then:

1) Ris Prüfer by Proposition 3.1(1.b);

2) Ris not Gaussian by Proposition 3.1(2.a) sinceA is not Gaussian (asA is non- Prüfer domain).

For enrich the literature with new examples of nonarithmetical Gaussian rings, we propose the next two examples.

Example 3.3. Let A M be a valuation domain, where M is its maximal ideal, and let R =A∝A/Mbe the trivial ring extension ofA byA/M. Then:

1) Ris Gaussian by Theorem 3.1(2.b) sinceA is a valuation domain;

2) Ris not arithmetical by Example 2.6(3).

Example 3.4. Let kbe a proper subﬁeld of a ﬁeld K, and letR =k∝K be the trivial ring extension of kbyK. Then:

1) Ris Gaussian by Bakkari and Mahdou (2006, Example 2.3(2.b));

2) R is not arithmetical by Bakkari and Mahdou (2006, Example 2.3(2.c)) sinceR is local.

Now we construct an arithmetical ringRsuch thatwdimR >1.

Example 3.5. LetK be a ﬁeld and R =K∝K be the trivial ring extension of K byK. Then:

1) Ris arithmetical;

2) wdimR= .

Proof. 1) Ris arithmetical by Bakkari and Mahdou (2006, Example 2.3(1.a)).

2) The ideal I R01is not ﬂat by the proof of Proposition 3.1(4). On the other hand, the exact sequence ofR-modules

0−→I−→R−→^u I−→0

whereua e=a e01=0 ashows thatfd_RI= . Hence, wdimR=

and this completes the proof.

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Let us see in the following example that even if we replace the ﬁeldK, in the above example, by a principal total ring of quotientsA, we don’t have in general R =A∝Ais Gaussian; in particular, it is not arithmetical.

The same example is a Prüfer non-Gaussian ring.

Example 3.6. Let A =/2ⁱ, wherei≥2 be an integer, and letR =A∝Abe the trivial ring extension ofAbyA. Then:

1) Ais a local principal total ring of quotients with maximal idealM=2A;

2) Ris a local total ring of quotients. In particular,Ris a Prüfer ring;

3) Ris not Gaussian. In particular,Ris not arithmetical.

Proof. 1) and 2) are clear since R is local with maximal ideal M ∝A and M∝A02ⁱ⁻¹=0_R.

It remains to show that R is not Gaussian. For that let f =2ⁱ⁻¹0+ 2ⁱ⁻¹1X ∈RX. We have f²=0 (and so C_Rf²=0) and C_Rf²=R02ⁱ⁻¹ =0R). Therefore,Ris not Gaussian and this completes the proof of Example 3.6.

The second application is devoted to the Nagata rings. Let A be a ring and R =AX=S⁻¹AX the localization of AX by S, where S is the multiplicative set inAXformed by all polynomialsfXsuch thatCf=A. By construction we have AX=A+XAX+C where C= ^fx_gx/fx gx∈AX d^ofX < d^ogx and Cgx=A (by Lam, 1978, Chapter IV, Proposition 1.4(1)), which implies thatA is a module retract of AX. The ring R =AX is called the Nagata ring.

See for instance Huckaba (1988) and Lam (1978).

By Section 2, we obtain the following proposition.

Proposition 3.7. LetAbe a ring,R =AXbe the Nagata ring. Then:

1) Assume thatE =Keris torsion-free. IfRis a Prüfer ring, then so isA;

2) IfR =AX is Gaussian, then so isA;

3) IfR =AX is arithmetical, then so isA.

Recall that a subset S of a ring R is called dense if AnnS=0. A ring R is called strongly Prüfer if every ﬁnitely generated dense ideal is locally principal.

Notice that the Nagata ring AX is Prüfer if and only if A is strongly Prüfer by Huckaba (1988, Theorem 18.10). For instance, a strongly Prüfer ring is a Prüfer ring by Huckaba (1988, Lemme 18.1).

Recall that a ringRsatisﬁesCH-property if each ﬁnitely generated ideal of Rhas a nonzero annihilator. It is clear that aCH-ring is strongly Prüfer ring. For instance, the trivial ring extensionR =A∝Eis aCH-ring (and so strongly Prüfer ring) for each local ringA M(whereM is its maximal ideal) and anA-moduleE such thatME=0.

Now we construct a new examples of non-Gaussian Prüfer rings, as shown below.

Example 3.8. Let A be a non-Gaussian CH-ring, and let R =AX be the Nagata ring. Then:

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1) Ris Prüfer;

2) Ris a not Gaussian.

Proof. 1) It is clear that A is strongly Prüfer ring since A is a CH-ring.

Therefore,Ris a Prüfer ring by Huckaba (1988, Theorem 18.10).

2) Ris not Gaussian by Proposition 3.6(2) sinceAis not Gaussian.

ACKNOWLEDGMENTS

The authors would like to express their sincere thanks for the referee for his/her helpful suggestions and comments, which have greatly improved this article.

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