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Trivial Ring Extensions Defined by Arithmetical-LikePropertiesAbdeslam Mimouni a , Mohammed Kabbour b & Najib Mahdou ba Department of Mathematics and Statistics , King Fahd University of Petroleum & Minerals ,Dhahran , Saudi Arabiab Department of Mathematics, Faculty of Science and Technology of Fez , University S. M.Ben Abdellah Fez , MoroccoPublished online: 23 Sep 2013.

To cite this article: Abdeslam Mimouni , Mohammed Kabbour & Najib Mahdou (2013) Trivial Ring Extensions Defined byArithmetical-Like Properties, Communications in Algebra, 41:12, 4534-4548, DOI: 10.1080/00927872.2012.705932

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Communications in Algebra®, 41: 4534–4548, 2013Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927872.2012.705932

TRIVIAL RING EXTENSIONS DEFINEDBY ARITHMETICAL-LIKE PROPERTIES

Abdeslam Mimouni1, Mohammed Kabbour2, and Najib Mahdou21Department of Mathematics and Statistics, King Fahd University of Petroleum& Minerals, Dhahran, Saudi Arabia2Department of Mathematics, Faculty of Science and Technology of Fez,University S. M. Ben Abdellah Fez, Morocco

In this article we investigate the transfer of the notions of elementary divisorring, Hermite ring, Bezout ring, and arithmetical ring to trivial ring extensions ofcommutative rings by modules. Namely, we prove that the trivial ring extension R �=A� B defined by extension of integral domains is an elementary divisor ring if and onlyif A is an elementary divisor ring and B = qf�A�; and R is an Hermite ring if and onlyif R is a Bezout ring if and only if A is a Bezout domain and qf�A� = B. We providenecessary and sufficient conditions for R = A� E to be an arithmetical ring when E

is a nontorsion or a finitely generated A−module. As an immediate consequences, weshow that A� A is an arithmetical ring if and only if A is a von Neumann regularring, and A�Q�A� is an arithmetical ring if and only if A is a semihereditary ring.

Key Words: Arithmetical ring and trivial ring extension; Bezout ring; Elementary divisor ring;Hermite ring; Valuation ring.

2010 Mathematics Subject Classification: 13D05; 13D02.

1. INTRODUCTION

All rings considered in this article are commutative with identity and allmodules are unital. A ring A is called a valuation ring if the set of all ideals of A istotally ordered by set inclusion, equivalently, for every elements x, y in A, x ∈ Ayor y ∈ Ax. This notion extends the notion of valuation domains to a ring with zero-divisors, and it is a particular class of uniserial modules.

Following Kaplansky [15], a ring A is said to be an elementary divisor ring ifevery matrix M over A is equivalent to a diagonal matrix, that is, there exist matricesP and Q which are invertible over A and such that PMQ is a diagonal matrix. In [17,Corollary 3.7], it is proved that a ring A is an elementary divisor ring if and only ifevery 2× 2 matrix over A is equivalent to a diagonal matrix. Noetherian elementarydivisor rings are just principal ideal rings, and the ring of entire functions is an

Received June 23, 2011; Revised May 15, 2012. Communicated by U. Walther.Address correspondence to Abdeslam Mimouni, Department of Mathematics and Statistics,

King Fahd University of Petroleum and Minerals, P. O. Box 278, Dhahran 31261, Saudi Arabia;E-mail: amimouni@kfupm.edu.sa

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TE DEFINED BY ARITHMETICAL-LIKE PROPERTIES 4535

example of a non-Noetherian (but Prüfer) elementary divisor ring (see [23]). Seekingfor elementary divisor rings outside the context of Noetherian rings, Kaplanskyproved that an elementary divisor ring A is Hermite (i.e., for every a, b in A, thereexist d, a′ and b′ in A such that a = da′, b = db′ and Aa′ + Ab′ = A), which is inturn a Bezout ring (i.e., every finitely generated ideal of A is principal). While itis well-known that a Bezout domain is Hermite ([15, Theorem 3.2]), the class ofBezout domains that are elementary divisor domains is unknown. However, in thecase of rings with zero-divisors, Gillman and Henriksen gave examples of a Bezoutring which is not Hermite and an Hermite ring which is not an elementary divisorring, see [6, Examples 3.4 and 4.11].

In 1932, Prüfer introduced and studied integral domains in which everynonzero finitely generated ideal is invertible [19], and later, in 1936 Krull [16] namedthese rings as Prüfer domains and stated equivalent conditions that make a domainPrüfer. The extension of this concept to rings with zero-divisors gave rise to manyclasses of Prüfer-like rings such as: semihereditary rings ( i.e., every finitely generatedideal is projective [4]); rings with weak global dimension ≤1 [7, 8]); arithmetical rings( i.e., every finitely generated ideal is locally principal [5, 13]); Gaussian rings ( i.e.,c�fg� = c�f�c�g� for any polynomials f , g with coefficients in A, where c�f� denotesthe content of f [21]); and Prüfer rings ( i.e., every finitely generated regular ideal isprojective [3, 10]).

We consider now the following properties on a ring A �

(1) A is a valuation ring;(2) A is an elementary divisor ring;(3) A is an Hermite ring;(4) A is a Bezout ring;(5) A is a semiheriditary ring;(6) w-dimA ≤ 1;(7) A is an arithmetical ring;(8) A is a Gausian ring;(9) A is a Prüfer ring.

Then we have the following implications: (1) ⇒ (2) ⇒ (3) ⇒ (4) ⇒ (7) ⇒ (8) ⇒ (9).The implications (1) ⇒ (2) ⇒ (3) ⇒ (4) are due to Kaplansky (see [15]), and wenoticed that Warfield [22, Theorem 3], and Larsen–Lewis–Shores [17, Theorem 3.8],proved separately that any valuation ring is an elementary divisor ring. Also we havethe implications (5) ⇒ (6) ⇒ (7), but all the above implications are not, in general,reversible, see for instance [6] and [9]. The diagram in Fig. 1 summarizes the pre-mentioned implications.

Let A be a ring and E an A-module. The trivial ring extension of A by E (alsocalled the idealization of E over A) is the ring R = A� E whose underlying set isA× E with multiplication given by �a� x��b� y� = �ab� ay + bx�. An element �a� x� ∈R is a unit if and only if a is. For the reader’s convenience, recall that if I is an idealof A and E′ is a submodule of E such that IE ⊆ E′, then I � E′ is an ideal of R;ideals of R need not be of this form. However, maximal ideals of R have the form�� E, where � is a maximal ideal of A. Considerable work, part of it summarizedin [1], Glaz’s book [7] and Huckaba’s book [12], has been concerned with trivialring extensions. These have proven to be useful in solving many open problems andconjectures for various contexts in (commutative and noncommutative) ring theory.

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Figure 1

The main purpose of this article is to study the transfer of the notions ofelementary divisor rings, Hermite rings, Bezout rings and arithmetical rings to someparticular classes of trivial ring extensions. Our motivation is to generalize well-known results and provide new and original families of examples of the abovenotions. It is worth mentioning that the transfer of some of the above 9 notions totrivial ring extensions were studied by different authors, namely we cite [1] and [2].Thus our findings come either as generalizations of well-known results or shed morelight on the structures of the rings and modules involved in this study.

Section 2 deals with the transfer of the notions of elementary divisor rings,Hermite rings, and Bezout rings in trivial ring extensions defined by ring extensions.Namely, we prove that for any extension of integral domains A ⊆ B, R �= A� B isan elementary divisor ring if and only if A is an elementary divisor ring and B =qf�A�; and R is an Hermite ring if and only if R is a Bezout ring if and only if A isa Bezout domain and B = qf�A� (Theorem 2.1).

Section 3 is devoted to the transfer of the notion of arithmetical rings. Mainly,we characterize when the trivial ring extension of a ring A by an A-module E isarithmetical when E is either a finitely generated or a non-torsion A-module. At thispoint we notice that a complete characterization of when a trivial ring extensionof a ring A by an arbitrary A-module is arithmetical is given in [1, Theorem 4.16].However, our approach is totally different and sheds more light on the structures ofthe ring A and the A-module E. Particulary, we prove that for a finitely generatedA-module E, R � A� E is an arithmetical ring if and only if if and only if for everymaximal ideal � of A, either A� is a valuation ring and E� = 0 or, A� is a fieldand E� is isomorphic to A� as an A�-vector space (Theorem 3.1). While if E is

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a non-torsion A-module, then A� E is an arithmetical ring if and only if A is areduced arithmetical ring and for every maximal ideal � of A, E� is isomorphic toA� as an A-module, where � is the unique minimal prime ideal contained in �. Inthis case � = �a ∈ A � �0 � a� � �� (Theorem 3.7). As an immediate applications, weobtain a generalization of [2, Theorem 2.1 (2)], that is, for an extension of integraldomains, A� B is an arithmetical ring if and only if A is a Prüfer domain and B =qf�A� (Corollary 3.8). Also we prove that for a ring A (which is not necessary anintegral domain), A� A is an arithmetical ring if and only if A is a von Neumannring (Corollary 3.5); and A�Q�A�, where Q�A� is the total quotient ring of A,is an arithmetical ring if and only if A is semihereditary (Theorem 3.9). We closethe article by extending [2, Theorem 2.1 (2)] to the trivial ring extension definedby a ring extension of non-integral domains. Particularly, we prove that if A hasa finite number of minimal prime ideals or Q�A� is a von Neuman regular ring,then A� B is an arithmetical ring if and only if A is a semihereditary ring and B =Q�A� (Corollary 3.13). Examples to illustrate the limit and scopes of our results areprovided.

2. ELEMENTARY DIVISOR RINGS, HERMITE RINGS AND BEZOUT RINGSIN TRIVIAL RING EXTENSIONS DEFINED BY RING EXTENSIONS

This section aims at characterizing when a trivial ring extension R = A�B defined by an extension of integral domains inherits any of the properties ofelementary divisor ring, Hermite ring, and Bezout ring. Let us denote by �n�X�,the set of all n× n matrices with entries from a set X. For any ring A, we will let�ln�A� denote the set of invertible matrices in �n�A�. If A is a ring, E is an A-module, and M = ��aij� eij�� ∈ �n�A� E�, we denote Ma = �aij� ∈ �n�A� and Me =�eij� ∈ �n�E�. Thus for every M ∈ �n�A� E�, M = Ma �Me. The product of twomatrices with entries from A� E� �Ma �Me��Na � Ne� is giving by

�Ma �Me��Na � Ne� = �MaNa�� �MaNe +MeNa�

Next, we state the main theorem of this section.

Theorem 2.1. Let A be an integral domain, B an extension of A, and let R = A� Bbe the trivial ring extension of A by B.

(1) R is an elementary divisor ring if and only if A is an elementary divisor ring andB = qf�A�.

(2) The following conditions are equivalent:

(i) R is an Hermite ring;(ii) R is a Bezout ring;(iii) A is a Bezout domain and B = qf�A�.

The proof of this theorem requires the following preparatory lemmas.

Lemma 2.2. Let A be a ring, and let E be an A-module.

(1) Let M ∈ �n�A� E�. Then M is invertible if and only if Ma is invertible.(2) If A� E is an elementary divisor ring, then so is A.

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Proof. (1) Let M = ��aij� eij�� ∈ �n�A� E�. The determinant of M , writtendetM , is the element of A� E given by

detM = ∑�∈�n

����n∏

i=1

�ai���i�� ei���i���

where �n denotes the set of all permutations on n letters and ���� denotes the signof �, for every � ∈ �n. The matrix M is invertible if and only if detM is a unit.But detM = �detMa� x� for some x ∈ E. Thus M is invertible if and only if Ma isinvertible.

(2) Assume that A� E is an elementary divisor ring, and let U ∈ �2�A�.Then the matrix U � 0 is equivalent to a diagonal matrix D in �2�A� E�. Thusthere exist P�Q ∈ �l2�A� E� such that P�U � 0�Q = D. We have the followingequalities:

D = �Pa � Pe��U � 0��Qa �Qe�

= �PaU � PeU��Qa �Qe�

= �PaUQa�� �PaUQe + PeUQa�

Thus PaUQa is a diagonal matrix. From part (1) of the proof, we deduce thatPa�Qa ∈ �l2�A�. Hence, U is equivalent to a diagonal matrix, as desired. �

Lemma 2.3. Let A be a ring and let E be an A-module. If A� E is a Bezout ring,then A is a Bezout ring and E is a divisible A-module.

Proof. Assume that A� E is a Bezout ring. Let I = a1A+ · · · + anA be a finitelygenerated ideal of A and let J be the finitely generated ideal of A� E given by J =�a1� 0��A� E�+ · · · + �an� 0��A� E�. Since A� E is a Bezout ring, J = �u� v��A�E� for some �u� v� ∈ J . Now, it is easy to check that I = uA as desired. It remains toprove that E is a divisible A-module. Let a be a non-zero-divisor element of A andx ∈ E. The ideal of A� E generated by �a� 0� and �0� x� is principal, and so thereexists �b� y� ∈ A� E such that �b� y��A� E� = �a� 0��A� E�+ �0� x��A� E�. Hencethere exist �� e�� ��� f�� ��� g�� � � h� ∈ A� E such that

�a� 0� = �b� y��� e�

�0� x� = �b� y���� f�

�b� y� = ��� g��a� 0�+ � � h��0� x�

Thus a = b, b = �a, �b = 0 and x = �y + bf . Since a is a non-zero-divisor elementof A, then so is b, and therefore � = 0. Hence x = bf = a��f�, and this completesthe proof of the lemma. �

Lemma 2.4. Let A ⊆ B be an extension of rings. If A� B is a Bezout ring, then B =Q�A�

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Proof. Assume that A� B is a Bezout ring. Let a be a non-zero-divisor element ofA. From Lemma 2.3, we deduce that 1 = ax for some x ∈ B (since B is a divisible A-module). Then 1

a∈ B and so Q�A� ⊆ B. Conversely, let x ∈ B and let I be the ideal

generated by �0� 1� and �0� x�. Since I is principal, there exists some y ∈ B such that

�0� 1��A� B�+ �0� x��A� B� = �0� y��A� B�. Therefore,

{�0� 1� = �a� e��0� y�

�0� x� = �b� f��0� y�for

some �a� e�� �b� f� ∈ A� B. Thus 1 = ay and x = by. It follows that a is a regularelement and x = b

a∈ Q�A�. Hence Q�A� = B, as desired. �

Proof of Theorem 2.1. . (1) The necessary condition follows from Lemma 2.2and Lemma 2.4.

Conversely, assume that A is an elementary divisor domain and B = qf�A�.We wish to show that R is an elementary divisor ring. Let M = Ma �Mb ∈ �2�R�.Two cases will be considered.

Case 1. Ma = 0. Put Mb = 1dM ′, where 0 �= d ∈ A and M ′ ∈ �2�A�. Since A is

an elementary divisor ring, M ′ is equivalent to a diagonal matrix with entries fromA. Hence there exist P�Q ∈ �l2�A� such that D = PMbQ is a diagonal matrix withentries from B. We then have the following equalities:

�P � 0�M�Q� 0� = �0� PMb��Q� 0� = 0� �PMbQ� = 0�D

But, by Lemma 2.2, P � 0 and Q� 0 are invertible matrices, and thus M isequivalent to a diagonal matrix.

Case 2. Ma �= 0 Let P1� Q1 ∈ �l2�A� such that P1MaQ1 =(

a1 00 a2

)for some

�0� 0� �= �a1� a2� ∈ A2. If a1 = 0, by multiplying the above equality on the left andright by the invertible matrix � 0 1

1 0 �, we obtain, P ′1MaQ

′1 =

(a2 00 0

) = D′1, where P ′

1 =� 0 11 0 � P1, and Q′

1 = Q1 �0 11 0 �, are invertible matrices. Thus without loss of generality,

we may assume that a1 �= 0. Put

P1MbQ1 =(x yz t

)� Q2 = Q1

(− xa1

− y

a1

0 0

)∈ �2�B�

and P2 =(

0 0− z

a1− t

a1

)P1 ∈ �2�B�. Substituting these expressions in

�P1 � P2�M�Q1 �Q2�, we obtain the following equalities:

�P1 � P2��Ma �Mb��Q1 �Q2� = �P1MaQ1�� �P1MaQ2 + P2MaQ1 + P1MbQ1�

=(a1 00 a2

)�

0 0

0�a1 − a2�t

a1

=�a1� 0� �0� 0�

�0� 0�(a2�

�a1 − a2�t

a1

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4540 MIMOUNI ET AL.

Thus there exist nonsingular matrices P = P1 � P2 and Q = Q1 �Q2 with entriesfrom R such that PMQ is a diagonal matrix. Consequently, R is an elementarydivisor ring, completing the proof of (1).

(2) �i� �⇒ �ii� is trivial and �ii� �⇒ �iii� follows from Lemma 2.3 andLemma 2.4. It remains to prove �iii� �⇒ �i�. Let �a� x�� �b� y� ∈ R. If a = b = 0, putx = x′

d� y = y′

dwhere x′� y′ ∈ A and 0 �= d ∈ A. Since A is a Bezout ring, there exist

a′� b′� e′ ∈ A such that x′ = a′e′

y′ = b′e′

e′ = x′ + �y′

for some � � ∈ A. Hence �0� x� = �a′� 0��0� e� and �0� y� = �b′� 0��0� e� where e =e′d. Also since �a′ + �b′ − 1�e′ = 0, a′ + �b′ = 1. Thus �� 0��a′� 0�+ ��� 0��b′� 0� =

�1� 0� and therefore �a′� 0�R+ �b′� 0�R = R.Now assume that �a� b� �= �0� 0�. Since A is an Hermite ring (as a Bezout

domain), there exist a1� b1� u� v ∈ A and 0 �= c ∈ A such thata = ca1

b = cb1

a1u+ b1v = 1

Thus �c� 0�(a1�

1cx) = �a� x� and �c� 0�

(b1�

1cy) = �b� y�. On the other hand,(

a1�1cx)R+ (

b1�1cy)R = R since �u� 0�

(a1�

1cx)+ �v� 0�

(b1�

1cy)is a unit. Hence R is

an Hermite ring, and this completes the proof of the theorem. �

Remark 2.5.

(1) The condition of being an integral domain cannot be removed in Theorem 2.1.Indeed, a trivial ring extension of a ring A by its total quotient ring Q�A� neednot be an arithmetical ring. In fact, a characterization of when such trivial ringextension is arithmetical is given in Theorem 3.9. It turns that this is equivalentto A being a semihereditary ring.

(2) The converse of the Lemma 2.3 is not true in general. Indeed, let K be a field andlet E be a K-vector space such that dimK E ≥ 2. Then the ideal of R �= K � Egenerated by �0� x� and �0� y�, where �x� y� is a free family of E, is not principalsince �0� x�R+ �0� y�R = 0� �Kx⊕ Ky�. Thus R is not a Bezout ring. However,K is trivially a Bezout ring and E is a divisible K−module.

(3) Let A be a ring, E an A-module, and let R = A� E be the trivial ring extensionof A by E. If R is an Hermite ring, then so is A. Indeed, let �a� b� ∈ A2. Thenthere exist �d� e�� �a′� x′�� �b′� y′� ∈ R such that

�a� 0� = �a′� x′��d� e�

�b� 0� = �b′� y′��d� e�

�a′� x′�R+ �b′� y′�R = R

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TE DEFINED BY ARITHMETICAL-LIKE PROPERTIES 4541

Thus a = a′db = b′da′A+ b′A = A�

and therefore, A is an Hermite ring as desired.

Next we give examples to illustrate Theorem 2.1. Recall that if A� E is avaluation ring, then A is a valuation ring.

Example 2.6. Let S be an elementary divisor domain, and let K be its quotientfield. Let K��x�� denotes the ring of formal power series over K in the indeterminatex. By [11, Example 1, p. 161], A = S + xK��x�� is an elementary divisor domain.Then A� K��x�� is an elementary divisor ring.

Example 2.7. Let A be an elementary divisor domain which is not a valuationdomain, for instance A = � (see [17, Proposition 2.7, p. 236] or [20, Corollaryp. 213]). Then R �= A� qf�A� satisfies the following statements:

(1) R is an elementary divisor ring;(2) R is not a valuation ring.

3. TRANSFER OF THE ARITHMETICAL PROPERTY

This section aims at characterizing when a trivial ring extension R �= A� Eof a ring A by an A-module E is an arithmetical ring, paying particular attentionto the cases where E is respectively a finitely generated A-module or a non-torsionA-module. It is worth noticing that Anderson and Winders [1, Theorem 4.16] gavea general characterization of when a trivial ring R = A� E is an arithmetical ring.Namely, they proved that R is an arithmetical ring if and only if E is an arithmeticalA-module (i.e., the lattice of submodules of E is distributive) and for each maximalideal � of A with E� �= 0, A� is a (valuation) domain and E� is a divisible A�-module. Several of the results of this section follow directly from their result. Ourmain contribution sheds more light on the structure of A� and E� when E is afinitely generated A-module; and the structure of A and E� when E is a non-torsionA-module. We start this section by noticing that it is easy to check that for any ringA and any A-module E, if R �= A� E is an arithmetical ring, then so is A (see also[2, Lemma 2.2]). Next, we state the first main theorem of this section.

Theorem 3.1. Let A be a ring, E a finitely generated A-module, and let R �= A� Ebe the trivial ring extension of A by E. Then R is an arithmetical ring if and only if forevery maximal ideal � of A, one of the two conditions is true:

(1) A� is a valuation ring and E� = 0;(2) A� is a field and E� is isomorphic to A� as an A�-vector space.

The proof of this theorem appeals to the following lemmas. A general versionof first lemma is due to Anderson and Winders, where a complete characterization

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of when A� E is a chained ring is given for arbitrary rings and modules. However,we list here a weaker version due to Kabbour and Mahdou and treats when a trivialring extensions is a valuation ring in the context of finitely generated or nontorsionA-module.

Lemma 3.2 ([1, Theorem 4.16] or [14, Theorem 2.1]). Let A be a ring and E annonzero A-module. Let R �= A� E be the trivial ring extension of A by E.

(1) Assume that E is a nontorsion A-module. Then R is a valuation ring if and only Ais a valuation domain and E is isomorphic to K �= qf�A�.

(2) Assume that E is a finitely generated A-module. Then R is a valuation ring if andonly if A is a field and E A.

The next lemma is due to Anderson–Winders ([1, Theorem 4.1]), and itshows how localization with a multiplicatively closed set behaves on the trivial ringextension R �= A� E. Recall that the maximal ideals of R �= A� E are of the form�� E where � is a maximal ideal of A ([1, Theorem 3.2]), and if M = �� E is amaximal ideal of R, where � is a maximal ideal of A, then RM is isomorphic, as aring, to A� � E�.

Lemma 3.3 ([1, Theorem 4.1]). Let A be a ring, E an A−module and R = A� ELet M be a maximal ideal of R�M = �� E for some maximal ideal � of A. Considerthe correspondence between the rings RM and A� � E�,

� � RM → A� � E� where �

(�a� x�

�s� y�

)=(as�sx − ay

s2

)

Then � is a ring isomorphism.

Proof of Theorem 3.1. For every maximal ideal � of A, E� is a finitely generatedA�-module. By Lemma 3.2,A� � E� is a valuation ring if and only if �A�� E�� satisfies(1) or (2). But R is arithmetical if and only if every localization at a maximal ideal Mis a valuation ring. The result is then an immediate consequence of Lemma 3.3. �

The next example illustrates Theorem 3.1 and provides a chain of arithmeticalrings whose union is an arithmetical ring. Recall that a ring A is said to be avon Neumann regular ring if for every x ∈ A, there is y ∈ A such that yx2 = x,equivalently, A� is a field for every maximal ideal � of A.

Example 3.4. Let K be a field of characteristic zero. We denote by A the ringof all eventually constant sequences of elements of K. For each n ∈ �, we definethe element en of the ring A by en�k� = �kn, where �kn is the Kronecker symbol.We also define the sequence u by u�k� = 1. Now, for each n ∈ �, let En be thefinitely generated ideal of A generated by e0� � en and Mn be the ideal of Agiven by Mn = A�u− en�. Note that Mn = �x ∈ A � x�n� = 0� and set M = K��� =�x ∈ A � ∃N ∈ ��∀k ≥ N � x�k� = 0�.

(1) A is a von Neumann regular ring.(2) �M�

⋃�Mn��n∈�� is the set of all maximal ideals of A.

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(3) For every n ∈ �, Rn �= A� En is an arithmetical ring.(4) R = ⋃

Rn = A�M is an arithmetical ring.

Proof. (1) For x ∈ A, we may define a sequence y by setting y�k� = 0 if x�k� = 0and y�k� = 1

x�k�if x�k� �= 0. It is easy to get successively that y ∈ A and that x2y = x.

Hence A is a von Neumann regular ring.

(2) Clearly, M is the ideal of A generated by all en. Let a ∈ A \M . Thenthere exist c ∈ K∗ and N ∈ �∗ such that a�k� = c for each k ≥ N . The element of Agiven by

N−1∑i=0

ei +(u−

N−1∑i=0

ei

)a

is invertible since all terms of this sequence are equal to c or 1. Hence M + Aa = A,and therefore, M is a maximal ideal of A. Now let b ∈ A\Mn. Then b�n� �= 0 and sou = u− en + 1

b�n�enb. Therefore,A = A�u− en�+ Ab. ThusMn is amaximal ideal ofA.

Conversely, let � be a maximal ideal of A. If for each n ∈ �� en ∈ �, thenM ⊆ �, and so � = M . If en �∈ � for some positive integer n, then Aen +� = A.Thus there exists �a� b� ∈ A×� such that u = aen + b. Multiplying this equality byu− en, we obtain u− en = b �u− en� ∈ �. Therefore, � = Mn, as desired.

(3) Fix n ∈ �, and let x be the sequence defined by x�k� = 0 if k ≤ n, andx�k� = 1 if k > n. It is easy to get the equality xEn = 0. Since x �∈ M , the localizationof En at the maximal ideal M is zero. On the other hand, let i ∈ �0� 1� � n�. Forevery j ∈ �0� 1� � n�� j �= i, we have eiej = 0. Then ei

∑0≤j≤n�j �=i Aej = 0, and hence

the localization of En at the maximal ideal Mi is AMi

ei1 , which is isomorphic to AMi

as an AMi-vector space since ei

1 �= 0. By the same way, we prove that the localizationof En at the maximal ideal Mi, where i > n, is zero. Consequently, Rn = A� En isan arithmetical ring (by Theorem 3.1).

(4) Since A is an elementary divisor ring, then it is Bezout, and so M isan arithmetical A-module. Now, for every maximal ideal � of A, since A is vonNeumann regular, A� is a field and thus M� is a divisible as an A�-module. By [1,Theorem 4.16], R is an arithmetical ring. �

The next corollary characterizes when a trivial extension of a ring A by itselfis arithmetical. It turns out that A is a von Neumann regular ring.

Corollary 3.5. Let A be a ring. Then A� A is an arithmetical ring if and only if A isa von Neumann regular ring.

The following corollary is an immediate consequence of the above corollary.It shows that for a trivial ring extension R =� A� A of an integral domain A byitself, all the pre-mentioned notions collapse and are equivalent to A being a field.

Corollary 3.6. Let A be an integral domain and R = A� A. Then the followingstatements are equivalent:

(1) R is a valuation ring;(2) R is an elementary divisor ring;

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4544 MIMOUNI ET AL.

(3) R is an Hermite ring;(4) R is a Bezout ring;(5) R is an arithmetical ring;(6) A is a field.

Now, we are ready to state our second main theorem. It characterizes whena trivial ring extension of a ring A by a nontorsion A-module E is an arithmeticalring.

Theorem 3.7. Let A be a ring, E a non-torsion A-module, and let R �= A� E bethe trivial ring extension of A by E. Then R is an arithmetical ring if and only if thefollowing statements hold:

(1) A is a reduced arithmetical ring;(2) For every maximal ideal� ofA, E� is isomorphic toA� as anA-module, where � is the

unique minimal prime ideal contained in�. In this case, � = �a ∈ A � �0 � a� � ��.

Proof. Notice first that R is an arithmetical ring if and only of for every maximalideal � of A, A� � E� is a valuation ring (Lemma 3.2). Let x ∈ E such that �0 �x� = 0, and let � be a maximal ideal of A. For each a

s∈ A�, we have the following

equivalences:

a

s

x

1= 0 ⇐⇒ ∃t ∈ A\�� tax = 0 ⇐⇒ ∃t ∈ A\�� ta = 0 ⇐⇒ a

s= 0

Hence the annihilator of x1 is zero, and therefore, E� is a nontorsion A�-module.

Next, from [14, Theorem 2.1], we deduce that R is an arithmetical ring if andonly if for every maximal ideal � of A, the following statements hold:

a) A� is a valuation domain.b) The A�-modules E� and qf �A�� are isomorphic.

By [18, Proposition 2.1], A� is an integral domain for every maximal ideal � of Aif and only if A is a reduced ring and every maximal ideal of A contains a uniqueminimal prime ideal. Therefore, R is an arithmetical ring if and only if A is a reducedarithmetical ring and for every maximal ideal � of A, E� is isomorphic to qf �A�� =A� as an A�- module, where � = {

a ∈ A � �0 � a� � �}. Finally, a mapping � �

E� → qf �A�� is an A�-isomorphism if and only if it is an A-isomorphism. �

Now, [2, Theorem 2.1] comes as an immediate application of Theorem 3.7.Thus, our next corollary restates when a trivial ring extension defined by extensionof integral domains is an arithmetical ring.

Corollary 3.8. Let A ⊆ B be an extension of domains, and let R = A� B be thetrivial ring extension of A by B. Then R is an arithmetical ring if and only if A is aPrüfer domain and B = qf�A�.

Proof. Since B is a nontorsion A-module, by Theorem 3.7, R is an arithmeticalring if and only if so is A and B� is isomorphic to A� as an A-module, where � is

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TE DEFINED BY ARITHMETICAL-LIKE PROPERTIES 4545

the unique minimal prime ideal contained in �, for every maximal ideal � of A.But since A and B are integral domains, A is a Prüfer domain and � = �0�. Thusqf�A� = A�0� = B� for each maximal ideal � of A. Hence B = qf�A�, completing theproof of the corollary. �

At this point, we note that if A is a domain and K = qf�A�, then A� K isan arithmetical ring if and only if A is a Prüfer domain, [2, Corollary 2.4]. Next,we state our third main theorem of this section. It is a generalization of the abovecorollary to any arbitrary ring.

Theorem 3.9. Let A be a ring. Then A�Q�A� is an arithmetical ring if and only ifA is a semihereditary ring.

Proof. First, we notice that a ring is semihereditary if and only if it is a reducedarithmetical ring and its full ring of quotient is a von Neumann regular ring, forinstance see [13, p. 120], or [8, Theorem 2.3].

⇐�� Assume that A is a semihereditary ring. We want to prove condition(2) of Theorem 3.7. For this, let � be a maximal ideal of A and � ={a ∈ A � �0 � a� � �

}be the unique minimal prime ideal contained in �.

Claim 1. For every non-zero-divisor element y of A and for every s ∈ A\�,sy �∈ �. Indeed, if sy ∈ �, then there exists v � � such that syv = 0. But since y is nota zero-divisor, sv = 0 ∈ �. Hence s ∈ � or v ∈ �, which is absurd.

Claim 2. For every as, a′

s′ ∈ A� and for every xy, x′

y′ ∈ Q�A�, if as= a′

s′ and xy=

x′y′ , then the elements ax

sy, a′x′

s′y′ are equal as elements of A�. Indeed, assume thatt�as′ − a′s� = 0 for some t ∈ A \� and �xy′ − x′y� = 0 for some non-zero-divisorelement of A. By Claim 1, t � � and t �axs′y′ − a′x′sy� = t�as′ − a′s�xy′ +t �xy′ − x′y�a′s = 0+ 0 = 0. Thus ax

sy= a′x′

s′y′ as elements of A�.

Now, consider the mapping � � A� ×Q�A� −→ A�,(

as� xy

)�→ ax

sy. By Claim 2,

� is well-defined, and clearly it is A-bilinear. Therefore, � induces a homomorphismf � A� ⊗A Q�A� → A� such that f

(as⊗ x

y

)= ax

sy. On the other hand, the map

� � Q�A�� −→ A� ⊗A Q�A�� where �(zs

)= 1

s⊗ z

is an A-module isomorphism.

Claim 3. f � � is an isomorphism. Indeed, if f � � ( zs

) = 0 for some zs∈

Q�A��, thenxys= 0 with z = x

y. Hence tx = 0 for some t ∈ A\� and x ∈ �. Therefore,

there exists u ∈ A\� such that ux = 0. Hence zs= 0 and thus ker f � � = 0. It

follows that f � � is injective. It remains to show that f is surjective. Let x ∈ Aand y � � = �a ∈ A� �0 � a� � ��. Since Q�A� is a von Neumann regular ring, thereexists a

b∈ Q�A� such that a

b

(y

1

)2 = y

1 . Hence 0 = ay2 − by = y�ay − b� and ay − b ∈�. Therefore, there exists t ∈ A\� such that tay = tb. Then �0 � tb� = �0 � t� ⊆ �,and hence tb �∈ �. It follows that

f(x1⊗ a

b

)= ax

b= atxy

tby= x

y

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4546 MIMOUNI ET AL.

Consequently, f � � is an A-module isomorphism. Finally, by Theorem 3.7, A�Q�A� is an arithmetical ring, as desired.

�⇒� Assume that A�Q�A� is an arithmetical ring. By Theorem 3.7, A isa reduced arithmetical ring since Q�A� is a nontorsion A-module. Therefore, it issufficient to prove that Q�A� is a von Neumann regular ring. Let a

b∈ Q�A�. Our aim

is to show that ab= z

(ab

)2for some z ∈ Q�A�. Let I (resp. J ) be the principal ideal

generated by(0� 1

b

)�resp. �a� 0��. The ideal I + J is again finitely generated. Using

condition II of [13, Theorem 3.], we get that ��I + J� � �I + J�� = �I � �I + J��+�J � �I + J��. But

{�I � �I + J�� = �I � J��J � �I + J�� = �J � I�

} Then R = �I � J�+ �J � I�, and therefore,

there exists �x1� z1� ∈ �I � J� such that �1� 0�− �x1� z1� ∈ �J � I�. Put�x1� z1��a� 0� =

(0�

1b

)�x2� z2�

��1� 0�− �x1� z1��

(0�

1b

)= �a� 0��x3� z3�

Then ax1 = ax3 = 0, az1 = x2b, and az3 = 1−x1

b. Thus a

b= a2z3 =

(ab

)2�b2z3�, as

desired. �

Our next examples illustrate Theorem 3.9. We show that A being a reducedarithmetical ring is not enough to get A�Q�A� an arithmetical ring.

Example 3.10. Let A be a semihereditary ring which is not a Bezout ring. ThenR = A�Q�A� is an arithmetical ring (by Theorem 3.9), but R is not a Bezout ring.

Example 3.11. Let �A�∈I be a family of semihereditary rings. We let Q =Q�A�, A = ∏

∈I A and Q = ∏∈I Q. The total ring of quotients of A is Q.

Since any arbitrary product of semihereditary rings is a semihereditary ring, Ais semihereditary. Therefore, A�Q is an arithmetical ring by Theorem 3.9. Forinstance, �n ��n is an arithmetical ring for every nonnegative integer n (since � isa semihereditary ring).

Example 3.12. Let A be the direct sum of An, with An = �/2� for each n ∈ �and define addition and multiplication componentwise in A. Let R = �× A, whereaddition is defined componentwise and multiplication is given by the formula

�m� a��m′� a′� = �mm′�ma′ +m′a+ aa′�

Let en =(�k�n

)kwith n ∈ � and let Pn = R�1� en�. By [18, Example 3], we get

successively that the full set of prime ideals of R is �m�× A� ∪ �Pn�, where mis a prime integer or m = 0, RPn

�/2� and that RM ��m� with M = m�× A.Thus R is a reduced arithmetical ring. We have Q�R� = ��2� × A (with twistedmultiplication). On the other hand, R is not a semihereditary ring since Q�R� is not avon Neumann regular ring. We conclude that R�Q�R� is not an arithmetical ring.

We do not know whether a trivial ring extension defined by a ring extensionof non-integral domains is arithmetical. However, we are able to extend [2,

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Theorem 2.1 (2)] to the case of non-integral domains but under the condition thateither A has a finite number of minimal primes or Q�A� is a von Neumann regularring.

Corollary 3.13. Let A ⊆ B be an extension of rings. If A has only finitely manyminimal prime ideals or Q�A� is a von Neumann ring, then A� B is an arithmeticalring if and only if A is a semihereditary ring and B = Q�A�.

Proof. Assume that A� B is an arithmetical ring. We need only show that B =Q�A�. Let a be a non-zero-divisor element of A and let � be a maximal ideal ofA. By Theorem 3.7, B� = A�, where � = �x ∈ A� �0 � x� � �� is the unique minimalprime ideal of A contained in �. Then au� = v� for some �u�� v�� ∈ B × �A\��.Since the ideal generated by each v�, where � ranges over all of Max�A�, is notcontained in any maximal ideal of A, there exists b ∈ B such that ab = 1. Weconclude that if A� B is an arithmetical ring the containment Q�A� ⊆ B alwayshold.

Conversely, let b ∈ B and let � be a maximal ideal of A. By Theorem 3.7, thereis some �r�� s�� ∈ A× �A \ ��, where � = �x ∈ A� �0 � x� � ��, such that s�b = r�.Let I be the ideal generated by all s� and let ��1� � �n� be the set of minimal primeideals of A. Then I �

⋃i

�i since I � �i for each i ∈ �1� � n�. But⋃

i �i is the set of

all zero-divisor elements of A ([18, Proposition 1.1]). It follows that I contains a non-zero-divisor element of A. Thus b ∈ Q�A� and so B = Q�A�. Finally, if Q�A� is a vonNeumann regular ring, by the previous part of the proof and [18, Proposition 1.4],B = Q�A�. The result now follows from the previous theorem. �

ACKNOWLEDGMENT

The authors would like to express their sincere thanks to the referee, whosecomments and corrections greatly improved this article.

This work is supported by KFUPM under Project FT1000002.

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