Research Article Generalized Malcev-Neumann …downloads.hindawi.com/archive/2015/595274.pdfResearch Article Generalized Malcev-Neumann Series Modules with the Beachy-Blair Condition

Research ArticleGeneralized Malcev-Neumann Series Modules withthe Beachy-Blair Condition

Mohamed A Farahat12

1Department of Mathematics and Statistics Faculty of Science Taif University PO Box 888 Al-Hawiyah Taif 21974 Saudi Arabia2Mathematics Department Faculty of Science Al-Azhar University PO Box 11884 Nasr City Cairo Egypt

Correspondence should be addressed to Mohamed A Farahat m farahat79yahoocom

Received 9 January 2015 Revised 9 March 2015 Accepted 10 March 2015

Academic Editor Andrei V Kelarev

Copyright copy 2015 Mohamed A Farahat This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We introduce a new class of extension rings called the generalizedMalcev-Neumann series ring 119877((119878 120590 120591))with coefficients in a ring119877 and exponents in a strictly ordered monoid 119878 which extends the usual construction of Malcev-Neumann series rings Ouyanget al in 2014 introduced the modules with the Beachy-Blair condition as follows A right 119877-module satisfies the right Beachy-Blaircondition if each of its faithful submodules is cofaithful In this paper we study the relationship between the right Beachy-Blaircondition of a right 119877-module119872

119877and its Malcev-Neumann series module extension119872((119878))

119877((119878120590120591))

1 Introduction

Throughout this paper 119877 denotes an associative ring withidentity (119878 sdot ⩽) is a strictly ordered monoid (ie (119878 ⩽) is anordered monoid satisfying the conditions that if 119904 lt 119904

1015840 then119904119905 lt 119904

1015840119905 and 119905119904 lt 119905119904

1015840 for 119904 1199041015840 119905 isin 119878) Recall that a subset119883 of (119878 ⩽) is said to be artinian if every strictly decreasingsequence of elements of 119883 is finite and that 119883 is narrowif every subset of pairwise order-incomparable elements of119883 is finite Suppose the two maps 120590 119878 rarr End(119877) and120591 119878 times 119878 rarr 119880(119877) (the group of invertible elements of119877) Let 119860 = 119877((119878 120590 120591)) denote the set of all formal sums119891 = sum

119909isin119878119886119909119909 such that supp(119891) = 119909 isin 119878 | 119886

119909= 0

is an artinian and narrow subset of 119878 with componentwiseaddition and the multiplication rule is given by

(sum

119909isin119878

119886119909119909)(sum

119910isin119878

119887119910119910)

= sum

119911isin119878

( sum

(119909119910)|119909119910=119911

119886119909120590119909(119887119910) 120591 (119909 119910)) 119911

(1)

for each sum119909isin119878

119886119909119909 and sum

119910isin119878119887119910119910 isin 119860 In order to ensure

the associativity it is necessary to impose two additionalconditions on 120590 and 120591 namely for all 119909 119910 119911 isin 119878

(i) 120590119909(120591(119910 119911))120591(119909 119910119911) = 120591(119909 119910)120591(119909119910 119911)

(ii) 120590119909120590119910= 120578(119909 119910)120590

119909119910 where 120578(119909 119910) denotes the auto-

morphism of 119877 defined by

120578 (119909 119910) (119903) = 120591 (119909 119910) 119903120591 (119909 119910)minus1

forall119903 isin 119877 (2)

It is now routine to check that 119860 = 119877((119878 120590 120591)) is a ringwhich is called the ring of generalizedMalcev-Neumann seriesWe can assume that the identity element of119860 is 1 this meansthat

1205901= Id119877 120591 (119909 1) = 120591 (1 119909) = 1 (3)

In this case 119903 997891rarr 1199031 is an embedding of 119877 as a subring into 119860For each 119891 isin 119860 0 we denote by 120587(119891) the set of

minimal elements of supp(119891) If (119878 le) is a strictly totallyordered monoid then supp(119891) is a nonempty well-orderedsubset of 119878 and 120587(119891) consists of only one element

Clearly the above construction generalizes the construc-tion of Malcev-Neumann series rings in case of 119878 = 119866

(an ordered group) which was introduced independently byMalcev and Neumann (see [1 2])

If the order le is the trivial order then 119860 = 119877((119878 120590 120591)) isthe usual crossed product ring 119877[119878 120590 120591] Also if the monoid119878 has the trivial order and 120591 is trivial then 119860 = 119877((S 120590 120591)) is

Hindawi Publishing CorporationAlgebraVolume 2015 Article ID 595274 4 pageshttpdxdoiorg1011552015595274

2 Algebra

the usual skewmonoid ring 119877[119878 120590] However if the monoid 119878has the trivial order and120590 is trivial then119860 = 119877((119878 120590 120591)) is theusual twistedmonoid ring119877[119878 120591] Finally if themonoid 119878 hasthe trivial order and 120590 and 120591 are trivial then 119860 = 119877((119878 120590 120591))

is the usualmonoid ring119877[119878] (see Sections 32 and 33 in [3])Moreover if 120572 is a ring endomorphism of 119877 set 119878 = Z

ge0

endowed with the trivial order Define 120590 119878 rarr End(119877) via120590(119909) = 120572

119909 for every 119909 isin Zge0

and 120591(119909 119910) = 1 for any 119909 119910 isin ZWe have 119860 = 119877((119878 120590 120591)) is the usual skew polynomial ring119877[119909 120572] However if le is the usual order then 119860 = 119877((119878 120590 120591))

is the usual skew power series ring 119877[[119909 120590]] If 120572 is a ringautomorphism of 119877 119878 = Z and le is the usual order then119860 = 119877((119878 120590 120591)) is the usual ring of skew Laurent power series119877[[119909 119909

minus1 120572]]

At the same time if we set also120590(119904) = Id119877isin End(119877) for all

119904 isin 119878 then it is easy to check that polynomial rings Laurentpolynomial rings formal power series rings and Laurentpower series rings are special cases of 119860 = 119877((119878 120590 120591))

If 119872119877is a unitary right 119877-module then the Malcev-

Neumann series module 119861 = 119872((119878)) is the set of allformal sums sum

119909isin119878119898119909119909 with coefficients in 119872 and artinian

and narrow supports with pointwise addition and scalarmultiplication rule is defined by

(sum

119909isin119878

119898119909119909)(sum

119910isin119878

119886119910119910)

= sum

119911isin119878

( sum

(119909119910)|119909119910=119911

119898119909120590119909(119886119910) 120591 (119909 119910)) 119911

(4)

where sum119909isin119878

119898119909119909 isin 119861 and sum

119910isin119878119886119910119910 isin 119860 One can easily

check that (i) and (ii) ensure that 119872((119878)) is a unitary right119860-module For each 120593 isin 119861 0 we denote by 120587(120593) the setof minimal elements of supp(120593) If (119878 le) is a strictly totallyordered monoid then supp(120593) is a nonempty well-orderedsubset of 119878 and 120587(120593) consists of only one element

Recall from Faith [4] that a ring 119877 is called a right zip ringand if the right annihilator r

119877(119883) of a subset 119883 sube 119877 is zero

then r119877(1198830) = 0 for a finite subset 119883

0of 119883 Although the

concept of zip rings was initiated by Zelmanowitz [5] it wasnot called so at that time

Recall from [6] that a right 119877-module119872119877is called a right

zip module provided that if the right annihilator of a subset119883 of119872

119877is zero then there exists a finite subset119883

0sube 119883 such

that r119877(1198830) = 0

According to Rodrıguez-Jorge [7] a ring 119877 satisfies theright Beachy-Blair condition if its faithful right ideals arecofaithful that is if 119868 is a right ideal of 119877 such that r

119877(119868)

vanishes then r119877(1198680) = 0 for a finite subset 119868

0of 119868 Clearly

a right zip ring is a right Beachy-Blair ringOuyang et al in [8] generalized the right Beachy-Blair

condition from rings into modules as follows A right 119877-module 119872

119877is called module with the Beachy-Blair condition

provided that if the right annihilator of a submodule 119873119877of


0sube 119873 such that

r119877(1198730) = 0

Themain aim of the present paper is to investigate condi-tions for theMalcev-Neumann seriesmodules119872((119878))

119877((119878120590120591))

to satisfy the right Beachy-Blair condition The proofs of ourresults obtained here are very similar to those obtained byOuyang et al in [8] and by Salem et al in [9]

2 Generalized Malcev-Neumann SeriesModules with the Beachy-Blair Condition

We start this section with the following notions and defini-tions

Let 119881 be a subset of119872119877 then

119881 ((119878))

= 120593 = sum

119909isin119878

119898119909119909 isin 119861 | 0 = 119898

119909isin 119881 119909 isin supp (120593)

(5)

Definition 1 A ring 119877 is called 119878-compatible if for all 119886 119887 isin 119877

and 119909 isin 119878 119886119887 = 0 if and only if 119886120590119909(119887) = 0

Definition 2 A right 119877-module119872119877is called 119878-compatible if

for each 119898 isin 119872 119886 isin 119877 and 119909 isin 119878 119898119886 = 0 if and only if119898120590119909(119886) = 0

Definition 3 A ring 119877 is called 119878-Armendariz if whenever119891119892 = 0 implies 119886

119909120590119909(119887119910) = 0 for each 119909 isin supp(119891) and

119910 isin supp(119892) where 119891 = sum119909isin119878

119886119909119909 and 119892 = sum

119910isin119878119887119910119910 are

elements of 119860

We extend the 119878-Armendariz concept to modules asfollows

Definition 4 A right 119877-module 119872119877is called 119878-Armendariz

if whenever 120593119891 = 0 implies 119898119909120590119909(119886119910) = 0 for each 119909 isin

supp(120593) and 119910 isin supp(119891) where 120593 = sum119909isin119878

119898119909119909 isin 119861 and

119891 = sum119910isin119878

119886119910119910 isin 119860

It is clear that 119877 is an 119878-Armendariz (119878-compatible) ringif and only if 119877

119877is an 119878-Armendariz (119878-compatible) module

For a subset 119880 of119872119877 we define r

119860(119880) as the set

r119860 (119880) = 119891 isin 119860 | (1199061) 119891 = 0 for each 119906 isin 119880 (6)

Lemma 5 Let 119872119877be a right 119877-module Then r

119860(119880) =

r119877(119880)((119878 120590 120591)) for any subset 119880 of119872

119877

Proof Let119891 = sum119904isin119878

119886119904119904 isin r119860(119880)Then for each 119906 isin 119880we have

(1199061)119891 = 0 Thus

0 = (1199061)(sum

119904isin119878

119886119904119904) = sum

119904isin119878

1199061205901(119886119904) 120591 (1 119904) 119904 = sum

119904isin119878

119906119886119904119904 (7)

which implies that 119906119886119904= 0 for each 119904 isin supp(119891) Hence

119886119904isin r119877(119880) for each 119904 isin supp(119891) So 119891 isin r

119877(119880)((119878 120590 120591)) and

r119860(119880) sube r

119877(119880)((119878 120590 120591))

On the other hand suppose that 119891 = sum119904isin119878

119886119904119904 isin r119877(119880)

((119878 120590 120591)) then 119886119904isin r119877(119880) for each 119904 isin supp(119891) Thus 119906119886

119904= 0

for each 119906 isin 119880 which implies that 1199061205901(119886119904)120591(1 119904) = 0 for each

119906 isin 119880 and 119904 isin supp(119891) Hence (1199061)119891 = 0 and 119891 isin r119860(119880) So

r119877(119880)((119878 120590 120591)) sube r

119860(119880) Therefore r

119860(119880) = r

119877(119880)((119878 120590 120591))

Algebra 3

When 119872119877= 119877119877we have the following consequence of

Lemma 5

Corollary 6 Consider r119860(119880) = r

119877(119880)((119878 120590 120591)) for any

subset 119880 of 119877

Note the following for 120593 = sum119909isin119878

119898119909119909 isin 119861 let C

120593= 119898119909|

119909 isin 119878 and for a subset 119881 sube 119872((119878)) we have C119881= cup120593isin119881

C120593

Lemma 7 Let 119872119877be an 119878-compatible and 119878-Armendariz 119877-

module Then

r119860 (119881) = r

119877(C119881) ((119878 120590 120591)) (8)

for any 119881 sube 119861

Proof Let119881 sube 119861 and 119879 = C119881= cup120593isin119881

C120593 = cup120593isin119881

119898119909| 119909 isin 119878

We show that r119860(119881) = r

119877(119879)((119878 120590 120591)) and it is enough to

show that r119860(120593) = r

119877(C120593)((119878 120590 120591)) for each 120593 = sum

119909isin119878119898119909119909 isin

119881 In fact let 119891 = sum119910isin119878

119886119910119910 isin r119860(120593) Then 120593119891 = 0 Since119872

119877

is an 119878 -Armendariz module119898119909119886119910= 0 for each 119909 isin supp(120593)

and 119910 isin supp(119891) Then 119886119910isin r119877(C120593) for each 119910 isin supp(119891)

Thus 119891 isin r119877(C120593)((119878 120590 120591)) and r

119860(120593) sube r

119877(C120593)((119878 120590 120591))

Now let 119891 = sum119910isin119878

119886119910119910 isin r119877(C120593)((119878 120590 120591)) Then 119886

119910isin r119877(C120593)

for each 119910 isin supp(119891) Hence 119898119909119886119910= 0 for each 119909 isin supp(120593)

and 119910 isin supp(119891) Since 119872119877is 119878-compatible it follows that

119898119909120590119909(119886119910) = 0 which implies that 119898

119909120590119909(119886119910)120591(119909 119910) = 0 for

each 119909 isin supp(120593) and 119910 isin supp(119891) Consequently

0 = sum

119911isin119878

( sum

(119909119910)|119909119910=119911

119898119909120590119909(119886119910) 120591 (119909 119910)) 119911 = 120593119891 (9)

So 119891 isin r119860(120593) and it follows that r

119877(C120593)((119878 120590 120591)) sube r

119860(120593) So

r119860 (119881) = ⋂

120593isin119881

r119860(120593) = ⋂

120593isin119881

r119877(C120593) ((119878 120590 120591))

= (⋂

120593isin119881

r119877(C120593)) ((119878 120590 120591))

= r119877 (119879) ((119878 120590 120591)) = r

119877(C119881) ((119878 120590 120591))

(10)

For a right 119877-module119872119877 we define

r119877(2119872) = r

119877 (119880) | 119880 sube 119872

r119860(2119861) = r

119860 (119881) | 119881 sube 119861

(11)

Lemma 5 gives us the map Π r119877(2119872) rarr r

119860(2119861) defined by

Π(119868) = 119868((119878 120590 120591)) for every 119868 isin r119877(2119872) Obviously Π is an

injective mapIn the following lemma we show thatΠ is a bijective map

if and only if119872119877is 119878-Armendariz

Lemma 8 Let119872119877be an 119878-compatible 119877-module The follow-

ing conditions are equivalent

(1) 119872119877is an 119878-Armendariz 119877-module

(2) Π r119877(2119872) rarr r

119860(2119861) defined byΠ(119868) = 119868((119878 120590 120591)) is

a bijective map

Proof (1)rArr(2)It is only necessary to show thatΠ is surjective Let119881 sube 119861

and 119879 = C119881 Since Π(r

119877(119879)) = r

119877(119879)((119878 120590 120591)) the proof of

this direction follows directly from Lemma 7(2)rArr(1)Let 119891 = sum

119910isin119878119886119910119910 isin 119860 and 120593 = sum

119909isin119878119898119909119909 isin 119861 such that

120593119891 = 0 Then 119891 isin r119860(120593) By assumption r

119860(120593) = 119879((119878 120590 120591))

for some right ideal 119879 of 119877 Hence 119891 isin 119879((119878 120590 120591)) whichimplies that 119886

119910isin 119879 sube r

119860(120593) for each 119910 isin supp(119891) So

120593(1198861199101) = 0 and we have that

0 = (sum

119909isin119878

119898119909119909) (119886

1199101) = sum

119909isin119878

119898119909120590119909(119886119910) 120591 (119909 1) 119909 (12)

for each 119909 isin supp(120593) and 119910 isin supp(119891) Thus 119898119909120590119909(119886119910) =

0 for each 119909 isin supp(120593) and 119910 isin supp(119891) So 119872119877is an 119878-

Armendariz module

Recall that a ring is reduced if it has no nonzero nilpotentelements Reduced rings have been studied for over forty-eight years (see [10]) In 2004 the reduced ring concept wasextended to modules by Lee and Zhou [11] as follows a right119877-module 119872

119877is reduced if for any 119898 isin 119872

119877and any 119886 isin 119877

119898119886 = 0 implies119898119877cap119872119886 = 0 Clearly if119872119877is reduced then

for all119898 isin 119872119877and 119886 isin 119877119898119886 = 0 implies119898119877119886 = 0 It is clear

that 119877 is a reduced ring if and only if 119877119877is a reduced module

Now we are able to prove the main result

Theorem 9 Let 119872119877

be a reduced 119878-compatible and 119878-Armendariz right 119877-module If 119872

119877satisfies the right Beachy-

Blair condition then 119861119860satisfies the right Beachy-Blair condi-

tion

Proof Suppose that a right 119877-module 119872119877satisfies the right

Beachy-Blair condition and 119869 is a right 119860-submodule of 119861such that r

119860(119869) = 0

From Lemma 8 we conclude that r119877(C119869)((119878 120590 120591)) =

Π(r119877(C119869)) = r119860(119869) = 0 Thus r

119877(C119869) = 0

Let C119869119877 denote the right 119877-submodule of 119872

119877generated

by C119869 Since C

119869sub C119869119877 we have r

119877(C119869119877) sub r

119877(C119869) = 0 Since

119872119877satisfies the right Beachy-Blair condition there exists a

finite subset

119883 =

119899119905

sum

119894=1

119902119905

119894119903119905

119894| 119902119905

119894isin C119869 119903119905

119894isin 119877 1 le 119905 le 119896 sub C

119869119877 (13)

such that r119877(119883) = 0 Let

1198830= 1199021

1 1199021

2 119902

1

1198991

1199022

1 1199022

2 119902

2

1198992

119902119896

1 119902119896

2 119902

119896

119899119896

(14)

Then1198830is a finite subset of C

119869 Nowwe will see that r

119877(1198830) =

0 Let 119886 isin r119877(1198830) then 119902

119905

119894119886 = 0 for 1 le 119894 le 119899

119905and 1 le 119905 le 119896

Since 119872119877is a reduced 119877-module then 119902

119905

119894119903119905

119894119886 = 0 for 1 le 119894 le

119899119905and 1 le 119905 le 119896 Then for each (sum

119899119905

119894=1119902119905

119894119903119905

119894) isin 119883 we have

(sum119899119905

119894=1119902119905

119894119903119905

119894)119886 = 0 Therefore 119886 isin r

119877(119883) = 0 and so r

119877(1198830) = 0

is proved

4 Algebra

For each 119902119905

119894isin 1198830 there exists an element 120593119905

119894isin 119869 such that

119902119905

119894isin C120593119905

119894

Let 119881 be a minimal subset of 119869 such that 120593119905119894isin 119881 for

each 119902119905119894isin 1198830 then119881 is a finite subset of 119869 and119883

0sub C119881 Thus

r119877(C119881) sub r119877(1198830) = 0 Now we show that r

119860(119881) = 0 Let the

contrary that is r119860(119881) = 0 and suppose that 119891 = sum

119910isin119878119887119910119910 isin

r119860(119881) 0 then 120593119891 = 0 for each 120593 = sum

119909isin119878119886119909119909 isin 119881 Let

119910 isin supp(119891) since119872119877is an 119878-Armendariz and 119878-compatible

module we have 119886119909119887119910= 0 for all 119886

119909isin C120593and each 120593 isin 119881

Hence 119887119910isin r119877(C119881) = 0 a contradiction Hence r

119860(119881) = 0

is proved Thus 119861119860satisfies the right Beachy-Blair condition


Theorem 9

Corollary 10 Suppose that 119877 is a reduced 119878-compatibleand 119878-Armendariz ring If 119877 satisfies the right Beachy-Blaircondition then 119860 satisfies the right Beachy-Blair condition

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The author would like to express deep gratitude to thereferee for hisher valuable suggestions which improved thepresentation of the paper

References

[1] A Malcev ldquoOn embedding of group algebras in a divisionalgebrardquo Doklady Akademii Nauk vol 60 pp 1499ndash1501 1948(Russian)

[2] B H Neumann ldquoOn ordered division ringsrdquoTransactions of theAmerican Mathematical Society vol 66 pp 202ndash252 1949

[3] A V Kelarev Ring Constructions and Applications WorldScientific River Edge NJ USA 2002

[4] C Faith ldquoAnnihilator ideals associated primes and Kasch-McCoy commutative ringsrdquoCommunications in Algebra vol 19no 7 pp 1867ndash1892 1991

[5] J M Zelmanowitz ldquoThe finite intersection property on anni-hilator right idealsrdquo Proceedings of the American MathematicalSociety vol 57 no 2 pp 213ndash216 1976

[6] C Zhang and J Chen ldquoZip modulesrdquo Northeastern Mathemat-ical Journal vol 24 pp 240ndash256 2008

[7] E Rodrıguez-Jorge ldquoRings with the Beachy-Blair conditionrdquoJournal of Algebra and Its Applications vol 11 no 1 Article ID1250006 2012

[8] L Ouyang J Liu and Y Xiang ldquoModules with the Beachy-Blairconditionrdquo Communications in Algebra vol 42 no 2 pp 853ndash871 2014

[9] R M Salem A M Hassanein and M A Farahat ldquoMalrsquocev-Neumann series over zip and weak zip ringsrdquo Asian-EuropeanJournal of Mathematics vol 5 no 4 2012

[10] G Renault ldquoAnneaux reduits non commutatifsrdquo Journal deMathematiques Pures et Appliquees vol 46 pp 203ndash214 1967

[11] T-K Lee and Y Zhou ldquoReduced modulesrdquo in Rings ModulesAlgebras and Abelian Groups vol 236 of Lecture Notes in Pureand Applied Mathematics pp 365ndash377 Marcel Dekker NewYork NY USA 2004

Submit your manuscripts athttpwwwhindawicom

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Algebra

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Stochastic AnalysisInternational Journal of

2 Algebra

the usual skewmonoid ring 119877[119878 120590] However if the monoid 119878has the trivial order and120590 is trivial then119860 = 119877((119878 120590 120591)) is theusual twistedmonoid ring119877[119878 120591] Finally if themonoid 119878 hasthe trivial order and 120590 and 120591 are trivial then 119860 = 119877((119878 120590 120591))

is the usualmonoid ring119877[119878] (see Sections 32 and 33 in [3])Moreover if 120572 is a ring endomorphism of 119877 set 119878 = Z

ge0

endowed with the trivial order Define 120590 119878 rarr End(119877) via120590(119909) = 120572

119909 for every 119909 isin Zge0

and 120591(119909 119910) = 1 for any 119909 119910 isin ZWe have 119860 = 119877((119878 120590 120591)) is the usual skew polynomial ring119877[119909 120572] However if le is the usual order then 119860 = 119877((119878 120590 120591))

is the usual skew power series ring 119877[[119909 120590]] If 120572 is a ringautomorphism of 119877 119878 = Z and le is the usual order then119860 = 119877((119878 120590 120591)) is the usual ring of skew Laurent power series119877[[119909 119909

minus1 120572]]

At the same time if we set also120590(119904) = Id119877isin End(119877) for all

119904 isin 119878 then it is easy to check that polynomial rings Laurentpolynomial rings formal power series rings and Laurentpower series rings are special cases of 119860 = 119877((119878 120590 120591))

If 119872119877is a unitary right 119877-module then the Malcev-

Neumann series module 119861 = 119872((119878)) is the set of allformal sums sum

119909isin119878119898119909119909 with coefficients in 119872 and artinian

and narrow supports with pointwise addition and scalarmultiplication rule is defined by

(sum

119909isin119878

119898119909119909)(sum

119910isin119878

119886119910119910)

= sum

119911isin119878

( sum

(119909119910)|119909119910=119911

119898119909120590119909(119886119910) 120591 (119909 119910)) 119911

(4)

where sum119909isin119878

119898119909119909 isin 119861 and sum

119910isin119878119886119910119910 isin 119860 One can easily

check that (i) and (ii) ensure that 119872((119878)) is a unitary right119860-module For each 120593 isin 119861 0 we denote by 120587(120593) the setof minimal elements of supp(120593) If (119878 le) is a strictly totallyordered monoid then supp(120593) is a nonempty well-orderedsubset of 119878 and 120587(120593) consists of only one element

Recall from Faith [4] that a ring 119877 is called a right zip ringand if the right annihilator r

119877(119883) of a subset 119883 sube 119877 is zero

then r119877(1198830) = 0 for a finite subset 119883

0of 119883 Although the

concept of zip rings was initiated by Zelmanowitz [5] it wasnot called so at that time

Recall from [6] that a right 119877-module119872119877is called a right

zip module provided that if the right annihilator of a subset119883 of119872


0sube 119883 such

that r119877(1198830) = 0

According to Rodrıguez-Jorge [7] a ring 119877 satisfies theright Beachy-Blair condition if its faithful right ideals arecofaithful that is if 119868 is a right ideal of 119877 such that r

119877(119868)

vanishes then r119877(1198680) = 0 for a finite subset 119868

0of 119868 Clearly

a right zip ring is a right Beachy-Blair ringOuyang et al in [8] generalized the right Beachy-Blair

condition from rings into modules as follows A right 119877-module 119872

119877is called module with the Beachy-Blair condition

provided that if the right annihilator of a submodule 119873119877of


0sube 119873 such that

r119877(1198730) = 0

Themain aim of the present paper is to investigate condi-tions for theMalcev-Neumann seriesmodules119872((119878))

119877((119878120590120591))

to satisfy the right Beachy-Blair condition The proofs of ourresults obtained here are very similar to those obtained byOuyang et al in [8] and by Salem et al in [9]

2 Generalized Malcev-Neumann SeriesModules with the Beachy-Blair Condition

We start this section with the following notions and defini-tions

Let 119881 be a subset of119872119877 then

119881 ((119878))

= 120593 = sum

119909isin119878

119898119909119909 isin 119861 | 0 = 119898

119909isin 119881 119909 isin supp (120593)

(5)

Definition 1 A ring 119877 is called 119878-compatible if for all 119886 119887 isin 119877

and 119909 isin 119878 119886119887 = 0 if and only if 119886120590119909(119887) = 0

Definition 2 A right 119877-module119872119877is called 119878-compatible if

for each 119898 isin 119872 119886 isin 119877 and 119909 isin 119878 119898119886 = 0 if and only if119898120590119909(119886) = 0

Definition 3 A ring 119877 is called 119878-Armendariz if whenever119891119892 = 0 implies 119886

119909120590119909(119887119910) = 0 for each 119909 isin supp(119891) and

119910 isin supp(119892) where 119891 = sum119909isin119878

119886119909119909 and 119892 = sum

119910isin119878119887119910119910 are

elements of 119860

We extend the 119878-Armendariz concept to modules asfollows

Definition 4 A right 119877-module 119872119877is called 119878-Armendariz

if whenever 120593119891 = 0 implies 119898119909120590119909(119886119910) = 0 for each 119909 isin

supp(120593) and 119910 isin supp(119891) where 120593 = sum119909isin119878

119898119909119909 isin 119861 and

119891 = sum119910isin119878

119886119910119910 isin 119860

It is clear that 119877 is an 119878-Armendariz (119878-compatible) ringif and only if 119877

119877is an 119878-Armendariz (119878-compatible) module

For a subset 119880 of119872119877 we define r

119860(119880) as the set

r119860 (119880) = 119891 isin 119860 | (1199061) 119891 = 0 for each 119906 isin 119880 (6)

Lemma 5 Let 119872119877be a right 119877-module Then r

119860(119880) =

r119877(119880)((119878 120590 120591)) for any subset 119880 of119872

119877

Proof Let119891 = sum119904isin119878

119886119904119904 isin r119860(119880)Then for each 119906 isin 119880we have

(1199061)119891 = 0 Thus

0 = (1199061)(sum

119904isin119878

119886119904119904) = sum

119904isin119878

1199061205901(119886119904) 120591 (1 119904) 119904 = sum

119904isin119878

119906119886119904119904 (7)

which implies that 119906119886119904= 0 for each 119904 isin supp(119891) Hence

119886119904isin r119877(119880) for each 119904 isin supp(119891) So 119891 isin r

119877(119880)((119878 120590 120591)) and

r119860(119880) sube r

119877(119880)((119878 120590 120591))

On the other hand suppose that 119891 = sum119904isin119878

119886119904119904 isin r119877(119880)

((119878 120590 120591)) then 119886119904isin r119877(119880) for each 119904 isin supp(119891) Thus 119906119886

119904= 0

for each 119906 isin 119880 which implies that 1199061205901(119886119904)120591(1 119904) = 0 for each

119906 isin 119880 and 119904 isin supp(119891) Hence (1199061)119891 = 0 and 119891 isin r119860(119880) So

r119877(119880)((119878 120590 120591)) sube r

119860(119880) Therefore r

119860(119880) = r

119877(119880)((119878 120590 120591))

Algebra 3


Lemma 5


119877(119880)((119878 120590 120591)) for any

subset 119880 of 119877


119898119909119909 isin 119861 let C

120593= 119898119909|


C120593


module Then

r119860 (119881) = r

119877(C119881) ((119878 120590 120591)) (8)

for any 119881 sube 119861


C120593 = cup120593isin119881

119898119909| 119909 isin 119878

We show that r119860(119881) = r


show that r119860(120593) = r

119877(C120593)((119878 120590 120591)) for each 120593 = sum

119909isin119878119898119909119909 isin


119886119910119910 isin r119860(120593) Then 120593119891 = 0 Since119872

119877



Thus 119891 isin r119877(C120593)((119878 120590 120591)) and r

119860(120593) sube r

119877(C120593)((119878 120590 120591))

Now let 119891 = sum119910isin119878

119886119910119910 isin r119877(C120593)((119878 120590 120591)) Then 119886

119910isin r119877(C120593)




119909120590119909(119886119910)120591(119909 119910) = 0 for


0 = sum

119911isin119878

( sum

(119909119910)|119909119910=119911

119898119909120590119909(119886119910) 120591 (119909 119910)) 119911 = 120593119891 (9)


119877(C120593)((119878 120590 120591)) sube r

119860(120593) So

r119860 (119881) = ⋂

120593isin119881

r119860(120593) = ⋂

120593isin119881

r119877(C120593) ((119878 120590 120591))

= (⋂

120593isin119881

r119877(C120593)) ((119878 120590 120591))

= r119877 (119879) ((119878 120590 120591)) = r

119877(C119881) ((119878 120590 120591))

(10)


r119877(2119872) = r

119877 (119880) | 119880 sube 119872

r119860(2119861) = r

119860 (119881) | 119881 sube 119861

(11)


119860(2119861) defined by







(2) Π r119877(2119872) rarr r

119860(2119861) defined byΠ(119868) = 119868((119878 120590 120591)) is

a bijective map


and 119879 = C119881 Since Π(r

119877(119879)) = r

119877(119879)((119878 120590 120591)) the proof of


119910isin119878119886119910119910 isin 119860 and 120593 = sum



119860(120593) = 119879((119878 120590 120591))


119910isin 119879 sube r


120593(1198861199101) = 0 and we have that

0 = (sum

119909isin119878

119898119909119909) (119886

1199101) = sum

119909isin119878

119898119909120590119909(119886119910) 120591 (119909 1) 119909 (12)



Armendariz module



119877and any 119886 isin 119877





Theorem 9 Let 119872119877




tion



119860(119869) = 0


Π(r119877(C119869)) = r119860(119869) = 0 Thus r

119877(C119869) = 0


119877generated

by C119869 Since C

119869sub C119869119877 we have r

119877(C119869119877) sub r

119877(C119869) = 0 Since


finite subset

119883 =

119899119905

sum

119894=1

119902119905

119894119903119905

119894| 119902119905

119894isin C119869 119903119905

119894isin 119877 1 le 119905 le 119896 sub C

119869119877 (13)

such that r119877(119883) = 0 Let

1198830= 1199021

1 1199021

2 119902

1

1198991

1199022

1 1199022

2 119902

2

1198992

119902119896

1 119902119896

2 119902

119896

119899119896

(14)



119877(1198830) =

0 Let 119886 isin r119877(1198830) then 119902

119905

119894119886 = 0 for 1 le 119894 le 119899

119905and 1 le 119905 le 119896


119905

119894119903119905

119894119886 = 0 for 1 le 119894 le


119899119905

119894=1119902119905

119894119903119905

119894) isin 119883 we have

(sum119899119905

119894=1119902119905

119894119903119905

119894)119886 = 0 Therefore 119886 isin r

119877(119883) = 0 and so r

119877(1198830) = 0

is proved

4 Algebra

For each 119902119905



119902119905

119894isin C120593119905

119894



0sub C119881 Thus


119860(119881) = 0 Let the


119910isin119878119887119910119910 isin


119909isin119878119886119909119909 isin 119881 Let





119860(119881) = 0



Theorem 9




Acknowledgment


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Algebra 3


Lemma 5


119877(119880)((119878 120590 120591)) for any

subset 119880 of 119877


119898119909119909 isin 119861 let C

120593= 119898119909|


C120593


module Then

r119860 (119881) = r

119877(C119881) ((119878 120590 120591)) (8)

for any 119881 sube 119861


C120593 = cup120593isin119881

119898119909| 119909 isin 119878

We show that r119860(119881) = r


show that r119860(120593) = r

119877(C120593)((119878 120590 120591)) for each 120593 = sum

119909isin119878119898119909119909 isin


119886119910119910 isin r119860(120593) Then 120593119891 = 0 Since119872

119877



Thus 119891 isin r119877(C120593)((119878 120590 120591)) and r

119860(120593) sube r

119877(C120593)((119878 120590 120591))

Now let 119891 = sum119910isin119878

119886119910119910 isin r119877(C120593)((119878 120590 120591)) Then 119886

119910isin r119877(C120593)




119909120590119909(119886119910)120591(119909 119910) = 0 for


0 = sum

119911isin119878

( sum

(119909119910)|119909119910=119911

119898119909120590119909(119886119910) 120591 (119909 119910)) 119911 = 120593119891 (9)


119877(C120593)((119878 120590 120591)) sube r

119860(120593) So

r119860 (119881) = ⋂

120593isin119881

r119860(120593) = ⋂

120593isin119881

r119877(C120593) ((119878 120590 120591))

= (⋂

120593isin119881

r119877(C120593)) ((119878 120590 120591))

= r119877 (119879) ((119878 120590 120591)) = r

119877(C119881) ((119878 120590 120591))

(10)


r119877(2119872) = r

119877 (119880) | 119880 sube 119872

r119860(2119861) = r

119860 (119881) | 119881 sube 119861

(11)


119860(2119861) defined by







(2) Π r119877(2119872) rarr r

119860(2119861) defined byΠ(119868) = 119868((119878 120590 120591)) is

a bijective map


and 119879 = C119881 Since Π(r

119877(119879)) = r

119877(119879)((119878 120590 120591)) the proof of


119910isin119878119886119910119910 isin 119860 and 120593 = sum



119860(120593) = 119879((119878 120590 120591))


119910isin 119879 sube r


120593(1198861199101) = 0 and we have that

0 = (sum

119909isin119878

119898119909119909) (119886

1199101) = sum

119909isin119878

119898119909120590119909(119886119910) 120591 (119909 1) 119909 (12)



Armendariz module



119877and any 119886 isin 119877





Theorem 9 Let 119872119877




tion



119860(119869) = 0


Π(r119877(C119869)) = r119860(119869) = 0 Thus r

119877(C119869) = 0


119877generated

by C119869 Since C

119869sub C119869119877 we have r

119877(C119869119877) sub r

119877(C119869) = 0 Since


finite subset

119883 =

119899119905

sum

119894=1

119902119905

119894119903119905

119894| 119902119905

119894isin C119869 119903119905

119894isin 119877 1 le 119905 le 119896 sub C

119869119877 (13)

such that r119877(119883) = 0 Let

1198830= 1199021

1 1199021

2 119902

1

1198991

1199022

1 1199022

2 119902

2

1198992

119902119896

1 119902119896

2 119902

119896

119899119896

(14)



119877(1198830) =

0 Let 119886 isin r119877(1198830) then 119902

119905

119894119886 = 0 for 1 le 119894 le 119899

119905and 1 le 119905 le 119896


119905

119894119903119905

119894119886 = 0 for 1 le 119894 le


119899119905

119894=1119902119905

119894119903119905

119894) isin 119883 we have

(sum119899119905

119894=1119902119905

119894119903119905

119894)119886 = 0 Therefore 119886 isin r

119877(119883) = 0 and so r

119877(1198830) = 0

is proved

4 Algebra

For each 119902119905



119902119905

119894isin C120593119905

119894



0sub C119881 Thus


119860(119881) = 0 Let the


119910isin119878119887119910119910 isin


119909isin119878119886119909119909 isin 119881 Let





119860(119881) = 0



Theorem 9




Acknowledgment


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









4 Algebra

For each 119902119905



119902119905

119894isin C120593119905

119894



0sub C119881 Thus


119860(119881) = 0 Let the


119910isin119878119887119910119910 isin


119909isin119878119886119909119909 isin 119881 Let





119860(119881) = 0



Theorem 9




Acknowledgment


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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Research Article Generalized Malcev-Neumann …downloads.hindawi.com/archive/2015/595274.pdfResearch Article Generalized Malcev-Neumann Series Modules with the Beachy-Blair Condition