Advances in Pure Mathematics, 2012, 2, 104108 http://dx.doi.org/10.4236/apm.2012.22014 Published Online March 2012 (http://www.SciRP.org/journal/apm) On PRegularity of Acts Akbar Golchin, Hossein Mohammadzadeh, Parisa Rezaei Department of Mathematics, University of Sistan and Baluchestan, Zahedan, Iran Email: agdm@math.usb.ac.ir Received September 27, 2011; revised December 17, 2011; accepted December 30, 2011 ABSTRACT By a regular act we mean an act that all its cyclic subacts are projective. In this paper we introduce Pregularity of acts over monoids and will give a characterization of monoids by this property of their right (Rees factor) acts. Keywords: PRegularity; Rees Factor Act 1. Introduction Throughout this paper will denote a monoid. We refer the reader to ([1]) and ([2]) for basic results, definitions and terminology relating to semigroups and acts over monoids and to [3,4] for definitions and results on flat ness which are used here. S S , A monoid is called left (right) collapsible if for every s zs zsS there exists such that zS zsz S ,, . A submonoid of a monoid is called weakly left collapsible if for all P sP zS the equality zsz implies that there exists an element such that . uPus , us SA monoid is called right (left) reversible if for every s S, there exist such that ,uvSus vs . us klK ,\ ,.Sx ySKxtytKxtyt v A right ideal K of a monoid S is called left stabilizing if for every , there exists such that and it is called left annihilating if, K lk k t If for all ,\ tSK :S S and all homomorphisms Ss StS , sft Kfsft then is called strongly left annihilating. K A right Sact satisfies Condition if for P as a s ,,A,aa , s S S implies the existence of such that and ,,uvaA ,aau a av ,aa A .usvs A right Sact A is called connected if for there exist 11 ,,, , nn tstS11 ,, n aaA and 111 12 22 1nn n asa t as at as at SF PF WPF WKF PWKF TKF such that We use the follow ing abbreviations: Strong flatness = ; Pullback flatness = ; Weak pullback flatness = ; Weak kernelflatness = ; Principal weak kernelflatness =; Translation kernelflatness = ; WP ; Weak homoflatness = Principal weak homoflatness = PWP WF PWF TF S ; Weak flatness = ; Principal weak flatness =; Torsion freeness = . 2. Characterization by PRegularity of Right Acts Definition 2.1. Let be a monoid. A right Sact is called Pregular if all cyclic subacts of A satisfy Condi tion P. We know that a right Sact is regular if every cy clic subact of is projective. It is obvious that every regular right act is Pregular, but the converse is no t true, for example if is a non trivial group, then is right reversible, and so by ([2, III, 13.7]), S is Pregular, but by ([2, III, 19.4]), S S S is not regular, since has no left zero element. S S S Theorem 2.1. Let be a monoid. Then: 1) is Pregular if and only if is right reversi ble. S SS 2) S is Pregular if and only if all principal right ideals of satisfy Condition P. is a right Sact and i iI, are subacts of 3) If , , then iI i is Pregular if and only if i is Pregular for every iI . 4) Every subact of a Pregular right Sact is Pregular. Proof. It is straightforward. q.e.d. Here we give a criterion for a right Sact to be P C opyright © 2012 SciRes. APM
A. GOLCHIN ET AL. 105 regular. Theorem 2.2. Let be a monoid and S a right S act. Then is Pregular if and only if for every aA and ,, yS ax ay,uv S implies that there exist such that and . aauav ux vy Proof. Suppose that is a Pregular right Sact and let , for and ax ay aA, ySaS. Then satis fies Condition . But PkerSa aS , and so by ([2, III, 13.4]), we are done. Conversely, we have to show that satisfies Con dition aS P for every a. Since AkeraS Sa , then it suffices to show that ker a S satisfies condition (P) and this is true by ([2, III, 13.4]) . q.e.d. We now give a characterization of monoids for which all right Sacts are Pregular. Theorem 2.3. For any monoid the following state ments are equivalent: S P P P x0S S 0, 1) All right Sacts are Pregular. 2) All finitely generated right Sacts are Pregular. 3) All cyclic right Sacts are Pregular. 4) All monocyclic right Sacts are Pregular. 5) All right Rees factor Sacts are Pregular. 6) S is a group or a group with a zero adjoined. Proof. Implications (1) (2) (3) (4) and (3) (5) are obvious. (4) (6). By assumption all monocyclic right S acts satisfy Condition , and so by ([2, IV, 9.9]), S is a group or a group with a zero adjoin ed. (5) (6). By assumption all right Rees factor Sacts satisfy Condition and again by ([2, IV, 9.9]), S is a group or a group with a zero adjoined. (6) (1). By ([2, IV, 9.9]), all cyclic right Sacts satisfy condition , and so by definition all right S acts are Pregular as required. q.e.d. Notice that freeness of acts does not imply Pregu larity, for if , with , then is free, but S is not Pregular, otherwise 0, 1,S2 x S SxS satis fies Condition as a cyclic subact of S, and so P ..0 xx, implies the existence of such that ,uv S xu xv and , and this is a contradiction. 0ux v Theorem 2.4. For any monoid the following state ments are equivalent: S 1) All right Sacts satisfying Condition E E are P regular. 2) All finitely generated right Sacts satisfying Condi tion are Pregular. 3) All cyclic right Sacts satisfying Condition E are Pregular. 4) All SF right Sacts are Pregular. 5) All SF finitely generated right Sacts are Pregular. 6) All SF cyclic right Sacts are Pregular. 7) All projective right Sacts are Pregular. 8) All finitely generated projective right Sacts are P regular. 9) All projective cyclic right Sacts are Pregular. 10) All projective generators in ActS are Pregular. 11) All finitely generated projective generators in Act S are Pregular. 12) All cyclic projective generators in ActS are P regular. 13) All free right Sacts are Pregular. 14) All finitely generated free right Sacts are Pregu lar. 15) All free cyclic right Sacts are Pregular. 16) All principal right ideals of S satisfy Condition P. ,, tz S 17) ,.zsztu vSzzuzvusvt S P Proof. Implications (1) (2) (3) (6) (9) (12) (15), (1) (4) (5) (6), (4) (7) (8) (9), (7) (10) (11) (12) and (10) (13) (14) (15) are obvious. (15) (16). As a free cyclic right Sact S is Pregular, and so by (2) of Theorem 2.1, all principal right ideals of S satisfy Condition . (17). By ([2, III, 13.10]), it is obvious. (16) (17) (1). Suppose the right Sact satisfies Condition Eax ayand let , for and aA , yS . Then there exist and such that aA uS aau and ux uy . Thus by assumption there exist tS uusut. and such that , xty ,aaua usas Therefore ,aa uautat xty, and so by Theorem 2.2, is Pregular. q.e.d . Notice that cofreeness does not imply Pregularity, otherwise every act is Pregular, since by ([2, II, 4.3]), every act can be embedded into a cofree act. But if 0,1,,Sx20,x with then as we saw before, is not Pregular, and so we have a contradiction. S S S Theorem 2.5. For any monoid the following state ments are equivalent: 1) All divisible right Sacts are Pregular. 2) All principally weakly injective right Sacts are P regular. 3) All fgweakly injective right Sacts are Pregular. 4) All weakly injective right Sacts are Pregular. 5) All injective right Sacts are Pregular. 6) All injective cogenerators in ActS are Pregular. 7) All cofree right Sacts are Pregular. 8) All right Sacts are Pregular. 9) S is a group or a group with a zero adjoined. Proof. Implications (1) (2) (3) (4) (5) (6) and (5) (7) are obvious. (6) (8). Suppose that is a right Sact, is an injective cogenerator in ActS and is an injective envelope of A (C exists by [2, III, 1 .23]). By ([5, Theo rem 2]), B C DBC D is an injective cogenerator in ActS, and so by assumption is Pregular. Since C, we have is Pregular. D, and so by Theorem 2.1, Copyright © 2012 SciRes. APM
A. GOLCHIN ET AL. 106 (7) (8). Let be a right Sact. Then by ([2, II, 4.3]), A can be embedded into a cofree right Sact. Since A is a subact of a cofree right Sact, by assumption A is a subact of a Pregular right Sact, and so by Theorem 2.1, is Pregular. (8) (9). By Theorem 2.3, it is obvious. x20x (8) (1). It is obvious. q .e.d. Theorem 2.6. Let S be a monoid. Then every strongly faithful right Sact is Pregular. Proof. By Theorem 2.2, it is obvious. q.e.d. Although strong faithfulness implies Pregularity, but faithfulness does not imply Pregularity, since every mo noid as an act is faithful, with 0, 1,S is faithful, but as we saw before, is not Pregular. Now see the following theorem. S S . Theorem 2.7. For any monoid S the following state ments are equivalent: 1) All faithfull right Sacts are Pregular. 2) All finitely generated faithfull right Sacts are P regular. 3) All faithfull right Sacts generated by at most two elements are Pregular. 4) S is a group or a group with a zero adjoined. Proof. Implications (1) (2) (3) are obvious. (3) (4). By Theorem 2.3, it suffices to show that every cyclic right Sact is Pregular. Thus we consider a cyclic right Sact and let SS aS aSSS Since S is faithful, S is faithful, also S is generated by at most two elements, thus by assumption S is Pregular. Since is a subact of aS S , by (4) of Theorem 2.1, is Pregular as required. aS (4) (1). By Theorem 2.3, it is obvious. q.e.d. S Since regularity does not imply flatness in general, Pregularity also does not imply flatness in general, but as the following theorem shows, for regular monoids P regularity implies flatness. Theorem 2.8. Let S be a regular monoid. Then every Pregular right Sact is flat. Proof. Suppose that S is a regular monoid, is a left Sact and S is a Pregular right Sact. Let in S am a m Mam for S and We show holds also in S ,aa a A m , SM . . mm Sm Sm Since in ama m S M , we have a tossing 11 2211 3 32 2 '' kk sm m smt m sm tm m tm 11 ,,, kk 11 1 12 22 1 kk k as a t as at as at of length , where k,, stt S . kS M 1,k , 1 1 ,, , kS aaA 1,,mm If then we have 11 11 11 . mm asa tmt m S1 asa t Since is regular, the equality 1 at atss implies that 1 111 , for 1 1 .Vs Since S is Pre gular, there exist S aA and uv such that ,S aa uav 1111 .vt s s and ut From the last equality we obtain 1 1111 111.umutm vtssm vtsm 11 msm Since , we get 11 , smm and so we have 111 11 1 11 1111 '' amassmassmat sm autsm avtsm avtsm aumaumam Sm Sm in S 2k.k 1 asa t . We now suppose that and that the required equality holds for every tossing of length less than From 11 we obtain equalities 11 1111 at atss for 11 Vs 111 asas tt and 1 for 1. Since 1 tVt S is Pregular, there exist 12 ,S aaA 12 ,, 12 ,vv S and uu such that 11111 ,aauav ut vtss 111111 2 12 111 ,.aauavus vstt and 222 2 21 12 22111112211 1 usmum ausaut usmutm 122111 112223 32 2 1 kk kkk usm utm ausat sm tm asatm tm 1.k Thus we have the fol lowing tossing of length 1 and of length From the tossing of length 1, we have 22 1122 aumausm S in M 1122 aumausm , and so we have 22 2122S in SumSusm 122111 11111 111,usmutm vtssmvtsmSm . Since 22 1122 aumausm we have S in Sm Sm 1k 1111 autmam . Also from the tossing of length , we have S in M . Thus we have 1111 autmam in 11 1SSut mSm 11111 1,utmvtsm Sm Since 1111 autmam we have , S in Sm Sm 2222112 2 1111 amaumauma usm autmam and so S Sm Sm in as required. q.e.d. Copyright © 2012 SciRes. APM
A. GOLCHIN ET AL. 107 3. Characterization by PRegularity of Right Rees Factor Acts In this section we give a characterization of monoids by P regularity of right Rees factor acts. Theorem 3.1. Let S be a monoid and S a right ideal of S. Then S is Pregular if and only if S SK S and S is right reversible or 1K S and all principal right ideals of S satisfy Condition .P Proof. Let S be a right ideal of S and suppose that S SK is Pregular. Then S satisfies Condition (P) If S SK S, then by ([2, III, 13.7]), S is right reversible, otherwise by ([2, III, 13.9]), 1K S, and so S SK S . Thus by (2) of Theorem 2.1, all principal right ideals of S satisfy Condition .P Conversely, suppose that S is a right ideal of If S .S S and S is right reversible, then by (1) of Theo rem 2.1, SS is Pregular. If SK 1 S K and all principal right ideals of satisfy Condition S P, then by (2) of Theorem 2.1, S SK S is Pregular. q.e.d. Although freeness of acts implies Condition P ,x20x in general, but notice that freeness of Rees factor acts does not imply Pregularity, for if with 0, 1,S , and S0, S then 0S SK S S S as a Rees fac tor act is free, but as we saw before, is not Pregu lar. S S SS Now let see the following theorem. Theorem 3.2. Let be a monoid and U be a property of Sacts implied by freeness. Then the follow ing statements are equivalent: 1) All right Rees factor Sacts satisfying property U regular. are P 2) All right Rees factor Sacts satisfying property U fy Condition P anither S contains no left zero or all principal right ideals of S satisfy Condition satis d e P. Proof. acts satisfyin (1) (2). By definiti Rees factor gon all right S property U satisfy Condition .P Suppose now that S containseft zero 0 z. Then 00S a l zS z is a right ideal of ,S and so SS S. by as p SK Since ). S S is free, S S Pregular, and so alrincipal right ideals of S satisfy Condition P. (2) (1Let is sumtion,l p S SK erty satisfies prop U for the rigideal S ht o Then by assumptionf S.S K satisfies Conditi PNow there are two cas follows: Case 1 S on . es as . S S. Then SS SK so by ([2, III, le, thus by , and 13.7]), S reversib(1) of Theorem 2.1, is right SS SK is Pregular. S Case 2. is a proper right ideal of S. Then by ([2, III, 13.9]), 1 S. Thus 0, S K z fo some 0 zSr , and so 0 z ption all prl right ides of S satisfy Condition P, that is is left zero. Thus by assuincipa al m SS SK S is Pregula r . q.e. d. y 3.1. For any monoiCorollar d S the following state mfactor Sacts satisfying Condition ents are equivalent: 1) All right Rees are Pregular. ) All WPF righ P 2t Rees factor Sacts are Pregular. PF SF roj lar. Pe right Rees factor Sacts are Pregular. deals of s 3) All right Rees factor Sacts are Pregular. 4) All right Rees factor Sacts are Pregular. 5) All pective right Rees factor Sacts are Pregu 6) All Rees factor projective generators in ActS are regular. 7) All fre 8) S contains no left zero or all principal right i Satisfy Condition .P P oof. By Theorem 3.2s obr, it ivious. q.e.d. ing state mees factor Sacts are Pregular. lat rl S Corollary 3.2. For any monoid S the follow ents are equivalent: 1) All WF right R 2) All fight Rees factor Sacts are Pregular. 3) S is not right reversible or no proper right idea , 2 S of S is left stabilizing, and if S contain , the all principal right ideals f S satisfy Condition Ks a left zerono .P Proof. Itw follos from Theorem 3.2, ([2, IV, 9.2]), and th . For any monoid S the following state mRees factor Sacts are Pregular. is righ at for Rees factor acts weak flatness and flatness are coinside. q.e.d. Corollary 3.3 ents are equivalent: 1) All PWF right 2) S t reversible, no proper right ideal S , 2Ks left zero a S of S is left stabilizing, and if S containa , thenll princip al right ideals of Ssatisfy Con dition .P Proo fof. It llows from Theorem 3.2, and ([2, IV, 9.7]). q. llary 3.4. For any monoid S the following state mes factor Sacts are Pregular. er tive mri S e.d. Coro ents are equivalent: 1) All TF right Re 2) EithS is a right reversible right cancella onoid or a ght cancellative monoid with a zero ad joined, and if S contains a left zero, then all principal right ideals of satisfy Condition .P Proof. It folls from Theorem 3.2, ad own([2, IV, 9.8]). q. ollary 3.5. For any monoid S the following st r Sacts satisfying Condition e.d. Cor atements are equivalent: 1) All right Rees facto P are Pregular. S is not right r W 2) eversible or no proper right ideal S , 2 S of S is left stabilizing and strongly left ihnd if S contains a left zero, then all prin cipal right ideals of satisfy Condition K ann ilating, aS .P Proof. It follows m Theorem 3.2, and fro([3, Proposi Copyright © 2012 SciRes. APM
A. GOLCHIN ET AL. Copyright © SciR APM 108 tio o monoid S the following state mfactor Sacts satisfying Condition P 2) S ie and no propeght ideal S 2012 es. n 3.26]). q.e.d. Corollary 3.6. Fis (2) (3) (4) are ob connected as a left Sact. Proof. Implications (1) r any ents are equivalent: 1) All right Rees vious. (1) (5). By Theorem 3.3, and ([4, Corollary 24]) it vio WP are Pregular. s right reversiblis obus. (2) (6). B r ri, 2 S of S is left stabilizing and left ann ihilating,d ains a left zero, then all principal right ideals of S satisfy Condition P. Proof. It follow y Theorem 3.3, and ([6, Proposition 8]) it vio K an if S cont s from Theorem 3.2 ([3, Corollary 3. nsider monoids over which Pregularity of R U be a prop er n a , and is obus. (4) (7). By Theorem 3.3, and ([6, Proposition 7]) it vio. By ([6, Proposition 28]), is obus. (4) (1) 27]). q.e.d. Here we co .PFPTKF Now if AS is a Pregu t is obvious that S Wlar right Rees factor Sact, then i satisfies Condi tion P, also by assumption S is,KF and so S ees factor acts implies other properties. Theorem 3.3. Let S be a monoid and T is W q.e.d. Corollary 3.10. .PF For any monoid S the following state m ty of Sacts implied by freeness. Thell Pregular right Rees factor Sacts satisfy property U if and only if S is not right reversible or S satisfies operty (U). Proof. Suppose that S is ht reversible. By (1) of ents are equivalent: 1) S is .WPF pr rig Theorem 2.1, SS SS atisfies pr is Pregular, and so by as sumption S soperty U. Conversel, suppose yS SK is egPrular for the right ideal S of S. Then thre two cases as follows: Cas. S ere a e 1 S. Then SS SK is Pregular, and S is right rso by (1) ofm 2.1, eversible. Thus by assumption Theore SS SK satisfies property (U). Case 2. S is a prideal of S. By Theooper t righ rem 3.1, 1K, and so SSS SK S. Ths uS SK is free, and so satisfies property . q.e. on U men all Pregu la ac from Teorem 3.3, and ([2, I, 5.23]). q. ollary 3.8. Let S be a monoid. Then all Pregular rig 2, III, 17.2 ]). q. ollary 3.9. Let S be a monoid. Then all Pregular rig ([2, III, 14 q. orem 3.4. For any monoid S the following state mt Rees factor Sacts are WPF. ke S 7) S is not right reveev,zS ker d Corollary 3.7. Let S be aoid. Th r right Rees factor Sts are free if and only if S is not right reversible or 1S. Proof. It followsh e.d. Cor ht Rees factor Sacts are projective if and only if S is not right reversible or S contains a left zero. Proof. It follows from Theorem 3. 3, and ([ e.d. Cor ht Rees factor Sacts are strongly flat if and only if S is not right reversible or S is left collapsible. Proof. It follows from Theorem 3. 3, and.3]). e.d The ents are equivalent: 1) All Pregular righ 2) All Pregular right Rees factor Sacts are WKF. 3) All Pregular right Rees factor Sacts are PWKF 4) All Pregular right Rees factor Sacts are TKF. 5) S is not right reversible or S is weakly left collapsible. 6) S is not right reversible or for every left ideal I of S, rf is connected for every homomorphism :. S fI S rsible or for ery 2) S is .WK F S ht revble and weakly left collapsible. S 3) is rigersi 4) is right reversible and for every left ideal of Sr fI 5) S is rit reversible and for every ,zS ker , kef is connected for every homomorphism :. SS S gh is ne (2) is obvious. concted as a left Sact. Proof. Implication (1) (1) (3). It is obvioby ([6, Corollary us 24]). (3) (4) (5). It is obvious by Theorem 3.4. (3) (4). I o bvious by ([6, Proposition 8]). q.e.dt is. lla ThCorory 3.11. Let S be a right reversible monoid. en S is WPF if ad only if S is .TKF ProoIt is ious that every PF rS n f. obvWight act is T.KF If S is ,TKF then by ([6osition 7]), for zS , Prop every , ker is lary connected as a left Sact, and so by Corol 3.10 S is .WPF q.e.d. REFERENCES [1] J. M. Howie, “roup Theory,” Cla Mikhalev, “Monoids, Acts Fundamentals of Semig rendon Press, Oxford, 1995. [2] M. Kilp, U. Knauer and A. and Categories,” W. de Gruyter, Berlin, 2000. doi:10.1515/9783110812909 [3] V. Laan, “Pullbacks and Flatness Properties of Acts,” perties of Acts I,” Tartu University Press, Tartu, 1999. [4] V. Laan, “Pullbacks and Flatness Pro Communications in Algebra, Vol. 29, No. 2, 2001, pp. 829 850. doi:10.1081/AGB100001547 [5] P. Normak, “Analogies of QFRing for Monoids. I,” Tar ks and tu Ülikooli Toimetised, Vol. 556, 1981, pp. 3846. [6] S. BulmanFleming, M. Kilp and V. Laan, “Pullbac Flatness Properties of Acts II,” Communications in Alge bra, Vol. 29, No. 2, 2001, pp. 851878. doi:10.1081/AGB100001548
