Journal of athematcs and Statstcs 7 (): 4448, 0 ISSN 5493644 00 Scence Publcatons odules n σ[] wth Chan Condtons on Small Submodules Al Omer Alattass Department of athematcs, Faculty of Scence, Hadramout Unversty of scence and Technology, P. O. Box 50663, ukalla, Yemen Abstract: Problem statement: Let be a rght module over a rng R. In ths artcle modules n σ[] wth chan condtons on small submodules are studed. Approach: Wth the help of known results about sngular, Artnan and Noetheran modules the technques of the proofs of our man results use the propertes of small, supplement and semmaxmal Results: odules n σ[] wth chan condtons on small are nvestgated, semmaxmal submodule s defned. Some Propertes of semmaxmal submodules are proved. As applcaton a new characterzaton of Artnan module n σ[] s obtaned n terms of small submodules and semmaxmal submodules, as well as small submodules and supplement Concluson/Recommendatons: Our results certanly generalzed several results obtaned earler. Key words: Small submodules, supplement submodules, chan condtons, sngular, supplemented module, fntely generated, unform dmenson, nonzero submodules, postve nteger INTRODUCTION ATERIALS AND ETHODS Throughout ths research, R denotes an assocatve rng wth unty and modules are untary rght R modules odr denotes the category of all rght R modules. Let be any R module. Any R module N s generated ( or generated by ) f there exsts an ( ) epmorphsm f: Λ N, for some ndexed set Λ. An R module N s sad to be subgenerated by f Ns somorphc to a submodule of an generated module. We denote by σ[] the full subcategory of the rght R modules whose objects are all rght Rmodules subgenerated by. Any module N σ [] s sad to be sngular f N L/K, for some L σ [] and K s Hence small submodules are the generalzaton of small submodules n the category odr Let L,K be two submodules of L s called a supplement of Kn f = L+K and L K L. L s called a supplement submodule of f L s a supplement of some submodule of. s called a supplemented module f every submodule of has a supplement n. If for every submodules L,K of wth =L+K there exsts a supplement N of L n such that N K, then s called an amply supplemented module. Now, let N σ [] and essental n L The class of all sngular modules s L,K N. L s called a supplement of K n N f closed under submodules, homohorphc mages and N=K+L and K L L. L s called a supplement drect sums. The concept of small submodule has been submodule of N f L s a supplement of some generalzed to small submodule by Zhou (000). submodule of N Ns called a supplemented Zhou called a submodule N of a module s small n ( notaton N ) f, whenever N+X= wth module f every submodule of N has a supplement. On the other hand N s called an /Xsngular, we have X= Ozcan and Alkan consder ths notaton n σ[] For a module N n amply supplemented module f for every σ[], Ozcan and Alkan (006) call a submodule L of N submodules L,K wth N= L+K there exsts a s small submodule, wrtten L N, n N f L supplement X of L such that X K. For the other defntons and notatons n ths study we refer to +K N for any proper submodule K of N wth N/K sngular. Clearly, f Ls small, then L s a small Anderson and Fuller (974) and Wsbauer (99). The propertes of small submodules that are submodule. lstedn n Zhou (000) Lemma.3 also hold n σ[]. 44
We wrte them for convenence Ozcan and Alkan, (006) lemma.3, Lemma.). Lemma.: Let N σ[]:. For modulesk and L wth, K L N, we have L N f and only f K N and L/K N/K. For submodules K and L of N, K+ L N f and only f K Nand L N 3. If K N,L σ[] and f:k Ls a homomorphsm, then f(k) L In partcular, f K N L, then K L 4. If K L N and K N, then K L Also Ozcan and Alkan (006) consder the followng submodule of a module N n σ[] Zhou (000). (N) = {K N: N / K s sngular smple } Lemma.: For any N n σ, (N) = {L N:L N}. The next Lemma s proven n Alattass (0). Lemma.3: Let N σ [] be supplemented. Then N/ (N) s semsmple. RESULTS AND DISCUSSION Theorem.: Let N σ []. Then (N) s Noetheran f and only f N satsfes ACC on small J. ath. & Stat., 7 (): 4448, 0 Theorem.3: Let N σ []. Then the followng are Proof: By lemma., every ascendng chan of small submodules of N s ascendng chan equvalent: submodules of (N). Hence the necessty s clear. Suffcency: Suppose to the contrary that (N) s a) (N) s Artnan. b) Every small submodule of N s Artnan. not Noetheran. Then there s a properly ascendng chan N N of submodules of (N). Let c) satsfes DDC on small submodules of N n Nand n N N, for each >. For each Proof: (a) (b). Ths s clear as every = j small j, let Kj = nr. Hence K j s fntely generated and submodules of N s a submodule of (N). = (b) (c). Ths s obvous. K j (N). So, by Lemma. and Lemma., (c) (a). By Anderson and Fuller (994), proposton Kj N, for each j. Hence K K. s a 0.0) t wll be suffce to show that every factor properly ascendng chan of small submodules of module of (N) s fntely cogenerated. For ths N. Ths mples N fals to satsfy ACC on small suppose that there exsts a factor module of (N) submodules, a contradcton. Thus (N) s Noetheran. whch s not fntely cogenerated. Then the set 45 Recall that a module s sad to have a unform dmenson n, where n s a nonnegatve nteger,f n s the maxmal number of summands n a drect sum of nonzero submodules of. In ths case we wrte u.dm = n and we say has a fnte unform dmenson. Theorem.: For any N σ [], the followng are equvalent: a) (N) has a fnte unform dmenson. b) Every small submodules of N has a fnte unform dmenson and there exsts a postve nteger n such that u.dml n, for any L N. c) N does not contan an nfnte drect sum of nonzero small submodules of N Proof: (a) (b). Ths s clear as any small submodule of N s contaned n (N). (b) (c). Assume that N N s an nfnte drect sum of nonzero small submodules of N. Then, by lemma., N N Nn+ Nand hence u.dm(n N N n+ ) n +, a contradcton to the hypothess. Hence (C) follows. (c) (a). Let N N be an nfnte drect sum of nonzero submodules of (N). For each, let n I be a nonzero element of N I Hence, by Lemmas. and., nr N. Thus nr nr s an nfnte drect sum of nonzero small submodules o f N Ths contradcts (C) and hence (N) has a fnte unform dmenson.
Λ= {L (N) : (N) / L s not fntely cogenerated} s nonempty. We show that Λ has a mnmal member. Let {L α } α Γ be a chan of submodules n Λ Consder the submodule L= L α. If L Λ, then α Γ (N) / L fntely cogenerated and so L= L α, for some a T a contradcton. Ths contradcton gves L Λ and we conclude that every chan of Λ has a lower bound n Λ. Hence, by Zorn's lemma, Λ has a mnmal member K. We clam that K N. Frst we show Soc( (N) / K) s not fntely generated. Let x (N) and x K. By lemmas.., xr N. Hence xr s Artnan. Ths mples (xr + K) / K s a nonzero Artnan as (xr + K) / K xr / (xr K). Therefore (xr + K) / K and hence (N) / K has an essental socle. Thus Soc( (N) / K) s not fntely generated Anderson and Fuller (000), Proposton 0.7. Now suppose that U s a submodules of N such that N = K+ Uwth N/U sngular. Let V be a submodule of (N), contanng K such that V/K= Soc( (N)/K). Then we have V = K + (U V). Suppose to the contrary that K U K. Then (N) / (K U) s fntely cogenerated. But J. ath. & Stat., 7 (): 4448, 0 V/K (K + (U V))/K (U V)/(K U) Soc( (N) / (K U)). So V/K s fntely generated, a contradcton. Ths contradcton gves K U = K and hence N=U Thus K N. Next we show V N. Suppose that W Nsuch that N=V+W wth N/W sngular. Then N / (K + W) = (U + W) / (K + W) U / (K + U W), mplyng that N/(K+ W) s semsmple. If N K+W then K+W N s contaned n a maxmal submodule Z of N Therefore N/Z s sngular smple. It follows that U (N) Z and so N=Z, a contradcton. Thus N=K+W whch wll mply N=W So V N. Therefore, by the hypothess, V and hence V/K s Artnan. The followng example explan that f every small submodule of N s Noetheran, then (N) need not be Noetheran. Example.4: Let R =, = and let N =, the (p ) Prufer P group. Hence N s an R module n fact N σ []. It s known that every submodule of N s Noetheran, but N s not Noetheran. oreover (N) N = Wang (007), Example.6. 46 Remark: If we look to a rng R as a module over t self and takng =R n.,.,.3 we get the results.3,.4,.5 n Wang (007) respectvely. Recall that a submodule N of an R module s called a semmaxmal submodule f N = N α, for some fnte set Λ wth N α and /N α sngular smple, for each α Λ. Here we consder ths defnton n the category σ []. Defnton.5: Let N σ[] and K N. K s called semmaxmal submodule of N f there s a fnte collecton {A α } α Λ of submodules of N such that K = and N/A α sngular smple for any A α α Λ α Λ. Snce any sngular module s sngular, any semmaxmal submodule of N σ [] s semmaxmal submodule of N. The next example gves a module wth a semmaxmal submodule whch s not semmaxmal submodue. Example.6: Let be a smple non projectve module. Then s sngular and not sngular Wsbauer (99). Hence the trval submodule s a semmaxmal submodule of but t s not semmaxmal submodule. Lemma.7: Let N σ []. Then:. (N) s contaned n any semmaxmal submodule of N. If N has DDC on the semmaxmal submodules, then N has a mnmal semmaxmal submodule Proof: The proof s standard and s omtted. Theorem.8: Let N σ []. Then the followng statements are equvalent: a) N s Artnan b) N satsfes DCC on small submodules and on semmaxmal submodules c) N satsfes DCC on small submodules and (N) s semmaxmal submodule d) N amply supplemented satsfes DCC on small submodules and suplementet Proof: (a) (b). Is obvous. α Λ
(b) (c). Let K be a mnmal semmaxmal submodule of N. We show that (N) = K. If (N) = N, then, by Lemma.7 (), N =(N) K and so (N) = K. Suppose that (N) N. By the defnton of (N) and Lemma.7 () t s suffce to show K L, for any submodule L of L wth N/Ls sngular smple. If L N such that N/L s sngular smple, then K Ls semmaxmal submodule of N Hence, by the mnmalty of K, K L= K and so K L. (c) (a). If N = (N), then N s Artnan by Theorem.3. Suppose that N (N). Then n = (N) = L, where N/L s sngular smple for each =, n Therefore N/ (N) s somorphc to a submodule of the fntely generated semsmple module = n N/L. Hence N/ (N) and so N s = Artnan. (d) (a). Suppose that N s an amply supplemented whch satsfes DCC on supplement submodules and small Then, by Theorem.3, (N) s Artnan and hence t s suffces to show N/ (N) s Artnan. N/ (N) s semsmple by Lemma.3. We clam that N/ (N) s Noetheran. Suppose that (N) N N s ascendng chan of submodules of N. We show by nducton there exsts descendng chan of submodules K K such that K s supplement N of n n for each. Snce N=N +N and N s amply supplemented, there exsts a supplement K of N n N Then N=N +K. Agan snce N=N +K,K, contans a supplement K of N n N. Now assume r and there s a descendng K K Kr of submodules such that K s supplementet of NI n N for each =,, r Hence N = Nr + Kr and so N = N + K. Agan snce Ns amply r+ r supplemented, we have a supplement K r + of N r + n N Proceedng n ths way we see that there exsts a descendng chan of submodules K K such that K s supplement of N n N for each. By the hypothess there exsts a postve nteger m such that Kn = K m, for each n m. Snce N=N +K J. ath. & Stat., 7 (): 4448, 0 47 and N K (N), N / (N) = N / (N) (K +(N) / (N). Thus Nn = N m, for each n m. Therefore N/ (N) s Noetheran and hence fntely generated. Thus N/ (N) s Artnan. Note: The condton N s amply supplemented n the statement (d) n Theorem.8 cannot be deleted (see the followng example). Example.9: Take RZ and =Z It s clear that σ[], satsfes DCC on supplement submodules and small submodules, but s not Artnan. The next corollary follows from the proof of (b) (c) n.8 and Lemma.7(). Corollary.9: If N satsfes one of the condtons of Theorem.8, then (N) s the least semmaxmal submodule of N. Corollary.0: The followng statements are equvalent for any R modulen. a) N s Artnan. b) N satsfes DCC on N small submodules and on semmaxmal N c) N satsfes DCC on N small submodules and N (N) N semmaxmal submodule. d) N s amply N supplemented satsfes DCC on N small submodules and N supplement e) N satsfes DCC on small submodules and on semmaxmal f) N satsfes DCC on small submodules and (N) s N semmaxmal submodule. g) N s amply supplemented satsfes DCC on small submodules and supplement Proof: (a) (b) (c) (d) s by takng =N n Theorem.8 and (a) (e) (f) (g) by takng =R n.8. Remark: The equvalence of (a,e,f,g) has been proved by Wang (007), Proposton.8 and Theorem (3.0) Then Theorem.8 s an extenson of such results. Corollary.: A fntely generated supplemented module N n σ[] s Artnan f and only f N satsfes DCC on small
J. ath. & Stat., 7 (): 4448, 0 Proof: The necessary part s trval. Suffcently part, suppose that N s a fntely generated supplemented module n σ[] satsfes DCC on small Then, by Lemma.3, N/ (N) s semsmple and hence t must be Artnan as N s fntely generated. By the hypothess and.3, (N) s Artnan. Thus N s Artnan. We end ths Artcle by showng that every factor module of a supplemented module that satsfes ACC on small submodules s also satsfes ACC on small Theorem.3: Let N σ[] be supplemented module. If N satsfes ACC on small submodules, then so does every factor modules of N. Proof. Let L N and let L /L L /L be an ascendng chan of a small submodules of N/L. Snce N s a supplemented module and L N, there exsts a submodule K of N such that N= L+K and L K K. Hence N/L (L+ K)/L K/L K. Let f:n/l K/L K be an somorphsm. Therefore for each, there exsts a submodule K of N contanng L K such that f(l /L) = K /K L. Hence, by Lemma., f(l /L) = K /K L K/L. Now we show that K N, for each. Suppose that X N such that N = K + X, wth N/X sngular. Then N/K L= K /K L + (X+ L K)/L K. But K /K L K/L and N/(X+ L K) s sngular. So N/K L = (X+ L K)/L K and hence N = (L K) + X. Therefore N=X. Thus we have a sendng chan K K of small submodules of N. Then, by the hypothess, there exsts a postve nteger n such that Kn = K n + =. Ths mples L/Ln = L/L n + =.Therefore N/L satsfes ACC on small small submodules s also has ACC on small submodules. () N s Artnan f and only f N satsfes DCC on small submodules and on semmaxmal submodules f and only f N amply supplemented satsfes DCC on small submodules and on supplement (v) If N s fntely generated supplemented, then N s Artnan f and N only f N satsfes DCC on small ACKNOWLEDGEENT The author s thankful for the facltes provded by department of mathematcs, at Unverst Tekonolog alaysa durng hs stay. REFERENCES Alattass, A., 0. On Supplemented and Lftng modules (submtted). Anderson, F.W. and K.R. Fuller, 974. Rngs and Categores of odules. st Edn., Sprngerverlage, New York, ISBN0: 0387900705, pp: 339. Ozcan, A.C. and. Alkan, 006. Semperfect modules wth respect to a preradcal. Comm. Alg., 34: 84 856. DOI: 0.080/009787050044593 Wang, Y., 007. small submodules and supplemented odules. Int. J. ath. ath. Sc., 007: 8. DOI: 0.55/007/583 Wsbauer, R. 99. Foundatons of odule and Rng Theory: A Handbook for Study and Research. st Edn., Gordon and Breach Scence Publshers, USA., ISBN0: 8848055, pp: 606. Zhou, Y.Q., 000. Generalzatons of Perfect, Semperfect, and semregular rngs. Alg. Coll., 7: 30538. CONCLUSION For any module N n σ[] we have obtaned a necessary and suffcent condtons for the sum of all small submodules of N to has a fnte unform dmenson. Also t s shown that () the sum of all small submodules of N s Noetheran (Artnan ) f and only f N satsfes ACC (DCC ) on small () Every factor module of a supplemented module n σ[] wth ACC on 48