International Journal of Algebra, Vol. 8, 2014, no. 5, HIKARI Ltd,

Internatonal Journal of Algebra, Vol. 8, 2014, no. 5, 229-238 HIKARI Ltd, www.m-hkar.com http://dx.do.org/10.12988/ja.2014.4212 On P-Duo odules Inaam ohammed Al Had Department of athematcs College of Educaton for pure scence/ Ibn-Al-Hatham Unversty of Baghdad, Baghdad, Iraq Rana Noor ajeed Department of athematcs College of Educaton for pure scence/ Ibn-Al-Hatham Unversty of Baghdad, Baghdad, Iraq Copyrght 2014 Inaam ohammed Al Had and Rana Noor ajeed. Ths s an open access artcle dstrbuted under the Creatve Commons Attrbuton Lcense, whch permts unrestrcted use, dstrbuton, and reproducton n any medum, provded the orgnal work s properly cted. Abstract In ths paper we ntroduce the concept P-dou module as a generalzaton of dou modules, where an R module s called a P-duo module f every pure submodule of s a fully nvarant. any results about ths concept are gven. Also some relatonshps between a P-dou module and other modules related wth t are presented. Keywords: P-duo modules, duo modules, weak duo modules, pure submodules. Introducton Throughout ths paper all rngs are commutatve rngs have an dentty and all modules are unttal. R s a rng and a left R-module. A submodule N of s called fully nvarant f f(n) N for every R-endomorphsm f of. Clearly 0 and are fully nvarant submodules of. The R-module s called a duo module f every submodule of s a fully nvarant. The rng R s a duo rng f t s duo as R-module. It s clearly that every rng s a duo rng. Harmanc and Smth n [1] ntroduced and studed weak duo modules, where an R-module s called

230 Inaam ohammed Al Had and Rana Noor ajeed a weak duo module provded every drect summand submodule of s fully nvarant. In ths paper we ntroduce purely duo module (brefly P-duo module) where an R- module s called a P-duo f every pure submodule of s fully nvarant where a submodule N of s called pure f I N = IN for each deal I of R, [2]. Equvalently, N s a pure submodule f I N = IN for each fntely generated deal I of R. It s well-known that every drect summand of a module s pure. Hence every P-duo module s weak duo. In ths paper we study P-duo module, and we gve some condtons under whch P-duo and weak duo are equvalent. Also we gve some relatonshps between P-duo module and other related modules. Defnton (1): An R-module s a P-duo module f every pure submodule of s a fully nvarant. Remarks and Examples (2): (1) It s clear that every duo module s P-duo and every P-duo s weakly duo, but the converse are not true n generals, for example: The Z-module Q s not duo, for the submodule Z of Q s not fully nvarant, x snce there exsts f:q Q defned by f(x) =, x Q and hence 2 1 f(z) = z Z 2. On the other hand, f s a dvsble module over a P.I.D R, then by [3,Cor.2.9, p.62] every pure submodule of s a drect summand. But Q has only two drect summands namely (0) and Q. Hence Q s a P-duo and a weakduo Z-module. Note that we clam that a weak-duo module need not be a P-duo module, but we have no example. (2) The Z-module Z 2 Z 2 s not weak-duo, snce there exsts f: Z 2 Z 2 Z 2 Z 2 defned by f(x, y) = (y, x), x, y Z2. So f N = Z 2 (0) (whch s a drect summand of Z 2 Z 2 ), then f(n) = (0) Z2 N. Also, Z 2 Z 2 s not a P-duo module and not a duo module. (3) The drect sum of P-duo modules may not be P-duo (see example two). (4) Every multplcaton module s a duo module, hence a P-duo module and a weakly duo module. (5) Every pure smple module (module wth only two pure submodules (0) and s a P-duo module, hence a weak-duo module. (6) A submodule of a P-duo module may not be a P-duo module as the followng example shows: Consder the vector space R 2 over R and let R=R R 2. Defne on R by:

On P-duo modules 231 (a,u) (b,v) = (ab,av + bu) for each a, b R and for each u, v R 2. Then R s a commutatve and so t s a duo R-module and so t s a P-duo module. However the submodule L=(0) R 2 whch s somorphc to R 2 s not a P-duo module, snce the submodule N= {(0,x):x R} s a pure submodule but t s not a fully nvarant submodule. Proposton (3): A drect summand of P-duo module s a P-duo module. Let N be a drect summand of a P-duo R-module. Then = N W for some W. Let K be a pure submodule of N and let f:n N be an R- homomorphsm module. Snce N s a drect summand, then N s a pure submodule n, hence K s a pure submodule n. f(x) x N Defne g: by: g(x) = 0 otherwse g s a well-defned R-homomorphsm. It follows that g(k) K, snce s a P- duo module and K s a pure submodule n. But g(k) = f(k). Hence f(k) K; that s K s a fully nvarant submodule of N. Thus N s a P-duo module. Recall that an R-module s called purely quas-njectve f for each pure submodule N of and for each f:n, there exsts an R-homomorphsm g: such that g = f where s the ncluson mappng [4]. f N g An R-module s called quas-projectve f for each R-module A and every epmorphsm g: and every R-homomorphsm f: N, there s f End() such that g f = f, [5] Proposton (4): Let be a P-duo R-module. Then (1) If s a purely quas-njectve, then every pure submodule of s a P-duo module. (2) If s a quas-projectve, then for any pure submodule K of, K s a P-duo R-module.

232 Inaam ohammed Al Had and Rana Noor ajeed (1) Let L be a pure submodule of N. Let f:n N. Snce N s a pure submodule n and s a purely quas-njectve module, there exsts g: such that g = f where s the ncluson mappng. Thus g (L) = g(l). But L s a pure submodule n N and N s a pure submodule n, mples L s a pure submodule n. Hence g(l) L. Also g (L) = f(l) = f(l). Thus g(l) = f(l) and so f(l) L. Therefore N s a P- duo module. f N N g (2) Let H K be a pure submodule of K and let g: be an R- K K homomorphsm. Let π : be the natural epmorphsm. Snce K K s quas-projectve, there exsts g*: such that g* = g. Hence g*(m) + K = g(m + K) for each m. But H K s a pure submodule n and K s a pure submodule n, K so that H s a pure submodule n. It follows that g*(h) H, snce s a P-duo module. H H Hence g ( ) g *(H) + K. Thus H s a P-duo module. K K K Proposton (5): Let be an R-module such that every cyclc submodule s pure. Then s a P-duo module f and only f for each f End() and for each m, there exsts r R such that f(m) = rm. ) Let f End(), m. Snce <m> s pure, then f(<m>) <m>. Hence the result s obtaned. ) The stated condton mples f(n) N for every f End(). It follows that s a duo module. Hence t s a P-duo module. Proposton (6): Let a module = 1 2 be a drect sum of submodules 1, 2 such that s a P-duo module. Then Hom( 1, 2 ) = 0.

On P-duo modules 233 Snce 1 s a drect sum of, 1 s a pure submodule n. But s a P- duo module, so 1 s a fully nvarant submodule n. Hence Hom( 1, 2 ) = 0 [1,lemma 1.9]. Theorem (7): Let an R-module = 1 2 be a drect sum of submodules 1, 2 such that ann 1 + ann 2 = R. Then s a P-duo module f and only f 1 and 2 are P-duo modules and Hom(, j ) = 0 for j,, j {1,2}. ) It follows by proposton 3 and proposton 6. ) Let N be a pure submodule of. Snce ann 1 + ann 2 =R, then by [6,Theorem 4.2] N = N 1 N 2 for some N 1 1, N 2 2. Hence N 1 s a pure submodule n 1 and N 2 s a pure submodule n 2. Let f: be an R-homomorphsm. Then f j f j : j j, j = 1,2, where j s the canoncal projecton and j s the ncluson map. Hence j f j (N j ) N j, j = 1,2, snce j (j = 1,2) s a P-duo module. oreover by hypothess k f j (N j ) = 0 for k j (k,j {1,2}). Then f(n) = f(n 1 ) + f(n 2 ) 2 ( f )(N ) 2 N N. Thus s a P-duo module. The followng s a well-known lemma. Lemma (8): Let be an R-module such that submodule of, then Theorem (9): Let a module N = (N ). I I j j j j j=1 j=1 =,. If N s a fully nvarant I = of submodules ( I). Then s a P-duo module f and only f (1) s a P-duo module for all I. (2) Hom(, j ) = 0 for all j,, j I. (3) N = (N ) for each pure submodule N of. I ) It follows by proposton 3, proposton 6 and lemma 8. ) Let N be a pure submodule of. By (3), N = (N ). Thus N s a pure submodule n. Let f:. For any j I. Consder the followng I j f j j j

234 Inaam ohammed Al Had and Rana Noor ajeed where j s the ncluson and j s the canoncal projecton. Hence j f j : j j and so j f j (N j ) N j for each j I. By condton (2), Hom(, j ) = 0 j. Hence f( (N )) ( j f j )(N j ) (N j I j ) = N. Thus s a P-duo module. j j I j I Recall that an R-module has the pure ntersecton property (brefly PIP) f the ntersecton of any two pure submodules s agan pure, [3, defnton 2.1,p.33]. Corollary (10): Let = wth for each I. Then s a P-duo module f the I followng condtons hold: (1) s P-duo for every fnte subset I of I. I (2) satsfes PIP. By (1), s a P-duo module for every I. Also j s a P-duo module for each j,, j I. Let N be any pure submodule of. Let x N, hence x = L, for some fnte subset I of I. Thus x N L. By condton (2), N L I s a pure submodule n. But N L L, so N L s a pure submodule n L. Snce L s a P-duo module by (1), N L s a fully nvarant submodule n L. Thus N L = [(N L) ] (N ). It follows that I I I x (N ). Thus N = (N ) I I theorem 9. x (N ) and so and hence s a P-duo module, by In [1], weak duo module satsfes summand sum property and summand ntersecton property, where A module s sad to satsfy summand sum property f K + L s a drect summand of whenever K and L are drect summands of. Also, satsfes the summand ntersecton property f K L s a drect summand of whenever K and L are drect summands of. We prove the followng: Proposton (11): Let be a P-duo module. If K s a drect summand of and L s a pure submodule of, then K L s a pure submodule of.

On P-duo modules 235 Snce K s a drect summand of, = K H for some H. Snce s a P-duo module and L s a pure submodule, then L s a fully nvarant. Hence L = (L K) (L H). Thus L K s a drect summand of L, so L K s a pure submodule n L. But L s a pure submodule n, hence L K s a pure submodule n. Proposton (12): Let be an R-module. Then satsfes (*), f and only f s a P-duo module, where satsfes (*) f for each pure submodules wth zero ntersecton, then ther sum s a fully nvarant submodule n. ) Let N be a pure submodule of. Let H = (0), then H s a pure submodule n and N H = (0). Hence by (*), N = N + H s a fully nvarant. Thus s a P-duo module. ) It s clear. To gve the next result, we gve the followng lemma: Lemma (13): Let be an R-module such that for each deal I of R and for each pure submodules N, W of such that I(N W) = IN IW. Then satsfes PIP. Let N and W be two pure submodules of. Then I N = IN and I W = IW. Hence I (N W) = (I N) W = IN W, also I (N W) = (I W) N = IW N. Hence IN W = IW N. On the other hand, I(N W) = IN IW. We clam that IN W = IN IW. Let x IN W = IW N. Hence x IN IW so IN W IN IW. The reverse ncluson s clear thus IN IW = IN W. Thus I (N W) = IN W = IN IW = I(N W). Therefore s satsfes PIP. Corollary (14): Let =, wth, I. Then s a P-duo module f the I followng condtons hold: (1) s a P-duo module for every fnte subset I of I. I (2) satsfes PIP. It follows drectly by corollary (10) and lemma (11). The followng remark s clear.

236 Inaam ohammed Al Had and Rana Noor ajeed Remark (15): Let be a semsmple R-module. Then the followng statements are equvalent: (1) s a duo-module. (2) s a P-duo module. (3) s a weak duo module. Recall that an R-module s called a fully stable module f every submodule N of s a stable submodule of, where N s called a stable submodule f f(n) N for any R-homomorphsm f:n,[6]. Note that every stable submodule s a fully nvarant submodule. Hence we get the followng result. Corollary (16): Let be a semsmple R-module. Then the followng statements are equvalent: (1) s a P-duo module. (2) s a duo module. (3) s a weak duo module. (4) s a fully stable module. Recall that an R-module s sad to be extendng (or CS), f every submodule of s essental n a drect summand of, [9]. Or equvalently, An R-module s extendng f and only f every closed submodule s a drect summand, [10,proposton 2.4, p.20]. Recall that an R-module s called a purely extendng module f every submodule of s essental n a pure submodule of [8]. It s clear every extendng module s a purely extendng module. A submodule N of an R-module s called closed f t has no proper essental extenson, [11]. Proposton (17): Let be a purely extendng R-module. Then the followng statements are equvalent: (1) s a P-duo module. (2) Every closed submodule of s a fully nvarant. (3) s a weak duo module. (1) (2) Let N be a closed submodule of. Snce s a purely extendng module, N s essental n a pure submodule (say H). Hence N = H, that s N s a pure submodule. It follows that N s a fully nvarant, snc s a P-duo module. (2) (3) Let N be a drect summand of. Hence N s a closed submodule of [11, exercse3, p.19] and so that N s a fully nvarant submodule of.

On P-duo modules 237 Corollary (18): Let be an extendng module over a PIR. Then the followng statements are equvalent: (1) s a P-duo module, (2) Every closed submodule of s a fully nvarant submodule of. (3) s a weak-duo module. (1) (2) (3) It follows by proposton 15. (3) (1) Let N be a pure submodule of. Snce R s a P.I.R then by [7,excersce 15,p.242], N s closed, and snce s extendng N s a drect summand and so by (3) N s a fully nvarant submodule of. Thus s a P-duo module. Recall that an R-module s sad to be regular f every submodule of s pure [12]. Proposton (19): Let be a regular R-module. Then the followng statements are equvalent: (1) s a P-duo module, (2) s a duo-module, (3) s a weak duo module. (1) (2) It s clear, snce every submodule of s pure. (2) (1) and (2) (3) are obvous. (3) (2) Let N. Snce s regular and s a weak duo. Hence by [13,proposton 2.2.10,p.44] s a fully stable, so every submodule of s stable. But every stable submodule s a fully nvarant. Thus s a duo module. It s well-known that every drect summand s pure but not conversely. Also t s clear that when every pure submodule s a drect summand, then P-duo module and weak duo module are equvalent. However under certan classes of modules, pure submodule s a drect summand, as the followng result, [3]. Lemma (20): Let be an R-module. Then (1) If s a prme njectve R-module, then every puure submodule s a drect summand [3,proposton 2.7,p.(61-62)]. (2) If s a dvsble module over a P.I.D., then every pure submodule s a drect summand [3,corollary 2.9,p.62]. (3) If s a fntely generated module over a Noetheran rng, then every pure submodule of s a drect summand [3,proposton 2.10,p.62]. (4) If s a Noetheran projectve R-module, then every pure submodule s a drect summand [3,proposton 2.11,p.63].

238 Inaam ohammed Al Had and Rana Noor ajeed Corollary (21): Let be an R-module. If s a prme njectve or s a dvsble over a P.I.D R, or s a fntely generated module over Noetheran rng or s a Noetheran projectve R-module, then s a P-duo module f and only f s a weak-duo. References 1. A.C.Özcan, A.Harmanc, Duo odules, Glasgow ath.j.48 (2006) 533-545. 2. F.W.Anderson and K.R.Fuller, Rngs and Categores of odules, Sprnger- Verlag, New York, Hedelberg-Berln, (1974). 3. B.H.Al-Bahraany, odules wth Pure Intersecton Property, Ph.D. Thess, Unversty of Baghdad, (2000). 4. ohanad Farhan Hamd, Purely Quas-Injectve odules,.sc. Thess, College of Scence, Al-ustansryah Unversty, (2007). 5. Anne Koehler, Quas-Projectve and Quas-Injectve odules, Pacfc Journal of athematcs, 1971, Vol.36, No.3. 6..S. Abas, On Fully Stable odules, Ph.D. Thess, College of Scence, Unversty of Baghdad, 1991. 7. T.Y.Lam, Lectures on odules and Rngs, Berkeley, Calforna Sprnger, 1998. 8. J.Clark, On Purely Extendng odules, The Proceedng of the Internatonal Conference on Abelan Groups and odules, 1999, 353-358. 9. anabu Harada, On odules wth Extendng Propertes, Harada.Osaka J.ath., (1982), Vol.19, pp.203-215. 10. Saad H. ohamed, Bruno J. uller, Contnuous and Dscrete odules, Combrdge Unversty Press, New York, Port Chester elbourne Sydney, (1990). 11. K.R.Goodearl, Rng Theory, Non Sngular Rngs and odules, arcel Dekker, Inc. New York and Basel, (1976). 12. D.J.Feldhous, Pure Theores, ath. Ann. 184, pp. 1-8, (1969). 13. Saad Abdulkadm Gatta Al-Saad, S-Extendng odules and Related Concepts, Ph.D. Thess, College of Scence, Al-ustansrya Unversty, 2007. Receved: February 11, 2014