مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 On Weakly Quasi-Prime Module M. A. Hassin Department of Mathematical, College of Basic Education, University of Al-Mustansriyah Received in: 3 April 2011 Accepted in: 18 October 2011 Abstract In this work we shall introduce the concept of weakly quasi-prime modules and give some properties of this type of modules. Key words: Prime module, quasi-prime module, weakly quasi-prime module. 1- Introduction Let R be a commutative ring with unity , and let M be an R-module, we introduce that an R-module M is called weakly quasi–prime module if annRM = annRrM for every r annRM, where annRM = {r: rR and rM = 0}. The main purpose of this work is to investigate the properties of weakly quasi-prime modules, and we give several characterizations of weakly quasi-prime modules. Recall that an R-module is called prime if annRM = annRN for every non-zero submodule N of M and annRM = {r: rR and rM = 0}, [1]. A submodule N of M is said to be prime if a m  N for a  R, m  M, then either m  N or a  [N:M] where [N:M] = {r: rR, rM  N}, [1], [2]. It was shown that in [1] M is prime module iff (0) is p rime submodule. The concept of quasi-prime module is introduced in [3] where an R-module M is quasi-prime module if annRN is prime ideal for every nonzero submodule N of M. If M is quasi-prime module then annRM = annRrM  r  annRM, [3]. But the converse is not true for example: Let M =  p Z as Z-module is not quasi-prime module since if N = <1/p 2 + z >   p Z . So annRN = p2z is not prime ideal in Z. But ann p Z = 0 and  r  0, let a  ann r p Z so a r p Z  = 0, so a r  ann p Z . a r = 0, but r  0 so a = 0 so ann r p Z = 0. Then ann p Z = ann r p Z . 2- Weakly Quasi-Prime Module In this section we introduce the concept of weakly quasi-prime module and give several results about it. 2.1 Definition: An R-module M is called weakly quasi-prime module (briefly W.q.p) if annRM = annRrM for every r ann RM. Recall that if R is an integral domain, an R-module M is said to be divisible iff rM = M for every nonzero element r in R, [4,p.35]. 2.2 Examples and Remarks: مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 1. If M is divisible over integral domain then M is W.q.p. 2. Every quasi-prime is W.q.p but the converse is not true (see the example in the introduction). 3. Z as Z-module is W.q.p module since annRZ = 0 = annR rZ,  r annRZ. 4. Z4 as Z-module is not W.q.p module Since annRZ4 = 4Z and annR2Z = annR( 2 ) = 2Z. Thus Z4 as Z-module is not W.q.p module. 5. Z6 as Z-module is not W.q.p module since annZ6 = 6Z and ann2Z6 = ann( 2 ) = 3Z, so annZ6 ann 2Z6. 6. Zn as Z-module is W.q.p module iff n is prime. 7. Let M = ZZp; p is prime number is W.q.p module since annM = ann rM = 0 for each r  ann(Z Zp). 8.  p Z is W.q.p module since ann  p Z = ann r  p Z = 0. 2.3 Note: Let M be W.q.p over integral domain in R. Then every divisible submodule of W.q.p module. Recall that a proper submodule N of M is called semi-prime submodule if every r  R, x  M, k  Z+, such that rkx  N, then rx  N, [4,p .50]. 2.4 Proposition: Let M be divisible and (0) submodule of M is semi-prime submodule, then the following statements are equivalent 1. M is prime module, 2. M is q.p module, 3. M is W.q.p module. Proof :(1) → (2), by [2,p10] (2) → (3), by [2,p20] (3) → (1) To prove M is prime module, i.e. to show that (0) is p rime submodule. Let rm = 0, r  R, m  M, to prove either m = 0 or r  annRM. Suppose r  annRM, so we must prove that m = 0. Since r  annRM, rM  0. Hence rM = M, because M is divisible. Thus m = rm1 for some m1  M. Since rm = r(rm1) = 0, that is r2m1 = 0 which implies that rm1 = 0, since (0) submodule of M is semi-prime. Thus m = 0. 2.5 Remark: The condition in proposition 2.4 is necessary as the following example shows:  p Z is not q.p since if N = 2 1 p + Z then ann N = p2Z is not prime ideal, but  p Z is W.q.p module (see the example in the introduction). 2.6 Theorem: Let M be a module over an integral domain R and every submodule of M is divisible then ann (rm) =ann (m), for each r  ann (m). Proof: Since (rm)  (m), so ann(m)  ann (rm) …(1) To prove ann (rm)  ann (m) Let x  ann (rm) so x (rm) = 0. Since every submodule of M is divisible, (rm) = (m) and so xm = 0 which implies x  ann (m). Thus ann(rm)  ann(m) …(2) From (1) and (2), we have ann (m) = ann(rm), for each r  ann (m). Recall that an R-module M is called multiplication R-module if for every submodule N of M , there exists an ideal I of R such that IM = N. مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 2.7 Theorem: Let M be multiplication W.q.p R-module. Then every submodule of M is W,.q.p module. Proof: Let N be submodule of M, since M is multiplication R-module, so N = IM; I be ideal of ring R. To prove N is W.q.p module. To prove annRN = annRrN,  r annRN since rN  N so annRN  annRrN …(1) To prove annRrN  annRN. Let x  annRrN so xrN = 0. Since M is multiplication so there exists an ideal I of R such that N = IM. Thus xrIM = 0; that is xI  annRrM = annRM, hence xIM = 0; so xN = 0 which implies x  annRN. Thus annRrN  annRN …(2) From (1) and (2) we have annRN = annRrN so N is W.q.p module. 2.8 Proposition: Let M be cyclic W.q.p R-module. Then M is q.p module. Proof: Let M be cyclic so there exist x  M; M = (x), let y  M, to prove annRy is prime ideal, so y = rx; r  R, let a, b  annRy, to prove either a  annRy or b  annRy. Since ab  annRy = annRrx, so abrx = 0. Suppose b  annRy = annRrx, i.e brx ≠ 0, so ab  annR(rx) = annR(x), since M is W.q.p module, so abx = 0 which implies that a annRbx= annR(x) (since M is W.q.p). Thus ax = 0 which implies rax = r.0 = 0 so a  ann (rx) which means a  annRy. 2.9 Theorem: Let M be cyclic R-module then the following statements are equivalent 1. M is prime module 2. annRM = annRIM; I ⊈ annRN 3. M is W.q.p module. Proof: To prove (1) → (2) It is clear by definition of prime submodules. (2)  (3) it is obvious. To prove (3)  (1), to prove M is prime module. By proposition (2.8) we have M is q.p module which implies that annRM is prime ideal, see [3,p.14] and by [3,p.8] we get M is a prime module. 2.10 Theorem: The direct sum of two W.q.p R-module is also W.q.p R-module. Proof: Let M = M 1  M 2 where M 1 and M 2 are two W.q.p module, to prove M is W.q.p module, i.e to prove annRM = annRrM, for all r  annRM. annRrM = annRr(M 1  M 2) = annR(rM 1  rM 2) , see [2, p.80] = annRrM 1  annRrM 2 , see [2, p.83] = annRM 1  annRM 2 , since M 1 and M2 are W.q.p = annR(M 1  M 2) = annRM 2.11 Corollary: Let M be an R-module if M is W.q.p module then for any positive integer n, M n is W.q.p module where M n is the direct sum of n copies of M . 2.12 Remark: مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 A direct summand of W.q.p module is need not be W.q.p module. For example: Let M = Z  Z4 so annRM = annRrM  r  annRM. But Z4 is not W.q.p module, (see remarks and examples (2.2(4)). 2.13 Theorem: Let M 1 ; M 2 then M1 is W.q.p iff M2 is W.q.p. Proof:  Let f: M 1 → M 2 be 1-1 and onto and homomorphisim and M2 is W.q.p. To prove M 1 = f – 1(M 2) is W.q.p module, that is to prove annRrf – 1(M 2)  annRf – 1(M 2); rannRf –1(M2), let x  annR r f – 1(M 2) so xrf – 1(M 2) = 0 and since f – 1is homomorphisim so f – 1(xrM 2) = f-1(0) and since f – 1 is 1-1 so xrM 2 = 0 which mean x  annRrM 2 but M 2 is W.q.p module and rannRM 2 then xM 2 = 0 which implies f – 1 (xM 2) = f – 1 (0), but f – 1 is homomorphisim so x f – 1 (M 2) = 0 implies x  annRf – 1 (M 2) so annRr f – 1(M 2)  annR f – 1(M 2) …(1) and since r f – 1(M 2)  f – 1(M 2 ), so annRf – 1(M 2)  annRr f – 1(M 2) …(2) From (1) and (2) we have annRf – 1(M 2) = annRrf – 1(M 2). So f – 1(M 2) is W.q.p module.  clearly. 2.14 Note: The condition "isomorphism" in theorem 2.13 is necessary as the following example shows Example: Let : Z  Z ⁄ (4) ; Z4, where Z is W.q.p, but Z4 is not W.q.p. It is known that, if M is an R-module and I is an ideal of R which is contained in annRM then M is R/I-module, by taking (r + 1)x = rx x  M, r  R, see [5,p.40]. Now, we give the following result. 2.15 Theorem: Let M be an R-module and let I be an ideal of R, which is contained in annRM. Then M is W.q.p R-module iff M is W.q.p R/I-module. Proof:  To prove M is W.q.p R/I-module, i.e. to prove annR/IM = annR/I(r + 1)M. Since (r + 1)M  M so annR/IM  annR/I(r + 1)M …(1) To prove annR/I(r + 1)M  annR/IM Let x  annR/I(r + 1)M so x(r + 1)M = 0, which implies (xr + 1)M = 0 so (xr)M = 0 (by definition), so x  annRrM = annRM (since M is W.q.p R-module). x  annR/IM (since I  annR/IM), so annR/I(r + 1)M  annR/IM …(2) From (1) and (2) we have annR/IM = annR/I(r + 1)M .  If M is W.q.p R/I-module then M is W.q.p R-module, i.e. to prove annRM = annRrM,  r  annRM. Since rM  M so annRM  annRrM …(1) To prove annRrM  annRM Let x  annRrM so (xr)M = 0 implies that (xr + 1)M = 0, so x(r + 1)M = 0, hence x  annR/I(r + 1)M = annR/IM (since M is W.q.p R/I-module). Thus x  annR/IM, which implies that x  annRM (since I  annRM), so annRrM  annRM …(2) مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 From (1) and (2) we have annRM = annRrM. So M is W.q.p module. Recall that a subset S of a ring R is called multiplicatively closed if 1  S and ab  S for every a, b  S. We know that every proper ideal P in R is prime if and only if R-P is multiplicatively closed, see [4,p.42]. Let M be a module on the ring R and S be a multiplicatively closed on R such that S  0 and let RS be the set of all fractional r/s where r  R and s  S and M S be the set of all fractional x/s where x  M, s  S; x1/s1 = x2/s2 if and only if there exists t  S such that t(s1x2 – s2x1) = 0. So, can make M S into RS-module by setting x/s + y/t = (tx + sy )/st, r/tx/s = rx/ts for every x, y  M and for every r  R, s, t  S. If S = R-P where P is a prime ideal we use M P instead of M S and RP instead of RS. A ring in which there is only one maximal ideal is called local ring, see [4,p.50], hence RP is often called the localization of R, similar M P is the localization of M at P. So we can define the two maps :R  RS, such that (r) = r /1, rR, :M  M S, such that (m) = m /1, mM, see [5,p.69]. Through this paper S – 1R and S – 1M represent RS and MS respectively. 2.16 Proposition: Let M be W.q.p R-module then S – 1M is W.q.p S – 1R-module for each multiplicatively closed set S of R. Proof: To prove 1 S Rann  S – 1 M = 1 S Rann  r/t S – 1 M  r t  1 S Rann  S – 1 M, since r/t S – 1 M  S – 1 M so 1 S Rann  S – 1 M  1 S Rann  r/t S – 1 M …(1) To prove 1 S Rann  r/t S – 1 M  1 S Rann  S – 1 M Let y/t '  1 S Rann  r/t S – 1 M so y/t 'r/t S – 1 M = 0 which implies that yr/tt 'S – 1 M = 0 where yr  M, tt '  S so yr/ tt 'S – 1 M = 0 which implies that yr/ tt 'M/S = 0 so yrM = 0. Hence y  annRrM = annRM. Since y  annRM so yM = 0. Thus yM/ts = 0 so y/tS – 1M = 0, y/t annRS – 1M, hence 1 S Rann  r/t S – 1 M  1 S Rann  S – 1 M …(2) From (1) and (2) we have 1 S Rann  S – 1M = 1 S Rann  r/t S – 1M, so S – 1M is W.q.p module. References: 1. AL-Bahraany, B., (1996), Note on Prime Modules and Pure Submodule, J.Science, 37, . 1431 – 1441. 2. Anderson, F.W. and Fuller,R.R., (1973), Rings and Categories of M module, University of Oregon. 3. Abdul Razak, H.M ., (1999), Quasi Prime Module and Quasi-Prime Submodule. M.Sc. thesis, University of Baghdad. 4. Sharpe, D.W. and Vamos, P., (1972), Injective Modules, Cambridge University, press. 5. Larsen, M.D. and Mccarl, P.J., (1971), Multiplicative Theory of Ideals, Academic Press, New York. مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 حول المودیوالت الشبه األولیه الضعیفة منتهى عبد الرزاق حسن قسم الریاضیات، كلیة التربیة االساسیة، الجامعة المستنصریة 2011 تشرین االول 18: قبل البحث في2011 نیسان 3: استلم البحث في الخالصة ُوقـد برهنـت بعـض الخـواص لهـذا النـوع مـن . ُ قدمت تعریف جدید وهو المودیوالت الشبه أولیه الـضعیفة في هذا العمل .المودیوالت . المودیول األولي ، المودیول الشبه األولي ، المودیول الشبه األولي الضعیف:الكلمات المفتاحیة