A Theorem on Unique Factorization Domains Analogue for Modules

Int. J. Contemp. Math. Sciences, Vol. 6, 2011, no. 32, 1589-1596 A Theorem on Unique Factorization Domains Analogue for Modules M. Roueentan Department of Mathematics Shiraz University, Iran m.rooeintan@yahoo.com S. Namazi Department of Mathematics Shiraz University, Iran namazi@shirazu.ac.ir Abstract All rings are assumed to be commutative with identity. We define and study the properties of integral modules, the generalization of integral domains. We also generalize basic results concerning the Unique Factorization Domains to modules. Mathematics Subject Classification: 13C13, 13C05, 13G05 Keywords: Prime submodules, Multiplication modules 1 Introduction Throughout this paper we consider only commutative rings with identity and modules which are unitary. An R-module M is called multiplication provided for each submodule N of M there exists an ideal I of R such that N = IM. (see [1], [2], [6]). The main goal of this paper is to define the concept of integral module, and to investigate its properties. Let M be an R-module, we say that M is an integral module if M = 0 or zero submodule is a prime submodule of M. The concept of integral module was introduced as namely prime module in [4]. In section 2, some results related to integral domains are generalized to integral modules. For example it is shown that the non-zero multiplication R-module M is integral module if and only if mm = 0 implies that m =0 or m = 0 where m, m M (for elements m, m of M the product of m and

1590 M. Roueentan and S. Namazi m is defined in [2]). Further it is shown that if M is a multiplication integral module, then there exists a bijection ϕ : Supp(M) Spec(M). In section 3, firstly we define the irreducible elements and prime elements of a multiplication R-module M and we show that every prime element of an integral multiplication module M is irreducible. Also 0 m M is irreducible if and only if Rm is maximal in the set S of all proper cyclic submodules of M. Finally we define a principal module and we show that every principal integral module is a unique factorization module. Let M be an R-module. There are three subsets of Spec(R) which depend on M; (1) N(M) ={P Spec(R) :PM M}. (2) V (ann(m)) = {P Spec(R) :ann(m) P }. (3) Supp(M) ={P Spec(R) :P ann(m), for some m M}. = {P Spec(R) :M P 0}. Lemma 1.1. The following statements are equivalent for a proper submodule N of a multiplication R-module M. (i) N is prime. (ii) ann( M ) is a prime ideal of R. N (iii) N = PM for some prime ideal P of R with ann(m) P. Proof. see [5]. Lemma 1.2. Let M be a multiplication module, then (i) Supp(M) N(M). (ii) If M 0 and P N(M) V (ann(m)), then (PM : M) =P. Proof. (i), (ii) hold by [6, Theorem 3.5 and Lemma 2.4] Definition 1.3. Let N = IM and K = JM for some ideals I and J of R. The product of two submodules N and K in M which is denoted by NK is defined by IJM. Further if m,m M and Rm = IM, Rm = JM then by mm, we mean the product of Rm and Rm in M. Lemma 1.4. Let N be a proper submodule of multiplication module M then N is prime if and only if KL N implies that K N or L N for each submodule K and L of M. Proof. [2, Theorem 3.16]. 2 Preliminary Notes Definition 2.1. Let M be an R-module then we say M is an integral module if M = 0 or its zero submodule is prime, i.e. the relation rm = 0 for m

Theorem on unique factorization domains 1591 M,r R implies that m =0orrM = 0 (i.e, r ann(m)) It is easy to check that every simple module is an integral module. Also if M is an R-module and N is a proper submodule of M, then N is a prime submodule of M if and only if M is an integral R-module. N Lemma 2.2. Let M be a non-zero multiplication R-module then M is an integral module if and only if mm = 0 implies that m =0orm =0(m, m M). Proof. See Lemma 1.4. Lemma 2.3. If M is an integral module, then Supp(M) =V (ann(m)). Proof. If M = 0, then Supp(M) =V (ann(m)) = φ. If M 0, then for every nonzero element m of M, ann(m) =ann(m) and hence Supp(M) = V (ann(m)). Lemma 2.4. Let N be a nonzero submodule of an integral module M, then (i) If N is finitely generated and JN = N for ideal J of R, then JM = M. (ii) If N is cyclic and IN = JN for ideals I,J of R, then IM = JM. Proof. (i) There exists t J such that 1 t ann(n) =ann(m). (ii) Consider N = Rm;0 m M. If IN = JN, then Im = Jm, thus for each i I, there is j J such that (i j)m = 0. Since m 0,thus i j ann(m). Consequently IM JM. Similarly JM IM and hence IM = JM. Corollary 2.5. Let m, m,m be elements of an integral multiplication R-module M. Ifm 0 and mm = mm, then Rm = Rm. For R-module M we denote the the set of all prime submodules of M by Spec(M). Theorem 2.6. Let M be an integral multiplication R-module. Then there is a one-to-one order-preserving corresponding; Supp(M) Spec(M). Proof. Suppose P Supp(M). By Lemma 2.4 and Lemma 1.2, PM is a P -prime submodule of M. If we define ϕ : Supp(M) Spec(M), by ϕ(p )=PM where P Supp(M), then ϕ is well-defined. Now define a map ψ : Spec(M) Supp(M) byψ(n) =(N : M) where N Spec(M). by Lemma 1.1 and Lemma 2.4, ψ is well-defined. By considering Theorem 1.2 (ii), ψ ϕ and ϕ ψ are identity maps. We know that being torsion-free is a local property (see [3, exercise 13.3]), that is, for a module M over an integral domain R the following statements

1592 M. Roueentan and S. Namazi are equivalent. (i) M is torsion-free. (ii) M P is torsion-free for all prime ideals P. (iii) M P is torsion-free for all maximal ideals P. We denote the subset of all torsion elements of M by T (M), and the set of its zero-divisors by Z(M). If M is an integral module, then T (M) 0 if and only if Z(M) 0. Theorem 2.7. Let M be an integral module over an integral domain R then the following statements are equivalent. (i) M is non torsion-free (T (M) 0). (ii) ann(m) 0 (iii) M is torsion module (T (M) =M) (iv) there exists a prime ideal P of R such that M P is a nonzero torsion R p -module. (v) There exists a maximal ideal P of R such that M P is a nonzero torsion R P -module. Proof. By definition of T (M), (i) (ii) (iii), that is, a nonzero integral module is torsion if and only if it is non torsion-free. Since for each prime ideal P, M P is an integral R P -module, (iii), (iv), (v) are equivalent. 3 Main Results Let M be an R-module. Following [2] a nonzero element u of M is said to be a unit provided that u is not contained in any maximal submodule of M. For R-module M with Max(M) φ, it is easy to check that M is a simple module if and only if every nonzero element of M is unit. If 0 M is either finitely generated or multiplication, then 0 u M is a unit if and only if Ru = M. Definition 3.1. Let M be an R-module. If N and K are submodules of M. We say that N divides K if there exists an ideal I of R such that K = IN, denoted by N K (see [1]). In particular for elements m and m of M we say that m divides m if Rm Rm. In this case we say that m is a divisor of m and we write m m. Clearly m m if and only if Rm Rm. Elements m,m of M are said to be associated if m m and m m. Lemma 3.2. Let M be a nonzero multiplication module. (i) The nonzero element m of M is a unit if and only if there exists the nonzero element m of M such that mm = M, equivalently, m divides every element of M.

Theorem on unique factorization domains 1593 (ii) M is a simple module if and only if for each non-zero element m of M there exists a non-zero element m of M such that m m = M. Definition 3.3. Let 0 m be an element of a multiplication R-module M. (i) We say that m M is irreducible provided that: (1) m is non-unit. (2) If Rm = R(sm ), s R and m M, then either m is a unit in M or R s + ann(m) is a unit in. ann(m) (ii) We say that m is prime provided that N = Rm is a prime submodule of M. By the above definition, it is clear that the associates of a prime (irreducible) are prime (irreducible). lemma 3.4. Let M be a multiplication R-module and m be a non-zero, non-unit element of M. (i) If m is a prime element and Rm (Rm )(Rm ), then Rm Rm or Rm Rm (m,m M). (ii) If the R-module N = Rm is multiplication, then the converse of (i) is true. Proof. Assume that Rm (Rm )(Rm ). By Lemma 1.4, we conclude that m m or m m. Conversely suppose N = Rm is a multiplication R-module, we prove that N is a prime submodule of M. Consider N 1 N 2 N, where N 1 and N 2 are submodules of M. IfN 2 N, then we prove that N 1 N. There exists x N 2 \N so for every y in N 1, (Rx)(Ry) N. There exists ideal I of R such that (Rx)(Ry) = IN = I(Rm). Therefore m x or m y. If m x, then x N and this is a contradiction so m y and y N. Theorem 3.5. Let M be a multiplication R-module and m be a non-zero, non-unit element of M then the following are equivalent: (i) m is prime. (ii) If r R and m M and m rm, then m m or m rm for every element m of M. Proof. (i) (ii) let m rm and m m,sorm Rm and R(rm ) Rm that is, m Rm and rm Rm. Thus r (Rm : M) and hence m rm for every m in M. (ii) (i) It is trivial by the definition of a prime submodule. Theorem 3.6. Let m be a non-zero element in a multiplication integral module M. Then m is irreducible if and only if Rm is maximal in the set S

1594 M. Roueentan and S. Namazi of all proper cyclic submodules of M. Proof. If m is irreducible, then Rm M. Consider N = Rm and N Rm S hence there exists r R such that m = rm and so Rm = R(rm ). Since m is irreducible there exists s R such that 1 rs ann(m). Therefore m = rsm and consequently Rm = Rm. Conversely, let Rm be maximal in S and Rm = R(sm ), where s R, m M. Suppose that Rm M. Since Rm Rm, Rsm = Rm = Rm.As m 0 and M is integral, thus there exists r R such that 1 rs ann(m). This implies that m is irreducible. Corollary 3.7. In a multiplication R-module M, the only divisors of an irreducible element m of M are its associates and the units of M. Theorem 3.8. Let M be a multiplication R-module. If M is integral module, then: (i) Every prime element of M is irreducible. (ii) If m is irreducible and Rm =(Rm )(Rm );(m,m M), then m or m is unit. Proof. (i) Let m be a prime element of M. IfRm = R(sm ), s R, m M, then m sm and hence m m or m sm, for each m M, by theorem 3.5. If m m, then Rm Rm and so Rm = Rm = R(sm ), thus there is r R such that 1 rs ann(m). If for each m M, m sm, then sm Rm = R(sm ). Since M is integral and s ann(m), M = Rm and m is a unit. (ii) consider Rm =(Rm )(Rm ) and m is a non-unit, we prove that m is a unit. Since Rm Rm, there exists s R such that m = sm and so Rm = R(sm ). Now if m is non-unit, then there exists r R such that 1 rs ann(m). Thus m = rsm = rm. Consequently, Rm = Rm so we have Rm =(Rm )(Rm) and by Lemma 2.4 (i), Rm = M. That is a contradiction. Definition 3.9. Let M be an R-module we say that M is a principal module if every submodule of M is cyclic. Definition 3.10. If M is a multiplication integral R-module, then we say that M is a unique factorization module provided that: (1) for each non-zero non-unit element m of M, the submodule Rm can be written as Rm =(Rm 1 ) (Rm n ), where m 1,,m n irreducible. (2) If Rm =(Rm 1 ) (Rm r ) and Rm =(Rm 1) (Rm n ) ; (m i,m j irreducible), then n = r and for each i, m i and m i are associates, apart from the order in which the factors occur. For example every simple module is a unique factorization module. Also

Theorem on unique factorization domains 1595 every cyclic module over a principal integral domain is principal. Theorem 3.11. Let M be a principal integral module. For a non-zero element m of M the following are equivalent: (i) Rm is a maximal submodule of M. (ii) Rm is a prime submodule of M. (iii) m is a prime element of M. (iv) m is an irreducible element of M. Proof. (i) (ii) (iii) are trivial. (iii) (iv) and (iv) (i) hold by Theorems 3.8 (i) and 3.6, respectively. Corollary 3.12. Let M be a principal integral module. (i) Every non-zero non-unit element m of M has an irreducible divisor. (ii) If m M is irreducible and Rm = N 1 N 2, then Rm = N 1 or Rm = N 2, where N 1 and N 2 are the submodules of M. Proof. Let m be a non-zero non-unit element of M. If m is irreducible, then this is trivial. Suppose m is not irreducible. By Theorem 3.6, N = Rm is not maximal and hence there exists a maximal submodule N = Rm such that N N. Therefore, by Theorem 3.11, m is an irreducible divisor of m. ii) Suppose m is irreducible and Rm = N 1 N 2. Clearly N 1 N 2 Rm. Thus by Theorem 3.11 and Lemma 1.4, we have N 1 Rm or N 2 Rm and hence N = N 1 or N = N 2. Theorem 3.13. Every principal integral module is a unique factorization module. Proof. Let m be a non-zero non-unit element of M. First we show that Rm = (Rx 1 ) (Rx n ), where each x i (1 i n) is irreducible. By Corollary 3.12 there exists irreducible element x 1 of M such that x 1 m, thus (Rx 1 )I = Rm for ideal I of R. Let Rx 1 = JM where J is an ideal of R, then Rm =(JM)(IM). If Rm = IM, then by Lemma 2.4, Rx 1 = M and this is a contradiction. Thus Rm IM. In this case, if IM = M, then Rm = Rx 1 and the claim is proved. So IM M and for some non-unit element m 1 of M, IM = Rm 1. By Corollary 3.12, there exists an irreducible element x 2 of M such that x 2 x 1 thus for some element m 2 of M we have Rm 1 =(Rx 2 )(Rm 2 ). Similarly, it is easy to check that Rm 1 Rm 2. If m 2 is unit, then Rm = (rx 1 )(Rx 2 ) and if for each i 3, m i is non-unit, then by continuing, we obtain an ascending chain Rm Rm 1 Rm 2 of the submodules of M. This is a contradiction, whence Rm =(Rx 1 ) (Rx n ), where for i =1,,n,x i is irreducible. Finally if (Rx 1 ) (Rx n )=Rm =(Ry 1 ) (Ry k ), (x i,y i, irreducible), then Rx 1 = Ry i for some y i, by theorem 3.14 and Lemma 1.4. Since x 1 is non-unit, it must be an associate of y i by Corollary 3.7. The proof

1596 M. Roueentan and S. Namazi of uniqueness is now completed by Corollary 2.5 and a routine inductively argument. References [1] M. M. Ali, D. J. Smith, Generalized GCD modules, Beitrage zur Algebra und Geometrie 46(2), (2005), 447-466. [2] R. Ameri, On the prime submodules of multiplication modules, IJMMs. (27), (2003), 1715-1724. [3] M. F. Atiyah and I-G macdonald, Introduction to commutative algebra, Addison-Wesley, 1969. [4] M. Behboodi, O. A. S. Karamzadeh, and H. Koohy, Modules whose certain submodules are prime, vietnam Jornal of mathematices 32(3), (2004), 303-317. [5] Z. A. El-Bast, P. F. Smith, Multiplication modules, Comm. Algebra 16(4), (1988), 755-779. [6] S. C. Lee, S. Kim and S. C. Chung, Ideals and submodules of multiplication modules, J. Korean math. Soc. 42(5), (2005), 933-948. Received: December, 2010