Some characterizations of strongly irreducible submodules in arithmetical and Noetherian modules
aa r X i v : . [ m a t h . A C ] J a n SOME CHARACTERIZATIONS OF STRONGLY IRREDUCIBLESUBMODULES IN ARITHMETICAL AND NOETHERIAN MODULES
REZA NAGHIPOUR ∗ AND MONIREH SEDGHI
Abstract.
The purpose of the present paper is to prove some properties of the stronglyirreducible submodules in the arithmetical and Noetherian modules over a commutativering. The relationship among the families of strongly irreducible submodules, irreduciblesubmodules, prime submodules and primal submodules is proved. Also, several newcharacterizations of the arithmetical modules are given. In the case when R is Noetherianand M is finitely generated, several characterizations of strongly irreducible submodulesare included. Among other things, it is shown that when N is a submodule of M suchthat N : R M is not a prime ideal, then N is strongly irreducible if and only if thereexist submodule L of M and prime ideal p of R such that N is p -primary, N $ L ⊆ p M and for all submodules K of M either K ⊆ N or L p ⊆ K p . In addition, we showthat a submodule N of M is strongly irreducible if and only if N is primary, M p isarithmetical and N = ( p M ) ( n ) for some integer n >
1, where p = Rad( N : R M ) with p Ass R R/ Ann R ( M ) and p M * N . As a consequence we deduce that if R is integraldomain and M is torsion-free, then there exists a strongly irreducible submodule N of M such that N : R M is not prime ideal if and only if there is a prime ideal p of R with p M * N and M p is an arithmetical R p -module. Introduction
Let R be a commutative ring with non-zero identity and let M be an arbitrary R -module. We say that a submodule N of M is a distributive submodule if for all submodules K, L of M , the following equivalent conditions are satisfied:(i) ( K + L ) ∩ N = ( K ∩ N ) + ( L ∩ N );(ii) ( K ∩ L ) + N = ( K + N ) ∩ ( L + N ) . Also, M is said to be distributive module if every submodule of M is a distributivesubmodule.We say that N is an irreducible submodule if N is not the intersection of two submodulesof M that properly contain it. It is easy to see that if an irreducible submodule N isdistributive, then for all submodules K, L of M the condition K ∩ L ⊆ N implies thateither K ⊆ N or L ⊆ N . These considerations motivated us to define a submodule N ofan R -module M to be strongly irreducible , if for all submodules K, L of M , the condition K ∩ L ⊆ N implies that either K ⊆ N or L ⊆ N . Key words and phrases.
Arithmetical module, irreducible submodule, multiplication module, stronglyirreducible submodule, prime submodule, primal submodule.2010
Mathematics Subject Classification : 13C05, 13E05. ∗ Corresponding author: e-mail: [email protected] (Reza Naghipour). ∗ AND MONIREH SEDGHI
The purpose of the present article is to introduce and examine some properties ofstrongly irreducible submodules. In particular, we relate the notions of strongly irre-ducible submodules, irreducible submodules, primal submodules and prime submodulesof an R -module M . Also a characterization of strongly irreducible submodules in an arith-metical module is given. Specifically, we show that a submodule N of an arithmetical R -module M is strongly irreducible if and only if N is a primal submodule if and onlyif N is an irreducible submodule. If moreover R is assumed to be Noetherian ring and M finitely generated, then N is strongly irreducible if and only if N is primary, M p isarithmetical and N = ( p M ) ( n ) for some integer n >
1, where p = Rad( N : R M ) with p Ass R ( R/ Ann R ( M )) and p M * N .We recall that an R -module M is said to be arithmetical module if M m is an uniserialmodule over R m , for each maximal ideal m of R , i.e., the submodules of M m are linearlyordered with respect to inclusion.A brief summary of the contents of this article will now be given. Let R be a com-mutative ring and let M be an arbitrary R -module. In Section 2, the notion of stronglyirreducible submodules are introduced, and some properties of them are considered. Atypical result in this direction, which is a generalization of the main result of Heinzer etal. (see [7, Theorem 2.6]) to strongly irreducible submodules, is the following: Theorem 1.1.
Let ( R, m ) be a quasi-local ring and let M be an R -module. Suppose that N is a strongly irreducible submodule of M such that N = N : M m . Then the followingconditions hold: (i) the submodule N : M m of M is cyclic, (ii) N = m ( N : M m ) , (iii) for each submodule K of M either K ⊆ N or N : M m ⊆ K . The proof of Theorem 1.1 is given in Theorem 2.11. Using this, it is shown that if N is a strongly irreducible m -primary submodule of a finitely generated module M over alocal Noetherian ring ( R, m ) with m M = N , then N = S { K | K is a submodule of M and K $ N : M m } , and N : M m = T { L | L is a submodule of M and N $ L } . In Section 3, we give some characterizations of arithmetical and distributive modules.We also establish two characterizations whenever a submodule of an arithmetical moduleover a commutative ring is strongly irreducible. More precisely, we shall prove:
Theorem 1.2.
Let M be an arithmetical R -module and let N be a submodule of M . Thenthe following statements are equivalent: (i) N is strongly irreducible. (ii) N is primal. (iii) N is irreducible. Finally, the main results of the Section 4, provide several characterizations for a finitelygenerated module M over a Noetherian ring R to have a strongly irreducible submodule.In this section, among other things, we shall show that: TRONGLY IRREDUCIBLE SUBMODULES 3
Theorem 1.3.
Let M be a finitely generated module over a Noetherian ring R and let N be a submodule of M . Then, N is strongly irreducible if and only if N is primary, M p is anarithmetical R p -module and N = ( p M ) ( n ) for some integer n > , where p = Rad( N : R M ) such that p M * N and p Ass R ( R/ Ann R ( M )) . The proof of Theorem 1.3 is given in Theorem 4.7. One of our tools for proving Theorem1.3 is the following:
Proposition 1.4.
Let M be a non-zero finitely generated module over a local (Noetherian)ring ( R, m ) such that deth R ( R/ Ann R ( M )) > and Γ m ( M ) = 0 . Let N be a stronglyirreducible submodule of M such that Rad( N : R M ) = m . Then Γ m ( M ) * N . Pursuing this point of view further we derive the following consequence of Theorem 1.3.
Corollary 1.5.
Let M be a non-zero finitely generated torsion-free module over a Noe-therian integral domain R . Then there exists a strongly irreducible submodule N of M such that N : R M is not prime ideal of R if and only if there is a prime ideal p of R with p M * N and M p is an arithmetical R p -module. Finally, using Theorem 1.1 we obtain the following proposition which gives us a char-acterization of a strongly irreducible submodule in a multiplication module over a com-mutative Noetherian ring.
Proposition 1.6.
Let M be a multiplication module over a Noetherian ring R , and sup-pose that N is a proper submodule of M such that the ideal N : R M of R is not prime.Then N is strongly irreducible if and only if there exist submodule L of M and prime ideal p of R such that N is p -primary, N $ L ⊆ p M and for all submodules K of M either K ⊆ N or L p ⊆ K p . Throughout this paper, R will always be a commutative ring with non-zero identityand a will be an ideal of R . For each R -module M and for any submodule N of M , thesubmodules S s ∈ S ( N : M s ) and S n ≥ (0 : M a n ) of M are denoted by S ( N ) and Γ a ( M )respectively, where S is a multiplicatively closed subset of R . In the case S = R \ S { p ∈ mAss R M/ a M } , for any integer m ≥
1, the submodule S ( a m M ) is denoted by ( a M ) ( m ) .The radical of a , denoted by Rad( a ), is defined to be the set { r ∈ R : r n ∈ a for some n ∈ N } . Finally, for any R -module L , we shall use Z R ( L ) (resp. mAss R L ) to denote the setof zerodivisors on L in R (resp. the set of minimal elements of Ass R L ).Let R be a Noetherian ring and let G be a non-zero finitely generated R -module. For p ∈ Supp( G ), the G -height of p , denoted by ht G p , is defined to be the supremum oflengths of chains of prime ideals of Supp( G ) terminating with p . We shall say that anideal a of R is G -proper if G/ a G = 0, and, when this is the case, we define the G -heightof a (written ht G a ) to be inf { ht G p : p ∈ Supp( G ) ∩ V ( a ) } , where V ( a ) denotes the set { p ∈ Spec( R ) : p ⊇ a } . Also, if ( R, m ) is local then we use depth R ( G ) to denote themaximum length of all G -sequences contained in m .A proper submodule P of an R -module M is said to be prime if whenever rx ∈ P for r ∈ R and x ∈ M , then x ∈ P or r ∈ ( P : R M ). (For more information about primesubmodules, see [10], [16]). REZA NAGHIPOUR ∗ AND MONIREH SEDGHI An R -module M is said to be a multiplication module if every submodule of M is of theform b M for some ideal b of R . One can easily check that M is a multiplication moduleif and only if, for all submodules K of M , we have K = ( K : R M ) M .For any unexplained notation and terminology we refer the reader to [5] or [11].2. Strongly irreducible submodules
Throughout this section, R will denote a commutative ring (with identity). The purposeof this section is to introduce the concept of strongly irreducible submodules. Severalproperties of them are considered. The main goals of this section are Theorem 2.11 andCorollary 2.12. We begin with Definition 2.1.
Let M be an R -module and let N be a submodule of M . We say that N is a strongly irreducible submodule of M if for every two submodules L and K of M ,the inclusion L ∩ K ⊆ N implies that either L ⊆ N or K ⊆ N .The first lemma shows that every strongly irreducible submodule in a Noetherian R -module M is primary. Lemma 2.2.
Let M be an R -module and let N be a strongly irreducible submodule of M .Then N is irreducible. In particular, if M is Noetherian, then N is a primary submoduleof M .Proof . Suppose that N is a strongly irreducible submodule of M . Let L and K besubmodules of M such that N = L ∩ K . Then L ∩ K ⊆ N . Since N is strongly submodule,it follows that either L ⊆ N or K ⊆ N , and so either L = N or K = N . Hence N isirreducible. Now, the second part follows from the proof of [11, Theorem 6.8]. (cid:3) In the next lemma we observe that every prime submodule in a multiplication R -module M is strongly irreducible. Lemma 2.3.
Let M be a multiplication R -module and let N be a prime submodule of M .Then N is strongly irreducible submodule.Proof . Let L and K be submodules of M such that L ∩ K ⊆ N . Then ( L ∩ K ) : R M ⊆ ( N : R M ), and so ( L : R M ) ∩ ( K : R M ) ⊆ ( N : R M ). As N : R M is a primeideal of R , it follows that either L : R M ⊆ N : R M or K : R M ⊆ N : R M . Hence,either ( L : R M ) M ⊆ ( N : R M ) M or ( K : R M ) M ⊆ ( N : R M ) M . Now, since M is amultiplication module, it follows that either L ⊆ N or K ⊆ N , as required. (cid:3) Remark 2.4.
Before bringing the next result we fix a notation, which is employed by P.Schenzel in [13] in the case M = R . Let S be a multiplicatively closed subset of R . Fora submodule K of M , we use S ( K ) to denote the submodule S s ∈ S ( K : M s ).In particular, for any ideal a of R , if S = R \ S { p ∈ mAss R M/ a M } then for every n ∈ N ; S ( a n M ) is denoted by ( a M ) ( n ) . Lemma 2.5.
Let M be an R -module and let N be a submodule of M . Let S be a multiplica-tively closed subset of R such that the submodule S − N in S − M is strongly irreducible.Then S ( N ) is also a strongly irreducible submodule of M . TRONGLY IRREDUCIBLE SUBMODULES 5
Proof . Let L and K be two submodules of M such that L ∩ K ⊆ S ( N ). Then S − ( L ∩ K ) ⊆ S − ( S ( N )), and so it is easy to see that S − ( L ) ∩ S − ( K ) ⊆ S − ( N ). Now,the hypothesis on S − N implies that either S − ( L ) ⊆ S − ( N ) or S − ( K ) ⊆ S − ( N ).Hence, either S − ( L ) ∩ M ⊆ S − ( N ) ∩ M or S − ( K ) ∩ M ⊆ S − ( N ) ∩ M ; and so either L ⊆ S ( N ) or K ⊆ S ( N ), as required. (cid:3) The following proposition shows that the notion of strongly irreducible submodulebehaves well under localization.
Proposition 2.6.
Let M be an R -module and let N be a strongly irreducible primarysubmodule of M . Let S be a multiplicatively closed subset of R such that Rad( N : R M ) ∩ S = ∅ . Then S − N is a strongly irreducible submodule of S − M .Proof . Suppose that Ω and Ω are two submodules of S − M such that Ω ∩ Ω ⊆ S − N .Then, in view of [14, Ex. 9.11], there exist submodules K and L of M such thatΩ = S − K and Ω = S − L . Hence S − K ∩ S − L ⊆ S − N , and so S ( K ) ∩ S ( L ) ⊆ S ( N ).Since N is primary and Rad( N : R M ) ∩ S = ∅ , one easily sees that S ( N ) = N . Whence S ( K ) ∩ S ( L ) ⊆ N . Now, as N is strongly irreducible, it follows that either S ( K ) ⊆ N or S ( L ) ⊆ N . Therefore, either S − ( S ( K )) ⊆ S − N or S − ( S ( L )) ⊆ S − N , and so either S − K ⊆ S − N or S − L ⊆ S − N , as required. (cid:3) Lemma 2.7.
Let p be a prime ideal of R and let M be an R -module. Suppose that N is a p -primary submodule of M such that N p is a strongly irreducible submodule of M p . Then N is a strongly irreducible submodule of M .Proof . Let L and K be two submodules of M such that L ∩ K ⊆ N . Then L p ∩ K p ⊆ N p .Since N p is strongly irreducible submodule, it follows that either L p ⊆ N p or K p ⊆ N p .Now, as N is p -primary submodule, it readily follows that either L ⊆ N or K ⊆ N , asrequired. (cid:3) The next lemma investigates how the strongly irreducible property behaves under thefaithfully flat extensions.
Lemma 2.8.
Assume that T is a commutative ring which is a faithfully flat R -algebra.Let M be an R -module and assume that N is a submodule of M such that N ⊗ R T is astrongly irreducible submodule of M ⊗ R T . Then N is a strongly irreducible submodule of M .Proof . Suppose that L and K are two submodules of M such that L ∩ K ⊆ N . Then,in view of [11, Theorem 7.4], ( L ⊗ R T ) ∩ ( K ⊗ R T ) ⊆ N ⊗ R T . Now, as N ⊗ R T is astrongly irreducible submodule of M ⊗ R T , it follows that either L ⊗ R T ⊆ N ⊗ R T or K ⊗ R T ⊆ N ⊗ R T . Now, if L ⊗ R T ⊆ N ⊗ R T , then we have( L + N ) /N ⊗ R T = (( L ⊗ R T ) + ( N ⊗ R T )) / ( N ⊗ R T ) = 0 , and so by the faithfully flatness of T over R we have L ⊆ N . A similar argument alsoshows that if K ⊗ R T ⊆ N ⊗ R T , then K ⊆ N . This completes the proof. (cid:3) REZA NAGHIPOUR ∗ AND MONIREH SEDGHI
Lemma 2.9.
Let M be an R -module and let U be a submodule of M . Assume that N isa strongly irreducible submodule of M containing U . Then N/U is a strongly irreduciblesubmodules of
M/U .Proof . Suppose that L and K are two submodules of M such that U ⊆ L ∩ K and K/U ∩ L/U ⊆ N/U . Then K ∩ L ⊆ N , and so as N is strongly irreducible it followsthat either K ⊆ N or L ⊆ N . Hence either K/U ⊆ N/U or L/U ⊆ N/U , as required. (cid:3)
Proposition 2.10.
Let M be an R -module and let N be a submodule of M . Then N isstrongly irreducible submodule if and only if for all cyclic submodules L and K of M thecondition K ∩ L ⊆ N implies that either K ⊆ N or L ⊆ N .Proof . One direction is clear. To prove the converse, suppose that T and S are twosubmodules of M such that T ∩ S ⊆ N and T * N . Then there exists y ∈ T such that y N . Now, for all x ∈ S , we have Rx ∩ Ry ⊆ N . According to hypothesis Rx ⊆ N ,and so S ⊆ N , as required. (cid:3) We are now ready to state and prove the main theorem of this section which providessome properties of a strongly irreducible submodule N of an arbitrary module M over aquasi-local ring ( R, m ) which is properly contained in N : M m . It is easy to see that if R is Noetherian and Rad( N : R M ) = m , then N is properlycontained in N : M m . This result plays an important role in the Section 4 where werestrict attention to the case R is a Noetherian ring. Theorem 2.11.
Let ( R, m ) be a quasi-local ring and let M be an R -module. Let N be astrongly irreducible submodule of M such that N = N : M m . Then (i) the submodule N : M m of M is cyclic, (ii) N = m ( N : M m ) , (iii) for each submodule K of M either K ⊆ N or N : M m ⊆ K .Proof . In order to show (i), in view of the hypothesis N = N : M m , there exists anelement x ∈ N : M m such that x N . It is enough for us to show that N : M m = Rx .To do this, let y be an arbitrary element of ( N : M m ) \ Rx . We claim that Rx ∩ Ry ⊆ N .To this end, set z ∈ Rx ∩ Ry . Then there exist elements a, b ∈ R such that z = ax = by .Now, if b is a unit in R , then y ∈ Rx , which is a contrdiction. Thus we may assume that b is not unit. Then b ∈ m , and so it follows from y ∈ N : M m that by ∈ N . That is z ∈ N ,and hence Rx ∩ Ry ⊆ N . Now, since N is strongly irreducible it follows from x N that Ry ⊆ N , i.e., y ∈ N . Therefore it follows that N : M m = N ∪ Rx , and so N : M m = N or N : M m = Rx . Consequently, as N : M m = N , it yields that N : M m = Rx , and so thesubmodule N : M m of M is cyclic.To prove (ii), in view of (i), we have N : M m = Rx , where x ∈ ( N : M m ) \ N . Hence m x ⊆ N , and so m ⊆ N : R x . As x N , it yields that m = N : R x . Thus m ( N : M m ) =( N : R x ) x . So it is enough for us to show that N ⊆ ( N : R x ) x . To do this, let y ∈ N .Then, as N $ ( N : M m ) = Rx , it follows that y = rx , for some r ∈ R . Since x N ,it yields that r is not unit and so r ∈ m . Hence y ∈ m x , and thus y ∈ ( N : R x ) x , asrequired. TRONGLY IRREDUCIBLE SUBMODULES 7
Finally, in order to prove (iii) suppose that K is an arbitrary submodule of M suchthat ( N : M m ) * K . We have to show that K ⊆ N . To do this, let y ∈ K . Since by part(i) N : M m = Rx , for some x ∈ ( N : M m ) \ N , and ( N : M m ) * K , it follows that x K .Moreover, Rx ∩ Ry ⊆ N . Because, if w ∈ Rx ∩ Ry , then there are elements r, s ∈ R suchthat w = rx = sy . Since y ∈ K and x K , it follows that r is not unit, and so r ∈ m .Hence w ∈ m x , i.e., w ∈ N . Now, as N is strongly irreducible we deduce that either Rx ⊆ N or Ry ⊆ N . As x N , it yields that y ∈ N , and hence K ⊆ N , as required. (cid:3) Before bringing the final result of this section we recall that a proper submodule N ofan R -module M is said to be sheltered if the set of submodules of M strictly containing N has a smallest member S , called the shelter of N (see [3, Exercise 18, p. 238]). Corollary 2.12.
Let ( R, m ) be a local (Noetherian) ring and let M be a finitely generated R -module. Suppose N is a strongly irreducible submodule of M such that Rad( N : R M ) = m . Then N is sheltered and its shelter is N : M m . Proof . Since N is a strongly irreducible and Rad( N : R M ) = m , it follows from Lemma2.2 that N is an m -primary submodule of M . Hence it is easy to see that N = N : M m .Now, the assertion follows from Theorem 2.11. (cid:3) Strongly irreducible submodule in Arithmetical modules
The first main result of this section, gives us several characterizations of an arithmeticalmodule over a commutative ring. Before stating that theorem, let us recall that an R -module M is called an arithmetical module if M m is uniserial R m -module for each maximalideal m of R , i.e., the submodules of M m are linearly ordered with respect to inclusion. Itis clear that every primary submodule of an arithmetical module is strongly irreducible. Theorem 3.1.
Let M be an R -module. Then the following statements are equivalent: (i) M is an arithmetical module. (ii) M is a distributive module. (iii) ( K + L ) : R N = ( K : R N ) + ( L : R N ) for all submodules K, L, N of M with N isfinitely generated. (iv) K : R ( L ∩ N ) = ( K : R N ) + ( L : R N ) for all submodules K, L, N of M with L and N are finitely generated. (v) Every finitely generated submodule of M is a multiplication module.Proof . (i) = ⇒ (ii): Let M be an arithmetical R -module and we show that the latticeof the submodules of M is distributive, i.e., for all submodules K, L, N of M , we have( K + L ) ∩ N = ( K + N ) ∩ ( L + N ) . To do this, in view of [1, Corollary 3.4 and Proposition 3.8], it is enough for us to showthat for all maximal ideals m of R , we have( K m + L m ) ∩ N m = ( K m + N m ) ∩ ( L m + N m ) . Since the submodules of M m are linearly ordered with respect to inclusion, it follows thateither K m ⊆ N m or N m ⊆ K m . Now, the desired result easily follows from this and themodular law (see [15, Proposition 1.2]). REZA NAGHIPOUR ∗ AND MONIREH SEDGHI (ii) = ⇒ (i): Let M be a distributive R -module. Then, it easily follows from [14, Ex.9.11] that, for all maximal ideals m of R , the R m -module M m is also distributive. Sowithout loss of generality we may assume that R is a quasi-local ring with the uniquemaximal ideal m , and we must show that the the submodules of M are linearly orderedwith respect to inclusion. To do this, suppose that the contrary is true, i.e., there existtwo submodules K and L of M such that K * L and L * K . Then there exist elements x, y ∈ M such that x ∈ K \ L and y ∈ L \ K . Now, since M is a distributive module, itfollows that R ( x + y ) ∩ Rx + R ( x + y ) ∩ Ry = R ( x + y ) ∩ ( Rx + Ry ) = R ( x + y ) , and so there are elements w , w ∈ M such that x + y = w + w , where w ∈ R ( x + y ) ∩ Rx and w ∈ R ( x + y ) ∩ Ry.
Therefore, there exist elements r, s, t ∈ R such that w = rx = s ( x + y ) and w = ty .Hence ( r − s ) x = sy , and so the elements r − s and s are not units. This means that r, s ∈ m . On the other hand, since x + y = rx + ty , it follows that (1 − r ) x = ( t − y ,and so 1 − r ∈ m , which is a contradiction. Consequently one of K ⊆ L and L ⊆ K musthold.(i) = ⇒ (iii): Let M be an arithmetical R -module and let K, L, N be submodules of M such that N is finitely generated. We show that( K + L ) : R N = ( K : R N ) + ( L : R N ) . To do this, in view of [1, Corollary 3.4 and Proposition 3.8], we may assume that R is aquasi-local ring. Then the submodules of M are linearly ordered with respect to inclusion.Hence, without loss of generality we may assume that K ⊆ L . Then( K + L ) : R N = L : R N and K : R N ⊆ L : R N. Now, the assertion follows.(iii) = ⇒ (i): In view of [1, Corollary 3.4 and Proposition 3.8], we may assume that R is a quasi-local ring with the unique maximal ideal m . To establish (i), suppose, onthe contrary, that M is not arithmetical, and seek a contradiction. Then, there exist twosubmodules K and L of M such that K * L and L * K . Thus, there exist elements x, y ∈ M such that x ∈ K \ L and y ∈ L \ K . By the hypothesis we have( Rx + Ry ) : R R ( x + y ) = ( Rx : R R ( x + y )) + ( Ry : R R ( x + y )) . hence R = ( Rx : R R ( x + y ))+( Ry : R R ( x + y )), and so 1 = a + b , where a ∈ ( Rx : R R ( x + y ))and b ∈ ( Ry : R R ( x + y )). Therefore, there exist elements r, s ∈ R such that ax + ay = rx and bx + by = sy . Hence ( a − r ) x = ay and bx = ( s − b ) y . Since x Ry and y Rx , itfollows that a, b ∈ m , so that 1 ∈ m , which is a contradiction.(i) = ⇒ (iv): Let M be an arithmetical R -module and suppose that K, L, N are submod-ules of M such that L and N are finitely generated. It is clear that K : R L ⊆ K : R ( L ∩ N )and K : R N ⊆ K : R ( L ∩ N ) , and so ( K : R L ) + ( K : R N ) ⊆ K : R ( L ∩ N ). Now, in orderto show the opposite inclusion, in view of [1, Proposition 3.8], it is enough for us to showthat, for all maximal ideals m of R ,( K : R ( L ∩ N ) / ( K : R L ) + ( K : R N )) m = 0 . TRONGLY IRREDUCIBLE SUBMODULES 9
To do this, we have( K : R ( L ∩ N )) m ⊆ K m : R m ( L ∩ N ) m = K m : R m ( L m ∩ N m ) , and in view of [1, Corollary 3.4 and Proposition 3.8],(( K : R L ) + ( K : R N )) m = ( K m : R m L m ) + ( K m : R m N m ) . Now, since the submodules of M m are linearly ordered (with respect to inclusion), we mayassume that L m ⊆ N m . Then( K : R ( L ∩ N )) m ⊆ ( K m : R m L m ) and (( K : R L ) + ( K : R N )) m = ( K m : R m L m )Therefore ( K : R ( L ∩ N )) m ⊆ (( K : R L ) + ( K : R N )) m . Now the assertion follows easily from [1, Corollary 3.4].(iv) = ⇒ (i): According to the definition we need to show that for every maximalideal m of R , the submodules of the R m -module M m are linearly ordered (with respect toinclusion). To this end, suppose that the contrary is true and look for a contradiction.Then, there exist two submodules K m and L m of M m (see [14, Ex. 9.11]) such that K m * L m and L m * K m , where K and L are submodules of M . Thus there are elements x, y ∈ M such that x/ ∈ K m \ L m and y/ ∈ L m \ K m . Now, by hypothesis (iv), we have R = ( Rx ∩ Ry : R Rx ∩ Ry ) = ( Rx ∩ Ry : R Rx )+( Rx ∩ Ry : R Ry ) = ( Ry : R Rx )+( Rx : R Ry ) . Hence R m = ( R m y : R m R m x ) + ( R m x : R m R m y ). But R m y : R m R m x and R m x : R m R m y areproper ideals in R m , we achieve a contradiction.Finally, the equivalence between (ii) and (v) follows from [2, Proposition 7]. (cid:3) The next result, which is the second main theorem of this section, gives us two char-acterizations of strongly irreducible submodules of the arithmetical modules in terms ofprimal and irreducible submodules. To this end, recall that a proper submodule N of an R -module M is called primal submodule of M if Z R ( M/N ), the set of zero-divisors of the R -module M/N , is an ideal of R (see [6, Section 2, P. 193]). Then, it is easy to see that p := Z R ( M/N ) is a prime ideal of R , called the adjoint prime ideal of N . Also, in thiscase we say that N is a p -primal submodule of M . Theorem 3.2.
Let M be an arithmetical R -module and let N be a submodule of M . Thenthe following statements are equivalent: (i) N is irreducible. (ii) N is strongly irreducible. (iii) N is primal.Proof . (i) = ⇒ (ii): Let N be an irreducible submodule of M and suppose that K and L are two submodules of M such that K ∩ L ⊆ N . Then, as M is arithmetical, it followsfrom Theorem 3.1 that N = N + ( K ∩ L ) = ( N + K ) ∩ ( N + L ) . Since N is irreducible, it yields that either N = N + K or N = N + L , and so either K ⊆ N or L ⊆ N , as required. ∗ AND MONIREH SEDGHI
The implication (ii) = ⇒ (i) follows from Lemma 2.2. In order to show (ii) = ⇒ (iii),suppose that N is a strongly irreducible and let a, b ∈ Z R ( M/N ). Then then there existelements x, y ∈ M \ N such that ax, by ∈ N , and so Rx ∩ Ry * N . Hence, there exists z ∈ Rx ∩ Ry such that z N . Thus, z = rx = sy for some elements r, s ∈ R , and so( a − b ) z = rax − sby ∈ N. Therefore ( a − b )( z + N ) = N , i.e., a − b ∈ Z R ( M/N ). Moreover, for every c ∈ R , wehave ac ( x + N ) = N , and so ac ∈ Z R ( M/N ). This shows that Z R ( M/N ) is an ideal of R ,and hence N is a primal submodule.(iii) = ⇒ (ii): Let N be a primal submodule of M . Then Z R ( M/N ) is a prime ideal of R ; say p := Z R ( M/N ). Suppose that S is the multiplication closed subset R \ p of R . Itis then easily seen that S ( N ) = N . Now, since the submodules of the R p -module M p arelinearly ordered with respect to inclusion, it follows that N p is a strongly irreducible sub-module of M p . Hence, in view of Lemma 2.5, N is also a strongly irreducible submoduleof M , as required. (cid:3) Strongly irreducible submodules in Noetherian modules
The purpose of this section is to give a characterization for a finitely generated module M over a Noetherian ring R to have a strongly irreducible submodule. The main goal isTheorem 4.7. To this end, as an application of Theorem 2.11, we first prove the followingproposition which is needed in the proof of that theorem. Proposition 4.1.
Let R be a Noetherian ring and let M be a finitely generated R -module.Suppose that N is a strongly irreducible submodule and assume that p M p = N p , where p = Rad( N : R M ) . Then the following conditions are hold: (i) The submodule ( N : M p ) p of M p is cyclic. (ii) N p = p ( N p : M p p R p ) . (iii) For each submodule K of M either K ⊆ N or ( N p : M p p R p ) ⊆ K p .Proof . First of all, we note that as N is a strongly irreducible submodule of M , it followsfrom Lemma 2.2 that N is a p -primary submodule of M . Moreover, in view of Proposition2.6, N p is a strongly irreducible submodule of M p . Also, since p = Rad( N : R M ) and R is Noetherian, it follows that there exists an integer n ≥ p n M ⊆ N .Then, it is easy to see that N p $ N p : M p p R p . Hence, it follows from Theorem 2.11 that N p = p ( N p : M p p R p ) and the submodule ( N p : M p p R p ) of M p is cyclic.Now, in order to show (iii), let K be an arbitrary submodule of M such that( N p : M p p R p ) " K p . Then, it follows from Theorem 2.11 that K p ⊆ N p , and henceas N is p -primary it is easy to see that K ⊆ N . (cid:3) The next result of this section investigates whenever N is a strongly irreducible sub-module of a module M over a local (Noetherian) ring ( R, m ), then N and N : M m arecomparable (under containment) to all submodules of M . Theorem 4.2.
Let ( R, m ) be a local (Noetherian) ring and let M be a finitely generated R -module. Suppose that N is a strongly irreducible submodule of M such that m M = N TRONGLY IRREDUCIBLE SUBMODULES 11 and
Rad( N : R M ) = m . Then N and N : M m are comparable by set inclusion to allsubmodules of M . In fact, N = S { K | K is a submodule of M and K $ N : M m } , and N : M m = T { L | L is a submodule of M and N $ L } . Proof . Let K be an arbitrary submodule of M . We must show that either N ⊆ K or K ⊆ N . To do this, in view of Theorem 2.11 either K ⊆ N or N : M m ⊆ K . Now, if K * N , then it follows that N : M m ⊆ K , and so N ⊆ K . That is N is comparableto all submodules of M . Also, if N : M m * K , it follows that K ⊆ N , and therefore K ⊆ N : M m . Thus N : M m is also comparable to all submodules of M . Now, weshow that N = S { K | K is a submodule of M and K $ N : M m } . To do this, if K is anarbitrary submodule of M such that K $ N : M m , then N : M m " K , and so it followsfrom Theorem 2.11 that K ⊆ N . Therefore S { K | K is a submodule of M and K $ N : M m } ⊆ N. On the other hand, in view of Lemma 2.2 and the hypothesis Rad( N : R M ) = m , wededuce that N is an m -primary submodule of M , so that there exists an integer k ≥ m k M ⊆ N . Hence we obtain that N $ N : M m , and so N ⊆ S { K | K is a submodule of M and K $ N : M m } . Finally, in order to show N : M m = T { L | L is a submodule of M and N $ L } , if L is an arbitrary submodule of M such that N $ L , then in view of Theorem 2.11 wehave N : M m ⊆ L . Hence N : M m ⊆ T { L | L is a submodule of M and N $ L } , Moreover, since N $ N : M m it follows that T { L | L is a submodule of M and N $ L } ⊆ N : M m ,and this completes the proof. (cid:3) As an application, we derive the following consequence of Theorem 4.2, which showsthat every strongly irreducible submodule N of finitely generated module M over aNoetherian ring R of dimension one, is a distributive submodule, whenever the ideal N : R M of R contains a regular element on M . Recall that a submodule N of an R -module M is called a distributive submodule if for all submodules L and K of M ,( K ∩ L ) + N = ( K + N ) ∩ ( L + N ) . Corollary 4.3.
Let R be a Noetherian ring and let M be a finitely generated R -modulesuch that dim M = 1 . Suppose that N is a strongly irreducible submodule of M such thatthe ideal N : R M contains a regular element on M . Then N is a distributive submodule.Proof . According to the definition it is enough to show that for all submodules L and K of M , we have ( K ∩ L ) + N = ( K + N ) ∩ ( L + N ) . ∗ AND MONIREH SEDGHI
To do this, it suffices to check the equation locally at each prime ideal p in Supp( M ).Now, if N : R M * p , then p Supp(
M/N ), and so N p = M p . Hence the equation clearlyholds in this case. We therefore may assume that N : R M ⊆ p . Then, it follows from N : R M * Z R ( M ) that ht M ( N : R M ) = ht M p = 1. Hence Rad( N : R M ) = p , andtherefore Rad( N p : R p M p ) = p R p . On the other hand, in view of Proposition 2.6 thesubmodule N p is strongly irreducible in M p . Consequently, without loss of generality wemay assume that ( R, p ) is local. Now, in view of Theorem 4.2, either K ⊆ N or N ⊆ K .If K ⊆ N , then( K ∩ L ) + N = N and ( K + N ) ∩ ( L + N ) = N ∩ ( L + N ) = N, and so the equation holds in this case. Also, if N ⊆ K then by the modular law (see [15,Proposition 1.2]), we have( K ∩ L ) + N = K ∩ ( L + N ) = ( K + N ) ∩ ( L + N ) , as required. (cid:3) The following proposition gives us a characterization of strongly irreducible submodulein a multiplication module over a commutative Noetherian ring.
Proposition 4.4.
Let R be a Noetherian ring and let M be a finitely generated multipli-cation R -module. Suppose that N is a proper submodule of M such that the ideal N : R M of R is not prime. Then N is strongly irreducible if and only if there exists a submodule L of M and a prime ideal p of R such that N $ L ⊆ p M and that N is p -primary andfor all submodules K of M either K ⊆ N or L p ⊆ K p .Proof . First, let N be a strongly irreducible submodule of M . Then it follows fromLemma 2.2 that N is primary, and so Rad( N : R M ) is a prime ideal of R . Let p =Rad( N : R M ), and put L = N : M p . Now, in view of Proposition 4.1, it is enough forus to show that N $ L ⊆ p M . To do this, as R is Noetherian it follows that thereexists an integer n ≥ p n M ⊆ N . Let t be the least integer n ≥ p n M ⊆ N . Thus p t M ⊆ N and p t − M * N (even if t = 1, simply because N = M ), andso it yields that N p $ N p : M p p R p . Hence N $ L . In order to show L ⊆ p M , since M is a multiplication R -module, there exists an ideal a of R such that L = N : M p = a M .Hence ap M ⊆ N . Now, as the ideal N : R M is not prime, it follows that p M * N , andso a ⊆ p , note that N is p -primary. Therefore a M ⊆ p M , i.e., L ⊆ p M .In order to show the converse, let L be a submodule of M and suppose that p is aprime ideal of R such that N $ L ⊆ p M and assume that N is p -primary. We show that N is a strongly irreducible submodule. To do this, let T and S be two submodules of M such that T ∩ S ⊆ N . Now, suppose contrary is true, i.e., T * N and S * N . Then, itfollows from hypothesis that L p ⊆ T p ∩ S p . Whence L p ⊆ N p , and so as N is p -primary iteasily follows that L ⊆ N , which is a contradiction. (cid:3) The following two propositions will serve to shorten the proof of the main theorem ofthis section.
TRONGLY IRREDUCIBLE SUBMODULES 13
Proposition 4.5.
Let ( R, m ) be a local (Noetherian) ring and let M be a non-zero finitelygenerated R -module such that deth R ( R/ Ann R ( M )) > and that Γ m ( M ) = 0 . Suppose N is a strongly irreducible submodule of M such that Rad( N : R M ) = m . Then Γ m ( M ) * N .Proof . In view of Proposition 4.1 the submodule N : M m of M is cyclic, and so thereexists w ∈ M such that N : M m = Rw . Now, in order to prove the claim, suppose thecontrary is true, that is Γ m ( M ) ⊆ N . Then Γ m ( M ) ⊆ Rw , and so there exists an ideal a of R such that Γ m ( M ) = a w . On the other hand, since Γ m ( M ) is finitely generated itfollows that there exists t ∈ N such that m t a w = 0. Now, since N ⊆ Rw and the R -module M/N has finite length, it follows that the R -module M/Rw has also finite length.Thus there exists l ∈ N such that m l M ⊆ Rw . Therefore m t + l a M ⊆ m t a w = 0, and so m t + l a ⊆ Ann R ( M ). Whence( m / Ann R ( M )) t + l ( a + Ann R ( M ) / Ann R ( M )) = 0 , and so ( a + Ann R ( M ) / Ann R ( M )) ⊆ Γ m ( R/ Ann R ( M )) . Now, since by hypotheses deth R ( R/ Ann R ( M )) >
0, it follows from [4, Lemma 2.1.1] thatΓ m ( R/ Ann R ( M )) = 0 , and so a ⊆ Ann R ( M ). Hence a w = 0, and so Γ m ( M ) = 0, whichis a contradiction. (cid:3) Proposition 4.6.
Let R be a Noetherian ring and let M be a finitely generated R -module.Suppose N is a strongly irreducible submodule of M such that p M * N and that p Ass R ( R/ Ann R ( M )) , where p = Rad( N : R M ) . Then, the ideal N p : R p M p of R p containsa regular element on M p .Proof . In view of Lemma 2.2 the submodule N of M is p -primary and hence it followsfrom p M * N that p M p * N p . So according to Proposition 2.6 it may be assumed that( R, p ) is a local ring, and we must show that the ideal b := ( N : R M ) contains a regularelement on M . To this end, in view of [4, Lemma 2.1.1], it is enough for us to show thatΓ b ( M ) = 0. Since Rad( b ) = p , it is suffices to establish that Γ p ( M ) = 0.In view of Proposition 4.1 either Γ p ( M ) ⊆ N or ( N : M p ) ⊆ Γ p ( M ). If Γ p ( M ) ⊆ N ,then the assertion follows from the proof of Proposition 4.5.Therefore it may be assumed that ( N : M p ) ⊆ Γ p ( M ). Then, there exists an in-teger n ≥ N ⊆ (0 : M p n ), and so p n N = 0. On the other hand, since p = Rad( N : R M ), there is an integer s ≥ p s M ⊆ N . Hence p s + t M ⊆ p t N = 0,and so p s + t M = 0. Therefore M has finite length. Now, it is easy to see that the R -module R/ Ann R ( M ) is Artinian, and so Ass R ( R/ Ann R ( M )) = { p } , which is a contradiction. (cid:3) Now we are prepared to prove the main theorem of this section, which gives a charac-terization for a finitely generated module M over a Noetherian ring R to have a stronglyirreducible submodule. Theorem 4.7.
Let R be a Noetherian ring and let M be a finitely generated R -module.Let N be a submodule of M . Then, N is strongly irreducible if and only if N is primary, ∗ AND MONIREH SEDGHI M p is an arithmetical R p -module and N = ( p M ) ( n ) for some integer n > , where p =Rad( N : R M ) such that p M * N and p Ass R ( R/ Ann R ( M )) .Proof . First, let N be a strongly irreducible submodule of M . Then, it follows fromLemma 2.2 that N is a primary submodule of M . Now, let p := Rad( N : R M ) and weshow that M p is an arithmetical R p -module and N = ( p M ) ( n ) for some integer n >
1. Todo this end, in view of Proposition 2.6 the submodule N p of M p is strongly irreducible.Moreover, it is easy to see that N = ( p M ) ( n ) if and only if N p = p n M p . Therefore,without loss of generality we may assume that ( R, p ) is local, and that ( p M ) ( n ) = p n M .Then, Proposition 4.6 shows that the ideal ( N : R M ) contains a regular element on M ,and according to Proposition 4.1 the submodule N : M p of M is cyclic. Hence, thereexists an element w ∈ M such that N : M p = Rw , and thus by Proposition 4.1 we have N = p w . Now, Since by the Krull intersection theorem, T k ≥ p k M = 0, and since w = 0,it follows that w ∈ p t M , where t is the greatest integer i such that w ∈ p i M . Then N = p w ⊆ p t +1 M . Now, we show that p t M is a cyclic submodule. To achieve this,suppose the contrary is true. Then p t M = p t +1 M ∪ Rw , and so there exists an element y ∈ p t M such that y p t +1 M ∪ Rw . Hence, it follows from N ⊆ p t +1 M that y N . Also,one easily sees from y p t +1 M ∪ Rw that w Ry . Next, we show that Rw ∩ Ry ⊆ N .To do this, let x ∈ Rw ∩ Ry . Then there exist elements a, b ∈ R such that x = aw = by .As w Ry , it follows that a ∈ p , and so aw ∈ p w . This shows that x ∈ N . Since N isstrongly irreducible, it yields that either Rw ⊆ N or Ry ⊆ N , which is a contradiction.Therefore p t M is a cyclic submodule of M . Hence, as p contains a regular element on M (note that N : R M contains a regular element on M and p = Rad( N : R M )), itfollows from [12, Theorem 2.3] that R/ Ann R ( M ) is a PID and that M ∼ = R/ Ann R ( M ).Consequently, every proper submodule of M is of the form p v M for some integer v ≥ M are linearly ordered with respect to inclusion,i.e., M is an arithmetical R -module and N = p n M for some integer n >
1, as required.In order to prove the converse, since M p is an arithmetical R -module, it follows thatevery submodule of M p is strongly irreducible. In particular N p is strongly irreduciblesubmodule of M p . Now, as N is a primary submodule of M , it follows from Lemma 2.5that N is strongly irreducible. (cid:3) Corollary 4.8.
Let R be a Noetherian integral domain and let M be a non-zero finitelygenerated torsion-free R -module. Then there exists a strongly irreducible submodule N of M with N : R M is not prime ideal of R if and only if there is a prime ideal p of R with p M * N and M p is an arithmetical R p -module.Proof . Let N be a strongly irreducible submodule of M such that N : R M is notprime ideal of R . Then, N : R M = 0 and in view of Lemma 2.2, N is primary. HenceRad( N : R M ) is a prime ideal of R . Say p = Rad( N : R M ). Then, since M is torsion-free, it is easy to see that p Ass R ( R/ Ann R ( M )), and so in view of Theorem 4.7 the R p -module M p is arithmetical. Moreover, as the ideal N : R M is not prime, one can easilycheck that p M * N . TRONGLY IRREDUCIBLE SUBMODULES 15
Conversely, let p be a prime ideal of R such that M p is an arithmetical R p -module.Then N := ( p M ) (2) is a p -primary submodule of M . Whence in view of Theorem 4.7, N is a strongly irreducible submodule of M . Also, the ideal N : R M is not a prime ideal in R , because if N : R M is a prime ideal, then N : R M = Rad( N : R M ) = Rad(( p M ) (2) : R M ) = p , and so p M ⊆ N , which is a contradiction. (cid:3) Before we state the final result of this paper, recall that the radical of a submodule N of an R -module M , denoted by rad M ( N ), is defined as the intersection of all primesubmodules containing N . Proposition 4.9.
Let R be a Noetherian ring and let M be a finitely generated R -module.Let N be a strongly irreducible submodule of M such that rad M ( N ) = N . Then N is aprime submodule of M .Proof . In view of Lemma 2.2 the submodule N of M is primary and hence Rad( N : R M )is a prime ideal of R . Hence by applying [9, Theorem 5] we deduce that ( N : R M ) is aprime prime ideal of R , say p := ( N : R M ), and so in view of [11, Theorem 6.6] we haveAss R ( M/N ) = { p } . Hence it follows from [14, Corollary 9.36] that Z R ( M/N ) = p . Now,it is easy to see that N is a prime submodule of M . (cid:3) Acknowledgments
The authors are deeply grateful to Professors L. J. Ratliff, Jr., and R. Nekooei, for readingof the original manuscript and valuable suggestions. Also, we would like to thank theInstitute for Research in Fundamental Sciences (IPM), for the financial support.
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Department of mathematics, university of tabriz, tabriz, iran, and School of Math-ematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box: 19395-5746,Tehran, Iran.
Email address : [email protected] (R. NAGHIPOUR) Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran.
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