Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 84 This work is licensed under a Creative Commons Attribution 4.0 International License. Strongly Maximal Submodules with A Study of Their Influence on Types of Modules Abstract Let S be a commutative ring with identity, and A is an S-module. This paper introduced an important concept, namely strongly maximal submodule. Some properties and many results were proved as well as the behavior of that concept with its localization was studied and shown. Keywords: Maximal submodule, regular module, regular ring, semi-simple module, prime module. 1.Introduction Along with this paper, S is a commutative ring with identity, and A is an S-module. A proper submodule N of an S-module A is named maximal if there exists a submodule D of A such that N⊊D⊆A, then D=A[1][11]. Equivalently, N is the maximal submodule in A if and only if A/N is a simple S-module [1][12]. Maximal submodules may not exist; for instance, the Z-module ℂ has no maximal submodules. The main goal of this paper is to introduce a new concept called strongly maximal submodule (for short, SM-submodule) where a proper non-zero submodule B of an S-module A is said to be strongly maximal submodule if and only if, for every non-zero ideal E of S implies A/E2B is a regular module. Field house is defined in [9], a pure submodule of the form: A submodule D of an S-module A is pure if IA⋂D = ID for every ideal I of S and Sahera introduced in [2], the definition of F-regular module, where module A is said to be F-regular if and only if every submodule of A is pure Every strongly maximal submodule is a maximal submodule, but the opposite does not true. This paper is divided into three sections. We reviewed some basic definitions and properties needed in our next work. Section three introduced the definition of strongly maximal submodule. Lots of properties and examples of this concept were shown. In section four, the Ibn Al Haitham Journal for Pure and Applied Science Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index Doi: 10.30526/35.1.2802 Article history: Received 24, August, 2021, Accepted 26, September, 2021, Published in January 2022. Fatima M. Mohialdeen Mathematics Department, College of Education Ibn Al-Haitham for pure Sciences, University of Baghdad, Iraq Fatima.mohy.fm@gmail.com Buthyna N. Shihab Mathematics Department, College of Education Ibn Al-Haitham for pure science, University of Baghdad, Iraq Dr.buthyna@yahoo.com https://creativecommons.org/licenses/by/4.0/ mailto:Fatima.mohy.fm@gmail.com mailto:Dr.buthyna@yahoo.com Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 85 behavior of strongly maximal submodules under localization was some characterized and results were proved. 2.Basic concepts and Results This part includes some well-known definitions, concepts, and results that are useful for us in our study of the next section. Proposition (2.1) Every submodule of the regular module is regular [2]. Proposition (2.2) An S-module A is cyclic if and only if it is isomorphic to a factor module of S [4]. Proposition (2.3) An S-module A is simple if and only if A ≅ S/E for some maximal ideal E of S [4] [1]. Definition (2.4) A submodule B of an S-module A is called prime if B ≠ A, and whenever tx ∈ B for t ∈ S and x ∈ A we have either t ∈ [B: A] or x ∈ B [5]. Definition (2.5) A submodule B of an S-module A is called a semimaximal submodule if and only if A/B is a semi-simple S-module [3, definition (2.1.1),p32]. Definition (2.6) Let B be a submodule of an S-module A , the closure of B is denoted by CL(B) = { x ∈ A : [B:(x)] essential in S }. It is clear that CL(B) is a submodule of A containing B. That is B ⊆ CL(B) [7] . S3: Strongly Maximal Submodules with its advantages In this section, the concept of strongly maximal submodule (for short, SM-submodule) was introduced, which was a generalization of the concept the strongly maximal ideal in a ring S. Several examples and properties were proved also a lot of characterizations, and different results were presented. Let us start with our basic definition. Definition (3.1) Let A be an S-module, and B be a non-zero proper submodule of A. Then B is named strongly maximal submodule (for short SM-submodule) if and only if, for every non-zero ideal E of S implies A/E2B is a regular module. Examples and Remarks (3.2) Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 86 1. All the following modules have no SM-submodules. (i) Z as a Z-module. (ii) Zp as a Z-module, p is a prime number. (iii) Zp as a Zp-module, p is a prime number. 2. Every simple S-module has no SM-submodule. But the opposite is not true and the following example shows that: The module A=Z4⊕Z as a Z-module. Since A has no SM- submodules , A is not a simple module. Also notice examples (ii) and (iii) in no.(1) . 3. It is important to note that it is not necessary that all modules contain SM-submodules; for example Zp∞ as- Z-module. Since, all the submodules of Zp∞ are of the form <1/pi + Z>, where p is a prime number and i= 0, 1, 2… Now, we write N = <1/pi +Z> and let E be an ideal of Z. If we take E= <1>, then, Zp∞/E2B = Zp∞/<1>2B = Zp∞/B ≅ Zp∞ is not a regular module and hence B = <1/pi + Z> is not SM- submodule of Zp∞. That is, Zp∞ has no SM-submodules. Also, we can give another example Z8 as a Z8-module that has no SM-submodules .Since <2> and <4> are not SM- submodules in Z8 . 4. The submodule <3 > of a Z6-module Z6 is an SM-submodule. To clarify, let E be an ideal of a ring Z6 . if E = <1> , then Z6/<1> 2 <3> = Z6/ <3> ≅Z3 is a regular module . if E = <2> , then Z6/<2> 2 <3 > = Z6/<0> ≅Z6 is a regular module . if E = <3 >, then Z6/ <3 > 2 <3 > = Z6/ <3 > ≅ Z3 is a regular module . Therefore <3 > is an SM-submodule of Z6 . In general, all submodules of Z6 as a Z6-module are SM-submodules. 5. In Z10 as a Z20-module, the submodule B = <5> is an SM-submodule. Since if we take E = <1> , <2> , <10> , <4> , <5> ,where E is ideal of Z20 , then Z20/E 2<5> is a regular module for all ideal E of Z20 . This ends the proof of example. 6. Consider Z4 as a Z-module. The submodule <2> is not SM-submodule of Z4. Since Z4/E 2<2> is not regular, E is an ideal of Z. To prove that, take E = <4>. Then, Z4/<4> 2 <2> ≅ Z4 is not regular module. 7. Let Z4 as a Z4-module. Then the submodule B = <2> is not an SM-submodule. Notice if E = <2> then Z4/E 2 <2> is not regular module. Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 87 8. Let B = <2> be a submodule of a Z20-module Z10 . Since Z10/ E 2B = Z10/<2> 2<2> ≅ Z8 is not a regular module, where E = <2> is an ideal of Z20. 9. Let A = Z6⊕Z3 be an Z12-module and B = <2>⊕<0> be a submodule of A. Then A/E 2B = Z6⊕Z3/E 2(<2>⊕<0>) is regular module where E = <1>, <2> , <3> , <4> and <6> be ideals of Z12 . Therefor B = <2>⊕<0> is an SM-Submodule of A = Z6⊕Z3 . 10. Consider A = Z6⊕Z as a Z-module. Then the submodule B = <3>⊕<2> of A is not SM- submodule. Since A/E2B = Z6⊕Z/<2> 2 (<3>⊕<2>) = Z6⊕Z/<0>⊕<8 > is not a regular module, where E = <2> is an ideal of Z. 11. Every SM-submodule is maximal but the opposite is not true and the following example shows that: The submodule <2> of a Z-module Z4 is a maximal submodule in Z4 but it is not a SM-submodule, see no.(6) . 12. A submodule of an SM-submodule is not necessary to be an SM-submodule, for example: A submodule <2> in a Z6-module Z6 is SM-submodule. See no.(4) , While <0> is a submodule of <2> and it is not SM-submodule . 13. The intersection of two SM-submodules is not condition to be SM-submodule, for example : The two submodules <2> , <3> in Z6-module Z6 are SM- Submodules but <2> ∩ <3> = <0> is not an SM-submodule 14. More generally, let {Bi}i=1 n be a finite collection of SM-submodules of an S-module A. Then ∩i=1 n Bi is not always SM-submodule. 15. The direct sum of two SM-submodules of an S-module A is not necessary to be an SM- submodule, for example: Let <2>, <3> be two SM-submodules of a Z6-module Z6, but <2>⊕<3>=Z6 is not an SM-submodule. 16. From the fact that every maximal submodule is a semimaximal by [3, Remarks and Examples (2.1.2), p32], and the fact no.(11), we obtain that every SM-submodule of an S- module A is a semimaximal while the converse is not true in general and the following shows that: Let 6Z be a submodule of a Z-module Z . Then, 6Z is a semimaximal submodule of Z. Since 6Z =2Z∩5Z where 2Z, 5Z are maximal submodules of Z. But 6Z is not an SM- submodule of Z. Since Z/(2Z)2(6Z) = Z/24Z ≅ Z24 is not a regular module . Proposition (3.3) Let B, D be two submodules of an S-module A with B⊆D. Then B is SM-submodule in D when B is SM-submodule in A. Proof: Suppose B is SM-submodule in A, then A/E2B is regular module for every non-zero ideal E of S. Since D/E2B is a submodule of A/E2B (Notice, E2B ⊆ B ⊆ D), and hence by proposition (2.1), D/E2B is a regular submodule of A/E2B. Thus B is SM-submodule in D. Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 88 Next, we will give an application to a proposition (3.3) Corollary (3.4) Let A be an S-module and B be a proper submodule of A. If B is an SM-submodule of [B AA : ] and [B AA : ] is an SM-submodule in A, then, B is an SM-submodule in A. Proof: It clear that B⊆ [B AA : ] ⊆ A. Therefore, by using proposition (3.3), we conclude that B is SM-submodule in A. Now, we will give the sufficient condition for a submodule to not be SM-submodule. Proposition (3.5) Let A be an S-module. If A is cyclic module (if A=Sx for some x∈A) and anns(x) is maximal ideal of S, then A has no SM-submodule. Proof: Since A= Sx for some x∈A, then A is isomorphic to a factor module of S by proposition (2.2). We can define f : S → A such that f(r) = rx . It is easily to show that f is well- define and epimorphisim, by the first fundamental theorem of isomorphism S/Ker f ≅ A. Next, Ker f ={r∈S : f(r) =0A } = {r∈S : rx=0A } =anns(x) That is S/anns(x) ≅A. Also, we have anns(x) is maximal ideal of S which implies S/anns(x) is simple and hence A is simple. Therefore, by examples and remarks ((3.2) No. (2)), we get the result. Next is the application of proposition (3.5) Corollary (3.6) If a non-zero prime and semi-simple S-module A, then A has no SM-submodule. Proof: Suppose that A is a prime and semi-simple module, then, we obtain A is simple module. To prove this, assume that A is not simple which implies A is a direct sum of simple S- modules. Then, there exists a simple module M1 and M2 which are a direct summand of A with M1 ≠ M2. M1≅ S/E, M2≅ S/D where E and D are maximal ideals of S, by proposition (2.3). But A is prime module, then anns(M1) = E= anns(A) and anns(M2) = D =anns(A).Thus E = D which implies that M1 = M2, and this is a contradiction. Hence, A is a simple module and by using proposition (3.5), we have A has no SM-submodule. As a direct of corollary (3.6), we have the following. Corollary (3.7) Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 89 Let P be a prime and semimaximal submodule of an S-module A. Then the quotient module by P has no SM-submodule. Proof: Since P is a semimaximal submodule of A, then by definition (2.5), we have A/P is a semi- simple S-module. On the other hand, P is a prime submodule of A, then, by [3, proposition (1.1.51)] we obtain that A/P is a prime S-module, and hence by corollary (3.6), then A/P is a simple S-module and hence A/P has no SM-submodule. S4: The behavior of SM-submodules under localization. Let K be a subset of a ring S, W is multiplication closed if the two condition hold: 1. I ∈ W . 2. xy ∈ W for every x , y ∈ W . We know that every proper ideal E in S is prime if and only if S-E is multiplicatively closed, see [8].If A is an S-module and W be a multiplicatively closed on S such that W≠<0> , then Sw be the set for all fractional r/w where r ∈S and w ∈W and Aw be the set of all fractional m/w where m ∈A and w ∈W . For m1, m2∈ A and w1,w2∈ W , m1/w1=m2/w2 if and only if ∃ t ∈ W such that t(w1m1-w2m2)= 0. Also, we can make Aw in to Sw-module by setting m1/w1+m2/w2 =(w2m1+w1m2)/w1w2 and (r/w1) (m1/w2) = rm1/w1w2 for every m1,m2 ∈A and every r ∈ S , w1,w2 ∈ W. If W =S-E where E is a prime ideal, we used AE instead of Aw and SE instead of Sw. If a ring has only one maximal ideal, then it is called a local ring. Hence SE is often called the localization of S at E, similar AE is the r/1,Ɐ r ∈ S and 𝜱 :A →Aw such that 𝜱(m)=m/1 ,Ɐm∈A. Furthermore, if B is a submodule of an S-module A and W be a multiplicatively closed in S, then Bw = {n/w: n ∈B, w∈W} be a submodule on Sw- module , see [8]. In this section we study the behavior of an SM-submodule under localization and several of results have been proved. The following lemma is needed in our next result. Lemma (4.1) [10] Let A be an S-module and B, L are two submodules of A. Then, B=L if and only if BP=LP for every maximal ideal P of S. The following proposition study the relationship between a module A and its locally and prove that they are equivalent. Proposition (4.2): Let A be an S-module and B is nonzero proper submodule of A. Then, BP is SM- submodule of an SP-submodule AP if and only if B is SM-submodule of an S-module A. Proof: Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 90 Suppose that B is nonzero proper submodule of A. We must prove that A/E2B is a regular S-module for every nonzero ideal E of S; that is, every submodule of A/E2B is pure. Let L/E2B be a submodule of A/E2B. It is clear that I(L/E2B) ⊇ I(A/E2B)⋂(L/E2B) where I is an ideal of S. Now, to prove I(L/E2B) ⊆ I(A/E2B)⋂(L/E2B). Let x ∈ I(L/E2B). Then x = ∑ ani=1 i(li+E 2B) . Therefore xs/s = (∑ ani=1 i(li+E 2B))s/s ∈ IP(LP/EP 2BP) but LP is SM-submodule in AP, then IP(LP/EP 2BP) = (IP(AP/EP 2BP))⋂(LP/EP 2BP) which implies xs/s ∈ IP(AP/EP 2BP)⋂(LP/EP 2BP) = ((IPAP+EP 2BP)/ EP 2BP)⋂(LP/EP 2BP )=(((IA)P+(E 2B)P)/(E 2B)P)⋂(LP/(E 2B)P) by [ 10]. And hence As/s ∈((IA+E2B)P)/(E 2B)P)⋂(LP/(E 2B)P)= ((IA+E 2B)/(E2B))P⋂(L/(E 2B))P = (((IA+E2B)/(E2B))⋂(L/(E2B)))P . Therefore x ∈ ((IA+E 2B)/(E2B))⋂(L/(E2B)) which implies x ∈ (I(A/(E2B))⋂(L/(E2B)) and hence I(L/E2B) ⊆ (I(A/(E2B)) ⋂ (L/(E2B)). Therefore I(L/E2B) = (I(A/(E2B))⋂(L/(E2B)). Thus L/E2B is pure submodule of A/E2B. This proves that A/E2B is regular and finally B is an SM-submodule of A. Conversely: - Suppose that B is an SM-submodule of A. To prove BP is SM-submodule of an Sp-module AP we must show that AP/EP 2BP is regular SP-module. It is clear that (IP(AP/EP 2BP))⋂(LP/EP 2BP) ⊆ IP(LP/EP 2BP). To prove IP(LP/EP 2BP) ⊆ (IP(AP/EP 2BP))⋂(LP/EP 2BP) . Let x/1 ∈ I and a/s ∈ IP(LP/EP 2BP) (xa/s + EP 2BP) = ∑ (b n i=0 i/si)(li/ti+EP 2BP 2) where si , ti ∉P and bi ∈I ,li ∈L . Put ci=siti . Therefore (xa/s + EP 2BP) = ((b1l1v1+b2l2v2+…+bnlnvn)/u)+ EP 2BP 2 where u = c1c2c3….cn and v1 = c2c3….cn , v2 = c1c3c4….cn , vn = c1c2c3….cn-1 . Thus there exist k∉ P such that kxau+E2B = k(b1l1v1+b2l2v2+…+bnlnvn)∈ I(L/E 2B) but L is pure submodule in A, that is I(L/E2B) = I(A/ E2B)⋂ (L/ E2B) (Since A/ E2B is regular S-module) and hence by (13,theorem (2.6)] we have (xa+ E2B) ∈ I(A/ E2B ⋂ L/ E2B). This leads us to write (xa/s+EP 2BP)∈(IP(AP/EP 2BP)⋂(LP/EP 2BP). This gives IP(LP/EP 2BP) ⊆ (IP(AP/EP 2BP))⋂(LP/EP 2BP) and AP/EP 2BP is a regular SP-module and finally , we obtain that BP is an SM-submodule of AP . Proposition (4.3): Let L, B be two finitely generated submodules of an S-module A. If LP, BP are SM- submodules of AP, then L⋂B is an SM-submodule of A. Proof: Since L, B are two finitely generated submodules of A, then by {10,p24}, [LP:BP] + [BP:LP] = SP for every maximal ideals P of S. Thus, LP⋂BP = LP or LP⋂BP = BP, but LP and BP are SM-submodules, then LP⋂BP is an SM-submodule, and we have LP⋂BP = (L⋂B)P. Therefore (L⋂B)P is an SM-submodule and by proposition(4.2), L⋂B is an SM-submodule of A. Proposition (4.4): Let L, B be two finitely generated submodules of an S-module A. Then, L+B is an SM- submodules of A, if LP ,BP are SM-submodules of an SP-module AP . Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 91 Proof: Let L, B be two finitely generated submodules of A. Then by {10, p24}, we have [LP:BP]+[BP:LP] = SP for every maximal ideal P of S. Let y1 ∈ [LP:BP] and y2 ∈ [BP:LP] such that y1+y2 =1 = unity of SP. Then, either y1 is a unit element or y2 is a unit element (Since SP is local ring). Therefore [LP:BP] = SP or [BP:LP] = SP and hence either LP ⊆ BP or BP ⊆ LP which implies LP + BP = LP or LP + BP = BP ,but LP , BP are SM-submodules of AP . Thus LP + BP is an SM-submodule and (L+B)P is an SM-submodule and by proposition(4.2) , L+B is an SM-submodule of A. 5.Conclusion The conclusion of this work is to study an important concept, namely strongly maximal submodule. 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