# Uniformly Primal Submodule over Noncommutative Ring

Authors: Lamis J. M. Abulebda
Journal: Journal of Mathematics
 (2020)

---

## Abstract

LetR be an associative ring with identity and M be a unitary right R-module. A submodule N of M is called a uniformly primal submodule provided that the subset B of R is uniformly not right prime to N, if there exists an element s∈M−N with sRB⊆N.The set adjN=r∈R|mRr⊆Nfor somem∈M is uniformly not prime to N.This paper is concerned with the properties of uniformly primal submodules. Also, we generalize the prime avoidance theorem for modules over noncommutative rings to the uniformly primal avoidance theorem for modules.


---

## Body

## 1. Introduction

Throughout this paper, all rings are associative with identity and all modules are unitary modules. For detailed description regarding rings and modules, interested readers are encouraged to go through the book of Kelarev et al. [1]. The concept of uniformly primal ideal has been introduced and studied by Barnes [2]. Let A be an ideal of R. The ideal B of R is uniformly not right prime to A, if there exists an element y∈R−A with yRB⊆A. A is called uniformly primal if adjA is uniformly not right prime to A where adjA=x∈R|yRx⊆Afor somey∈R−A. The prime avoidance theorem for rings with identity [3] states that if an ideal I of a ring is contained in a union of a finite number of prime ideals P1,P2,…,Pn, then I must be contained in Pk for some k∈1,2,…,n. Karamzadeh [4] generalizes the prime avoidance theorem for any ring that is not necessarily commutative. The aim of Section 1 is to generalize the prime avoidance theorem for rings over noncommutative rings to the uniformly primal avoidance theorem over noncommutative rings.The concept of uniformly primal submodules has been introduced and studied by Dauns in [5]. A submodule N of M is called a uniformly primal submodule provided that the set adjN=x∈R|mRr⊆Nfor somem∈M is uniformly not prime to N, where the subset B of R is uniformly not right prime to N if there exists an element s∈M−N with sRB⊆N. In particular, a number of papers concerning primal submodules have been studied by various authors (see, for example, [6–10]). In Section 2, we give some basic results about uniformly primal submodules and show that N1,N2,…,Nn is a finite collection of uniformly primal submodules of an R-module M with adjNj=Pj for every j and Nj:M⊄adjNk whenever j≠k. Then, S′=M−∪i=1nNi is an S-system subset of M, where S=R−∪i=1nPi. Also, we study the prime avoidance theorem for modules over noncommutative rings and generalize it to the uniformly primal avoidance theorem for modules.


## 2. Uniformly Primal Ideal

The concept of primal ideals over noncommutative has been introduced and studied by Fuchs [11].Definition 1.
Let A be an ideal ofR. The adjoint of A is the set of all elements of R that are not right prime to A and denoted by adjA. In other words, adjA=x∈R|yRx⊆Afor somey∈R−A.Definition 2.
The idealA of R is said to be primal if adjA forms an ideal of R. In this case, the adjoint of A will also be called the adjoint ideal of A.Definition 3.
The idealB of R is uniformly not right prime to A if there exists an element y∈R−A with yRB⊆A.Definition 4.
An idealA of a ring R is said to be uniformly primal if adjA is uniformly not right prime to A.Proposition 1.
(see [2]).If A is a uniformly primal ideal in R, thenadjA is a prime ideal of R.Proposition 2.
(see [3]). IfP,P1,andP2 are ideals of R such that P⊆P1∪P2, then either P⊆P1 or P⊆P2. The definition of efficient union of ideals was introduced in the rings that are commutative (see [12]). We give a generalization to it in rings that are not necessary commutative as follows.Definition 5.
LetP,P1,P2,…,Ps be ideals of a ring R. The covering P⊆P1∪P2∪⋯∪Ps of P is called efficient if P is not contained in the union of any s−1 of the idealsPi′s. Analogously, we shall say P=P1∪P2∪⋯∪Ps is an efficient union if none of the Pi′s may be excluded. Any cover or union consisting of ideals of R can be reduced to an efficient one, called an efficient reduction, by deleting any unnecessary terms. The following very important lemma is based on McCoy over commutative rings (see [3]).Lemma 1.
(see [3]).LetP=P1∪P2∪⋯∪Psbe an efficient union of ideals where s > 2. Then, ∩j≠kPj=∩j=1sPj for all k∈1,2,…,s. As an application, we obtain following corollary.Corollary 1.
LetP⊆P1∪P2∪⋯∪Ps be an efficient cover of ideals of a ring R where s > 2. Then, P∩∩j≠kPj⊆Pk for all k∈1,2,…,s.Proof.
SinceP⊆P1∪P2∪⋯∪Psis an efficient covering, P⊆P∩P1∪P∩P2∪⋯∪P∩Ps is an efficient union. Now, by Lemma 1, P∩∩j≠kPj=∩j≠kP∩Pj⊆Pk. To prove the uniformly primal avoidance theorem for rings, we need the following result on the uniformly primal ideal.Proposition 3.
LetP⊆P1∪P2∪⋯∪Ps be an efficient covering consisting of ideals where s > 2. If adjPi⊄adjPk for every i≠j, then no Pk for k∈1,2,…,s is a uniformly primal ideal of R.Proof.
Suppose that somePkis uniformly primal ideal. SinceP⊆P1∪P2∪⋯∪Ps is an efficient covering, there exists an element ak∈P−Pk. If i≠j, then adjPi⊄adjPk, so there exists rj∈adjPj such that rj∉adjPk. Since Pk is a uniformly primal ideal, then by Proposition 1, adjPk is a prime ideal of R. Therefore, r=∏i≠kri∈adjPj, but r∉adjPk. Consequently, ekRr⊆P∩Pj for every k≠j, but ekRr⊄Pk, which contradicts the fact that P∩∩j≠kPj=∩j≠kP∩Pj⊆Pk (by Corollary 1). Therefore, no Pk is a uniformly primal. Now, we will give the proof of the main theorem of this section.Theorem 1.
(uniformly primal avoidance theorem of rings). LetP1,P2,…,Ps be a finite number of ideals of a ring R and P be an ideal of R such that P⊆P1∪P2∪⋯∪Ps. Assume that at least two of the Pi′s are not uniformly primal and that adjPi⊄adjPj whenever i≠j. Then, P⊆Pk for some k∈1,2,…,s.Proof.
For the given coveringP⊆P1∪P2∪⋯∪Ps, let P⊆Pα1∪Pα2∪⋯∪Pαmbe its efficient reduction. Then, 1≤m≤s and m≠2 where if m=2, then by Proposition 2, P⊆P1 or P⊆P2. If m>2, then there exists at least one Pαrto be uniformly primal ideal. By Proposition 3, this is impossible as adjPi⊄adjPj if i≠j. Hence, m=1, so P⊆Pk for some k.


## 3. Uniformly Primal Submodule

The concept of primal submodules has been introduced and studied by Dauns in [5].Definition 6.
LetM be an R-module and N be a submodule of M. For any a∈R, the submodule m∈M|mRa⊆N is denoted by Na−1. Analogously, for a subset A of R, NA−1  = m∈M|mRA⊆N where RA=mra:|r∈R,a∈A.Definition 7.
LetM be an R-module and N be a submodule of M. The element a∈R is right prime to N if Na−1=N, i.e., if mRa⊆Nwherem∈M implies m∈N. The element a∈R is not right prime to N if Na−1≠N, i.e., there exists an element m∈M−N with mRa⊆N, since N⊆Na−1. A subset A of R is not right prime to N if for any a∈A, a is not right prime to N. In this case, we say that A is pointwise not right prime to N. The subset A of R is uniformly not right prime to N if there exists an element s∈M−N with sRA⊆N, i.e., A is uniformly not right prime to N if and only if NA−1≠N.Definition 8.
LetM be an R-module and N be a submodule of M. The adjoint of N is the set of all elements of R that are not right prime to N and denoted by adjN. In the other words, adjN=r∈R|mRr⊆Nforsomem∈M.Definition 9.
LetM be an R-module. A proper submodule N of M is said to be primal if adj (N) forms an ideal of R. In this case, the adjoint of N will also be called the adjoint ideal of N.Proposition 4.
LetN be a submodule of an R-module M. If adjN is uniformly not prime to N, then adjN is an ideal of R, and as a consequence, N is a primal submodule.Definition 10.
LetM be an R-module. A proper submodule N of M is said to be uniformly primal if adjN is uniformly not prime to N.Proposition 5.
(see [8]). LetM be an R-module. If N is a uniformly primal submodule of M, then adjN is a prime ideal of R. In the following propositions, we show the behavior of a primal submodule under isomorphism.Proposition 6.
Letf:M⟶M′ be a module isomorphism. If N is a primal submodule of M, then fadjN=adjfN.Proof.
Leta∈fadjN. Since f is a module isomorphism, then f−1a∈adjN so that there exist s∈M−Nsuch that sRf−1a⊆N. Since f is a module isomorphism, then fsRa=fsRf−1a⊆fN, but s∈M−Nso that fs∈M′−fN which implies that a∈adjfN. Thus, fadjN⊆adjfN. Now, let a∈fadjN so that there exist t∈M′−fN such that tRa⊆fN. Since f is a module isomorphism, then f−1tRf−1a⊆fN=f−1tRa⊆f−1fN=N, but t∈M′−fN so that f−1t∈M−Nwhich implies that f−1a∈adjN. That is, a∈fadjN. Thus, fadjN⊆adjfN.Proposition 7.
Letf:M⟶M′be a module isomorphism. If N is a primal submodule of M, then fN is a primal submodule of M′.Proof.
Leta,b∈adjfN. Since f is a module isomorphism, then by Proposition 6, we have f−1a,f−1b∈f−1adjfN=adjf−1fN=adjN. But N is a primal submodule; then, f−1a+b=f−1a+f−1b∈adjN. Also, since f is a module isomorphism, we have a+b∈fadjN. Hence, a+b∈adjfN by Proposition 6, where fN is a primal submodule of M′.Proposition 8.
LetN and L be proper submodules of an R-module M and I be an ideal of R. If LI⊆N, then either L⊆Nor I⊆adjN.Proof.
AssumeL⊄N; then, there is l∈L−N. For each a∈I, lRa⊆LI⊆Nwhile l∉N; thus, a∈adjN. Callialp and Takir introduced the following definition (see [13]).Definition 11.
LetN,N1,N2,…,Nn be submodules of an R-module M. The covering N⊆N1∪N2∪⋯∪Nn of N is called efficient if N is not contained in the union of any n−1 of the submodulesNi′s. Analogously, we shall say N⊆N1∪N2∪⋯∪Nnis an efficient union if none of the Ni′s may be excluded.Proposition 9.
IfL and N are submodules of an R-module M, thenL∩N:M=L:M∩N:M.Proof.
Leta∈L∩N:M. Then, Ma⊆L∩N. Thus, Ma⊆L and Ma⊆N; hence, a∈L:M∩N:M. Therefore, L∩N:M⊆L:M∩N:M. Now, if b∈L:M∩N:M, then b∈L:Mandb∈N:M so that Mb⊆L and Mb⊆N; thus, Mb⊆L∩N and then b∈L∩N:M. Therefore, L:M∩N:ML∩N:M.Proposition 10.
(see [13]).LetN⊆N1∪N2∪⋯∪Nn be an efficient cover of submodules of an R-module M where n>2. Then, N∩​∩j≠kNj⊆Nkfor all k∈1,2,…,n. Now, by using Propositions 9 and 10 we will prove the following lemma and theorem.Lemma 2.
LetN⊆N1∪N2∪⋯∪Nn be an efficient cover of submodules of an R-module M where n>2; then, for all j∈1,2,…,n, ∩i≠jNi:M⊆adjNj.Proof.
Letj∈1,2,…,n. Put ∩i≠jNi:M=Pj. By Proposition 9, ∩i≠jNi:M=Pj. So, MPj⊆∩i≠jNi, and thus NPj⊆∩i≠jNi. But NPj⊆N; then, NPj⊆N∩∩i≠jNi⊆Nj by Proposition 10. This implies either N⊆Nj or Pj⊆adjNj. But N⊄Nj, and this implies that∩i≠jNi:M⊆adjNj.Theorem 2.
LetN be a submodule of an R-module M. If N1,N2,...,Nn are submodules of M such that N⊆N1∪N2∪⋯∪Nnand ∩i≠jNi:M⊄adjNj for all j=1,2,…,n except possibly for at most two of the j’s, then N⊆Nkfor some k∈1,2,…,n.Proof.
For the given coveringN⊆N1∪N2∪⋯∪Nn, let N⊆Nα1∪Nα2∪⋯∪Nαmbe its efficient reduction. Then, 1≤m≤n and m≠2. If m>2, then there exists at least one Nij satisfying ∩i≠jNαi:M⊆adjNαj which is contradiction to Lemma 2. Hence, m=1, so N⊆Nα1=Nk for some k∈1,2,…,n.
Bland in [14] proved the following result.Corollary 2.
LetP be a prime ideal of a ring R and suppose that A and B are ideals in R. If A∩B⊆P, then either A⊆PorB⊆P. The following corollary follows immediately from Proposition 5 and Corollary 2.Corollary 3.
LetN1,N2,…,Nn be uniformly primal submodules. Then, the following two conditions are equivalent:(a)
∩i≠jNi:M⊄adjNj(b)
Ni:M⊄adjNjwheneveri≠j
Now, the main theorem of this section is uniformly primal avoidance theorem which follows immediately from Proposition5, Theorem 2, and Corollary 3.Theorem 3.
(uniformly primal avoidance theorem for modules).Let N be a submodule of an R-module M.N1,N2,…,Nn are submodules of M such that N⊆N1∪N2∪⋯∪Nn. Assume that at most two of the Ni′s are not uniformly primal and Ni:M⊄adjNjwheneveri≠j; then, N⊆Nkfor some k∈1,2,…,n.Corollary 4.
(see [5]).LetM be an R-module. If N is a prime submodule of M, then N is primal. Since every prime submodule is uniformly primal by Corollary 4, then the uniformly primal avoidance theorem is a generalization of the prime avoidance theorem for modules. Now, we will recall the concept of S-system subsets of modules, which was introduced in [11] (also see [13,15]). Then, we will prove some results on the S-system and uniformly primal submodule.Definition 12.
A nonempty subsetS of a ring R is said to be an m-system if for any a,b∈S, there exists r∈R such that arb∈S.Definition 13.
LetM be an R-module and S be an m-system. A nonempty subset N of R-module M is said to be a S-system if for any a∈S and m∈N, there exists r∈R such that mra∈N.Proposition 11.
LetM be an R-module and N be a uniformly primal submodule of M. Then, M−N is an S-system where S=R−adjN.Proof.
Since N is a uniformly primal submodule ofM, by Proposition 5, adjN is a prime ideal of R. Let a∈S=R−adjN and t∈M−N, so tRa⊄N. Therefore, M−N is an S-system.Proposition 12.
LetP1,P2,…,Pn be a finite number of prime ideals in a ring R and S=R−∪i=1nPi. Then, S is an m-system subset of R.Proof.
Leta,b∈S and assume on the contrary that aRb⊄S; thus, aRb⊆∪i=1nPi. Then, RaRbR⊆∪i=1nPi. Hence, by the prime avoidance theorem for rings (see [3] and [4]), we have RaRRbR=RaRbR⊆Pi for some i=1,2,…,n. Since Pi is prime, then either RaR⊆Pi or RbR⊆Pi. If RaR⊆Pi, then a∉S which is a contradiction. Similarly, if RbR⊆Pi, then there exists r∈R such that arb∈S. Therefore, S is an m-system subset of R.Proposition 13.
LetN1,N2,…,Nn be a finite collection of uniformly primal submodules of an R-module M with adjNj=Pj for every j and Nj:M⊄adjNkwheneverj≠k. Then, S∗=M−∪i=1nNi is an S-system subset of M, where S=R−∪i=1nPi.Proof.
By Proposition12, S is an m-system subset of R, so to prove S∗is an S-system, let a∈S and m∈S∗ and assume on the contrary that mRa⊄S∗; thus, mRa⊆∪i=1nNi. Then, mRaR⊆∪i=1nNi, so by Theorem 3 (uniformly primal avoidance theorem for modules), we have mRaR⊆Niand mRa⊆Ni for some i=1,2,…,n. Since Niis uniformly primal, then m∈Nior a∈adjNi. If m∈Ni, then m∉S∗ which is a contradiction. If a∈adjNi, then a∉S. Then, there exists r∈R such that mra∈S∗. Therefore, S∗is an S-system subset of M.



---

*Source: 1593253-2020-10-22.xml*