Hindawi Publishing Corporation Algebra Volume 2015, Article ID 183930, 6 pages http://dx.doi.org/10.1155/2015/183930 Research Article On π-Absorbing Primary Elements in Lattice Modules Sachin Ballal and Vilas Kharat Department of Mathematics, Savitribai Phule Pune University, Pune 411 007, India Correspondence should be addressed to Sachin Ballal; [email protected] Received 18 December 2014; Accepted 31 March 2015 Academic Editor: Andrei V. Kelarev Copyright © 2015 S. Ballal and V. Kharat. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Let πΏ be a πΆ-lattice and let π be a lattice module over πΏ. Let π : π β π be a function. A proper element π β π is said to be π-absorbing primary if, for π₯1 , π₯2 , . . . , π₯π β πΏ and π β π, π₯1 π₯2 β β β π₯π π β€ π and π₯1 π₯2 β β β π₯π π β° π(π) together imply βπ, for some π β {1, 2, . . . , π}. We study some basic properties of π-absorbing π₯1 π₯2 β β β π₯π β€ (π : 1π ) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ π primary elements. Also, various generalizations of prime and primary elements in multiplicative lattices and lattice modules as π-absorbing elements and π-absorbing primary elements are unified. 1. Introduction A lattice πΏ is said to be complete, if, for any subset π of πΏ, we have β¨π, β§π β πΏ. Since every complete lattice is bounded, 1πΏ (or 1) denotes the greatest element and 0πΏ (or 0) denotes the smallest element of πΏ. A complete lattice πΏ is said to be a multiplicative lattice, if there is a defined binary operation ββ β called multiplication on πΏ satisfying the following conditions: (1) π β π = π β π, for all π, π β πΏ, (2) π β (π β π) = (π β π) β π, for all π, π, π β πΏ, (3) π β (β¨πΌ ππΌ ) = β¨πΌ (π β ππΌ ), for all π, ππΌ β πΏ, (4) π β 1 = π, for all π β πΏ. Henceforth, π β π will be simply denoted by ππ. An element π =ΜΈ 1 of a multiplicative lattice πΏ is said to be prime if ππ β€ π implies either π β€ π or π β€ π, for π, π β πΏ. Radical of an element π β πΏ is denoted byβπ and is defined asβπ = β¨{π₯ β πΏ | π₯π β€ π, for some π β Z+ }. An element π of a complete lattice πΏ is said to be compact if π β€ β¨πΌ ππΌ implies π β€ β¨ππ=1 ππΌπ , where π β Z+ . The set of all compact elements of a lattice πΏ is denoted by πΏ β . By a πΆ-lattice we mean a multiplicative lattice πΏ with a multiplicatively closed set π of compact elements which generates πΏ under join. A complete lattice π is said to be a lattice module over a multiplicative lattice πΏ, if there is a multiplication between elements of π and πΏ, denoted by ππ for π β πΏ and π β π, which satisfies the following properties: (1) (ππ)π = π(ππ), (2) (β¨πΌ ππΌ )(β¨π½ ππ½ ) = β¨πΌ,π½ (ππΌ ππ½ ), (3) 1πΏ π = π, (4) 0πΏ π = 0π, for all π, π, ππΌ β πΏ and for all π, ππ½ β π, where 1π denotes the greatest element of π and 0π denotes the smallest element of π. For π β π and π β πΏ, denote (π : π) = β¨{π β π : ππ β€ π}. For π, π β πΏ, (π : π) = β¨{π₯ β πΏ : ππ₯ β€ π} and for π΄, π΅ β π, (π΄ : π΅) = β¨{π₯ β πΏ : π₯π΅ β€ π΄}. For π β π, βπ = β¨{π₯ β πΏ : π₯π 1π β€ π} for some positive integer π and it is also denoted byβ(π : 1π). For π β π, we define π β π =β(π : 1π)1π. An element π β π is said to be weak join principal if it satisfies the following identity πβ¨(0π : π) = (ππ : π) for all π β πΏ. A lattice module π over a multiplicative lattice πΏ is called a multiplication lattice module if for π β π there exists an element π β πΏ such that π = π1π. An element π =ΜΈ 1π in π is said to be prime if ππ β€ π implies π β€ π or π1π β€ π, that is, π β€ (π : 1π) for every π β πΏ and π β π. An element π β π is called compact if π β€ β¨πΌ π΄ πΌ implies π β€ π΄ πΌ1 β¨ π΄ πΌ2 β¨ β β β β¨ π΄ πΌπ for some πΌ1 , πΌ2 , . . . , πΌπ . If each element of π is a join of principal (compact) elements of 2 π, then π is called principally generated lattice (compactly generated lattice). CΜ§allΔ±alp et al. [1] studied the concepts of weakly prime and almost prime elements in multiplicative lattices as extensions of, respectively, weakly prime and almost prime ideals in commutative rings. An element π =ΜΈ 1 in πΏ is said to be weakly prime if 0 =ΜΈ ππ β€ π implies either π β€ π or π β€ π and almost prime if ππ β€ π and ππ β° π2 implies either π β€ π or π β€ π for π, π β πΏ. In [2], the authors generalized these concepts, respectively, to weakly primary and almost primary elements in multiplicative lattices. A proper element π β πΏ is said to be weakly primary if 0 =ΜΈ ππ β€ π implies either π β€ π or π β€βπ and almost primary if ππ β€ π and ππ β° π2 implies either π β€ π or π β€βπ for π, π β πΏ. The concept of prime elements in multiplicative lattices is further generalised to 2-absorbing and weakly 2-absorbing elements in multiplicative lattices by Jayaram et al. [3]. An element π < 1 in πΏ is said to be 2-absorbing if πππ β€ π implies either ππ β€ π or ππ β€ π or ππ β€ π and is said to be weakly 2-absorbing element if 0 =ΜΈ πππ β€ π implies either ππ β€ π or ππ β€ π or ππ β€ π, for π, π, π β πΏ. Joshi and Ballal [4] defined the concept of π-prime elements in multiplicative lattices as a generalization of prime elements. An element π < 1 of a multiplicative lattice πΏ is said to be an π-prime if we can express it as meet of at most π primes, where π is a positive integer. In [5], the authors defined the concepts of π-absorbing, weakly π-absorbing, and π-almost π-absorbing elements in multiplicative lattices as generalizations of, respectively, 2absorbing, weakly 2-absorbing, and almost prime elements, where π β₯ 2. An element π < 1 of a multiplicative lattice πΏ is called π-absorbing if π₯1 π₯2 π₯3 β β β π₯π+1 β€ π implies π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π+1 β€ π and called weakly π-absorbing if, for π₯1 , π₯2 , π₯3 , . . . , π₯π+1 β πΏ, 0 =ΜΈ π₯1 π₯2 π₯3 β β β π₯π+1 β€ π implies π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π+1 β€ π, π β {1, 2, . . . , π}, and π₯1 , π₯2 , π₯3 , . . . , π₯π+1 β πΏ. An element π < 1 of a multiplicative lattice πΏ is called π-almost π-absorbing if π₯1 π₯2 π₯3 β β β π₯π+1 β€ π and π₯1 π₯2 π₯3 β β β π₯π+1 β° ππ together imply π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π+1 β€ π for π β {1, 2, . . . , π}, π₯1 , π₯2 , π₯3 , . . . , π₯π+1 β πΏ. Manjarekar and Bingi [6] unified the theory of generalizations of prime and primary elements in multiplicative lattice as π-prime and π-primary elements. Let π : πΏ β πΏ be a function. A proper element π β πΏ is said to be π-prime if ππ β€ π and ππ β° π(π) implies either π β€ π or π β€ π and it is said to be π-primary if ππ β€ π and ππ β° π(π) implies either π β€ π or π β€βπ, for π, π β πΏ. As a generalization of primary ideals in commutative rings, Badawi et al. [7] introduced the concept of 2-absorbing primary ideals. In this paper, we extend the various generalizations of prime ideals and primary ideals in commutative rings to lattice modules. Also, we unify various generalizations of prime and primary elements in multiplicative lattices and lattice modules, respectively, as π-absorbing elements and π-absorbing primary elements. For basic concepts and terminologies of lattice modules, one may refer to [8β11] and for multiplicative lattices, one may refer to [12β14]. Algebra 2. π-Absorbing Primary Elements We introduce the concepts of π-absorbing elements and π-absorbing primary elements in lattice modules which generalizes, respectively, the concepts of prime and primary elements in multiplicative lattices and lattice modules (see [1β 3, 5, 6]). Essentially, we have the following definitions, where π is a lattice module over a multiplicative lattice πΏ and π β₯ 1. Definition 1. A proper element π β π is said to be πabsorbing if π₯1 π₯2 β β β π₯π π β€ π implies π₯1 π₯2 β β β π₯π β€ (π : 1π) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ π for π β {1, 2, . . . , π}, where π₯1 , π₯2 , . . . , π₯π β πΏ and π β π. Definition 2. Let π : π β π be a function. A proper element π β π is said to be π-absorbing if π₯1 π₯2 β β β π₯π π β€ π and π₯1 π₯2 β β β π₯π π β° π(π) together imply π₯1 π₯2 β β β π₯π β€ (π : 1π) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ π for π β {1, 2, . . . , π}, where π₯1 , π₯2 , . . . , π₯π β πΏ and π β π. Definition 3. A proper element π β π is said to be πabsorbing primary if π₯1 π₯2 β β β π₯π π β€ π implies π₯1 π₯2 β β β π₯π β€ π (π : 1π) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ β π for π β {1, 2, . . . , π}, where π₯1 , π₯2 , . . . , π₯π β πΏ and π β π. Definition 4. Let π : π β π be a function. A proper element π β π is said to be π-absorbing primary if π₯1 π₯2 β β β π₯π π β€ π and π₯1 π₯2 β β β π₯π π β° π(π) together imply π π₯1 π₯2 β β β π₯π β€ (π : 1π) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ β π for π β {1, 2, . . . , π}, where π₯1 , π₯2 , . . . , π₯π β πΏ and π β π. Let π be a lattice module over a multiplicative lattice πΏ. For a map ππΌ : π β π, we have the following. (1) π0 : π(π) = 0 defines weakly π-absorbing primary elements of π. (2) π2 : π(π) = (π : 1π)π defines almost π-absorbing primary elements of π. (3) ππ+1 (π β₯ 1) : π(π) = (π : 1π)π π defines π-almost π-absorbing primary elements of π. π (4) ππ : π(π) = β§β π=1 (π : 1π ) π defines π-absorbing primary elements of π. Remark 5. It follows immediately from the definition that any π-absorbing primary element of π is a π-absorbing primary. However, the converse does not necessarily holds. In fact, we have the following theorem in which the converse is true under certain condition. Theorem 6. Let π be a lattice module over a πΆ-lattice πΏ. Then every π-absorbing primary element π β π with (π : 1π)π π β° π(π) is π-absorbing primary. Proof. Suppose that (π : 1π)π π β° π(π) and π₯1 π₯2 β β β π₯π π β€ π, for π₯1 , π₯2 , . . . , π₯π β πΏ, π β π. If π₯1 π₯2 β β β π₯π π β° π(π), then as π is π-absorbing primary, we have π₯1 π₯2 β β β π₯π β€ (π : 1π) π or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ β π for some π β {1, 2, . . . , π}, which concludes that π is π-absorbing primary. Algebra 3 Now, suppose that π₯1 π₯2 β β β π₯π π β€ π(π). We assume that π₯1 π₯2 β β β π₯πβπ (π : 1π)π π β€ π(π), for all π β {1, 2, . . . , π β 1}. If π₯1 π₯2 β β β π₯πβπ (π : 1π)π π β° π(π), then there exists π1 , π2 , . . . , ππ β€ (π : 1π) such that π₯1 π₯2 β β β π₯πβπ π1 π2 β β β ππ π β° π(π). Hence, π₯1 π₯2 β β β π₯πβπ (π₯πβπ+1 β¨ π1 )(π₯πβπ+2 β¨ π2 ) β β β (π₯π β¨ ππ )π β€ π and π₯1 π₯2 β β β π₯πβπ (π₯πβπ+1 β¨ π1 )(π₯πβπ+2 β¨ π2 ) β β β (π₯π β¨ ππ )π β° π(π). Since π is π-absorbing primary, π₯1 π₯2 β β β π₯πβπ (π₯πβπ+1 β¨ π1 ) β β β (π₯π β¨ ππ ) β€ (π : 1π) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π π₯πβπ (π₯πβπ+1 β¨ π1 )(π₯πβπ+2 β¨ π2 ) β β β (π₯π β¨ ππ )π β€ β π for some π β {1, 2, . . . , π β 1} and so π₯1 π₯2 β β β π₯πβ1 π₯π+1 π β β β π₯πβπ π₯πβπ+1 π₯πβπ+2 β β β π₯π π β€ β π for some π β {1, 2, . . . , π β 1} or π₯1 π₯2 β β β π₯π β€ (π : 1π). β Similarly, we assume that, for {π1 , π2 , . . . , ππβπ } {1, 2, . . . , π}, 1 β€ π β€ π, π₯π1 π₯π2 β β β π₯ππβπ (π : 1π)π π β€ π(π). Also, we assume that π₯π1 π₯π2 β β β π₯ππβπ (π : 1π)π π β€ π(π), 1 β€ π β€ π, because if π₯π1 π₯π2 β β β π₯ππβπ (π : 1π)π π β° π(π), β° π(π), where then π₯π1 π₯π2 β β β π₯ππβπ π1 π2 β β β ππ π π1 π2 π3 β β β ππ β€ (π : 1π) and so π₯1 π₯2 β β β π₯πβπ (π₯πβπ+1 β¨ π π1 )(π₯πβπ+2 β¨ π2 ) β β β (π₯π β¨ ππ )(π β¨ π) β€ β π and π₯1 π₯2 β β β π₯πβπ (π₯πβπ+1 β¨ π1 )(π₯πβπ+2 β¨ π2 ) β β β (π₯π β¨ ππ )(π β¨ π) β° π(π). Now, since π is π-absorbing primary, π₯1 π₯2 β β β π₯π β€ π (π : 1π) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ β π, for some π β {1, 2, . . . , π β 1}. Also, (π : 1π)π π β° π(π) implies that there exist π1 , π2 , . . . , ππ β€ (π : 1π) with π1 π2 β β β ππ π β° π(π). We have (π₯1 β¨ π1 )(π₯2 β¨ π2 ) β β β (π₯π β¨ ππ )(π β¨ π) β€ π and (π₯1 β¨ π1 )(π₯2 β¨ π2 ) β β β (π₯π β¨ππ )(πβ¨π) β° π(π). Now, π is π-absorbing primary; therefore, π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯πβπ π₯πβπ+1 π₯πβπ+2 β β β π₯π π β€ π, for some π β {1, 2, . . . , π β 1} or π₯1 π₯2 β β β π₯π β€ (π : 1π) and consequently π is π-absorbing primary. From the above theorem, it follows that if π is a π-primary element of π that is not π-absorbing primary, then (π : 1π)π π β€ π(π). Corollary 7. Let π be a lattice module over a πΆ-lattice πΏ. If π β π is weakly π-absorbing primary that is not π-absorbing primary, then (π : 1π)π π = 0. Theorem 8. Let π be a multiplication lattice module over πΆlattice πΏ and let π β π. Then the following holds. (1) Let π1 , π2 : π β π be two functions with π1 β€ π2 , that is, π1 (π) β€ π2 (π) for each π β π. Then π is π2 -absorbing primary if it is π1 -absorbing primary. (2) Consider the following statements. (a) π is π-absorbing primary. (b) π is weakly π-absorbing primary. (c) π is π-absorbing primary. (d) π is π-almost π-absorbing primary. (e) π is almost π-absorbing primary. Then (a) β (b) β (c) β (d) β (e). (3) π is π-absorbing primary if and only if it is π-almost π-absorbing primary for π β₯ 1. Proof. (1) Suppose that π β π is π1 -absorbing primary and also suppose that π₯1 π₯2 β β β π₯π π β€ π and π₯1 π₯2 β β β π₯π π β° π2 (π) for π₯1 , π₯2 , . . . , π₯π β πΏ and π β π. Since π1 (π) β€ π2 (π) for each π β π, we have π₯1 π₯2 β β β π₯π π β° π1 (π). It follows from the fact π₯1 π₯2 β β β π₯π π β€ π, π₯1 π₯2 β β β π₯π π β° π1 (π), and π is π1 -absorbing primary that π₯1 π₯2 β β β π₯π β€ (π : 1π) or π π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ β π for some π β {1, 2, . . . , π} and so, π is π2 -absorbing primary. (2) By definition, every π-absorbing primary element is weakly π-absorbing primary and therefore (a) β (b) holds. (b) β (c) Suppose that π is weakly π-absorbing primary and also suppose that π₯1 π₯2 β β β π₯π π β€ π and π₯1 π₯2 β β β π₯π π β° π β§β π=1 (π : 1π ) π, for π₯1 , π₯2 , . . . , π₯π β πΏ and π β π. Then π₯1 π₯2 β β β π₯π π =ΜΈ 0π. By the assumption, π₯1 π₯2 β β β π₯π β€ (π : 1π) π or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ β π for some π β {1, 2, . . . , π} and so, π is π-absorbing primary. (c) β (d) Suppose that π is π-absorbing primary and also suppose that π₯1 π₯2 β β β π₯π π β€ π and π₯1 π₯2 β β β π₯π π β° (π : 1π)π π for π₯1 , π₯2 , . . . , π₯π β πΏ, π β π, and π β₯ 1. Then π π₯1 π₯2 β β β π₯π π β€ π and π₯1 π₯2 β β β π₯π π β° β§β π=1 (π : 1π ) π. Since π is π-absorbing primary, it follows that π₯1 π₯2 β β β π₯π β€ π (π : 1π) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ β π for some π β {1, 2, . . . , π} and so, π is π-almost π-absorbing primary. The statement (e) is a particular case of the statement (d) for π = 1 and therefore (d) β (e) holds. (3) Suppose that π β π is π-almost π-absorbing primary for π β₯ 1 and also suppose that π₯1 π₯2 β β β π₯π π β€ π and π π₯1 π₯2 β β β π₯π π β° β§β π=1 (π : 1π ) π for π₯1 , π₯2 , . . . , π₯π β πΏ and π β π. Then π₯1 π₯2 β β β π₯π π β€ π and π₯1 π₯2 β β β π₯π π β° (π : 1π)π π for some π β₯ 1. Since π is π-almost πabsorbing primary, we have π₯1 π₯2 β β β π₯π β€ (π : 1π) or π π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ β π for some π β {1, 2, . . . , π}. Consequently, π is π-absorbing primary. The converse follows from (c) β (d). Corollary 9. Let π be a lattice module over a πΆ-lattice πΏ. If π β π is π-absorbing primary, where π β€ ππ+2 , then π is πabsorbing primary. Proof. Suppose that π β π is π-absorbing primary. If π is π-absorbing primary, then, by Theorem 8, it is π-absorbing primary. Now, if π is not π-absorbing primary, then, by Theorem 6, (π : 1π)π π β€ π(π) β€ ππ+2 (π) = (π : 1π)π+1 π. Consequently, π(π) = (π : 1π)π π, for each π β₯ π. And, by Theorem 8(3), π is π-absorbing primary. Lemma 10. Let πΏ be a multiplicative lattice. If π β πΏ is πabsorbing primary with π(βπ) = βπ(π), then βπ is also πabsorbing. β€ Proof. (I) Suppose that π₯1 π₯2 β β β π₯π+1 βπ and π₯1 π₯2 β β β π₯π+1 β° π(βπ) = βπ(π), for π₯1 , π₯2 , . . . , π₯π+1 β πΏ. If π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π+1 β° βπ for π β {1, 2, . . . , π}, then we have (π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π+1 )π β° π for any positive integer π and for π β {1, 2, . . . , π}. (II) Now, since π₯1 π₯2 β β β π₯π+1 β€ βπ, there exists a positive π integer π such that (π₯1 π₯2 β β β π₯π+1 )π = π₯1π π₯2π β β β π₯π+1 β€ π. 4 Algebra π‘ (III) As π₯1 π₯2 β β β π₯π+1 β° βπ(π), we have π₯1π‘ π₯2π‘ β β β π₯π+1 β° π(π) for a positive integer π‘. From the assumption and (I), (II), and (III), it follows that (π₯1 π₯2 β β β π₯π )π = π₯1π π₯2π β β β π₯ππ β€ βπ. Consequently, π2 1 β π1 β€ πβπ 1 and π2 β€ βπ2 ; π₯1 π₯2 β β β π₯π β€ ββπ and therefore βπ is π-absorbing primary. π₯ β€β(π1 : 1π1 ) and π¦ β€β(π2 : 1π2 ); It is easy to observe that if (πΏ 1 , β§1 , β¨1 , β1 ) and (πΏ 2 , β§2 , β¨2 , β2 ) are multiplicative lattices then πΏ 1 × πΏ 2 is also a multiplicative lattice with componentwise meet, join, and multiplication. Also, if π1 and π2 are lattice modules over multiplicative lattices πΏ 1 and πΏ 2 , respectively, then π1 × π2 is a lattice module over πΏ 1 × πΏ 2 with componentwise meet, join, and multiplication given by (π, π)(π1 , π2 ) = (ππ1 , ππ2 ), where (π, π) β πΏ 1 × πΏ 2 and (π1 , π2 ) β π1 × π2 . Lemma 11. Let π = π1 × π2 and πΏ = πΏ 1 × πΏ 2 , where ππ is a lattice module over πΆ-lattice πΏ π , for π = 1, 2. Then ((π1 : 1π1 ), (π2 : 1π2 )) = ((π1 , π2 ) : (1π1 , 1π2 )), for π1 β π1 and π2 β π2 . Proof. Let (π₯, π¦) β πΏ β . Now, (π₯, π¦) β€ ((π1 : 1π1 ), (π2 : 1π2 )): β π₯ β€ (π1 : 1π1 ) and π¦ β€ (π2 : 1π2 ), β π1 β€ π₯1π1 and π2 β€ π¦1π2 for some π₯, π¦ β πΏ with β π₯π‘ β€ (π1 : 1π1 ) and π¦π β€ (π2 : 1π2 ) for some positive integers π‘ and π. Choose π = π‘ + π. Then π₯π β€ (π1 : 1π1 ) and π¦π β€ (π2 : 1π2 ): β (π₯, π¦)π = (π₯π , π¦π ) β€ ((π1 : 1π1 ), (π2 : 1π2 )); β (π₯, π¦) β€β((π1 : 1π1 ), (π2 : 1π2 )); β (π₯, π¦)(1π1 , 1π2 ) β€ β((π1 : 1π1 ), (π2 : 1π2 )) (1π1 , 1π2 ); β (π1 , π2 ) β€ (π₯, π¦)(1π1 , 1π2 ) β€ β((π1 : 1π1 ), (π2 : 1π2 ))(1π1 , 1π2 ) = β(π1 , π2 ). π Consequently, π 1 2 π1 , πβ π2 ) β€ β (π1 , π2 ). (πβ (II) β π₯1π1 β€ π1 and π¦1π2 β€ π2 , From (I) and (II), result follows. β (π₯, π¦)(1π1 , 1π2 ) = (π₯1π1 , π¦1π2 ) β€ (π1 , π2 ), Theorem 13. Let π = π1 × π2 and πΏ = πΏ 1 × πΏ 2 , where ππ is a lattice module over πΆ-lattice πΏ π , for π = 1, 2, and let π : π β π be a function. If π1 β π1 is a weakly π-absorbing primary such that (0, 1π2 ) β€ π(π1 , 1π2 ), then (π1 , 1π2 ) β π1 × π2 is π-absorbing primary. β (π₯, π¦) β€ (π1 , π2 ) : (1π1 , 1π2 ). Lemma 12. Let π = π1 × π2 and πΏ = πΏ 1 × πΏ 2 , where ππ is a π lattice module over πΆ-lattice πΏ π , for π = 1, 2. Then β(π 1 , π2 ) = π1 π2 ( βπ1 , βπ2 ), for π1 β π1 and π2 β π2 . π Proof. Let (π1 , π2 ) β πβ with (π1 , π2 ) β€ β(π 1 , π2 ) = β€ β(π1 , π2 ) : (1π1 , 1π2 )(1π1 , 1π2 ). Then (π1 , π2 ) (π, π)(1π1 , 1π2 ), where (π, π) β€β(π1 , π2 ) : (1π1 , 1π2 ). Now, (π, π) β€β(π1 , π2 ) : (1π1 , 1π2 ) β (π, π)π β€ (π1 , π2 ) : (1π1 , 1π2 ) for a positive integer π: β ππ β€ (π1 : 1π1 ) and ππ β€ (π2 : 1π2 ) (by Lemma 11), β (π, π) β€ (β(π1 : 1π1 ),β(π2 : 1π2 )), β (π, π)(1π1 , 1π2 ) β€ (β(π1 : 1π1 ),β(π2 : 1π2 )) (1π1 , 1π2 ) = (β(π1 : 1π1 )1π1 ,β(π2 : 1π2 )1π2 ), π2 1 β (π, π)(1π1 , 1π2 ) β€ (πβπ 1 , βπ2 ). Consequently, π 1 2 β (π1 , π2 ) β€ (πβ π1 , πβ π2 ) . (I) 1 Next, let (π1 , π2 ) β πβ be such that (π1 , π2 ) β€ (πβπ 1, βπ2 ): π2 β π1 β€β(π1 : 1π1 )1π1 and π2 β€β(π2 : 1π2 )1π2 ; Proof. Let (π1 , π1 ), (π2 , π2 ), . . . , (ππ , ππ ) β πΏ and (π1 , π2 ) β π = be such that (π1 , π1 )(π2 , π2 ) β β β (ππ , ππ )(π1 , π2 ) (π1 π2 β β β π2 π1 , π1 π2 β β β π2 π2 ) β€ (π1 , 1π2 ) and (π1 , π1 )(π2 , π2 ) β β β (ππ , ππ )(π1 , π2 ) = (π1 π2 β β β ππ π1 , π1 π2 β β β ππ π2 ) β° π(π1 , 1π2 ). Since (0, 1π2 ) β€ π(π1 , 1π2 ), we have (π1 , π1 )(π2 , π2 ) β β β (ππ , ππ )(π1 , π2 ) = (π1 π2 β β β ππ π1 , π1 π2 β β β ππ π2 ) β° (0, 1π2 ) and so 0 =ΜΈ π1 π2 β β β ππ π1 β€ π1 . Since π1 β π1 is weakly π-absorbing primary, we have π1 π2 β β β ππ β€ 1 (π1 : 1π1 ) or π1 π2 β β β ππβ1 ππ+1 β β β ππ π1 β€ πβπ 1 , for some π β {1, 2, . . . , π}. This implies that (π1 , π1 )(π2 , π2 ) β β β (ππ , ππ ) = (π1 π2 β β β ππ , π1 π2 β β β ππ ) β€ ((π1 : 1π1 ), (1π2 : 1π2 )) = ((π1 , 1π2 ) : (1π1 , 1π2 )) or (π1 , π1 )(π2 , π2 ) 1 β β β (ππβ1 , ππβ1 )(ππ+1 , ππ+1 ) β β β (ππ , ππ )(π1 , π2 ) β€ (πβπ 1 , 1π2 ) = π β(π 1 , 1π2 ) for some π β {1, 2, . . . , π}, by Lemmas 11 and 12. Therefore, (π1 , 1π2 ) is π-absorbing primary element of π1 × π2 . Theorem 14. Let π be a lattice module over a πΆ-lattice πΏ and π β π. Then π is π-absorbing primary if and only if (π : π₯1 π₯2 β β β π₯πβ1 π) = (π : π₯1 π₯2 β β β π₯πβ1 1π) or (π : π₯1 π₯2 β β β π₯πβ1 π) = (π(π) : π₯1 π₯2 β β β π₯πβ1 π) or (π : π₯1 π₯2 β β β π₯πβ1 π) = (βπ : π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯πβ1 π) for some π β {1, 2, . . . , π β 1}, for π₯1 , π₯2 , . . . , π₯πβ1 β πΏ and π β π with π π₯1 π₯2 β β β π₯πβ1 π β° β π. Algebra Proof. Suppose that π β π is π-absorbing primary and π π₯1 π₯2 β β β π₯πβ1 π β° β π, for π₯1 , π₯2 , . . . , π₯πβ1 β πΏ and π β π. Let π β πΏ β be such that π β€ (π : π₯1 π₯2 β β β π₯πβ1 π) which essentially implies that π₯1 π₯2 β β β π₯πβ1 ππ β€ π. We have the following two cases. Case 1. If ππ₯1 π₯2 β β β π₯πβ1 π β° π(π), then as π is πabsorbing primary, we have π₯1 π₯2 β β β π₯πβ1 π β€ (π : 1π) or π π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯πβ1 ππ β€ β π for some π β {1, 2, . . . , π β π 1}. Therefore, π β€ (π : π₯1 π₯2 β β β π₯πβ1 1π) or π β€ (β π : π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯πβ1 π) for some π β {1, 2, . . . , π β 1}. Case 2. If ππ₯1 π₯2 β β β π₯πβ1 π β€ π(π), then π β€ (π(π) : π₯1 π₯2 β β β π₯πβ1 π). So (π : π₯1 π₯2 β β β π₯πβ1 π) β€ (π(π) : π₯1 π₯2 β β β π₯πβ1 π). Now, from Cases 1 and 2, it follows that (π : π₯1 π₯2 β β β π₯πβ1 π) = (π : π₯1 π₯2 β β β π₯πβ1 1π) or (π : π₯1 π₯2 β β β π₯πβ1 π) = (π(π) : π₯1 π₯2 β β β π₯πβ1 π) or (π : π₯1 π₯2 β β β π₯πβ1 π) = π (β π : π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯πβ1 π) for some π β {1, 2, . . . , πβ1}. Conversely, suppose that π₯1 π₯2 β β β π₯π π β€ π and π₯1 π₯2 β β β π₯π π β° π(π), for π₯1 , π₯2 , . . . , π₯π β πΏ and π β π. If π π₯1 π₯2 β β β π₯πβ1 π β€ β π, then the result is obvious. So, suppose π that π₯1 π₯2 β β β π₯πβ1 π β° β π. Now, (π : π₯1 π₯2 β β β π₯πβ1 π) = (π : π₯1 π₯2 β β β π₯πβ1 1π) or (π : π₯1 π₯2 β β β π₯πβ1 π) = (π(π) : π₯1 π₯2 β β β π₯πβ1 π) or (π : π π₯1 π₯2 β β β π₯πβ1 π) = (β π : π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯πβ1 π) for some π β {1, 2, . . . , π β 1}. Since π₯1 π₯2 β β β π₯π π β€ π, we have π₯π β€ (π : π₯1 π₯2 β β β π₯πβ1 π). But π₯π β° (π(π) : π₯1 π₯2 β β β π₯πβ1 π) and so π π₯1 π₯2 β β β π₯πβ1 π₯π β€ (π : 1π) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ β π for some π β {1, 2, . . . , π}. Consequently, π is π-absorbing primary. Theorem 15. Let π = π1 × π2 × β β β × ππ and πΏ = πΏ 1 × πΏ 2 × β β β ×πΏ π , where each ππ is a compactly generated lattice module over a πΆ-lattice πΏ π , for π β {1, 2, . . . , π}. Let π : π β π such that π(π) = (π1 (π1 ), π2 (π2 ), . . . , ππ (ππ )), where ππ β ππ , ππ : ππ β ππ , π β {1, 2, . . . , π}, and π = (π1 , π2 , . . . , ππ ) is πabsorbing primary. Then ππ is a ππ -absorbing primary element of ππ , for each π with ππ =ΜΈ 1ππ . Proof. Let ππ =ΜΈ 1ππ , ππ β ππ , and π₯1 , π₯2 , . . . , π₯π β πΏ π be such that π₯1 π₯2 β β β π₯π ππ β€ ππ and π₯1 π₯2 β β β π₯π ππ β° ππ (ππ ). Thus (1, . . . , 1, π₯1 , 1, . . . , 1)(1, . . . , 1, π₯2 , 1, . . . , 1) β β β (1, . . . , 1, π₯π , 1, = (0, . . . , 0, π₯1 π₯2 β β β π₯π ππ , . . . , 1)(0, . . . , 0, ππ , 0, . . . , 0) 0, . . . , 0) β€ π, and (0, . . . , 0, π₯1 π₯2 β β β π₯π ππ , 0, . . . , 0) β° π(π). As π is π-absorbing primary, (1, 1, . . . , π₯1 π₯2 β β β π₯π , 0, . . . , 0) β€ (π : 1π) or (1, 1, . . . , π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π ππ , 0, . . . , 0) β€ π β π. Now, by Lemmas 11 and 12, we have π₯1 π₯2 β β β π₯π β€ (ππ : π 1ππ ) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π ππ β€ βπ π for π β {1, 2, . . . , π} and consequently, ππ is ππ -absorbing primary element of ππ , for each π. Corollary 16. Let π = π1 × π2 × β β β × ππ and πΏ = πΏ 1 × πΏ 2 × β β β ×πΏ π , where each ππ is a compactly generated lattice module over a πΆ-lattice πΏ π , for π β {1, 2, . . . , π}. If π = (π1 , π2 , . . . , ππ ) β π is ππ -absorbing primary, where ππ β ππ , then ππ β ππ is ππ -absorbing primary with ππ =ΜΈ 1ππ (π, π β₯ 2). 5 Proof. We have ππ (π) = (π : 1π)πβ1 π = ((π1 : 1π1 )πβ1 π1 , (π2 : 1π2 )πβ1 π2 , . . . , (ππ : 1ππ )πβ1 ππ ) = (ππ (π1 ), ππ (π2 ), . . . , ππ (ππ )) and the result follows from Theorem 15. Theorem 17. Let π be a lattice module over a πΆ-lattice πΏ. Suppose that π1π β π is a weak join principal element with π1π =ΜΈ 1π and (0π : π1π) β€ βπ. Then π1π is π-almost π-absorbing primary, π β₯ 1, if and only if it is π-absorbing primary. Proof. Suppose that π1π β π is π-almost π-absorbing primary and π₯1 π₯2 β β β π₯π π β€ π1π, for π₯1 , π₯2 , . . . , π₯π β πΏ and π β π. If π₯1 π₯2 β β β π₯π π β° (π1π : 1π)π π1π, then π₯1 π₯2 β β β π₯π β€ (π1π : 1π) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ π βπ1 π for some π β {1, 2, . . . , π} and therefore π1π is πabsorbing primary. Now, suppose that π₯1 π₯2 β β β π₯π π β€ (π1π : 1π)π π1π. Since π₯1 π₯2 β β β π₯π π β€ π1π, we have (π₯1 β¨ π)π₯2 β β β π₯π π β€ π1π. If (π₯1 β¨ π)π₯2 β β β π₯π π β° (π1π : 1π)π π1π, then it follows from the fact (π₯1 β¨ π)π₯2 β β β π₯π π β€ π1π, (π₯1 β¨ π)π₯2 β β β π₯π π β° (π1π : 1π)π π1π, and π1π is π-almost π-absorbing primary that π₯1 π₯2 β β β π₯π β€ (π1π : 1π) or π₯1 π₯2 β β β π₯πβ1 π₯π+1 β β β π₯π π β€ π βπ1 π for some π β {1, 2, . . . , π} and we are done. So assume that (π₯1 β¨ π)π₯2 β β β π₯π π β€ (π1π : 1π)π π1π. Then as π₯1 π₯2 β β β π₯π π β€ (π1π : 1π)π π1π, we have ππ₯2 β β β π₯π π β€ (π1π : 1π)π π1π. Next, ππ₯2 β β β π₯π π β€ (π1π : 1π)π π1π and π1π is weak join principal; together they imply that π₯2 β β β π₯π β€ ((π1π : 1π)π π1π : π1π) = (π1π : 1π)π β¨ (0π : π1π) β€ π β¨ (0π : π1π) β€ βπ. Consequently, π π₯2 β β β π₯π π β€βππ β€βπ1π = βπ1 π which implies that π1π is π-absorbing primary. The converse follows from Theorem 8(2). Note. The results pertaining to π-absorbing elements are essentially the corollaries to the respective results of πabsorbing primary elements, as such the results of πabsorbing elements are the immediate consequences of results of π-absorbing primary elements in this paper. Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. Acknowledgment This research work is an outcome of the project supported by Board of College and University Development, Savitribai Phule Pune University, Pune. References [1] F. CΜ§allΔ±alp, C. Jayaram, and UΜ. 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