Palestine Journal of Mathematics

Vol. 5(Special Issue: 1) (2016) , 127–135 © Palestine Polytechnic University-PPU 2016

ON φ-2-ABSORBING ELEMENTS IN MULTIPLICATIVE LATTICES Ece Yetkin Celikel, Emel A. Ugurlu and Gulsen Ulucak

Communicated by Ayman Badawi

MSC 2010 Classifications: 03G10, 03G99.

Keywords and phrases: φ-prime element, φ-primary element, 2-absorbing element, weakly 2-absorbing element, φ-2- absorbing element. Abstract In this paper, we introduce the concept of φ-2-absorbing elements in multiplicative lattices. Let φ : L L be a function. We will say a proper element q of L to be a → ∪ {∅} φ-2-absorbing element of L if whenever a,b,c L with abc q and abc  φ(q) implies either ab q or ac q or bc q. We give some basic∈ properties and≤ establish some characterizations of φ≤-2-absorbing≤ elements≤ in some special lattices.

1 Introduction

Several authors have studied various extensions of prime and primary ideals. A. Badawi [6] introduced the concept of 2-absorbing ideals in a commutative ring with identity, which is a gen- eralization of prime ideals. A. Badawi and A.Y. Darani [5] studied weakly 2-absorbing ideals which are generalizations of weakly prime ideals [3]. Weakly prime elements in multiplica- tive lattices are studied in [10]. The concepts of 2-absorbing primary and weakly 2-absorbing primary ideals of commutative rings are studied in [7] and [8]. The concepts of 2-absorbing, weakly 2-absorbing, 2-absorbing primary and weakly 2-absorbing primary elements in mul- tiplicative lattices are studied in [16] and [11] as generalizations of prime and weakly prime elements. Later, the concepts of φ-prime, φ-primary ideals are recently introduced in [12], [9], and generalizations of these are studied in [17]. In this work, our aim is to extend the concepts of 2-absorbing elements to φ-2-absorbing elements and investigate some characterizations in some special lattices. Throughout this paper R denotes a commutative ring with identity and L(R) denotes the lattice of all ideals of R. An element a of L is said to be compact if whenever a aα ≤α ∨∈I implies a aα for some finite subset I0 of I. A multiplicative lattice, we mean a complete ≤α∈ ∨I0 lattice L with the least element 0L and compact greatest element 1L, on which there is defined a commutative, associative, completely join distributive product for which 1L is a multiplicative identity. Throughout this paper L denotes a multiplicative lattice and L∗ denotes the set of all compact elements of L. Bya C-lattice we mean a (not necessarily modular) multiplicative lattice which is generated under joins by a multiplicatively closed subset C of compact elements. We note that in a C-lattice, a finite product of compact elements is again compact. An element a L is said to be idempotent if a = a2. For any a L, L/a = b L : a b is a multiplicative∈ ∈ { ∈ ≤ } lattice with the multiplication c d = cd a. An element a L is said to be proper if a< 1L. A proper element p of L is said to◦ be prime∨ if ab p implies∈ either a p or b p. C-lattices can ≤ ≤ ≤ be localized. For any prime element p of L, Lp denotes the localization at F = xǫC : x  p . { } If 0L is prime, then L is said to be a domain. A proper element p is called as φ-prime if ab p ≤ and ab  φ(p) implies either a p or b p for a,b L. Ina C-lattice, an element p is φ-prime ≤ ≤ ∈ if and only if ab p and ab  φ(p) implies either a p or b p for all a,b L∗ by [17]. An ≤ ≤ ≤ ∈ element m < 1L is said to be maximal in L if m < x 1L implies x = 1L. It can be easily shown that maximal elements are prime. For a,b L, we≤ denote (a : b) = x L : xb a . For a L, we define √a = p L : p is prime∈ and a p . Note that∨{ in∈ a C-lattice≤ L}, ∈ ∧{ ∈ ≤ } n + √a = p L : a p is a minimal prime over a = x L∗ : x a for some n Z . ∧{ ∈ ≤ } ∨{ ∈ ≤ ∈ } A proper element q is said to be primary if ab q implies either a q or b √q for every ≤ ≤ ≤ 128 Ece Yetkin Celikel, Emel A. Ugurlu and Gulsen Ulucak pair of elements a,b L. A proper element q is said to be φ-primary if for every a,b L with ∈ ∈ ab q and ab  φ(q) implies either a q or b √q. A proper element q of L is said to be a 2-absorbing≤ element if whenever a,b,c≤ L with≤abc q implies either ab q or bc q or ac q. ∈ ≤ ≤ ≤ ≤A multiplicative lattice is called a Noether lattice if it is modular, principally generated (every element is a join of some principal elements) which satisfies the ascending chain condition. A Noether lattice L is local if it contains precisely one maximal prime. If L is a Noether lattice and 0L is prime, then L is said to be a Noether domain. In[18], J. F. Wells studied the restricted cancellation law of a Noether lattice. An element a in a Noether lattice L satisfies the restricted cancellation law if ab = ac = 0L implies b = c for any a,b,c L. 6 ∈

2 φ-2-absorbing elements

Definition 2.1. Let φ : L L be a function and q L be a proper element. Then q is said → ∪{∅} ∈ to be a φ-2-absorbing element of L whenever if a,b,c L with abc q and abc  φ(q) implies either ab q or ac q or bc q. ∈ ≤ ≤ ≤ ≤ We can define the following special functions φα as follows: Let q be a φα-2-absorbing element of L. Then we say φ∅(q) = q is a 2-absorbing element, ∅ ⇒ φ0(q) = 0 q is a weakly 2-absorbing element, 2 ⇒ φ2(q) = q q is an almost 2-absorbing element, ⇒ ... n φn(q) = q q is an n-almost 2-absorbing element for n> 2, ∞ n ⇒ φω(q) = q q is a ω-2-absorbing element. ∧n=1 ⇒ Throughout this paper, φ denotes a function defined from L to L . Since for an element ∪ {∅} a L with a q but a  φ(q) implies that a  q φ(q), there is no loss generality in assuming ∈ ≤ ∧ that φ(q) q. We henceforth make this assumption. For any two functions ψ1, ψ2 : L L , ≤ → ∪{∅} we say ψ1 ψ2 if ψ1(a) ψ2(a) for each a L. Thus clearly we have the following order: ≤ ≤ ∈ φ∅ φ0 φω ... φn 1 φn ... φ2 φ1. ≤ ≤ ≤ ≤ + ≤ ≤ ≤ ≤ Lemma 2.2. Let q be a proper element of L and ψ1, ψ2 : L L be two functions with → ∪ {∅} ψ1 ψ2. If q is a ψ1-2-absorbing element of L, then q is a ψ2-2-absorbing element of L. ≤ Proof. Suppose that q is a ψ1-2-absorbing element of L and a,b,c L such that abc q and ∈ ≤ abc  ψ2(q). Since abc q and abc  ψ1(q), we are done. ≤ Hence we have the following relations among the concepts mentioned in Definition 2.1: Theorem 2.3. Let q be a proper element of L. Then (i) q is a 2-absorbing element of L q is a weakly 2-absorbing element of L q is a ω- 2-absorbing element of L q is⇒ an (n + 1)-almost 2-absorbing element of L⇒ q is an n-almost 2-absorbing element⇒ of L for all n 2 q is an almost 2-absorbing element⇒ of L. ≥ ⇒ (ii) q is a φ-prime element of L q is a φ-2-absorbing element of L. ⇒ (iii) A proper element q of L is an idempotent element q is a ω-2-absorbing element of L and q is an n-almost 2-absorbing element of L for all n⇒ 2. ≥ (iv) q is an n-almost 2-absorbing element of L for all n 2 q is a ω-2-absorbing element of L. ≥ ⇔ Proof. (i) It is clear from Lemma 2.2. (ii) Suppose that a,b,c L with abc q, abc φ(q) and ab q. Hence c q as q is a φ-prime element of L. Thus∈ac q or bc ≤q, we are6≤ done. 6≤ ≤ (iii) Suppose that q is an idempotent≤ ≤ element of L. Then q = qn for all n > 0, and so ∞ n φω(q) = n=1q = q. Thus q is a ω-2-absorbing element of L. Finally, q is an n-almost 2- absorbing∧ element for all n 2 from (i). ≥ on φ-2-absorbing elements 129

∞ n m (iv) Let a,b,c L with abc q but abc  n=1q . Hence abc q but abc  q for some m 2. Since q is∈n-almost 2-absorbing≤ for all∧n 2, this implies≤ either ab q or bc q or ac ≥ q, we are done. The converse is clear from (i).≥ ≤ ≤ ≤ Theorem 2.4. Let q be a φ-2-absorbing element of L. If φ(q) is a 2-absorbing element of L, then q is 2-absorbing.

Proof. Let abc q for some a,b,c L. If abc φ(q), then we have either ab q or ac q or bc q as q is φ-2-absorbing.≤ Suppose∈ that abc 6≤φ(q). Hence we conclude that≤ either ab ≤φ(q) or ac≤ φ(q) or bc φ(q). Since φ(q) q, we≤ are done. ≤ ≤ ≤ ≤ Definition 2.5. Let q be a φ-2-absorbing element of L and a,b,c L. If abc φ(q) but ab  q, ∈ ≤ bc  q, ac  q, then (a,b,c) is called a φ-triple zero of q.

Remark 2.6. If q is a φ-2-absorbing element of L which is not 2-absorbing, then there exists (a,b,c) a φ-triple zero of q for some a,b,c L. ∈ Lemma 2.7. Let q be a φ-2-absorbing element of L and suppose that (a,b,c) is a φ-triple zero of q for some a,b,c L. Then ∈ (i) abq, bcq, acq φ(q). ≤ (ii) aq2, bq2, cq2 φ(q). ≤ (iii) q3 φ(q). ≤ Proof. (i) Suppose that abq  φ(q). Then ab(c q)  φ(q). Since ab  q and q is φ-2-absorbing, we have a(c q) q or b(c q) q. So ac q∨or bc q, which contradicts with our hypothesis. Thus abq ∨φ(q)≤. Similarly∨ one≤ can easily show≤ that≤bcq φ(q) and acq φ(q). ≤ ≤ ≤ (ii) Suppose that aq2  φ(q). Hence we have a(b q)(c q)  φ(q) by (i). So we conclude either a(b q) q or a(c q) q or (b q)(c q) ∨q. Thus∨ either ab q or ac q or bc q, a contradiction.∨ ≤ Therefore∨aq2 ≤ φ(q). Similarly∨ ∨ it can≤ be easily verified≤ that bq2,≤ cq2 φ(q≤). ≤ ≤ (iii) Assume that q3  φ(q). Then we have (a q)(b q)(c q) q but (a q)(b q)(c q)  φ(q) by (i) and (ii). Since q is φ-2-absorbing, (∨a q∨)(b q∨) ≤q or (a ∨q)(c ∨ q) ∨ q or (b q)(c q) q, so we conclude ab q or ac∨ q∨or bc≤ q, a contradiction.∨ ∨ ≤ Thus q3 ∨ φ(q).∨ ≤ ≤ ≤ ≤ ≤ Now we can give a condition for a φ-2-absorbing element to be a 2-absorbing element of L.

Corollary 2.8. Let q be a proper element of L. Then the following statements hold:

(i) If q is a φ-2-absorbing element of L such that q3  φ(q), then q is a 2-absorbing element of L. (ii) Let L be a C-lattice. If q is a φ-2-absorbing element of L that is not a 2-absorbing, then √q = pφ(q). Proof. (i) The proof is clear by Remark 2.6 and Lemma 2.7 (iii). (ii) Since q is not a 2-absorbing element of L, q3 φ(q) by Lemma 2.7 (iii). Hence √q ≤ ≤ pφ(q). Since φ(q) q is always hold, we get √q = pφ(q). ≤ Recall from [13] that an element e L is said to be principal, if it satisfies the dual identities (i) a be = ((a : e) b)e and (ii) ((ae∈ b) : e)=(b : e) a. Elements satisfying the identity (i) are∧ called meet principal∧ and elements∨ satisfying the identity∨ (ii) are called join principal. If the both identities are satisfied, then e is said to be a principal element of L. Note that by [13, Lemma 3.3 and Lemma 3.4], a finite product of principal elements of L is again principal. If every element of L can be written as a join of some principal elements of L, then L is said to be join principally generated lattice.

Theorem 2.9. Let L be a join principally generated C-lattice and a,b,c be proper join principal elements of L. Then abc is a φ-2-absorbing element of L if and only if abc = φ(abc). 130 Ece Yetkin Celikel, Emel A. Ugurlu and Gulsen Ulucak

Proof. Suppose that abc is a φ-2-absorbing element of L. Assume that abc = φ(abc). Then we have either ab abc or ac abc or bc abc. Without loss generality we6 may assume that ≤ ≤ ≤ ab abc. Since ab is principal, we conclude that 1L = (abc : ab) = c (0L : ab). Observe ≤ ∨ that (0L : ab) = 1L. Indeed, if (0L : ab) = 1L, then abc = 0L φ(abc), a contradiction. 6 ≤ Since (0L : ab) = 1L, c J(L) we conclude that 1L = c (0L : ab), a contradiction. Thus abc = φ(abc). The6 converse≤ part is clear. 6 ∨

Theorem 2.10. Let L be a local Noether domain. If q is an φn-2-absorbing element of L for all n > 2, then q is a 2-absorbing element of L.

Proof. Let abc q for some a,b,c L. If abc φn(q), then we have either ab q or bc q ≤ ∈ 6≤ ∞ n≤ ≤ or ac q as q is φn-2-absorbing. So suppose that abc φn(q). Since q = 0L, from ≤ ≤ ∧n=1 Corollary 3.3 of [13], we conclude that abc 0L. Thus a 0L or b 0L or c 0L as L is a domain, so clearly ab q or bc q or ac q≤. ≤ ≤ ≤ ≤ ≤ ≤ We remind to the reader that for any a L, L/a = b L : a b is a multiplicative lattice with multiplication c d = cd a. ∈ { ∈ ≤ } ◦ ∨ Theorem 2.11. Let q be a proper element of L. Then the following statements hold:

(i) q is a φ-2-absorbing element of L if and only if q is a weakly 2-absorbing element of L/φ(q). (ii) q is a φ-prime element of L if and only if q is a weakly prime element of L/φ(q). (iii) q is a φ-primary element of L if and only if q is a weakly primary element of L/φ(q).

Proof. (i) If φ(q) = , then there is nothing to prove. Thus assume that φ(q) = . Let φ(q) = (a φ(q)) (b φ(q∅)) (c φ(q)) = abc φ(q) q for some a,b,c L.6 Then∅ abc 6q, ∨ ◦ ∨ ◦ ∨ ∨ ≤ ∈ ≤ but abc  φ(q). Hence either ab q or bc q or ac q. So (a φ(q)) (b φ(q)) q or (b φ(q)) (c φ(q)) q or (a ≤φ(q)) (c≤ φ(q)) ≤q. Therefore∨ q is a◦ weakly∨ 2-absorbing≤ element∨ of◦L/φ∨(q). ≤ ∨ ◦ ∨ ≤ Conversely, let abc q and abc  φ(q) for some a,b,c L. Then φ(q) =(a φ(q)) (b φ(q)) (c φ(q)) q≤. Hence (a φ(q)) (b φ(q)) ∈q or (b φ(q)) 6 (c ∨φ(q)) ◦ q or∨ (a φ◦(q))∨ (c φ(≤q)) q. Thus ab∨ q or ◦bc ∨q or ac ≤ q. Similarly∨ one◦ can easily∨ prove≤ (ii) and∨ (iii). ◦ ∨ ≤ ≤ ≤ ≤

Corollary 2.12. Let q be a proper element of L and n 2. Then ≥ n (i) q is a φn-2-absorbing element of L if and only if q is a weakly 2-absorbing element of L/q . n (ii) q is a φn-prime element of L if and only if q is a weakly prime element of L/q . n (iii) q is a φn-primary element of L if and only if q is a weakly primary element of L/q .

n Proof. Since φn(q) = q , the proof is clear by Theorem 2.11.

Corollary 2.13. Let q be a φ-2-absorbing element of L such that φ φ3. Then ≤ (i) q is a φn-2-absorbing element of L for every n 3. ≥ (ii) q is a φω-2-absorbing element of L.

Proof. Suppose that q is a 2-absorbing element of L. Hence (i) and (ii) are clear. (i) Assume that q is not a 2-absorbing element of L. Thus q3 φ(q) by Lemma 2.7 (iii). 3 3 3 n ≤ Then we have q φ(q) q as φ φ3. This follows q = q = φ(q) for every n 3, so we are done. ≤ ≤ ≤ ≥ (ii) Let abc q and abc  ∞ qn. Then abc  qn for some n 2. If n 3, then it is clear ≤ ∧n=1 ≥ ≥ from (i). So suppose that n = 2. Hence abc  q2 which implies that abc  q3, so from (i) the result is obtained.

Theorem 2.14. Let x and y be two proper elements of L such that x y and let n 2. If y is a ≤ ≥ φn-2-absorbing element of L, then y is a φn-2-absorbing element of L/x. on φ-2-absorbing elements 131

Proof. Suppose that y is a φn-2-absorbing element of L. Assume that (a x) (b x) (c x) = ∨ ◦ ∨ ◦ ∨ abc x y and (a x) (b x) (c x) = abc x  yn for some a,b,c L. As y L/x, then ∨ ≤ ∨ ◦ ∨ ◦ ∨ ∨ ∈ ∈ yn = y y y ... y = yn x. Since x y and abc x  yn = yn x, then we have abc y and ◦ ◦ ◦ ◦ ∨ ≤ ∨ ∨ ≤ abc  yn. Thus ab y or ac y or bc y. Hence (a x) (b x) y or (a x) (c x) y ≤ ≤ ≤ ∨ ◦ ∨ ≤ ∨ ◦ ∨ ≤ or (b x) (c x) y which means that y is a φn-2-absorbing primary element of L/x. ∨ ◦ ∨ ≤

Corollary 2.15. Let x and y be two proper elements of L such that x y. If y is a φω-2-absorbing ≤ element of L, then y is a φω-2-absorbing element of L/x.

Proof. Similar to the proof of Theorem 2.14.

Definition 2.16. Let x be a proper element of L/q such that q x. Then x is called a φq-2- ≤ absorbing element of L/q if whenever a,b,c L/q with abc x and abc  φ(x) q implies ab x or ac x or bc x. ∈ ≤ ∨ ≤ ≤ ≤ Theorem 2.17. Let p and q be two elements of L with q p< 1. If p is a φ-2-absorbing element ≤ of L, then p is a φq-2-absorbing element of L/q.

Proof. Let (a q) (b q) (c q) p and abc q = (a q) (b q) (c q)  φ(p) q ∨ ◦ ∨ ◦ ∨ ≤ ∨ ∨ ◦ ∨ ◦ ∨ ∨ for some a,b,c L. Hence abc p and abc  φ(p). Since p is φ-2-absorbing element of L, we conclude that ab∈ p or ac p or≤bc p. So we get (a q) (b q) p or (a q) (c q) p or (b q) (c q≤) p. ≤ ≤ ∨ ◦ ∨ ≤ ∨ ◦ ∨ ≤ ∨ ◦ ∨ ≤ Theorem 2.18. Let p and q be two proper elements of L such that q φ(p). Then the following statements are equivalent: ≤

(i) p is a φ-2-absorbing element of L.

(ii) p is a φq-2-absorbing element of L/q.

n (iii) p is a φqn -2-absorbing element of L/q .

Proof. (i) (ii): It is clear by Theorem 2.17. (ii) (iii):⇒ Let n 1. Since q φ(p), we have qn q φ(p). Suppose that (a qn) (b ⇒ ≥ ≤ ≤ ≤ ∨ ◦ ∨ qn) (c qn) p and (a qn) (b qn) (c qn)  φ(p) qn for some a,b,c L. Hence ◦ ∨ ≤ ∨ ◦ ∨ ◦ ∨ ∨ ∈ abc  φ(p). Since q φ(p) and abc  φ(p), we have abc  q. Thus (a q) (b q) (c q) p ≤ ∨ ◦ ∨ ◦ ∨ ≤ and (a q) (b q) (c q)  φ(p) q. Since p is a φq- 2-absorbing element of L/q, one can conclude∨ that◦ ab∨ p◦or ac∨ p or bc ∨ p. Thus ab qn p or ac qn p or bc qn p (in L/qn). ≤ ≤ ≤ ∨ ≤ ∨ ≤ ∨ ≤ (iii) (i): Suppose that abc p and abc  φ(p) for some a,b,c L. Since qn φ(p), we ⇒ ≤ ∈ ≤ have abc  qn. As qn φ(p) p, we get (a qn) (b qn) (c qn) = abc qn p and n n ≤ n ≤ n ∨ ◦ ∨ ◦ ∨ ∨ ≤n (a q ) (b q ) (c q )  φ(p) q . Since p is a φqn -2-absorbing element of L/q , one can∨ conclude◦ ∨ that ab◦ p∨or ac p or bc∨ p. ≤ ≤ ≤ Corollary 2.19. Let p and q be two proper elements of L. Suppose that q is not a weakly 2- absorbing element of L. The following statements are equivalent:

(i) p is a φ-2-absorbing element of L.

3 (ii) p is a φp3 -2-absorbing element of L/p . n (iii) p is a φpn -2-absorbing element of L/p for every n 3. ≥ Proof. Suppose that p is not a weakly 2-absorbing element of L. Hence p is not a 2-absorbing element of L. So we conclude p3 φ(q) by Lemma 2.7. Thus we are done by Theorem 2.18. ≤ Definition 2.20. Let q be a proper element of L and n 2. We call q as an n-potent 2-absorbing if whenever a,b,c L with abc qn, then ab q or bc≥ q or ac q. ∈ ≤ ≤ ≤ ≤ Theorem 2.21. Let q be an n-almost 2-absorbing element for some n 2. If q is k-potent 2- absorbing for some k n, then q is a 2-absorbing element. ≥ ≤ 132 Ece Yetkin Celikel, Emel A. Ugurlu and Gulsen Ulucak

Proof. Suppose that q is an n-almost 2-absorbing element. Let abc q for some a,b,c L. ≤ ∈ If abc  qk, then abc  qn. It implies either ab q or bc q or ac q as q is an n-almost 2-absorbing element. If abc qk, then we obtain≤ the same result≤ as q is≤ k-potent 2-absorbing, so we are done. ≤

In the following theorems, we obtain some conditions under which a φ-2-absorbing element of L is a 2-absorbing element of L. Let J(L) = m L : m is a maximal element of L . ∧{ ∈ } Theorem 2.22. Let L be a Noether domain. Then an element q of L with q J(L) is a 2- ≤ absorbing element of L if and only if q is a φn-2-absorbing element of L for all n 2. ≥

Proof. If q is 2-absorbing, then q is φn-2-absorbing by Theorem 2.3. Conversely, suppose that k q is φn-2-absorbing for all n 2 and let a,b,c L with abc q. If abc  q for some k 2, we have either ab q or bc ≥q or ac q. So suppose∈ that abc≤ qn for all n 2. Since≥L is ≤ ≤ ≤∞ n ≤ ≥ a Noether domain, we conclude abc q = 0L by Corollary 1.4 in [4]. Since 0L is prime, ≤ ∧n=1 we get either a = 0L or b = 0L or c = 0L. Without loss generality suppose that a = 0L. This implies that ab = 0L q which completes the proof. ≤ Theorem 2.23. Let L be a Noether lattice. Let q be a non-zero non-nilpotent proper element of L satisfying the restricted cancellation law. Then q is a φ-2-absorbing element of L for some φ φn and for all n 2 if and only if q is a 2-absorbing element of L. ≤ ≥ Proof. Assume that q is a 2-absorbing element of L. Then q is a φ-2-absorbing element of L for all φ. Thus q is φ-2-absorbing for some φ φn and for all n 2. ≤ ≥ Conversely assume that q is a φ-2-absorbing element of L for some φ φn for all n 2. ≤ ≥ Hence q is a φn-2-absorbing element of L for all n 2 by Lemma 2.2. Let abc q for some a,b,c in L. Here there are two cases: ≥ ≤ Case 1: Let abc  qn for some n 2. Then by hypothesis we get ab q or bc q or ac q. Case 2: Let abc qn for all n 2.≥ We have that a(b q)(c q) = abc≤ abq acq≤ aq2 ≤ q. ≤ ≥ ∨ ∨ ∨ ∨ ∨ ≤ If a(b q)(c q)  qn, then a(b q) q or a(c q) q or (b q)(c q) q. It follows that either ∨ab q∨or bc q or ac ∨q. If≤a(b q)(c∨ q)≤ qn, then∨ a(b ∨q)(c≤ q) = abc abq acq aq≤2 qn q≤2. By[18,≤ Lemma 1.11],∨ we get∨ ab≤ q and ac q∨. Thus ∨q is a 2-absorbing∨ ∨element∨ of L≤. ≤ ≤ ≤

Remark 2.24. Let L = L1 L2 ...Ln where L1, L2, ..., Ln are multiplicative lattices (n 1) × × ≥ and let φ = ψ1 ψ2 ... ψn where ψi : Li Li (i = 1, ..., n) be a function. Let × × × → ∪ {∅} a = (a1,a2, ..., an) be an element of L. Observe that if ψi(ai) = for some i = 1, ..., n, then ∅ there is no element of φ(a) and vice versa. Thus φ(a) = if and only if ψi(ai) = for some i = 1, ..., n. ∅ ∅

Lemma 2.25. Let L = L1 L2 where L1,L2 are two multiplicative lattices. Let φ = ψ1 ψ2, × × where ψi : Li Li (i = 1, 2) is a function. Then q1 is a 2-absorbing element of L1 if and → ∪ {∅} only if q =(q1, 1L2 ) is a 2-absorbing element of L.

Proof. Suppose that q1 is a 2-absorbing element of L1 and (a, 1L )(b, 1L )(c, 1L ) q for some 2 2 2 ≤ elements (a, 1L ), (b, 1L ), (c, 1L ) of L. Then abc q1 which implies that either ab q1 or 2 2 2 ≤ ≤ bc q1 or ac q1. It follows (a, 1L2 )(b, 1L2 ) q or (b, 1L2 )(c, 1L2 ) q or (a, 1L2 )(c, 1L2 ) q. Thus≤ q is a 2-absorbing≤ element of L. ≤ ≤ ≤

Conversely suppose that q =(q1, 1L2 ) is a 2-absorbing element of L but assume that q1 is not a 2-absorbing element of L1. Hence there exists a,b,c L1 with abc q1 but neither ab q1 ∈ ≤ ≤ nor bc q1 nor ac q1. Thus we conclude (a, 1L )(b, 1L )(c, 1L ) q but (a, 1L )(b, 1L ) q ≤ ≤ 2 2 2 ≤ 2 2 6≤ and (b, 1L )(c, 1L ) q and (a, 1L )(c, 1L ) q, a contradiction. 2 2 6≤ 2 2 6≤

Theorem 2.26. Let L = L1 L2 where L1,L2 are two multiplicative lattices. Let φ = ψ1 ψ2, × × where ψi : Li Li (i = 1, 2) is a function. Then the following statements hold: → ∪ {∅}

(i) If qi is a proper element of Li with ψi(qi) = qi (i = 1, 2), then q = (q1, q2) is a φ-2- absorbing element of L. on φ-2-absorbing elements 133

(ii) If q1 is a ψ1-2-absorbing element of L1, and ψ2(1L2 ) = 1L2 , then q = (q1, 1L2 ) is a φ-2- absorbing element of L.

(iii) If q2 is a ψ2-2-absorbing element of L2 and ψ1(1L1 ) = 1L1 , then q = (1L1 , q2) is a φ-2- absorbing element of L.

Proof. (i) If ψ1(q1) = q1 and ψ2(q2) = q2, then there is no such an element (a,b) which satisfies (a,b) (q1, q2) and (a,b)  φ(q1, q2)=(q1, q2), so we are done. ≤ (ii) Suppose that ψ1(q) = . Then q =(q1, 1L ) is a φ-2-absorbing element of L by Lemma ∅ 2 2.25. So assume that ψ1(q) = and q1 is ψ1-2-absorbing element of L1. Let a = (a1,a2), 6 ∅ b = (b1,b2) and c = (c1,c2) such that abc q and abc  φ(q). Hence a1b1c1 q1 and ≤ ≤ a1b1c1  ψ1(q1), this implies that either a1b1 q1 or b1c1 q1 or a1c1 q1. Thus either ab q or bc q or ac q. ≤ ≤ ≤ ≤ (iii)≤ This can≤ be easily obtained similar to (ii).

Theorem 2.27. Let L = L1 L2 where L1,L2 are two multiplicative lattices and φ = ψ1 ψ2, × × where ψi : Li Li (i = 1, 2) is a function such that ψ2(1L ) = 1L . Let q1 be a proper −→ ∪ {∅} 2 6 2 element of L1 and q =(q1, 1L2 ). Then the following statements are equivalent:

(i) (q1, 1L2 ) is a φ-2-absorbing element of L.

(ii) (q1, 1L2 ) is a 2-absorbing element of L.

(iii) q1 is a 2-absorbing element of L1.

Proof. If ψ1(q1) = and ψ2(1 ) = , then φ(q) = by Remark 2.24. So we are done from ∅ L2 ∅ ∅ Lemma 2.25. Thus assume that ψ1(q1) = or ψ2(1 ) = . 6 ∅ L2 6 ∅ (i) (ii): Assume that q = (q1, 1L ) is a φ-2-absorbing element of L. Then q1 is a ψ1- ⇒ 2 2-absorbing element of L1. Indeed, if q1 is not a ψ1-2-absorbing element of L1, then there exist a,b,c in L1 such that abc q1 and abc  ψ1(q1) but ab  q1 and bc  q1 and ac  ≤ q1. Then (abc, 1L )=(a, 1L )(b, 1L )(c, 1L ) q and (abc, 1L )=(a, 1L )(b, 1L )(c, 1L )  2 2 2 2 ≤ 2 2 2 2 (ψ1(p1), ψ2(1L )) = φ(q). This implies (ab, 1L )=(a, 1L )(b, 1L ) q or 2 2 2 2 ≤ (bc, 1L )=(b, 1L )(c, 1L ) q or (ac, 1L )=(a, 1L )(c, 1L ) q, which means ab q1 or 2 2 2 ≤ 2 2 2 ≤ ≤ bc q1 or ac q1, a contradiciton. Thus q1 is a ψ1-2-absorbing element of L1. ≤ ≤ If q1 is a 2-absorbing element of L1, then it is clear. Assume that q1 is not a 2-absorbing element of L1. Hence q1 has a ψ1-triple-zero (x,y,z) for some x,y,z in L1 by Remark 2.6. Since ψ2(1L ) = 1L , then we get (xyz, 1L )=(x, 1L )(y, 1L )(z, 1L ) q and (xyz, 1L ) = 2 6 2 2 2 2 2 ≤ 2 (x, 1L )(y, 1L )(z, 1L )  φ(q). Therefore (x, 1L )(y, 1L ) q or (y, 1L )(z, 1L ) q or 2 2 2 2 2 ≤ 2 2 ≤ (x, 1L )(z, 1L ) q. So we get xy q1 or yz q1 or xz q1, a contradiction. Thus q1 is 2 2 ≤ ≤ ≤ ≤ a 2-absorbing element of L1. Consequently, (q1, 1L2 ) is a 2-absorbing element of L. (ii) (iii): It can be easily shown similar to the argument in (i) (ii). (iii)⇒ (i): It is clear. ⇒ ⇒

Lemma 2.28. Let L = L1 L2 L3 where L1,L2,L3 are C-lattices. Let φ = ψ1 ψ2 ψ3, × × × × where ψi : Li Li (i = 1, 2, 3) is a function with ψi(1Li ) = 1Li . If q =(q1, q2, q3) is a φ-2-absorbing−→ element∪ of {∅}L, then either q = φ(q) or q is a 2-absorbing6 element of L.

Proof. If φ(q) = , then we are done. So assume φ(q) = . Suppose that q = φ(q). Hence there is an element (a,b,c∅ ) L with (a,b,c) q but (a,b,c)6 ∅φ(q). So 6 ∈ ≤ 6≤ (a,b,c)=(a, 1L , 1L )(1L , b, 1L )(1L , 1L ,c) q implies that either (a, 1L , 1L )(1L , b, 1L ) 2 3 1 3 1 2 ≤ 2 3 1 3 ≤ q or (1L1 , b, 1L3 )(1L1 , 1L2 ,c) q or (a, 1L2 , 1L3 )(1L1 , 1L2 ,c) q. Without loss of generality as- ≤ ≤ 3 sume that (a, 1L2 , 1L3 )(1L1 , b, 1L3 ) q. Then q3 = 1L3 which means that q φ(q). Thus q is a 2-absorbing element of L by Corollary≤ 2.8. 6≤

Theorem 2.29. Let L = L1 L2 L3 where L1,L2,L3 are C-lattices. Let φ = ψ1 ψ2 ψ3, × × × × where ψi : Li Li (i = 1, 2, 3) is a function with ψi(1Li ) = 1Li . If q = φ(q), then the followings are−→ equivalent:∪ {∅} 6 6

(i) q is a φ-2-absorbing element of L. (ii) q is a 2-absorbing element of L. 134 Ece Yetkin Celikel, Emel A. Ugurlu and Gulsen Ulucak

(iii) q is in one of the following type:

I) q =(1L1 , q2, q3), where q2 is a prime element of L2 and q3 is a prime element of L3.

II) q =(q1, 1L2 , q3), where q1 is a prime element of L1 and q3 is a prime element of L3.

III) q =(q1, q2, 1L3 ), where q1 is a prime element of L1 and q2 is a prime element of L2.

IV) For some i 1, 2, 3 , qi is a 2-absorbing element of Li and qj = 1lj for every j 1, 2, 3 i .∈ { } ∈{ }\{ } Proof. (i) (ii): If φ(q) = and q is a φ-2-absorbing element, then obviously q is a 2-absorbing ⇒ ∅ element of L. So assume that φ(q) = . Let q = (q1, q2, q3) be a φ-2-absorbing element of L, then q is a 2-absorbing element of L6 by∅ Lemma 2.28. (ii) (iii): Suppose that q is a 2-absorbing element of L. Since q = φ(q), there is a com- ⇒ 6 pact element of L such that (a1,a2,a3) q and (a1,a2,a3) φ(q). Since (a1,a2,a3) = ≤ 6≤ (a1, 1L , 1L )(1L ,a2, 1L )(1L , 1L ,a3) and q is φ-2-absorbing, we have (a1,a2, 1L ) q or 2 3 1 3 1 2 3 ≤ (1L ,a2,a3) q or (a1, 1L ,a3) q. This means that either q1 = 1L or q2 = 1L or q3 = 1L . 1 ≤ 2 ≤ 1 2 3 Case I. Suppose that q = (1L , q2, q3) where q2 = 1L and q3 = 1L . We show that q2 is 1 6 2 6 3 a prime element of L2. Let xy q2. Hence (1L , x, 1L )(1L , 1L , q3)(1L , y, 1L ) q and it ≤ 1 3 1 2 1 3 ≤ implies that (1L , x, 1L )(1L , 1L , 0L ) q or (1L , x, 1L )(1L , y, 1L ) q or 1 3 1 2 3 ≤ 1 3 1 3 ≤ (1L , 1L , 0L )(1L , y, 1L ) q. Since q3 is proper, we get 1 2 3 1 3 ≤ (1L , xy, 1L )=(1L , x, 1L )(1L , y, 1L ) q. Thus x q2 or y q2, which shows that q2 is 1 3 1 3 1 3 6≤ ≤ ≤ prime. By the similar argument one can easily show that q3 is a prime element of L3. Case II. q =(q1, 1L , q3), where q1 = 1L and q3 = 1L and Case III. q =(q1, q2, 1L ), where 2 6 1 6 3 3 q1 = 1L and q2 = 1L can be easily obtained similar to Case I. 6 1 6 2 Case IV. Without loss of generality suppose that q = (q1, 12, 1L3 ) where q1 is a proper ele- ment of L1. Let x1x2x3 q1 for some x1, x2, x3 L1. Then ≤ ∈ (x1x2x3, 1L , 0L )= (x1, 1L , 0L )(x2, 1L , 0L )(x3, 1L , 0L ) q and (x1x2x3, 1L , 0L ) 2 3 2 3 2 3 2 3 ≤ 2 3 6≤ φ(q). Since q is φ-2-absorbing, we have either (x1x2, 1L , 0L ) q or (x2x3, 1L , 0L ) q 2 3 ≤ 2 3 ≤ or (x1x3, 1L , 0L ) q. So x1x2 q1 or x2x3 q1 or x2x3 q1. 2 3 ≤ ≤ ≤ ≤ (iii) (i): Suppose that q2 and q3 are prime elements of L2 and L3, respectively and q = ⇒ (1L , q2, q3). Let (a1,a2,a3), (b1,b2,b3), (c1,c2,c3) L such that 1 ∈ (a1,a2,a3)(b1,b2,b3)(c1,c2,c3) q and (a1,a2,a3)(b1,b2,b3)(c1,c2,c3) φ(q). Assume that ≤ 6≤ (a1,a2,a3)(b1,b2,b3) q. Hence a2b2 q2 or a3b3 q3. Without loss of generality we may 6≤ 6≤ 6≤ suppose that a2b2 q2 and a3b3 q3. Since q2 is prime, we have c2 q2, which implies that 6≤ ≤ ≤ (a1,a2,a3)(c1,c2,c3) q, we are done. ≤ If q = (q1, 12, 1L3 ) where q1 is a 2-absorbing element of L1, then it can be seen that q is a 2-absorbing element of L. Thus q is a φ-2-absorbing element of L.

Theorem 2.30. Let L = L1 L2 L3 where L1,L2,L3 are multiplicative lattices. Let φ = ψ1 × × × ψ2 ψ3, where ψi : Li Li (i = 1, 2, 3) is a function. If a proper a =(a1,a2,a3) L is × −→ ∪{∅} ∈ a φ-2-absorbing element, then ψi(ai) = or ψi(ai) = ai (i = 1, 2, 3) for every proper element ∅ ai of Li.

Proof. Assume on the contrary that ψ1(a1) = a1 and ψ1(a1) = for some proper element 6 6 ∅ a1 L1. Put a = (a1, 0L , 0L ). Hence (a1, 1L , 1L )(1L , 0L , 1L )(1L , 1L , 0L ) a, but ∈ 2 3 2 3 1 2 3 1 2 3 ≤ (a1, 1L , 1L )(1L , 0L , 1L )(1L , 1L , 0L ) φ(a). Since a is a φ-2-absorbing element, we con- 2 3 1 2 3 1 2 3 6≤ clude either (a1, 1L , 1L )(1L , 0L , 1L ) a or (1L , 0L , 1L )(1L , 1L , 0L ) a or 2 3 1 2 3 ≤ 1 2 3 1 2 3 ≤ (a1, 1L , 1L )(1L , 1L , 0L ) a. It follows 1L a3 or 1L a1 or 1L a2, which are contra- 2 3 1 2 3 ≤ 3 ≤ 1 ≤ 2 ≤ dictions. Thus ψi(ai) = ai (i = 1, 2, 3) for every proper element ai of Li.

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Author information Ece Yetkin Celikel, Gaziantep University, Department of Mathematics, 27310, Gaziantep, Turkey.

E-mail:  Emel A. Ugurlu, Marmara University, Department of Mathematics, Ziverbey, Goztepe, Istanbul, Turkey.

E-mail:  Gulsen Ulucak, Gebze Technical University, Department of Mathematics, 141 41400 Kocaeli, Turkey.

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