c Allerton Press, Inc., 2009. ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2009, Vol. 53, No. 2, pp. 41–58. 

Ideal Extensions of Lattices* N. Kehayopulu1** 1

University of Athens, Panepistimiopolis, GR-157 84, Athens, Greece Received November 23, 2006

Abstract—Following the well-known Schreier extension of groups, the (ideal) extension of semigroups (without order) have been first considered by A. H. Clifford in Trans. Amer. Math. Soc. 68 (1950), with a detailed exposition of the theory in the monographs of Clifford–Preston and Petrich. The main theorem of the ideal extensions of ordered semigroups has been considered by Kehayopulu and Tsingelis in Comm. Algebra 31 (2003). It is natural to examine the same problem for lattices. Following the ideal extensions of ordered semigroups, in this paper we give the main theorem of the ideal extensions of lattices. Exactly as in the case of semigroups (ordered semigroups), we approach the problem using translations. We start with a lattice L and a lattice K having a least element, and construct (all) the lattices V which have an ideal L which is isomorphic to L and the Rees quotient V |L is isomorphic to K. Conversely, we prove that each lattice which is an extension of L by K can be so constructed. An illustrative example is given at the end. DOI: 10.3103/S1066369X09020042 Key words and phrases: translation, inner translation, (ideal) extension of a lattice.

INTRODUCTION The extension problem for groups is as follows: Given two groups H and K construct all groups G which have a normal subgroup N which is isomorphic to H and the quotient G/N of G by N is isomorphic to K. G is the well-known Schreier extension (or simply the extension) of H by K. Following the Schreier extension of groups, the ideal extensions of semigroups have been considered by A. H. Clifford in [1]. A detailed exposition of the ideal extensions of semigroups can be found in [2, 3]. The main theorem of the ideal extension of semigroups is as follows: Given a semigroup S and a semigroup Q with zero such that S ∩ Q∗ = ∅ (where Q∗ = Q \ {0}), construct all the semigroups V which have an ideal S  which is isomorphic to S and the Rees quotient V |S  is isomorphic to Q. To avoid confusion, for the Rees quotient we use the notation V |S instead of the usual one V /S. Extensions of weakly reductive semigroups, strict and pure extensions, retract extensions, dense extensions, equivalent extensions have been also considered in [3]. Ideal extensions of totally ordered semigroups have been studied in [4, 5], those of topological semigroups in [6, 7]. We are often interested in building more complex semigroups, lattices, ordered sets, ordered or topological semigroups and this can be sometimes achieved by constructing the ideal extensions. Ideal extensions of ordered sets have been considered in [8]. The retract and the equivalent extensions of ordered sets have been considered in [9, 10]. For the ideal extensions of ordered semigroups we refer to [11]. As in semigroups (without order), for ordered semigroups we approach the problem using left and right translations. We start with an ordered semigroup S and an ordered semigroup Q with zero (to avoid trivialities Q must have at least one nonzero element) such that S ∩ Q∗ = ∅ and construct all the ordered semigroups V having an ideal S  which is isomorphic to S and the Rees quotient V |S  is isomorphic to Q. Conversely, we prove that each ordered semigroup which is an extension of S by Q can be so constructed. Since the problem of the ideal extensions has been already considered for semigroups, for ordered semigroups, and ordered sets, it is quite natural to examine the same problem for lattices. The aim of this paper is to construct the ideal extensions of lattices. We give the main theorem of such extensions and an illustrative example ∗ **

The text was submitted by the author in English. E-mail: [email protected].

41

42

KEHAYOPULU

to the theorem. Exactly as in the case of semigroups, we approach the problem using translations. In case of semigroups or ordered semigroups we use both left and right translations. Unlike in semigroups and ordered semigroups we approach the problem using only one kind of translation (similar to the left one) which we just call translation. We start with a lattice L and a lattice K having a least element such that L ∩ K ∗ = ∅ (K ∗ = K \ {0}), and construct (all) the lattices V which have an ideal L which is isomorphic to L and the Rees quotient V |L is isomorphic to K. Conversely, we prove that each lattice which is an extension of L by K can be so constructed. The results of this paper have been announced without proofs in [12]. Later, some authors were interested in a detailed exposition of the results in [12] which are explained in detail in the present paper. The paper is inspired by semigroups. The aim of this paper is to give an analog of the ideal extensions of semigroups (or ordered semigroups) for lattices. We present the corresponding concepts and the analogous results. We do not use papers related with the lattice theory in the references, though the concept of translation has been already defined in case of lattices many years ago (see, for example [13–16]). 1. DEFINITIONS AND LEMMAS We are going to use the well-known concept of ideal [17]. A nonempty subset L of a lattice V is called an ideal of V if 1) a, b ∈ L implies a ∨ b ∈ L and 2) a ∈ L and V  b ≤ a imply b ∈ L. Equivalently, if 1) a, b ∈ L implies a ∨ b ∈ L and 2) if a ∈ V and b ∈ L imply a ∧ b ∈ L. Definition 1. Let L be a lattice. A mapping λ : L → L is called a translation of L if λ(x ∧ y) = λ(x) ∧ y for all x, y ∈ L. We denote by T (L) the set of all translations of L. Clearly, if λ is a translation of L, then λ(x) ≤ x for all x ∈ S. Lemma 1. Let L be a lattice. We define an operation “∧” on T (L) as follows: ∧ : T (L) × T (L) → T (L) | (λ1 , λ2 ) → λ1 ∧ λ2 , where λ1 ∧ λ2 : L → L | x → λ1 (x) ∧ λ2 (x). Then the set T (L) endowed with the operation “∧” is a lower semilattice. Furthermore, for each λ1 , λ2 ∈ T (L), we have λ1 ∧ λ2 = λ1 λ2 = λ2 λ1 , where λ1 λ2 : L → L | x → λ1 (λ2 (x)). Proof. One can prove that the operation “∧” on T (L) is well defined, it is idempotent, commutative and associative and so the set T (L) endowed with the operation “∧” is a lower semilattice. Now let λ1 , λ2 ∈ T (L) and x ∈ L. Then (λ1 ∧ λ2 )(x) := λ1 (x) ∧ λ2 (x) = λ1 (x ∧ λ2 (x)) (since λ1 ∈ T (L)) = λ1 (λ2 (x)) (since λ2 ∈ T (L)). Thus λ1 ∧ λ2 = λ1 λ2 . By symmetry, we have λ2 ∧ λ1 = λ2 λ1 . We have λ1 λ2 = λ2 λ1 because the operation “∧” is commutative. Lemma 2. Let (V, ∨, ∧) be a lattice and let L be an ideal of V . We consider the set V |L := V \ L ∪ {0} where 0 is an arbitrary element of L (V \ L is the complement of L to V ). We define operations “” and “ ” on V |L as follows:  : V |L × V |L → V |L | (x, y) → x  y, where ⎧ x ∨ y, if x, y ∈ V \ L; ⎪ ⎪ ⎪ ⎨ x, if x ∈ V \ L, y = 0; x  y := ⎪ y, if y ∈ V \ L, x = 0; ⎪ ⎪ ⎩ 0, if x = y = 0, RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

IDEAL EXTENSIONS OF LATTICES

43

: V |L × V |L → V |L | (x, y) → x y, where  x ∧ y, if x ∧ y ∈ V \ L; x y := 0 otherwise. Then the set V |L endowed with the operations “” and “ ” is a lattice and the element 0 is the least element in V |L. Proof. (A) We first prove that the operations “ ” and “ ” are well defined and they are commutative. To prove that the operation “” is associative, we have to consider the cases: 1. x, y, z ∈ V \ L, 2. x, y ∈ V \ L, z = 0, 3. y, z ∈ V \ L, x = 0, 4. z, x ∈ V \ L, y = 0, 5. x ∈ V \ L, y = z = 0, 6. y ∈ V \ L, z = x = 0, 7. z ∈ V \ L, x = y = 0, 8. x = y = 0. Consider case 1. Cases 2–8 can be proved similarly. 1. Let x, y, z ∈ V \ L. Then x  y := x ∨ y ∈ V \ L and (x  y)  z := (x  y) ∨ z = (x ∨ y) ∨ z. In addition, y  z := y ∨ z ∈ V \ L and x  (y  z) := x ∨ (y  z) = x ∨ (y ∨ z). Thus we have (x  y)  z = x  (y  z). (B) For each x ∈ V |L, we have x 0 = 0 x = 0. Really, let x ∈ V |L. Since x ∈ V , 0 ∈ L and L is an ideal of V , we also have x ∧ 0 ∈ L. Since x ∧ 0 ∈ / V \ L, we have x 0 := 0. Similarly, 0 x = 0. (C) The operation “ ” is associative. Really, let x, y, z∈V |L. Since x, y, z∈V , we have x ∧ y ∧ z∈V . 1. Let x ∧ y ∧ z ∈ V \ L. Then, since L is an ideal of V , we have x ∧ y ∈ / L and y ∧ z ∈ / L. Since x ∧ y, y ∧ z ∈ V \ L, we have x y := x ∧ y and y z := y ∧ z. Since (x ∧ y) ∧ z ∈ V \ L, we have (x ∧ y) z := (x ∧ y) ∧ z. Then (x y) z = (x ∧ y) ∧ z. Since x ∧ (y ∧ z) ∈ V \ L, we have x (y ∧ z) := x ∧ (y ∧ z). Hence x (y z) = x ∧ (y ∧ z). Therefore we have (x y) z = x (y z). 2. Let x ∧ y ∧ z ∈ / V \ L. We consider the following cases: / V \ L, 2.1. Let x ∧ y, y ∧ z ∈ V \ L. Then x y := x ∧ y and y z := y ∧ z. Since (x ∧ y) ∧ z ∈ we have (x ∧ y) z := 0, therefore (x y) z = 0. Since x ∧ (y ∧ z) ∈ / V \ L, we have x (y ∧ z) := 0, therefore x (y z) = 0. Hence we have (x y) z = x (y z). 2.2. Let x ∧ y ∈ V \ L and y ∧ z ∈ / V \ L. Then x y := x ∧ y and y z := 0; (x y) z = (x ∧ y) z and x (y z) = x 0 = 0. Since (x ∧ y) ∧ z ∈ / V \ L, we have (x ∧ y) z := 0. Thus we have (x y) z = x (y z). 2.3. Let x ∧ y ∈ / V \ L and y ∧ z ∈ V \ L. The proof is similar to that of 2.2. 2.4. Let x ∧ y, y ∧ z ∈ / V \ L. Since x y := 0 and y z := 0, we have (x y) z = 0 z = 0 and x (y z) = x 0 = 0. (D) For all x, y ∈ V |L we have x (x  y) = x. Really, let x, y ∈ V |L. 1. Let x, y ∈ V \ L, then V |L  x  y := x ∨ y and x (x  y) = x (x ∨ y). Since x ∧ (x ∨ y) = x ∈ V \ L, we have x (x ∨ y) := x ∧ (x ∨ y) = x. Hence x (x  y) = x. 2. Let x ∈ V \ L, y = 0, then V |L  x  y := x, and x (x  y) = x x. Since V \ L  x = x ∧ x, we have x x = x ∧ x = x. Hence x (x  y) = x. 3. Let x = 0, y ∈ V \ L, then V |L  x  y := y. Hence x (x  y) = x y = 0 y = 0 = x. 4. Let x = y = 0, then V |L  x  y := 0, and x (x  y) = x 0 = 0 = x. (E) For each x, y ∈ V |L, we have x  (x y) = x. Really, let x, y ∈ V |L. Clearly, x ∧ y ∈ V . 1. Let x ∧ y ∈ V \ L. Then V \ L  x y := x ∧ y and x  (x y) = x  (x ∧ y). If x = 0, then x y = 0 y = 0 ∈ L, which is impossible. Thus we have x ∈ V \ L. Since x, x ∧ y ∈ V \ L, we have x  (x ∧ y) := x ∨ (x ∧ y) = x. Therefore x  (x y) = x. 2. Let x ∧ y ∈V / \ L. Since x y := 0∈V |L, we have x  (x y) = x  0. If x∈V \ L, then x  0 := x, and x  (x y) = x. If x = 0, then x  (x y) = 0  0 = 0 = x. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

44

KEHAYOPULU

Lemma 3. Let (V, ∧, ∨) be a lattice and let L be an ideal of V . For each v ∈ V we consider the mapping μv : L → L | x → v ∧ x. The mapping μv is a translation of L. Proof. Since L is an ideal of V , the mapping μv is well defined. It is a translation of L. Really, if x, y ∈ L, then μv (x ∧ y) = v ∧ (x ∧ y) = (v ∧ x) ∧ y = μv (x) ∧ y.



In this paper we use the following denotations. Notation 1. Let (L, ∧, ∨) be a lattice and t ∈ L. We denote by λt the translation of L defined as follows: λt : L → L | x → t ∧ x. It is called the inner translation of L with respect to t. Notation 2. Let (L, ∧, ∨) be a lattice and t ∈ L. We denote by π the mapping of L into T (L) defined as follows: π : L → T (L) | t → λt . Remark. The mapping π is (1–1). Really, if x, y ∈ L and λx = λy , then λx (x) = λy (x) and λx (y) = λy (y). Hence x = x ∧ x = y ∧ x and x ∧ y = y ∧ y = y, so x = y. For a lattice L we denote by 0L (or just by 0) the least element of L. Definition 2. Let L be a lattice, let K be a lattice with 0, and L ∩ K ∗ = ∅, where K ∗ = K \ {0}. A lattice V is called an ideal extension (or just an extension) of L by K if there exists an ideal L of V such that L is isomorphic to L and the Rees quotient V |L is isomorphic to K (in symbols, L ≈ L , V |L ≈ K). In what follows we denote by A(L, K ∗ ) the set of all mappings of L into K ∗ . A mapping f of A into B is usually denoted by f : A → B. Unless otherwise is stated, we denote the order on a lattice L by “≤L ” and we do the operations of supremum and infimum on L by “∨L ” and “∧L ”, respectively. We denote the supremum and the infimum on V |L by “” and “ ”, respectively. 2. THE MAIN RESULT Let us adduce the main theorem of the (ideal) extensions of lattices. Given a lattice L and a lattice K having a least element, we construct (all) the lattices V which are (ideal) extensions of L by K. Conversely, we prove that each lattice V which is an extension of L by K can be so constructed. Theorem. Let L be a lattice, let K be a lattice with 0 and L ∩ K ∗ = ∅. Assume that there exist mappings θ1 : K ∗ → T (L) | a → λa , θ2 : K ∗ → A(L, K ∗ ) | a → ρa , f : {(a, b) | a, b ∈ K ∗ , a ∧K b = 0} → L, which satisfy the following conditions: (C1) θ1 (a)θ1 (b) = λf (a,b) for all a, b ∈ K ∗ , a ∧K b = 0; (C2) θ1 (a ∧K b) = θ1 (a)θ1 (b) for all a, b ∈ K ∗ , a ∧K b = 0; (C3) ρa∨K b (c) = ρa (c) ∨K b for all a, b ∈ K ∗ , c ∈ L; c (b)

(C4) ρc (a ∨L b) = ρρ

(a) for all a, b ∈ L, c ∈ K ∗ ;

(C5) ρa (λa (b)) = a for all a ∈ K ∗ , b ∈ L; b (a)

(C6) λρ

(a) = a for all a ∈ L, b ∈ K ∗ ;

(C7) ρa (f (a, b)) = a for all a, b ∈ K ∗ , a ∧K b = 0. Let V := L ∪ K ∗ . We define operations “∨” and “∧” on V as follows: RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

IDEAL EXTENSIONS OF LATTICES

45

∨ : V × V → V | (a, b) → a ∨ b, where ⎧ a ∨L b, ⎪ ⎪ ⎪ ⎨ b ρ (a), a ∨ b := ⎪ ρa (b), ⎪ ⎪ ⎩ a ∨K b,

if a, b ∈ L; if a ∈ L, b ∈ K ∗ ; if a ∈ K ∗ , b ∈ L; if a, b ∈ K ∗ ,

(S1)

if a, b ∈ L; if a ∈ L, b ∈ K ∗ ; if a ∈ K ∗ , b ∈ L; if a, b ∈ K ∗ , a ∧K b = 0; if a, b ∈ K ∗ , a ∧K b =  0.

(P 1)

(S2) (S3) (S4)

∧ : V × V → V | (a, b) → a ∧ b, where ⎧ ⎪ a ∧L b, ⎪ ⎪ ⎪ ⎪ b ⎪ ⎨λ (a), a ∧ b := λa (b), ⎪ ⎪ ⎪ f (a, b), ⎪ ⎪ ⎪ ⎩ a ∧K b,

(P 2) (P 3) (P 4) (P 5)

Then (V, ∨, ∧) is a lattice and it is an extension of L by K. Conversely, let V be an extension of L by K. Then there exist mappings θ1 : K ∗ → T (L) | a → λa , θ2 : K ∗ → A(L, K ∗ ) | a → ρa , f : {(a, b) | a, b ∈ K ∗ , a ∧K b = 0} → L, such that conditions (C1)–(C7) mentioned in the first part of the theorem are satisfied. Furthermore, the set L ∪ K ∗ endowed with the operations “∨” and “∧” defined in the first part of the theorem is a lattice, and L ∪ K∗ ≈ V . Proof. (A) The operation “∨” is well defined. Really, let a, b ∈ V . If a, b ∈ L then by (S1) we have a ∨ b := a ∨L b ∈ L ⊆ V . If a ∈ L, b ∈ K ∗ then by (S2) we have a ∨ b := ρb (a). Since b ∈ K ∗ , ρb is a mapping of L into K ∗ . Since a ∈ L, we have ρb (a) ∈ K ∗ ⊆ V , hence a ∨ b ∈ V . If a ∈ K ∗ , b ∈ L then by (S3) we have a ∨ b := ρa (b). Since a ∈ K ∗ , we have ρa ∈ A(L, K ∗ ). Since b ∈ L, we have ρa (b) ∈ K ∗ ⊆ V . If a, b ∈ K ∗ then by (S4) we have a ∨ b := a ∨K b. If a ∨K b = 0, then we have a = b = 0 (because 0 is the zero of K), which is impossible. Thus a ∨ b ∈ K ∗ ⊆ V . In a similar way we prove that if a, b, c, d ∈ V such that (a, b) = (c, d), then a ∨ b = c ∨ d. (B) The operation “∨” on V is commutative. Really, if a, b ∈ L, then a ∨ b := a ∨L b = b ∨L a := b ∨ a. If a ∈ L, b ∈ K ∗ then by (S2) we have a ∨ b := ρb (a) and by (S3) we have b ∨ a := ρb (a), thus a ∨ b = b ∨ a. If a ∈ K ∗ , b ∈ L, by symmetry we get a ∨ b = b ∨ a. If a, b ∈ K ∗ , then a ∨ b := a ∨K b = b ∨K a := b ∨ a. (C) (a ∨ b) ∨ c = a ∨ (b ∨ c) for all a, b, c ∈ V . Really, there are eight cases. 1. Let a, b, c ∈ L. Then a ∨ b := a ∨L b ∈ L(⊆ V ), b ∨ c := b ∨L c ∈ L(⊆ V ), and (a ∨ b) ∨ c = (a ∨L b) ∨ c := (a ∨L b) ∨L c (by (S1)) = a ∨L (b ∨L c) = a ∨L (b ∨ c) := a ∨ (b ∨ c) (by (S1)). 2. Let a, b ∈ L, c ∈ K ∗ . Since a, b ∈ L, we have a ∨ b := a ∨L b ∈ L. Since a ∨ b ∈ L and c ∈ K ∗ , by (S2) we have (a ∨ b) ∨ c := ρc (a ∨ b). Since a ∨ b = a ∨L b ∈ L and ρc is a mapping of L into K ∗ , we have ρc (a ∨ b) = ρc (a ∨L b). Thus we have (a ∨ b) ∨ c = ρc (a ∨L b). Since b ∈ L, c ∈ K ∗ , by (S2) we have b ∨ c := ρc (b) ∈ K ∗ . Since a ∈ L, b ∨ c ∈ K ∗ , by (S2) we have a ∨ (b ∨ c) := ρb∨c (a). Since c c c b ∨ c = ρc (b) ∈ K ∗ , we have ρb∨c = ρρ (b) . Since ρb∨c , ρρ (b) : L → K ∗ , ρb∨c = ρρ (b) and a ∈ L, we c c have ρb∨c (a) = ρρ (b) (a). Then we have a ∨ (b ∨ c) = ρρ (b) (a). On the other hand, by (C4) we have c ρc (a ∨L b) = ρρ (b) (a). 3. Let b, c ∈ L, a ∈ K ∗ . Since a ∈ K ∗ , b ∈ L, by (S3) we have a ∨ b := ρa (b) ∈ K ∗ . Since a ∨ b ∈ K ∗ , c ∈ L, by (S3), we have (a ∨ b) ∨ c := ρa∨b (c). Since a ∨ b = ρa (b) ∈ K ∗ , we have ρa∨b = a a a a ρρ (b) . Since ρa∨b , ρρ (b) : L → K ∗ , ρa∨b = ρρ (b) and c ∈ L, we have ρa∨b (c) = ρρ (b) (c). Thus we RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

46

KEHAYOPULU

have (a ∨ b) ∨ c = ρρ (b) (c). Since b, c ∈ L, we have b ∨ c := b ∨L c ∈ L. Since a ∈ K ∗ , b ∨ c ∈ L, by (S3) we have a ∨ (b ∨ c) := ρa (b ∨ c). Since ρa : L → K ∗ and b ∨ c = b ∨L c ∈ L, we have ρa (b ∨ c) = ρa (b ∨L c) = ρa (c ∨L b). Thus we have a ∨ (b ∨ c) = ρa (c ∨L b). Since c, b ∈ L, a ∈ K ∗ , by (C4), we a have ρa (c ∨L b) = ρρ (b) (c). 4. Let c, a ∈ L, b ∈ K ∗ . Since a ∈ L, b ∈ K ∗ , by (S2), we have a ∨ b := ρb (a) ∈ K ∗ . Since a ∨ b ∈ K ∗ and c ∈ L, by (S3), we have (a ∨ b) ∨ c := ρa∨b (c). Since a ∨ b = ρb (a) ∈ K ∗ , we have b b b b ρa∨b = ρρ (a) . Since ρa∨b , ρρ (a) : L → K ∗ , ρa∨b = ρρ (a) , and c ∈ L, we have ρa∨b (c) = ρρ (a) (c). b Thus we have (a ∨ b) ∨ c = ρρ (a) (c). Since b ∈ K ∗ , c ∈ L, by (S3) we have b ∨ c := ρb (c) ∈ K ∗ . Since a ∈ L, b ∨ c ∈ K ∗ , by (S2) we have a ∨ (b ∨ c) := ρb∨c (a). Since b ∨ c = ρb (c) ∈ K ∗ , we have b b b b ρb∨c = ρρ (c) . Since ρb∨c , ρρ (c) : L → K ∗ , ρb∨c = ρρ (c) , and a ∈ L, we have ρb∨c (a) = ρρ (c) (a). Since b c, a ∈ L, b ∈ K ∗ , by (C4) we have ρb (c ∨L a) = ρρ (a) (c). Since a, c ∈ L, b ∈ K ∗ , by (C4) we have b ρb (a ∨L c) = ρρ (c) (a). Since ρb : L → K ∗ and c ∨L a = a ∨L c, we get ρb (c ∨L a) = ρb (a ∨L c). Thus we have a

b (a)

(a ∨ b) ∨ c = ρρ

b (c)

(c) = ρb (c ∨L a) = ρb (a ∨L c) = ρρ

(a) = ρb∨c (a) = a ∨ (b ∨ c).

5. Let a ∈ L, b, c ∈ K ∗ . Since a ∈ L, b ∈ K ∗ , by (S2) we have a ∨ b := ρb (a) ∈ K ∗ . Since a ∨ b, c ∈ K ∗ , by (S4) we have (a ∨ b) ∨ c := (a ∨ b) ∨K c. Since a ∨ b = ρb (a) ∈ K, we have (a ∨ b) ∨K c = ρb (a) ∨K c. Hence we have (a ∨ b) ∨ c = ρb (a) ∨K c. Since b, c ∈ K ∗ , by (S4), we have b ∨ c := b ∨K c ∈ K ∗ . Since a ∈ L, b ∨ c ∈ K ∗ , by (S2) we have a ∨ (b ∨ c) := ρb∨c (a). Since b ∨ c = b ∨K c ∈ K ∗ , we have ρb∨c = ρb∨K c . Then we get ρb∨c (a) = ρb∨K c (a) because ρb∨c , ρb∨K c : L → K ∗ and a ∈ L. Thus we have a ∨ (b ∨ c) = ρb∨K c (a). On the other hand, since b, c ∈ K ∗ and a ∈ L, by (C3), we have ρb∨K c (a) = ρb (a) ∨K c. 6. Let c, a ∈ K ∗ , b ∈ L. Then a ∨ b := ρa (b) ∈ K ∗ (by (S3)) and (a ∨ b) ∨ c := (a ∨ b) ∨K c = ρa (b) ∨K c (by (S4)). In addition, b ∨ c := ρc (b) ∈ K ∗ (by (S2)) and a ∨ (b ∨ c) := a ∨K (b ∨ c) = a ∨K ρc (b) (by (S4)). Since a, c ∈ K ∗ , b ∈ L, by (C3) we have ρa∨K c (b) = ρa (b) ∨K c. Since c, a ∈ K ∗ , b ∈ L, by (C3) we have ρc∨K a (b) = ρc (b) ∨K a. Since a ∨K c = c ∨K a ∈ K ∗ , we have ρa∨K c = ρc∨K a , then ρa∨K c (b) = ρc∨K a (b). Hence we have (a ∨ b) ∨ c = ρa (b) ∨K c = ρa∨K c (b) = ρc∨K a (b) = ρc (b) ∨K a = a ∨K ρc (b) = a ∨ (b ∨ c). 7. Let c ∈ L, a, b ∈ K ∗ ,. Then a ∨ b := a ∨K b ∈ K ∗ (by (S4)), (a ∨ b) ∨ c := ρa∨b (c) = ρa∨K b (c) (by (S3)), b ∨ c := ρb (c) ∈ K ∗ (by (S3)), and a ∨ (b ∨ c) := a ∨K (b ∨ c) = a ∨K ρb (c) (by (S4)). Since a ∨K b = b ∨K a ∈ K ∗ , we have ρa∨K b = ρb∨K a , and ρa∨K b (c) = ρb∨K a (c). Since b, a ∈ K ∗ , c ∈ L, by (C3) we have ρb∨K a (c) = ρb (c) ∨K a. Thus we have (a ∨ b) ∨ c = ρa∨K b (c) = ρb∨K a (c) = ρb (c) ∨K a = a ∨K ρb (c) = a ∨ (b ∨ c). 8. Let a, b, c ∈ K ∗ . Then a ∨ b := a ∨K b ∈ K ∗ , (a ∨ b) ∨ c := (a ∨ b) ∨K c = (a ∨K b) ∨K c, b ∨ c := b ∨K c ∈ K ∗ and a ∨ (b ∨ c) = a ∨K (b ∨ c) = a ∨K (b ∨K c) = (a ∨K b) ∨K c. (D) The operation “∧” is well defined. Really, let a, b ∈ V . If a, b ∈ L then, by (P1), a ∧ b := a ∧L b ∈ L ⊆ V . If a ∈ L, b ∈ K ∗ then, by (P2), a ∧ b := λb (a) ∈ L ⊆ V . If a ∈ K ∗ , b ∈ L then, by (P3), a ∧ b := λa (b) ∈ L ⊆ V . Let a, b ∈ K ∗ . If a ∧K b = 0 then, by (P4), a ∧ b := f (a, b) ∈ L ⊆ V . If a ∧K b = 0 then, by (P5) we have a ∧ b := a ∧K b ∈ K ∗ ⊆ V . In a similar way we prove that if a, b, c, d ∈ V , a = c, b = d, then a ∧ b = c ∧ d. (E) The operation “∧” is commutative. Really, let a, b ∈ V . If a, b ∈ L, then a ∧ b := a ∧L b. Since b, a ∈ L, we have b ∧ a := b ∧L a = a ∧L b. Thus a ∧ b = b ∧ a. If a ∈ L, b ∈ K ∗ then, by (P2), a ∧ b := RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

IDEAL EXTENSIONS OF LATTICES

47

λb (a). By (P3), b ∧ a := λb (a). Thus a ∧ b = b ∧ a. If a ∈ K ∗ , b ∈ L, then, by (P3), a ∧ b := λa (b). By (P2), b ∧ a := λa (b). Thus a ∧ b = b ∧ a. Let a, b ∈ K ∗ . (a) Let a ∧k b = 0. By (P4) we have a ∧ b := f (a, b). Since b, a ∈ K ∗ , b ∧K a = 0, again by (P4) we have b ∧ a := f (b, a). On the other hand, f (a, b) = f (b, a). Really, since a, b ∈ K ∗ , a ∧K b = 0, by (C1) we have θ1 (a)θ1 (b) = λf (a,b) . Since b, a ∈ K ∗ , b ∧K a = 0, by (C1) we have θ1 (b)θ1 (a) = λf (b,a) . Since θ1 (a)θ1 (b) are translations of L, by Lemma 1 we have θ1 (a)θ1 (b) = θ1 (b)θ1 (a). Thus we have λf (a,b) = λf (b,a) . Since the mapping π : L → T (L) | t → λt is (1–1), f (a, b), f (b, a) ∈ L, and λf (a,b) = λf (b,a) , we have f (a, b) = f (b, a). Hence we have a ∧ b = b ∧ a. (b) Let a ∧k b = 0. By (P5) we have a ∧ b := a ∧K b. Since b, a ∈ K ∗ and b ∧K a = 0, by (P5) we have b ∧ a := b ∧K a. Since a ∧K b = b ∧K a, we have a ∧ b = b ∧ a. (F) The operation “∧” is associative. Really, there are eight cases. 1. Let a, b, c ∈ L. Then a ∧ b := a ∧L b ∈ L, b ∧ c := b ∧L c ∈ L, (a ∧ b) ∧ c := (a ∧ b) ∧L c = (a ∧L b) ∧L c, a ∧ (b ∧ c) := a ∧L (b ∧ c) = a ∧L (b ∧L c). Since (a ∧L b) ∧L c = a ∧L (b ∧L c), we have (a ∧ b) ∧ c = a ∧ (b ∧ c). 2. Let a, b ∈ L, c ∈ K ∗ . Then a ∧ b := a ∧L b ∈ L, b ∧ c := λc (b) ∈ L, (a ∧ b) ∧ c := λc (a ∧ b) = ∧L b), a ∧ (b ∧ c) := a ∧L (b ∧ c) = a ∧L λc (b). On the other hand,

λc (a

λc (a ∧L b) = λc (b ∧L a) = λc (b) ∧L a (since λc ∈ T (L)) = a ∧L λc (b). 3. Let b, c ∈ L, a ∈ K ∗ . Then a ∧ b := λa (b) ∈ L, b ∧ c := b ∧L c ∈ L, (a ∧ b) ∧ c := (a ∧ b) ∧L c = λa (b) ∧L c, a ∧ (b ∧ c) := λa (b ∧ c) = λa (b ∧L c). Since λa ∈ T (L), we have λa (b ∧L c) = λa (b) ∧L c. 4. Let c, a ∈ L, b ∈ K ∗ . Then a ∧ b := λb (a) ∈ L, b ∧ c := λb (c) ∈ L, (a ∧ b) ∧ c := (a ∧ b) ∧L c = λb (a) ∧L c, a ∧ (b ∧ c) := a ∧L (b ∧ c) = a ∧L λb (c), On the other hand, λb (a) ∧L c = λb (a ∧L c) (since λb ∈ T (L)) = λb (c ∧L a) = λb (c) ∧L a (since λb ∈ T (L)) = a ∧L λb (c). 5. Let a ∈ L, b, c ∈ K ∗ . Then a ∧ b := λb (a) ∈ L and by Lemma 1, (a ∧ b) ∧ c := λc (a ∧ b) = λc (λb (a)) := (λc λb )(a). 5.1. Let b ∧K c = 0. Then b ∧ c := f (b, c) ∈ L and a ∧ (b ∧ c) := a ∧L (b ∧ c) = a ∧L f (b, c) = f (b, c) ∧L a = λf (b,c) (a). On the other hand, since b, c ∈ K ∗ and b ∧K c = 0, by (C1) we have θ1 (b)θ1 (c) = λf (b,c) . Then λc λb = λb λc = λf (b,c) (see Lemma 1), and (λc λb )(a) = λf (b,c) (a). 5.2. Let b ∧K c = 0. Then b ∧ c := b ∧K c ∈ K ∗ , and a ∧ (b ∧ c) := λb∧c (a) = λb∧K c (a). On the other hand, (λc λb )(a) = λb∧K c (a). Really, since b, c ∈ K ∗ and b ∧K c = 0, by (C2) we have θ1 (b ∧K c) = θ1 (b)θ1 (c). Then λb∧K c = λb λc = λc λb (see Lemma 1) and (λc λb )(a) = λb∧K c (a). 6. Let b ∈ L, c, a ∈ K ∗ . Then a ∧ b := λa (b) ∈ L, b ∧ c := λc (b) ∈ L, (a ∧ b) ∧ c := λc (a ∧ b) = := (λc λa )(b), a ∧ (b ∧ c) := λa (b ∧ c) = λa (λc (b)) := (λa λc )(b). Since λc , λa ∈ T (L), we c a have λ λ = λa λc (see Lemma 1), then (λc λa )(b) = (λa λc )(b).

λc (λa (b)) 7.

Let c ∈ L, a, b ∈ K ∗ .

Then b ∧ c := λb (c) ∈ L and a ∧ (b ∧ c) := λa (b ∧ c) = λa (λb (c)) :=

(λa λb )(c). 7.1. Let a ∧K b = 0. Then a ∧ b := f (a, b) ∈ L and (a ∧ b) ∧ c := (a ∧ b) ∧L c = f (a, b) ∧L c. On the other hand, (λa λb )(c) = f (a, b) ∧L c. Really, since a, b ∈ K ∗ and a ∧K b = 0, by (C1) we have θ1 (a)θ1 (b) = λf (a,b) , consequently, λa λb = λf (a,b) and (λa λb )(c) = λf (a,b) (c) := f (a, b) ∧L c. 7.2. Let a ∧K b = 0. Then a ∧ b := a ∧K b ∈ K ∗ and (a ∧ b) ∧ c := λa∧b (c) = λa∧K b (c). On the other hand, λa∧K b (c) = (λa λb )(c). Really, since a, b ∈ K ∗ and a ∧K b = 0, by (C2) we have θ1 (a ∧K b) = θ1 (a)θ1 (b). Then λa∧K b = λa λb , and λa∧K b (c) = (λa λb )(c). 8. Let a, b, c ∈ K ∗ . Consider the following cases. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

48

KEHAYOPULU

8.1. Let a ∧K b = b ∧K c = 0. Then a ∧ b := f (a, b) ∈ L, b ∧ c := f (b, c) ∈ L, (a ∧ b) ∧ c := λc (a ∧ b) = λc (f (a, b)) ∈ L, a ∧ (b ∧ c) := λa (b ∧ c) = λa (f (b, c)) ∈ L. We have λλc (f (a,b)) = λλa (f (b,c)) . Really, let x ∈ L. Then λλc (f (a,b)) (x) := λc (f (a, b)) ∧L x := λc (f (a, b) ∧L x) (since λc ∈ T (L)) := λc (λf (a,b) (x)). Since a, b ∈ K ∗ and a ∧K b = 0, by (C1) we have λf (a,b) = θ1 (a)θ1 (b) = λa λb . Hence, since λa λb ∈ T (L), we have λc (λf (a,b) (x)) = λc ((λa λb )(x)) := (λc (λa λb )(x). Thus we have λλc (f (a,b)) (x) = (λc (λa λb ))(x). Similarly, we get λλa (f (b,c)) (x) = (λa (λb λc ))(x). On the other hand, by Lemma 1, we have λc (λa λb ) = (λa λb )λc (since λa λb ∈ T (L)) = (λa ∧ λb ) ∧ λc = λa ∧ (λb ∧ λc ) = λa (λb λc ) (see Lemma 1) . Since the mapping π : L → T (L) | t → λt is (1–1), we have λc (f (a, b)) = λa (f (b, c)). 8.2. Let a ∧K b = 0 and b ∧K c = 0. Then a ∧ b := f (a, b) ∈ L, (a ∧ b) ∧ c := λc (a ∧ b) = c λ (f (a, b)) ∈ L, b ∧ c := b ∧K c ∈ K ∗ . Since a ∧K (b ∧ c) = a ∧K (b ∧K c) = (a ∧K b) ∧K c = 0, we have a ∧ (b ∧ c) := f (a, b ∧ c) = f (a, b ∧K c). Since the mapping π is (1–1), suffice it to prove that λλc (f (a,b)) = λf (a,b∧K c) . Let now x ∈ L. Then λλc (f (a,b)) (x) := λc (f (a, b)) ∧L x = λc (f (a, b) ∧L x) (since λc ∈ T (L)) = λc (λf (a,b) (x)). Since a, b ∈ K ∗ and a ∧K b = 0, by (C1) we get λf (a,b) = θ1 (a)θ1 (b) = λa λb . Then λλc (f (a,b)) (x) = λc (λf (a,b) (x)) = λc ((λa λb )(x)) = (λc (λa λb ))(x). Thus we have λλc (f (a,b)) = λc (λa λb ). Since a, b ∧K c ∈ K ∗ and a ∧K (b ∧K c) = 0, by (C1) we get λf (a,b∧K c) = θ1 (a)θ1 (b ∧K c) = λa θ1 (b ∧K c). Since b, c ∈ K ∗ and b ∧K c = 0, by (C2) we have θ1 (b ∧K c) = θ1 (b)θ1 (c) = λb λc . Then λf (a,b∧K c) = λa (λb λc ). On the other hand, λc (λa λb ) = λa (λb λc ). 8.3. Let a ∧K b = 0, b ∧K c = 0. The proof is similar to that of 8.2. 8.4. Let a ∧K b = 0, b ∧K c = 0. Then a ∧ b := a ∧K b ∈ K ∗ , b ∧ c := b ∧K c ∈ K ∗ . Clearly, (a ∧ b) ∧K c = (a ∧K b) ∧K c ∈ K. (a) Let (a ∧ b) ∧K c = 0. Then (a ∧ b) ∧ c := f (a ∧ b, c) = f (a ∧K b, c). Since a ∧K (b ∧ c) = a ∧K (b ∧K c) = (a ∧K b) ∧K c = (a ∧ b) ∧K c = 0, we have a ∧ (b ∧ c) := f (a, b ∧ c) = f (a, b ∧K c). On the other hand, λf (a∧K b,c) = λf (a,b∧K c) . Really, since a ∧K b, c ∈ K ∗ and (a ∧K b) ∧K c = 0 by (C1), we have λf (a∧K b,c) = θ1 (a ∧K b)θ1 (c) = θ1 (a ∧K b)λc . Since a, b ∈ K ∗ and a ∧K b = 0 by (C2), we have θ1 (a ∧K b) = θ1 (a)θ1 (b) = λa λb . Hence we have λf (a∧K b,c) = (λa λb )λc . In a similar way, by (C1) and (C2) we get λf (a,b∧K c) = (λa λb )λc . On the other hand, (λa λb )λc = λa (λb λc ) (see Lemma 1). (b) Let (a ∧ b) ∧K c = 0. Then (a ∧ b) ∧ c := (a ∧ b) ∧K c = (a ∧K b) ∧K c. Since a ∧K (b ∧ c) = a ∧K (b ∧K c) = (a ∧K b) ∧K c = (a ∧ b) ∧K c, we have a ∧K (b ∧ c) = 0. Then a ∧ (b ∧ c) := a ∧K (b ∧ c) = a ∧K (b ∧K c). (G) a ∧ (a ∨ b) = a for all a, b ∈ V . Really, if a, b ∈ L, then a ∨ b := a ∨L b ∈ L and a ∧ (a ∨ b) := a ∧L (a ∨ b) = a ∧L (a ∨L b) = a. If a ∈ L, b ∈ K ∗ , then a ∨ b := ρb (a) ∈ K ∗ and a ∧ (a ∨ b) := b λa∨b (a) = λρ (a) = a by (C6). If a ∈ K ∗ , b ∈ L, then a ∨ b := ρa (b) ∈ K ∗ . On the other hand, a ∧K (a ∨ b) = a ∧K ρa (b) = a ∧K ρa∨K a (b) = a ∧K (ρa (b) ∨K a) = a = 0 by (C3). Hence a ∧ (a ∨ b) := a ∧K (a ∨ b) = a. If a, b ∈ K ∗ , then a ∨ b := a ∨k b ∈ K ∗ . Since a ∧K (a ∨ b) = a ∧K (a ∨K b) = a = 0, we have a ∧ (a ∨ b) = a ∧K (a ∨ b) = a. (H) a ∨ (a ∧ b) = a for all a, b ∈ V . Really, if a, b ∈ L, then a ∧ b := a ∧L b ∈ L and a ∨ (a ∧ b) := a ∨L (a ∧ b) = a ∨L (a ∧L b) = a. If a ∈ L, b ∈ K ∗ , then a ∧ b := λb (a) ∈ L and a ∨ (a ∧ b) := a ∨L (a ∧ b) = a ∨L λb (a). In addition, a ∨L λb (a) = a. Really, since λb is a translation of L, we have a ∨L λa (a) = a ∨L λb (a ∧L a) = a ∨L (λb (a) ∧L a) = a. If a ∈ K ∗ , b ∈ L, then a ∧ b := λa (b) ∈ L and a ∨ (a ∧ b) := ρa (a ∧ b) = ρa (λa (b)) = a by (C5). Let a, b ∈ K ∗ . If a ∧K b = 0, then a ∧ b := f (a, b) ∈ RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

IDEAL EXTENSIONS OF LATTICES

49

L, and a ∨ (a ∧ b) := ρa (a ∧ b) = ρa (f (a, b)) = a by (C7). If a ∧K b = 0, then a ∧ b := a ∧K b ∈ K ∗ and a ∨ (a ∧ b) := a ∨K (a ∧ b) = a ∨K (a ∧K b) = a. (I) L is an ideal of V . Indeed, if a, b ∈ L, then a ∨ b := a ∨L b ∈ L. Let now a ∈ V , b ∈ L. If a ∈ L, then a ∧ b := a ∧L b ∈ L; if a ∈ K ∗ , then a ∧ b := λa (b) ∈ L. (J) (K, ∨K , ∧K ) ≈ (V |L, , ). Really, since V := L ∪ K ∗ and L ∩ K ∗ = ∅, we have V \ L = K ∗. Consider the mapping  g : K → V |L  x →



(∗)

x, if x ∈ K ∗ ; 0V |L , if x = 0K .

Due to (∗) the mapping g is well defined, it is (1–1) and onto mapping. The mapping g is a homomorphism. Really, g(x ∨K y) = g(x)  g(y) for all x, y ∈ K. Really, let x, y ∈ K. 1. Let x, y ∈ K ∗ . Then x ∨K y ∈ K ∗ , g(x) := x, g(y) := y, g(x ∨K y) := x ∨K y. Since x, y ∈ ∗ K = V \ L, we have x ∨ y = x ∨K y (by (S4)) and x  y = x ∨ y (see Lemma 2). Then g(x ∨K y) = g(x)  g(y). 2. Let x ∈ K ∗ , y = 0K . Then g(x) := x, g(y) := 0V |L , x ∨K y = x, and g(x ∨K y) = g(x) = x. Since x ∈ K ∗ = V \ L ⊆ V |L and 0V |L is the least element of V |L, we have x = x  0V |L . Then we have g(x ∨K y) = x = x  0V |L = g(x)  g(y). 3. Let x = 0K , y ∈ K ∗ . The proof is similar to that of case 2. 4. Let x = y = 0K . Then g(x) := 0V |L , g(y) := 0V |L , x ∨K y = 0K , and g(x ∨K y) = g(0K ) := 0V |L . Thus we have g(x ∨K y) = 0V |L = 0V |L  0V |L = g(x)  g(y). g(x ∧K y) = g(x) g(y) for all x, y ∈ K. Really, let x, y ∈ K. 1) Let x, y ∈ K ∗ , Then g(x) := x, g(y) := y. 1.1. Let x ∧K y = 0K . Then g(x ∧K y) = g(0K ) := 0V |L . By (P4) we have x ∧ y := f (x, y) ∈ L. Since x ∧ y ∈ / V \ L, we have x y := 0V |L (see Lemma 2). Then g(x ∧K y) = 0V |L = x y = g(x) g(y). 1.2. Let x ∧K y = 0K . Then g(x ∧K y) = x ∧K y. By (P5) we have x ∧ y := x ∧K y ∈ K ∗ = V \ L. Since x ∧ y ∈ V \ L, we have x y = x ∧ y (see Lemma 2). Then g(x ∧K y) = x ∧k y = x ∧ y = x y = g(x) g(y). 2) Let x ∈ K ∗ , y = 0K . Then g(x) := x, g(y) := 0V |L , x ∧K y = 0K , and g(x ∧K y) = g(0K ) := 0V |L . Since x ∈ K ∗ = V \ L(⊆ V |L), we have x 0V |L = 0V |L . Then we have g(x ∧K y) = 0V |L = x 0V |L = g(x) g(y). 3) Let x = 0K , y ∈ K ∗ . The proof is similar to that of case 2. 4) Let x = y = 0V |L . Then g(x) = g(y) = g(x ∧K y) = 0V |L . Hence g(x ∧K y) = 0V |L = 0V |L 0V |L = g(x) g(y). The converse statement: Let (V, ∨V , ∧V ) be an extension of (L, ∨L , ∧L ) by (K, ∨K , ∧K ). Then there exist an ideal I of V , an isomorphism φ : L → I, and an isomorphish ψ : K → V |I. (I) Consider the mapping h : K ∗ → V |I \ {0V |I } | x → ψ(x). 1. The mapping h is well defined. Really, let x ∈ K ∗ . Then h(x) := ψ(x) ∈ V |I. Let ψ(x) = 0V |I . Since 0V |I ∈ V |I and ψ is onto, there exists t ∈ K such that ψ(t) = 0V |I . Since ψ is a homomorphism, we have ψ(0K ) = ψ(0K ∧K t) = ψ(0K ) ψ(t) = ψ(0K ) 0V |I = 0V |I . Since ψ(x) = ψ(0K ) and ψ is (1–1), we have x = 0K , which is impossible. If x, y ∈ K ∗ , x = y, then h(x) := ψ(x), h(y) := ψ(y) and ψ(x) = ψ(y). 2. The mapping h is (1–1). This is because ψ is an (1–1) mapping. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

50

KEHAYOPULU

3. The mapping h is onto. Really, let y ∈ V |I \ {0V |I }. Since y ∈ V |I and ψ is onto, there exists x ∈ K such that ψ(x) = y. If x = 0K then, since ψ is an isomorphism, we have y = ψ(0K ) = 0V |I , which is impossible. Thus we have x ∈ K ∗ and h(x) := ψ(x). (II) Consider the mapping θ1 : K ∗ → T (L) | a → λa := φ−1 μh(a) φ, where μh(a) : I → I | x → h(a) ∧V x. The mapping θ1 is well defined. Really, let a ∈ K ∗ . Since V is a lattice, I is an ideal of V , and h(a) ∈ V |I \ {0V |I } = V \ I ⊆ V , the mapping μh(a) is a translation of I (see Lemma 3). Since φ : L → I, μh(a) : I → I, and φ−1 : I → L, the mapping φ−1 μh(a) φ : L → L is well defined. Furthermore, it is a translation of L. Really, let x, y ∈ L. Then (φ−1 μh(a) φ)(x ∧L y) = φ−1 (μh(a) φ(x ∧L y)) = φ−1 (μh(a) (φ(x) ∧I φ(y))) (since φ is a homomorphism) = φ−1 (μh(a) φ(x) ∧I φ(y)) (since μh(a) ∈ T (I)) = φ−1 (μh(a) φ(x)) ∧L φ−1 (φ(y)) (φ−1 is a homomorphism) = (φ−1 μh(a) φ)(x) ∧L y. Let now a, b ∈ K ∗ be such that a = b. Then h(a) = h(b), μh(a) = μh(b) (see Lemma 3), therefore φ−1 μh(a) φ = φ−1 μh(b) φ. (III) Consider the mapping θ2 : K ∗ → A(L, K ∗ ) | a → ρa , where ρa : L → K ∗ | x → h−1 (h(a) ∨V φ(x)). The mapping θ2 is well defined. Really, let a ∈ K ∗ and x ∈ L. Then h(a) ∈ V |I \ {0V |I } = V \ I ⊆ V , φ(x) ∈ I, h(a) ∨V φ(x) ∈ V . If h(a) ∨V φ(x) ∈ I then, since V  h(a) ≤V h(a) ∨V φ(x) ∈ I and I is an ideal of V , we have h(a) ∈ I, which is impossible. Thus we have h(a) ∨V φ(x) ∈ V \ I = V |I \ {0V |I }; consequently, h−1 (h(a) ∨V φ(x)) ∈ K ∗ . If x, y ∈ L, x = y, then φ(x) = φ(y), h(a) ∨V φ(x) = h(a) ∨V φ(y), and h−1 (h(a) ∨V φ(x)) = h−1 (h(a) ∨V φ(y)). So the mapping ρa is well defined. Let now a, b ∈ K ∗ be such that a = b. Then h(a) = h(b), h(a) ∨V φ(x) = h(b) ∨V φ(x), h−1 (h(a) ∨V φ(x)) = h−1 (h(b) ∨V φ(x)), whence ρa = ρb . (IV) Consider the mapping f : {(a, b) | a, b ∈ K ∗ , a ∧K b = 0} → L | (a, b) → φ−1 (h(a) ∧V h(b)). The mapping f is well defined. Really, let a, b ∈ K ∗ , a ∧K b = 0. Since a, b ∈ K ∗ , we have h(a), h(b) ∈ V |I \ {0V |I } = V \ I ⊆ V . Then h(a) ∧V h(b) ∈ V . On the other hand, h(a) ∧V h(b) ∈ I. Really, h(a) h(b) = ψ(a) ψ(b) = ψ(a ∧K b) = ψ(0K )0V |I (since ψ is a homomorphism). / V \ I. Since h(a) ∧V h(b) ∈ I, we have By Lemma 2 we have h(a) ∧V h(b) ∈ −1 φ (h(a) ∧V h(b)) ∈ L. Let a, b, c, d ∈ K ∗ be such that a ∧K b = c ∧K d = 0, a = c, b = d. Then f (a, b) := φ−1 (h(a) ∧V h(b)) and f (c, d) := φ−1 (h(c) ∧V h(d)). Since h(a) = h(c), h(b) = h(d), we have f (a, b) = f (c, d). Let us show that conditions (C1)–(C7) mentioned in the first part of the Theorem are satisfied. (C1) Let a, b ∈ K ∗ , a ∧K b = 0. Then θ1 (a)θ1 (b) = λf (a,b) . Really, since a, b ∈ K ∗ , a ∧K b = 0, we have f (a, b) := φ−1 (h(a) ∧V h(b)). Since a, b ∈ K ∗ , we have θ1 (a) := φ−1 μh(a) φ, θ1 (b) := φ−1 μh(b) φ, and θ1 (a)θ1 (b) = φ−1 μh(a) μh(b) φ. On the other hand, φ−1 μh(a) μh(b) φ = λf (a,b) . Really, let x ∈ L. Then (φ−1 μh(a) μh(b) φ)(x) := φ−1 (μh(a) (μh(b) φ(x))) = φ−1 (μh(a) (h(b) ∧V φ(x))) = φ−1 (h(a) ∧V (h(b) ∧V φ(x))) = φ−1 ((h(a) ∧V h(b)) ∧V φ(x)) = φ−1 ((h(a) ∧V h(b)) ∧I φ(x)) (since h(a) ∧V h(b) ∈ I, φ(x) ∈ I) = φ−1 (h(a) ∧V h(b)) ∧L φ−1 (φ(x)) = φ−1 (h(a) ∧V h(b)) ∧L x := λφ−1 (h(a)∧V h(b)) (x) = λf (a,b) (x). (C2) Let a, b∈K ∗ , a ∧K b=0. Then θ1 (a ∧K b) = θ1 (a)θ1 (b). Really, since a, b∈K ∗ , a ∧K b=0, we have θ1 (a) := φ−1 μh(a) φ, θ1 (b) := φ−1 μh(b) φ, θ1 (a ∧K b) := φ−1 μh(a∧K b) φ. On the other hand, RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

IDEAL EXTENSIONS OF LATTICES

51

φ−1 μh(a∧K b) φ = φ−1 μh(a) μh(b) φ. Really, let x ∈ L. Then (φ−1 μh(a∧K b) φ)(x) := φ−1 (μh(a∧K b) φ(x)). Since a ∧K b ∈ K ∗ and φ(x) ∈ I, we have μh(a∧K b) φ(x) := h(a ∧K b) ∧V φ(x). Therefore, (φ−1 μh(a∧K b) φ)(x) = φ−1 (h(a ∧K b) ∧V φ(x)). On the other hand, (φ−1 μh(a) μh(b) φ)(x) := φ−1 (μh(a) (μh(b) φ(x))) = φ−1 (μh(a) (h(b) ∧V φ(x))) = φ−1 (h(a) ∧V (h(b) ∧V φ(x))) = φ−1 ((h(a) ∧V h(b)) ∧V φ(x)). Moreover, h(a ∧K b) = h(a) ∧V h(b). Really, since a, b, a ∧K b ∈ K ∗ , we have h(a) := ψ(a) ∈ V |I \ {0V |I },

h(b) := ψ(b) ∈ V |I \ {0V |I },

h(a ∧K b) := ψ(a ∧K b) ∈ V |I \ {0V |I }. Then 0V |I = h(a ∧K b) = ψ(a ∧K b) = ψ(a) ψ(b) (since ψ : K → V |I is an isomorphism) = h(a) h(b). Since h(a), h(b) ∈ V |I and h(a) h(b) = 0V |I , by Lemma 2 we have h(a) h(b) = h(a) ∧V h(b). Consequently, h(a ∧K b) = h(a) ∧V h(b). (C3) Let a, b ∈ K ∗ , c ∈ L. Then ρa∨K b (c) = ρa (c) ∨K b. Really, since a, a ∨K b ∈ K ∗ and c ∈ L, we have ρa∨K b (c) := h−1 (h(a ∨K b) ∨V φ(c)) ∈ K ∗ , ρa (c) := h−1 (h(a) ∨V φ(c)) ∈ K ∗ . On the other hand, h−1 (h(a ∨K b) ∨V φ(c)) = h−1 (h(a) ∨V φ(c)) ∨K b. Really, suffice it to prove that h(a ∨K b) ∨V φ(c) = h(h−1 (h(a) ∨V φ(c)) ∨K b). We have h(h−1 (h(a) ∨V φ(c)) ∨K b) := ψ(h−1 (h(a) ∨V φ(c)) ∨K b) = ψ(h−1 (h(a) ∨V φ(c)))  ψ(b) (since ψ is a homomorphism) = h(h−1 (h(a) ∨V φ(c)))  h(b). Since h(h−1 (h(a) ∨V φ(c))) ∈ V \ I and h(b) ∈ V \ I, by Lemma 2 we have h(h−1 (h(a) ∨V φ(c)))  h(b) = h(h−1 (h(a) ∨V φ(c))) ∨V h(b). Then h(h−1 (h(a) ∨V φ(c)) ∨K b) = h(h−1 (h(a) ∨V φ(c))) ∨V h(b) = (h(a) ∨V φ(c)) ∨V h(b) = (h(a) ∨V h(b)) ∨V φ(c). Since h(a), h(b) ∈ V \ I, by Lemma 1 we have h(a) ∨V h(b) = h(a)  h(b). Hence we get h(h−1 (h(a) ∨V φ(c)) ∨K b) = (h(a) ∨V h(b)) ∨V φ(c) = (h(a)  h(b)) ∨V φ(c) = (ψ(a)  ψ(b)) ∨V φ(c) = ψ(a ∨K b) ∨V φ(c) (since ψ is a homomorphism) = h(a ∨K b) ∨V φ(c) (since a ∨K b ∈ K ∗ ). (C4) Let a, b ∈ L, c ∈ K ∗ . Then ρc (a ∨L b) = ρa (a). Really, we have ρc (a ∨L b) := h−1 (h(c) ∨V φ(a ∨L b)) ∈ K ∗ , ρc (b) := h−1 (h(c) ∨V φ(b)) ∈ K ∗ , and K ∗  ρρ

c (b)

(a) := h−1 (h(ρc (b)) ∨V φ(a)) = h−1 (h(h−1 (h(c) ∨V φ(b))) ∨V φ(a)) := h−1 ((h(c) ∨V φ(b)) ∨V φ(a)).

Suffice it to prove that h(c) ∨V φ(a ∨L b) = (h(c) ∨V φ(b)) ∨V φ(a). We have h(c) ∨V φ(a ∨L b) = h(c) ∨V (φ(a) ∨I φ(b)) = h(c) ∨V (φ(a) ∨V φ(b)) (since φ(a), φ(b) ∈ I ⊆ V ) = (h(c) ∨V φ(b)) ∨V φ(a). (C5) Let a ∈ K ∗ , b ∈ L. Then ρa (λa (b)) = a. Really, since a ∈ K ∗ and b ∈ L, we have L  λa (b) := = φ−1 (μh(a) φ(b)) = φ−1 (h(a) ∧V φ(b)), therefore,

(φ−1 μh(a) φ)(b)

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

52

KEHAYOPULU

ρa (λa (b)) := h−1 (h(a) ∨V φ(λa (b))) = h−1 (h(a) ∨V φ(φ−1 (h(a) ∧V φ(b)))) = h−1 (h(a) ∨V (h(a) ∧V φ(b))) = h−1 (h(a)) = a. (C6) Let a ∈ L, b ∈ K ∗ . Then λρ h−1 (h(b) ∨V φ(a)) ∈ K ∗ , and

b (a)

b (a)

λρ

(a) := φ−1 (μh(ρ

b (a))

:= a.

Really, since a ∈ L, b ∈ K ∗ , we have ρb (a) :=

φ(a)) = φ−1 (h(ρb (a)) ∧V φ(a))

= φ−1 (h(h−1 (h(b) ∨V φ(a))) ∧V φ(a)) = φ−1 ((h(b) ∨V φ(a)) ∧V φ(a)) = φ−1 (φ(a)) = a. (C7) Let a, b ∈ K ∗ , a ∧K b = 0. Then ρa (f (a, b)) = a. Really, since a, b ∈ K ∗ , a ∧K b = 0, we have f (a, b) := φ−1 (h(a) ∧V h(b)) ∈ L, therefore, ρa (f (a, b)) := h−1 (h(a) ∨V φ(f (a, b))) = h−1 (h(a) ∨V φ(φ−1 (h(a) ∧V h(b)))) = h−1 (h(a) ∨V (h(a) ∧V h(b))) = h−1 (h(a)) = a. From the second part of the theorem, the set L ∪ K ∗ endowed with the operations “∨” and “∧” on V is a lattice and it is an extension of L by K. Furthermore, (L ∪ K ∗ , ∨, ∧) ≈ (V, ∨V , ∧V ). To prove this fact, we consider the mapping:  ϕ(a), if a ∈ L; g : L ∪ K∗ → V | a → h(a), if a ∈ K ∗ . One can easily prove that the mapping g is well defined, it is (1–1) and an onto mapping. Let us show that the mapping g is a homomorphism. 1. For all a, b ∈ L ∪ K ∗ we have g(a ∨ b) = g(a) ∨V g(b). Really, let a, b ∈ L ∪ K ∗ . Consider the following cases. Let a, b ∈ L, then g(a) := φ(a) ∈ I, g(b) := φ(b) ∈ I. Since a, b ∈ L, by (S1) we have a ∨ b := a ∨L b ∈ L, therefore, g(a ∨ b) = g(a ∨L b) := φ(a ∨L b) = φ(a) ∨L φ(b) (since φ is a homomorphism) = φ(a) ∨V φ(b) (since φ(a), φ(b) ∈ I ⊆ V ) = g(a) ∨V g(b). Let a ∈ L, b ∈ K ∗ . Then g(a) := φ(a), g(b) := h(b). Moreover, by (S2) we have a ∨ b := ρb (a) ∈ K ∗ , so g(a ∨ b) = g(ρb (a)) := h(ρb (a)). Since ρb (a) := h−1 (h(b) ∨V φ(a)), we have g(a ∨ b) = h(b) ∨V φ(a) = g(b) ∨V g(a). Let a ∈ K ∗ , b ∈ L. Then g(a) := h(a), g(b) := φ(b) and, by (S3) we have a ∨ b := ρa (b) := h−1 (h(a) ∨V φ(b)) ∈ K ∗ . Then g(a ∨ b) := h(a ∨ b) = h(a) ∨V φ(b) = g(a) ∨V g(b). Let a, b ∈ K ∗ , then g(a) := h(a), g(b) := h(b), a ∨ b := a ∨K b ∈ K ∗ (by (S4)), and g(a ∨ b) = g(a ∨K b) := h(a ∨K b). On the other hand, h(a ∨K b) = h(a)  h(b). Really, since a ∨K b ∈ K ∗ , we have h(a ∨K b) := ψ(a ∨K b) = ψ(a)  ψ(b) (since ψ is a homomorphism) = h(a)  h(b) (since a, b ∈ K ∗ ). Since h(a), h(b) ∈ V |I \ {0V |I } = V \ I, by Lemma 2 we get h(a)  h(b) = h(a) ∨V h(b). Therefore, g(a ∨ b) = h(a ∨K b) = h(a)  h(b) = h(a) ∨V h(b) = g(a) ∨V g(b). 2. For all a, b ∈ L ∪ K ∗ we have g(a ∧ b) = g(a) ∧V g(b). Really, let a, b ∈ L ∪ K ∗ . Consider the following cases. If a, b ∈ L, then g(a) := φ(a) ∈ I, g(b) := φ(b) ∈ I, a ∧ b := a ∧L b ∈ L (by (P2)), and g(a ∧ b) = g(a ∧L b) := φ(a ∧L b) = φ(a) ∧I φ(b) (since φ is a homomorphism) = φ(a) ∧V φ(b) (since φ(a), φ(b) ∈ I ⊆ V ) = g(a) ∧V g(b). RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

IDEAL EXTENSIONS OF LATTICES

53

Let a ∈ L, b ∈ K ∗ . Then g(a) := φ(a), g(b) := φ(b) and by (P2), a ∧ b := λb (a) = (φ−1 μh(b) φ)(a) = φ−1 (μh(b) φ(a)) = φ−1 (h(b) ∧V φ(a)) ∈ L. Then we have g(a ∧ b) := φ(a ∧ b) = h(b) ∧V φ(a) = g(b) ∧V g(a) = g(a) ∧V g(b). K ∗,

b ∈ L can be proved similarly. The case a ∈ Let a, b ∈ K ∗ . Then g(a) := h(a), g(b) := h(b). We consider the cases. (a) Let a ∧K b = 0. By (P4) we have a ∧ b := f (a, b) ∈ L. Then g(a ∧ b) = g(f (a, b)) := φ(f (a, b)). Since f (a, b) := φ−1 (h(a) ∧V h(b)), we have g(a ∧ b) = h(a) ∧V h(b) = g(a) ∧V g(b). (b) Let a ∧K b = 0. By (P5) we have a ∧ b := a ∧K b ∈ K ∗ . Then g(a ∧ b) = g(a ∧K b) := h(a ∧K b). Since a, b, a ∧K b ∈ K ∗ , we have h(a ∧K b) = h(a) ∧V h(b) (cf. the proof of (C2)). Hence g(a ∧ b) = h(a) ∧V h(b) = g(a) ∧V g(b). Let us give an illustrative example of this theorem. Example. Consider the sets L and K defined as follows: L := {(0, 0), (1, 2), (2, 1)} ∪ {(2, t) | t ∈ R, 2 ≤ t < 3}, K := {(0, 0), (0, 4), (2, 3), (3, 0), (3, 4)}. (R is the set of real numbers). The sets L, K with the relation “≤” on R2 defined by: (x, y) ≤ (z, t) ⇔ x ≤ z, y ≤ t are lattices, the element (0, 0) is the least element of K and L ∩ K ∗ = ∅, where K ∗ := K \ {(0, 0)}. The least element (0, 0) of K is also denoted by 0. Their diagrams are depicted in Fig. 1 and 2.

(I) Consider the mapping:

 i , θ1 : K ∗ → T (L) | a → λa := L λ(0,0)

if a = (2, 3) or a = (3, 4); otherwise.

(We denote by iL the identity mapping on L.) The mapping θ1 is well defined. Really, let a ∈ K ∗ . If a = (2, 3) or a = (3, 4), then λa := iL ∈ T (L). If a = (0, 4) or a = (3, 0), then λa := λ(0,0) ∈ T (L) (cf. Notation 1). Analogously, if a, b ∈ K ∗ and a = b, then λa = λb . (II) Consider the mapping θ2 : K ∗ → A(L, K ∗ ) | a → ρa , where ρa is defined as follows: If a = (2, 3), then ρa : L → K ∗ | x → a. If a = (2, 3), then  a, if x = (0, 0); ρa : L → K ∗ | x → (3, 4), if x = (0, 0). RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

54

KEHAYOPULU

The mapping θ2 is well defined. Really, 1. Let a ∈ K ∗ . Then ρa ∈ A(L, K ∗ ). Really, 1.1. Let a = (2, 3). If x ∈ L, then ρa (x) := a ∈ K ∗ . If x, y ∈ L, x = y, then ρa (x) := a := ρa (y). 1.2. Let a = (2, 3). If x = (0, 0), then ρa (x) := a ∈ K ∗ . If x = (0, 0), then ρa (x) := (3, 4) ∈ K ∗ . Let x, y ∈ L, x = y. If x = y = (0, 0), then ρa (x) := a := ρa (y). If x = y = (0, 0), then ρa (x) := (3, 4) := ρa (y). 2. Let a, b ∈ K ∗ , a = b. Let us prove that ρa = ρb . Really, let x ∈ L. Then 2.1. If a = b = (2, 3), then ρa (x) := a = b := ρb (x). 2.2. Let a = b = (2, 3). If x = (0, 0), then ρa (x) := a = b := ρb (x). If x = (0, 0), then ρa (x) := (3, 4) := ρb (x). (III) Consider the mapping f : {(a, b) | a, b ∈ K ∗ , a ∧k b = (0, 0)} → L | (a, b) → (0, 0). The mapping f is well defined. Really, if a, b ∈ K ∗ , a ∧K b = (0, 0), then f (a, b) := (0, 0) ∈ L. If a, b, c, d ∈ K ∗ , a ∧K b = c ∧K d = (0, 0), then f (a, b) := (0, 0) := f (c, d). (IV) Conditions (C1)–(C7) mentioned in the Theorem are satisfied. (C1) Let a, b ∈ K ∗ , a ∧K b = (0, 0). Let us prove that θ1 (a)θ1 (b) = λf (a,b) . By hypothesis, f (a, b) := (0, 0), f (b, a) := (0, 0). On the other hand, θ1 (a)θ1 (b) = λ(0,0) . Consider the following cases: 1. a = (0, 4), b = (2, 3), 2. a = (0, 4), b = (3, 0), 3. a = (2, 3), b = (0, 4), 4. a = (2, 3), b = (3, 0), 5. a = (3, 0), b = (0, 4), 6. a = (3, 0), b = (2, 3). 1. Let a = (0, 4), b = (2, 3). Then θ1 (a) := λ(0,0) , θ1 (b) := iL , θ1 (a)θ1 (b) = λ(0,0) iL = λ(0,0) . 2. Let a = (0, 4), b = (3, 0). Then θ1 (a) := λ(0,0) , θ1 (b) := λ(0,0) , θ1 (a)θ1 (b) = λ(0,0) λ(0,0) = λ(0,0) ∧ λ(0,0) = λ(0,0) (cf. Lemma 1). 3. Let a = (2, 3), b = (0, 4). Then by item 1, θ1 (a)θ1 (b) = θ1 (b)θ1 (a) = λf (b,a) = λf (a,b) . 4. Let a = (2, 3), b = (3, 0). Then θ1 (a) := iL , θ1 (b) := λ(0,0) , and θ1 (a)θ1 (b) = iL λ(0,0) = λ(0,0) . The proofs of items 5 and 6 are similar. (C2) Let a, b ∈ K ∗ , a ∧K b = (0, 0). Let us prove that θ1 (a ∧K b) = θ1 (a)θ1 (b). In fact, if a = b, then θ1 (a ∧K b) = θ1 (a ∧K a) = θ1 (a) = θ1 (a)θ1 (a) (cf. Lemma 11) = θ1 (a)θ1 (b). Let a = b. Consider the following cases: 1. a = (0, 4), b = (3, 4), 2. a = (2, 3), b = (3, 4), 3. a = (3, 0), b = (3, 4), 4. a = (3, 4), b = (0, 4), 5. a = (3, 4), b = (2, 3), 6. a = (3, 4), b = (3, 0). 1. Let a = (0, 4), b = (3, 4). Then a ∧K b = (0, 4), θ1 (a) := λ(0,0) , θ1 (b) := iL , θ1 (a ∧K b) := λ(0,0) = λ(0,0) iL = θ1 (a)θ1 (b). 2. Let a = (2, 3), b = (3, 4). Then a ∧K b = (2, 3), θ1 (a) := iL , θ1 (b) := iL , θ1 (a ∧K b) := iL = iL iL = θ1 (a)θ1 (b). 3. Let a = (3, 0), b = (3, 4). Then a ∧K b = (3, 0), θ1 (a) := λ(0,0) , θ1 (b) := iL , θ1 (a ∧K b) := λ(0,0) = λ(0,0) iL = θ1 (a)θ1 (b). 4. Let a = (3, 4), b = (0, 4). Then by item 1, θ1 (a ∧K b) = θ1 (b ∧K a) = θ1 (b)θ1 (a) = θ1 (a)θ1 (b). The proofs of items 5 and 6 are similar. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

IDEAL EXTENSIONS OF LATTICES

(C3) Let a, b ∈ K ∗ , c ∈ L. a ∨K b ∈ K ∗ .

55

In fact, since a, b ∈ K ∗ , we have

Then ρa∨K b (c) = ρa (c) ∨K b.

1. Let a ∨K b = (2, 3). Then ρa∨K b (c) = a ∨K b = (2, 3). Since a ∈ K ∗ , a = (2, 3), we have ρa (c) := a = (2, 3). Since K ∗  b ≤ a ∨K b = (2, 3), we have b = (2, 3). On the other hand, ρa (c) ∨K b = (2, 3) ∨K (2, 3) = (2, 3). 2. Let a ∨K b = (2, 3). 2.1. Let c = (0, 0). Then ρa∨K b (c) := a ∨K b. Then we have the following case: 2.1.1. If a = (2, 3), then ρa (c) := a and ρa (c) ∨K b = a ∨k b. Hence ρa∨K b (c) = ρa (c) ∨K b. 2.1.2. If a = (2, 3) then c = (0, 0), we have ρa (c) := a, and ρ(c) ∨K b = a ∨K b. Therefore = ρa (c) ∨K b.

ρa∨K b (c)

2.2. Let c = (0, 0). Then ρa∨K b (c) := (3, 4). 2.2.1. Let a = (2, 3). If b = (2, 3), then a ∨K b = (2, 3), which is impossible. Thus we have b = (2, 3). Then ρa (c) := a = b. Since ρa (c), b ∈ K ∗ , we have ρa (c) ∨K b = (3, 4). Thus we have ρa∨K b (c) = ρa (c) ∨K b. 2.2.2. Let a = (2, 3). Since c = (0, 0), we have ρa (c) := (3, 4) ≥ b. Then ρa (c) ∨K b = (3, 4). Therefore ρa∨K b (c) = ρa (c) ∨K b. (C4) Let a, b ∈ L, c ∈ K ∗ . Then ρc (a ∨L b) = ρρ

c (b)

1. Let c = (2, 3). Then

ρc (b)

:= c,

c ρρ (b) (a)

=

ρc (a)

(a). Really, we obtain the following assertion. := c, ρc (a ∨L b) := c.

2. Let c = (2, 3). c (b)

2.1. Let b = (0, 0). Then ρc (b) := c, ρρ

(a) = ρc (a), a ∨L b = a, and ρc (a ∨L b) = ρc (a). c (b)

2.2. Let b = (0, 0). Then ρc (b) := (3, 4), ρρ

(a) = ρ(3,4) (a).

2.2.1. If a = (0, 0), then ρ(3,4) (a) := (3, 4), a ∨L b = b = (0, 0), ρc (a ∨L b) := (3, 4). 2.2.2. If a = (0, 0), then ρ(3,4) (a) := (3, 4), a ∨L b = (0, 0) and ρc (a ∨L b) = (3, 4). (C5) Let a ∈ K ∗ , b ∈ L. Then ρa (λa (b)) = a. Let us prove this equality. 1. Let a = (2, 3). Then λa (b) := iL (b) = b ∈ L and ρa (λa (b)) := a. 2. Let a = (2, 3). 2.1. Let a = (3, 4). Then λa (b) := iL (b) = b, ρa (λa (b)) = ρa (b). If b = (0, 0), then ρa (b) := a. If b = (0, 0), then ρa (b) := (3, 4) = a. 2.2. Let a = (3, 4). Then λa := λ(0,0) and λa (b) = λ(0,0) (b) = (0, 0) ∧L b = (0, 0). Since a = (2, 3) and λa (b) = (0, 0), we have ρa (λa (b)) := a. (C6) Let a ∈ L, b ∈ K ∗ . Let us prove that λρ

b (a)

ρb (a)

1. If b = (2, 3), then ρb (a) := b, and λ

(a) = a.

(a) = λb (a) := a.

2. Let b = (2, 3). b (a)

2.1. Let a = (0, 0). Since λρ b (a)

λρ

b (a)

(a) = λρ

is a translation of L, we have b (a)

(a ∧L a) = λρ

b (a)

2.2. Let a = (0, 0). Then ρb (a) := (3, 4), λρ K ∗,

b (a)

(a) ∧L a = λρ

(a) ∧L (0, 0) = (0, 0) = a.

b (a)

:= iL and λρ

(a) = a.

a ∧K b = (0, 0). Then = a. In order to prove this assertion, consider (C7) Let a, b ∈ two cases. 1. Let a = (2, 3). Since f (a, b) ∈ L, we have ρa (f (a, b)) = a. ρa (f (a, b))

2. Let a = (2, 3). Since a, b ∈ K ∗ , a ∧K b = (0, 0), we have L  f (a, b) := (0, 0). Then, since a = (2, 3), we have ρa (f (a, b)) = a. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

56

KEHAYOPULU

According to this theorem, the set V := L ∪ K ∗ with the operations “∨” and “∧” on V defined below is an extension of L by K: ⎧ (S1) a ∨L b, if a, b ∈ L; ⎪ ⎪ ⎪ ⎨ b ∗ ρ (a), if a ∈ L, b ∈ K ; (S2) a∨b:= a ∗ ⎪ if a ∈ K , b ∈ L; ρ (b), ⎪ (S3) ⎪ ⎩ a ∨K b, if a, b ∈ K ∗ , (S4) ⎧ (P 1) ⎪ a ∧L b, if a, b ∈ L; ⎪ ⎪ ⎪ ⎪ b ⎪ (P 2) if a ∈ L, b ∈ K ∗ ; ⎨λ (a), ∗ a ∧ b : = λa (b), if a ∈ K , b ∈ L; (P 3) ⎪ ⎪ ∗ ⎪ ⎪f (a, b), if a, b ∈ K , a ∧K b = 0; (P 4) ⎪ ⎪ ⎩ ∗ a ∧K b, if a, b ∈ K , a ∧K b = 0. (P 5) Let a ∈ L, b ∈ K ∗ . If b = (2, 3), then ρb (a) := b = (2, 3). Let b = (2, 3). If a = (0, 0), then ρb (a) := b. If a = (0, 0), then ρb (a) := (3, 4). Let a ∈ K ∗ , b ∈ L. If a = (2, 3), then ρa (b) := a = (2, 3). Let a = (2, 3). If b = (0, 0), then a ρ (b) := a. If b = (0, 0), then ρa (b) := (3, 4). Let a ∈ L, b ∈ K ∗ . If b = (2, 3) or b = (3, 4), then λb (a) := iL (a) = a. If b = (0, 4) or b = (3, 0), then b λ (a) := λ(0,0) (a) = (0, 0) ∧L a = (0, 0). Let a ∈ K ∗ , b ∈ L. If a = (2, 3) or a = (3, 4), then λa (b) := iL (b) = b. If a = (0, 4) or a = (3, 0), then a λ (b) := λ(0,0) (b) = (0, 0) ∧L b = (0, 0). If a, b ∈ K ∗ , a ∧K b = 0, then f (a, b) := (0, 0). As a consequence, the operations “∨” and “∧” take the form ⎧ ⎪ a ∨L b, if a, b ∈ L; ⎪ ⎪ ⎪ ⎪ ⎪ (2, 3), if a ∈ L, b = (2, 3); ⎪ ⎪ ⎪ ⎪ b, if L  a = (0, 0), b = (2, 3); ⎪ ⎪ ⎪ ⎨ (3, 4), if L  a = (0, 0), b = (2, 3); a∨b := ⎪(2, 3), if a = (2, 3), b ∈ L; ⎪ ⎪ ⎪ ⎪ ⎪ a, if a = (2, 3), L  b = (0, 0); ⎪ ⎪ ⎪ ⎪ ⎪ (3, 4), if a = (2, 3), L  b = (0, 0); ⎪ ⎪ ⎩ a ∨K b, if a, b ∈ K ∗ , ⎧ ⎪ a ∧ b, if a, b ∈ L; ⎪ ⎪ L ⎪ ⎪ a, if a ∈ L, b = (2, 3); ⎪ ⎪ ⎪ ⎪ ⎪ a, if a ∈ L, b = (3, 4); ⎪ ⎪ ⎪ ⎪ ⎪ (0, 0), if a ∈ L, b = (0, 4); ⎪ ⎪ ⎪ ⎪ ⎪(0, 0), if a ∈ L, b = (3, 0); ⎨ a ∧ b : = b, if a = (2, 3), b ∈ L; ⎪ ⎪ ⎪ b, if a = (3, 4), b ∈ L; ⎪ ⎪ ⎪ ⎪ ⎪ (0, 0), if a = (0, 4), b ∈ L; ⎪ ⎪ ⎪ ⎪ ⎪ (0, 0), if a = (3, 0), b ∈ L; ⎪ ⎪ ⎪ ⎪ ⎪ (0, 0), if a, b ∈ K ∗ , a ∧K b = 0; ⎪ ⎪ ⎩ a ∧K b, if a, b ∈ K ∗ , a ∧K b = 0. Finally, note that RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

IDEAL EXTENSIONS OF LATTICES

57

1. x ≤V (2, 3) for each x ∈ L. This takes place because x ∨ (2, 3) = (2, 3) for all x ∈ L. 2. (0, 0) ≤V (0, 4) and (0, 0) ≤V (3, 0). Really, (0, 0) ∨ (0, 4) = (0, 4) since (0, 4) = (2, 3), and (0, 0) ∨ (3, 0) = (3, 0) if (3, 0) = (2, 3). 3. x ∨ (0, 4) = (3, 4) and x ∨ (3, 0) = (3, 4) for all x ∈ L, x = (0, 0). Really, let x ∈ L, x = (0, 0). 1. Since (0, 4) = (2, 3), we have x ∨ (0, 4) = (3, 4). 2. Since (3, 0) = (2, 3), we have x ∨ (3, 0) = (3, 4). The diagram of V is the following:

REFERENCES 1. A. H. Clifford, “Extensions of Semigroups,” Trans. Amer. Math. Soc. 68, 165–173 (1950). 2. A. H. Clifford and G. B. Preston, The Algebraic Theory of Semigroups (American Math. Soc., Providence, Rhode Island, 1977) Mathematical Surveys 1, No. 7. 3. M. Petrich, Introduction to Semigroups (Merrill Research and Lecture Series. Charles E. Merrill Publishing Co. Columbus, Ohio, 1973). 4. A. J. Hulin, “Extensions of Ordered Semigroups,” Semigroup Forum 2 (4), 336–342 (1971). 5. A. J. Hulin, “Extensions of Ordered Semigroups,” Czechoslovak Math. J. 26 (1), 1–12 (1976). 6. Fr. T. Christoph (jr.), “Ideal Extensions of Topological Semigroups,” Canad. J. Math. 22 (6), 1168–1175 (1970). 7. J. A. Hildebrant, “Ideal Extensions of Compact Reductive Semigroups,” Semigroup Forum 25 (3–4), 283– 290 (1982). 8. N. Kehayopulu, “Ideal Extensions of Ordered Sets,” Int. J. Math. and Math. Sci., No. 53 (2847–2861) (2004). 9. N. Kehayopulu, J. S. Ponizovskii, and K. P. Shum, “Retract Extensions of Ordered Sets,”Zap. Nauchn. Semin. POMI 321 (205–212) (2005). 10. N. Kehayopulu and K. P. Shum, “Equivalent Extensions of Ordered Sets,” Algebras Groups Geom. 20 (4), 387–402 (2003). 11. N. Kehayopulu and M. Tsingelis, “The Ideal Extensions of Ordered Semigroups” Comm. Algebra 31 (10), 4939–4969 (2003). 12. N. Kehayopulu and P. Kiriakuli, “The Ideal Extensions of Lattices,” Simon Stevin 64 (1), 51–60 (1990). ´ ¨ 13. G. Szasz, “Die Translationen der Halbverbande,” Acta Sci. Math. 17, 165–169 (1956). ¨ ´ and J. Szendrei, “Uber ¨ 14. G. Szasz die Translationen der Halbverbande,” Acta Sci. Math. 18, 44–47 (1957). ´ ¨ 15. G. Szasz, “Translationen der Verbande,” Acta Fac. Rer. Nat. Univ. Comenianae Math. 5, 449–453 (1961). RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009

58

KEHAYOPULU

¨ ¨ 16. M. Kolibiar, “Bemerkungen uber Translationen der Verbande,” Acta Fac. Rer. Nat. Univ. Comenianae Math. 5, 455–458 (1961). 17. G. Birkhoff, Lattice Theory (Providence, Rhode Island, 1967), American Math. Soc. Colloquium Publ., Vol. 25.

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 53 No. 2 2009