IDEAL EXTENSIONS OF LATTICES

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isomorphic to Q. To avoid confusion, for the Rees quotient we use the ...... Charles E. ... [7] Hildebrant J.A. Ideal extensions of compact reductive semigroups ...
IDEAL EXTENSIONS OF LATTICES NIOVI KEHAYOPULU

Abstract Following the well known Schreier’s 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 L0 which is isomorphic to L and the Rees quotient V |L0 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.

1. 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’s extension (or simply the extension) of H by K. Following the Schreier’s 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 2000 Mathematics Subject Classification: 06B05 Keywords: Translation, inner translation, (ideal) extension of a lattice L by a lattice K (K having a least element).

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V which have and ideal S 0 which is isomorphic to S and the Rees quotient V |S 0 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], of topological semigroups in [6,7]. We are often interested in building more complex semigroups, lattices, ordered sets, ordered or topological semigroups out of some of ”simpler” structure 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 0 which is isomorphic to S and the Rees quotient V |S 0 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. Following the ideal extensions of ordered semigroups considered in [11], 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 to the theorem. Exactly as in the case of semigroups (ordered 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 L0 which is isomorphic to L and the Rees quotient V |L0 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 the analogous of the ideal extensions of semigroups (or ordered semigroups) for lattices. We present the corresponding concepts and the analogous results. We do

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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]). 2. 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) If a ∈ L and V 3 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 . Since the operation ∧” is commutative, we have λ1 λ2 = λ2 λ1 .

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Lemma 2. Let (V, ∨, ∧) be a lattice and L 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 ” t ” and ” u ” on V |L as follows: t : V |L × V |L → V |L | (x, y) → x t y, where

   x∨y     x x t y :=  y      0

if x, y ∈ V \L if x ∈ V \L, y = 0 if y ∈ V \L, x = 0 if x = y = 0.

u : V |L × V |L → V |L | (x, y) → x u y, where

( x u y :=

x ∧ y if x ∧ y ∈ V \L 0

otherwise.

Then the set V |L endowed with the operations ”t” and ”u” is a lattice and the element 0 is the least element in V |L. Proof. (A) We first prove that the operations ” u ” and ” u ” are well defined and they are commutative. To prove that the operation ” t ” 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. Let us prove 1. Conditions 2–8 can be proved similarly. 1. Let x, y, z ∈ V \L. Then xty := x∨y ∈ V \L and (xty)tz := (xty)∨z = (x∨y)∨z. Also y t z := y ∨ z ∈ V \L and x t (y t z) := x ∨ (y t z) = x ∨ (y ∨ z). Thus we have (x t y) t z = x t (y t z). (B) For each x ∈ V |L, we have x u 0 = 0 u x = 0. Indeed: Let x ∈ V |L. Since x ∈ V , 0 ∈ L and L is an ideal of V , we have x ∧ 0 ∈ L. Since x ∧ 0 ∈ / V \L, we have x u 0 := 0. Similarly, 0 u x = 0. 4

(C) The operation ” u ” is associative. In fact: 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 u y := x ∧ y and y u z := y ∧ z. Since (x ∧ y) ∧ z ∈ V \L, we have (x ∧ y) u z := (x ∧ y) ∧ z. Then (x u y) u z = (x ∧ y) ∧ z. Since x ∧ (y ∧ z) ∈ V \L, we have x u (y ∧ z) := x ∧ (y ∧ z). Then x u (y u z) = x ∧ (y ∧ z). Therefore we have (x u y) u z = x u (y u z). 2. Let x ∧ y ∧ z ∈ / V \L. We consider the following cases: 2.1. Let x ∧ y, y ∧ z ∈ V \L. Then x u y := x ∧ y and y u z := y ∧ z. Since (x ∧ y) ∧ z ∈ / V \L, we have (x ∧ y) u z := 0, then (x u y) u z = 0. Since x ∧ (y ∧ z) ∈ / V \L, we have x u (y ∧ z) := 0, then x u (y u z) = 0. Hence we have (x u y) u z = x u (y u z). 2.2. Let x ∧ y ∈ V \L and y ∧ z ∈ / V \L. Then x u y := x ∧ y and y u z := 0. Then (x u y) u z = (x ∧ y) u z and x u (y u z) = x u 0 = 0. Since (x ∧ y) ∧ z ∈ / V \L, we have (x ∧ y) u z := 0. Thus we have (x u y) u z = x u (y u 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 u y := 0 and y u z := 0, we have (x u y) u z = 0 u z = 0 and x u (y u z) = x u 0 = 0. (D) For each x, y ∈ V |L, we have x u (x t y) = x. In fact: Let x, y ∈ V |L. Then 1. Let x, y ∈ V \L. Then V |L 3 x t y := x ∨ y, so x u (x t y) = x u (x ∨ y). Since x ∧ (x ∨ y) = x ∈ V \L, we have x u (x ∨ y) := x ∧ (x ∨ y) = x. Hence x u (x t y) = x. 2. Let x ∈ V \L, y = 0. Then V |L 3 x t y := x, and x u (x t y) = x u x. Since V \L 3 x = x ∧ x, we have x u x = x ∧ x = x. Hence x u (x t y) = x. 3. Let x = 0, y ∈ V \L. Then V |L 3 x t y := y. Hence x u (x t y) = x u y = 0 u y = 0 = x. 4. Let x = y = 0. Then V |L 3 x t y := 0, and x u (x t y) = x u 0 = 0 = x. (E) For each x, y ∈ V |L, we have x t (x u y) = x. In fact: Let x, y ∈ V |L. Clearly x ∧ y ∈ V . 1. Let x ∧ y ∈ V \L. Then V \L 3 x u y := x ∧ y, and x t (x u y) = x t (x ∧ y). If x = 0, then x u y = 0 u y = 0 ∈ L which is impossible. Thus we have x ∈ V \L. Since x, x ∧ y ∈ V \L, we have x t (x ∧ y) := x ∨ (x ∧ y) = x. Therefore x t (x u y) = x. 2. Let x ∧ y ∈ / V \L. Since x u y := 0 ∈ V |L, we have x t (x u y) = x t 0. If x ∈ V \L, then x t 0 := x, and x t (x u y) = x. If x = 0, then x t (x u y) = 0 t 0 = 0 = x.

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Lemma 3. Let (V, ∧, ∨) be a lattice and L an ideal of V. For each v ∈ V , we consider 5

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. Indeed, if x, y ∈ L, then µv (x ∧ y) = v ∧ (x ∧ y) = (v ∧ x) ∧ y = µv (x) ∧ y.

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In this paper we use the following notations: Notation 1. Let (L, ∧, ∨) be a lattice and t ∈ L. We denote by λt the translation of L defined by 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 by π : L → T (L) | t → λt . Remark 1. The mapping π is (1–1). Indeed, 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, K 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 L0 of V such that L0 is isomorphic to L and the Rees quotient V |L0 is isomorphic to K (in symbol, L ≈ L0 , V |L0 ≈ K). In the following we denote by A(L, K ∗ ) the set of all mappings of L into K ∗ . A mapping f of A into B, as usually, is denoted by f : A → B. Unless otherwise is stated, we denote the order on a lattice L by ” ≤L ” and the operations of supremum and infimum on L by ” ∨L ” and ” ∧L ”, respectively. We denote the supremum and the infimum on V |L by ” t ” and ” u ”, respectively. 3. The main result Here we give 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 6

(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, K a lattice with 0 and L ∩ K ∗ = ∅. Suppose 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 each a, b ∈ K ∗ , a ∧K b = 0 (C2) θ1 (a ∧K b) = θ1 (a)θ1 (b) for each a, b ∈ K ∗ , a ∧K b 6= 0 (C3) ρa∨K b (c) = ρa (c) ∨K b for each a, b ∈ K ∗ , c ∈ L c (b)

(C4) ρc (a ∨L b) = ρρ

(a) for each a, b ∈ L, c ∈ K ∗

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

(C6) λρ

(a) = a for each a ∈ L, b ∈ K ∗

(C7) ρa (f (a, b)) = a for each a, b ∈ K ∗ , a ∧K b = 0. Let V := L ∪ K ∗ . We define operations ” ∨ ” and ” ∧ ” on V as follows: ∨ : V × V → V | (a, b) → a ∨ b, where

   a ∨L b     ρb (a) a ∨ b :=  ρa (b)      a ∨K b

if a, b ∈ L

(S1)

if a ∈ L, b ∈ K ∗

(S2)

K ∗,

(S3)

if a ∈

b∈L

if a, b ∈ K ∗

(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

if a, b ∈ L

(P 1)

if a ∈ L, b ∈ K ∗

(P 2)

K ∗,

(P 3)

if a ∈

if a, b ∈

b∈L

K ∗,

a ∧K b = 0

(P 4)

if a, b ∈ K ∗ , a ∧K b 6= 0

(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 7

θ2 : K ∗ → A(L, K ∗ ) | a → ρa f : {(a, b) | a, b ∈ K ∗ , a ∧K b = 0} → L such that conditions (C1)−(C7) mentioned on the first part of the theorem are satisfied. Furthermore, the set L ∪ K ∗ endowed with the operations ” ∨ ” and ” ∧ ” defined on the first part of the theorem is a lattice, and L ∪ K ∗ ≈ V . Proof. (A) The operation ” ∨ ” is well defined. Indeed: 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, since 0 is the zero of K, we have a = b = 0 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. Indeed: 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 . Indeed: 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 b ∨ c = ρc (b) ∈ K ∗ , c (b)

we have ρb∨c = ρρ ρb∨c (a)

=

we have

c ρρ (b) (a).

ρc (a

. Since ρb∨c , ρρ

c (b)

: L → K ∗ , ρb∨c = ρρ

Then we have a ∨ (b ∨ c) =

∨L b) =

c ρρ (b) (a).

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c ρρ (b) (a).

c (b)

and a ∈ L, we have

On the other hand, by (C4),

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 = ρρ ρa∨b (c)

=

a (b)

a ρρ (b) (c).

. Since ρa∨b , ρρ

a (b)

: L → K ∗ , ρa∨b = ρρ

Thus we have (a ∨ b) ∨ c =

a ρρ (b) (c).

a (b)

and c ∈ L, we have

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 have a (b)

ρa (c ∨L b) = ρρ

(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 ∗ , b (a)

we have ρa∨b = ρρ ρa∨b (c)

=

b ρρ (a) (c).

. Since ρa∨b , ρρ

b (a)

: L → K ∗ , ρa∨b = ρρ

Thus we have (a∨b)∨c =

b ρρ (a) (c).

b (a)

Since b ∈

and c ∈ L, we have

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 ρb∨c = ρρ b (c)

and a ∈ L, we have ρb∨c (a) = ρρ ρb (a)

ρb (c ∨L a) = ρ Since

ρb

:L→

K∗

b (c)

. Since ρb∨c , ρρ

b (c)

: L → K ∗ , ρb∨c = ρρ

(a). Since c, a ∈ L, b ∈ K ∗ , by (C4), we have b (c)

(c). Since a, c ∈ L, b ∈ K ∗ , by (C4), we have ρb (a ∨L c) = ρρ and c ∨L a = a ∨L c we get

(a ∨ b) ∨ c = ρρ

b (a)

b (c)

ρb (c

∨L a) =

ρb (a

(a).

∨L c). Thus we have 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, since ρb∨c , ρb∨K c : L → K ∗ and a ∈ L, we get ρb∨c (a) = ρb∨K c (a). 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)). Also 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 9

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. Indeed: 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 6= 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. Indeed: 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 := 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 ∗ . Then (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). Indeed: 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 10

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 6= 0. By (P5), we have a ∧ b := a ∧K b. Since b, a ∈ K ∗ and b ∧K a 6= 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. In fact: 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) = λc (a ∧L b). a ∧ (b ∧ c) := a ∧L (b ∧ c) = a ∧L λc (b). On the other hand, 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). 11

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) (cf. Lemma 1), and (λc λb )(a) = λf (b,c) (a). 5.2. Let b ∧K c 6= 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). Indeed: Since b, c ∈ K ∗ and b ∧K c 6= 0, by (C2), we have θ1 (b ∧K c) = θ1 (b)θ1 (c). Then λb∧K c = λb λc = λc λb (cf. 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)) := (λc λa )(b), a ∧ (b ∧ c) := λa (b ∧ c) = λa (λc (b)) := (λa λc )(b). Since λc , λa ∈ T (L), we have λc λa = λa λc (cf. Lemma 1), then (λc λa )(b) = (λa λc )(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. Indeed: Since a, b ∈ K ∗ and a ∧K b = 0, by (C1), we have θ1 (a)θ1 (b) = λf (a,b) , then λa λb = λf (a,b) , and (λa λb )(c) = λf (a,b) (c) := f (a, b) ∧L c. 7.2. Let a ∧K b 6= 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). Indeed: Since a, b ∈ K ∗ and a ∧K b 6= 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).

12

8. Let a, b, c ∈ K ∗ . We consider the following cases: 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)) . Indeed: 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 . Then, 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 ) (cf. 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 6= 0. Then a ∧ b := f (a, b) ∈ L, (a ∧ b) ∧ c := λc (a ∧ b) = λc (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), it is enough 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)). 13

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 6= 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 6= 0, b ∧K c = 0. The proof is similar to that of 8.2. 8.4. Let a ∧K b 6= 0, b ∧K c 6= 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) . Indeed: 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 6= 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),(C2), we get λf (a, b∧K c) = (λa λb )λc . On the other hand, (λa λb )λc = λa (λb λc ) (cf. Lemma 1). (b) Let (a ∧ b) ∧K c 6= 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) 6= 0. Then a ∧ (b ∧ c) := a ∧K (b ∧ c) = a ∧K (b ∧K c). (G) a ∧ (a ∨ b) = a for all a, b ∈ V . Indeed: 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) := λa∨b (a) = λρ (C6). If a ∈

K ∗,

b ∈ L, then a ∨ b :=

ρa (b)

∈

K ∗.

b (a)

= a by

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 6= 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 6= 0, we have a ∧ (a ∨ b) = a ∧K (a ∨ b) = a. 14

(H) a ∨ (a ∧ b) = a for each a, b ∈ V . In fact: 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). Besides, a ∨L λb (a) = a. Indeed, 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) ∈ L, and a ∨ (a ∧ b) := ρa (a ∧ b) = ρa (f (a, b)) = a by (C7) . If a ∧K b 6= 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. Then: 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, t, u). In fact: Since V := L ∪ K ∗ and L ∩ K ∗ = ∅, we have V \L = K ∗

(∗)

We consider the mapping: ( g : K → V |L | x →

x

if x ∈ K ∗

0V |L if x = 0K .

Because of (∗), the mapping g is well defined, it is an (1–1) and onto mapping. The mapping g is a homomorphism. In fact: g(x ∨K y) = g(x) t g(y) for each x, y ∈ K. Indeed: Let x, y ∈ K. Then 1. Let x, y ∈ K ∗ . Then x ∨K y ∈ K ∗ , g(x) := x, g(y) := y, g(x ∨K y) := x ∨K y. Since 15

x, y ∈ K ∗ = V \L, we have x ∨ y = x ∨K y (by (S4)) and x t y = x ∨ y (cf. Lemma 2). Then g(x ∨K y) = g(x) t 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 t 0V |L . Then we have g(x ∨K y) = x = x t 0V |L = g(x) t g(y). 3. Let x = 0K , y ∈ K ∗ . The proof is similar to that of 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 t 0V |L = g(x) t g(y). g(x ∧K y) = g(x) u g(y) for each x, y ∈ K. Indeed: Let x, y ∈ K. Then 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 u y := 0V |L (cf. Lemma 2). Then g(x ∧K y) = 0V |L = x u y = g(x) u g(y). 1.2. Let x ∧K y 6= 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 u y = x ∧ y (cf. Lemma 2). Then g(x ∧K y) = x ∧k y = x ∧ y = x u y = g(x) u 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 u 0V |L = 0V |L . Then we have g(x ∧K y) = 0V |L = x u 0V |L = g(x) u g(y). 3. Let x = 0K , y ∈ K ∗ . The proof is similar to that of 2. 4. Let x = y = 0V |L . Then g(x) = g(y) = g(x ∧K y) = 0V |L . Thus we have g(x ∧K y) = 0V |L = 0V |L u 0V |L = g(x) u g(y). The converse statement: Let (V, ∨V , ∧V ) be an extension of (L, ∨L , ∧L ) by (K, ∨K , ∧K ). Then, there exists an ideal I of V , an isomorphism φ : L → I and an isomorphish ψ : K → V |I. (I) We consider the mapping: h : K ∗ → V |I\{0V |I } | x → ψ(x). 16

1. The mapping h is well defined. Indeed: 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 . Then, since ψ is a homomorphism, we have ψ(0K ) = ψ(0K ∧K t) = ψ(0K ) u ψ(t) = ψ(0K ) u 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 a (1–1) mapping. 3. h is onto. Indeed: 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) We 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. Indeed: Let a ∈ K ∗ . Since V is a lattice, I an ideal of V and h(a) ∈ V |I\{0V |I } = V \I ⊆ V , the mapping µh(a) is a translation of I (cf. 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. Indeed: 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)) (since φ−1 is a homomorphism) = (φ−1 µh(a) φ)(x) ∧L y. Let now a, b ∈ K ∗ such that a = b. Then h(a) = h(b), µh(a) = µh(b) (cf. Lemma 3), thus φ−1 µh(a) φ = φ−1 µh(b) φ. (III) We 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. Indeed: 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, 17

since V 3 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 }, then 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 ∗ 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)), that is ρa = ρb . (IV) We 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. In fact: 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. Indeed: h(a) u h(b) = ψ(a) u ψ(b) = ψ(a ∧K b) = ψ(0K ) = 0V |I (since ψ is an isomorphism) . By Lemma 2, we have h(a) ∧V h(b) ∈ / V \I. Since h(a) ∧V h(b) ∈ I, we have φ−1 (h(a) ∧V h(b)) ∈ L. Let a, b, c, d ∈ K ∗ 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). Conditions (C1)–(C7) mentioned on 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) . In fact: 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) . Indeed: 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). 18

(C2) Let a, b ∈ K ∗ , a ∧K b 6= 0. Then θ1 (a ∧K b) = θ1 (a)θ1 (b). In fact: Since a, b ∈ K ∗ , a ∧K b 6= 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, φ−1 µh(a∧K b) φ = φ−1 µh(a) µh(b) φ. Indeed: 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). Thus we have (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). Indeed: 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 we have 0V |I 6= h(a ∧K b) = ψ(a ∧K b) = ψ(a) u ψ(b) (since ψ : K → V |I is an isomorphism) = h(a) u h(b). Since h(a), h(b) ∈ V |I and h(a) u h(b) 6= 0V |I , by Lemma 2, we have h(a) u h(b) = h(a) ∧V h(b). Therefore we have 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. In fact: 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. Indeed: It is enough 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))) t ψ(b) (since ψ is an homomorphism)

=

h(h−1 (h(a) ∨V φ(c))) t h(b). 19

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))) t h(b) = h(h−1 (h(a) ∨V φ(c))) ∨V h(b). Then we have 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) t 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) t h(b)) ∨V φ(c) = (ψ(a) t 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). In fact: 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 c (b)

K ∗ 3 ρρ

(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)). It is enough 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. In fact: Since a ∈ K ∗ and b ∈ L, we have L 3 λa (b) := (φ−1 µh(a) φ)(b) = φ−1 (µh(a) φ(b)) = φ−1 (h(a) ∧V φ(b)), then 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. 20

b (a)

(C6) Let a ∈ L, b ∈ K ∗ . Then λρ

:= a. In fact:

Since a ∈ L, b ∈ K ∗ , we have ρb (a) := h−1 (h(b) ∨V φ(a)) ∈ K ∗ , and b (a)

λρ

b (a))

(a) : = φ−1 (µh(ρ

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. Indeed: Since a, b ∈ K ∗ , a ∧K b = 0, we have f (a, b) := φ−1 (h(a) ∧V h(b)) ∈ L, then 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 first part of the theorem, the set L ∪ K ∗ , with the operations ” ∨ ” and ” ∧ ” on V defined in the first part of the theorem, is a lattice and it is an extension of L by K. Furthermore, (L ∪ K ∗ , ∨, ∧) ≈ (V, ∨V , ∧V ). In fact: We consider the mapping: ( g : L ∪ K∗ → V | a →

ϕ(a)

if a ∈ L

h(a)

if a ∈ K ∗ .

One can easily prove that the mapping g is well defined, it is (1–1) and an onto mapping. The mapping g is a homomorphism. In fact: 1. For each a, b ∈ L ∪ K ∗ , we have g(a ∨ b) = g(a) ∨V g(b). Indeed: Let a, b ∈ L ∪ K ∗ . We 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, thus we have 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)), 21

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) t h(b). Indeed: Since a ∨K b ∈ K ∗ , we have h(a ∨K b) := ψ(a ∨K b) =

ψ(a) t ψ(b) (since ψ is a homomorphism)

=

h(a) t h(b) (since a, b ∈ K ∗ ).

Since h(a), h(b) ∈ V |I\{0V |I } = V \I, by Lemma 2, we get h(a) t h(b) = h(a) ∨V h(b). Therefore, we have g(a ∨ b) = h(a ∨K b) = h(a) t h(b) = h(a) ∨V h(b) = g(a) ∨V g(b). 2. For each a, b ∈ L ∪ K ∗ , we have g(a ∧ b) = g(a) ∧V g(b). Indeed: Let a, b ∈ L ∪ K ∗ . We 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). 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). The case a ∈ K ∗ , b ∈ L can be proved similarly. Let a, b ∈ K ∗ . Then g(a) := h(a), g(b) := h(b). We consider the cases: 22

(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 6= 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)). Thus we have g(a ∧ b) = h(a) ∧V h(b) = g(a) ∧V g(b).

2

As an illustrative example of the Theorem we give the following Example. We 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 usual 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 the following:

23

(I) We consider the mapping: ( ∗

a

θ1 : K → T (L) | a → λ :=

iL

if a = (2, 3) or a = (3, 4)

λ(0,0)

otherwise.

(We denote by iL the identity mapping on L). The mapping θ1 is well defined. In fact: 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. also Notation 1). Similarly, if a, b ∈ K ∗ and a = b, then λa = λb . (II) We 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 6= (2, 3), then ( ρa

: L→

K ∗|

x→

a

if x = (0, 0)

(3, 4) if x 6= (0, 0).

The mapping θ2 is well defined. In fact: 1. Let a ∈ K ∗ . Then ρa ∈ A(L, K ∗ ). Indeed: 1.1. Let a = (2, 3). Then: If x ∈ L, then ρa (x) := a ∈ K ∗ . If x, y ∈ L, x = y, then ρa (x) := a := ρa (y). 1.2. Let a 6= (2, 3). Then: If x = (0, 0), then ρa (x) := a ∈ K ∗ . If x 6= (0, 0), then ρa (x) := (3, 4) ∈ K ∗ . Let x, y ∈ L, x = y. Then: If x = y = (0, 0), then ρa (x) := a := ρa (y). If x = y 6= (0, 0), then ρa (x) := (3, 4) := ρa (y). 2. Let a, b ∈ K ∗ , a = b. Then ρa = ρb . Indeed: Let x ∈ L. Then 2.1. If a = b = (2, 3), then ρa (x) := a = b := ρb (x). 2.2. Let a = b 6= (2, 3). Then: If x = (0, 0), then ρa (x) := a = b := ρb (x). If x 6= (0, 0), then ρa (x) := (3, 4) := ρb (x). (III) We 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. Indeed: 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. 24

(C1) Let a, b ∈ K ∗ , a ∧K b = (0, 0). Then θ1 (a)θ1 (b) = λf (a,b) . In fact: By hypothesis, we have f (a, b) := (0, 0), f (b, a) := (0, 0). On the other hand, θ1 (a)θ1 (b) = λ(0,0) . Indeed: We consider the 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 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 5 and 6 are similar. (C2) Let a, b ∈ K ∗ , a ∧K b 6= (0, 0). Then 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 1) = θ1 (a)θ1 (b). Let a 6= b. We consider the 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 , 25

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 1, θ1 (a ∧K b) = θ1 (b ∧K a) = θ1 (b)θ1 (a) = θ1 (a)θ1 (b). The proofs of 5 and 6 are similar. (C3) Let a, b ∈ K ∗ , c ∈ L. Then ρa∨K b (c) = ρa (c) ∨K b. In fact: Since a, b ∈ K ∗ , we have a ∨K b ∈ K ∗ . 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 ∗ 3 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 6= (2, 3). Then 2.1. Let c = (0, 0). Then ρa∨K b (c) := a ∨K b. Then we have the following: 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 6= (2, 3) then, since c = (0, 0), we have ρa (c) := a, and ρ(c) ∨K b = a ∨K b. Thus ρa∨K b (c) = ρa (c) ∨K b. 2.2. Let c 6= (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 6= (2, 3). Then ρa (c) := a 6= 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 6= (2, 3). Then, since c 6= (0, 0), we have ρa (c) := (3, 4) ≥ b. Then ρa (c) ∨K b = (3, 4). Thus ρa∨K b (c) = ρa (c) ∨K b. c (b)

(C4) Let a, b ∈ L, c ∈ K ∗ . Then ρc (a ∨L b) = ρρ 1. Let c = (2, 3). Then ρc (b) := c, ρρ

c (b)

(a). In fact:

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

2. Let c 6= (2, 3). Then 2.1. Let b = (0, 0). Then ρc (b) := c, ρρ

c (b)

(a) = ρc (a), a ∨L b = a, and

ρc (a ∨L b) = ρc (a). 2.2. Let b 6= (0, 0). Then ρc (b) := (3, 4), ρρ

c (b)

(a) = ρ(3,4) (a).

2.2.1. If a = (0, 0), then ρ(3,4) (a) := (3, 4), a∨L b = b 6= (0, 0), ρc (a∨L b) := (3, 4). 2.2.2. If a 6= (0, 0), then ρ(3,4) (a) := (3, 4), a∨L b 6= (0, 0), and ρc (a∨L b) = (3, 4). (C5) Let a ∈ K ∗ , b ∈ L. Then ρa (λa (b)) = a. In fact: 1. Let a = (2, 3). Then λa (b) := iL (b) = b ∈ L, and ρa (λa (b)) := a. 2. Let a 6= (2, 3). Then 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 6= (0, 0), then ρa (b) := (3, 4) = a. 26

2.2. Let a 6= (3, 4). Then λa := λ(0,0) , and λa (b) = λ(0,0) (b) = (0, 0) ∧L b = (0, 0). Since a 6= (2, 3) and λa (b) = (0, 0), we have ρa (λa (b)) := a. b (a)

(C6) Let a ∈ L, b ∈ K ∗ . Then λρ 1. If b = (2, 3), then

ρb (a)

(a) = a. In fact: b (a)

:= b, and λρ

(a) = λb (a) := a.

2. Let b 6= (2, 3). Then 2.1. Let a = (0, 0). Since λρ b (a)

λρ

b (a)

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

(a) = λρ = λ

ρb (a)

(a ∧L a) = λρ

b (a)

(a) ∧L a

(a) ∧L (0, 0) = (0, 0) = a. b (a)

2.2. Let a 6= (0, 0). Then ρb (a) := (3, 4), λρ

:= iL , and λρ

b (a)

(a) = a.

(C7) Let a, b ∈ K ∗ , a ∧K b = (0, 0). Then ρa (f (a, b)) = a. In fact: 1. Let a = (2, 3). Since f (a, b) ∈ L, we have ρa (f (a, b)) = a. 2. Let a 6= (2, 3). Since a, b ∈ K ∗ , a ∧K b = (0, 0), we have L 3 f (a, b) := (0, 0). Then, since a 6= (2, 3), we have ρa (f (a, b)) = a. By the Theorem, the set V := L ∪ K ∗ with the operations ” ∨ ” and ” ∧ ” on V defined below is an extension of L by K.    a ∨L b     ρb (a) a ∨ b :=  ρa (b)      a ∨K b   a ∧L b     b    λ (a) a ∧ b := λa (b)     f (a, b)     a ∧K b

if a, b ∈ L

(S1)

if a ∈ L, b ∈ K ∗

(S2)

K ∗,

(S3)

if a ∈

b∈L

if a, b ∈ K ∗

(S4)

if a, b ∈ L

(P 1) K∗

(P 2)

b∈L

(P 3)

if a ∈ L, b ∈ if a ∈

K ∗,

if a, b ∈ K ∗ , a ∧K b = 0 if a, b ∈

K ∗,

a ∧K b 6= 0

(P 4) (P 5)

Let a ∈ L, b ∈ K ∗ . If b = (2, 3), then ρb (a) := b = (2, 3). Let b 6= (2, 3). Then: If a = (0, 0), then ρb (a) := b. If a 6= (0, 0), then ρb (a) := (3, 4). Let a ∈ K ∗ , b ∈ L. If a = (2, 3), then ρa (b) := a = (2, 3). Let a 6= (2, 3). Then: If b = (0, 0), then ρa (b) := a. If b 6= (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 λb (a) := λ(0,0) (a) = (0, 0) ∧L a = (0, 0). 27

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 λa (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 ” ∧ ” are the following:                  a ∨ b :=

a ∨L b

if a, b ∈ L

(2, 3)

if a ∈ L, b = (2, 3)

b

if L 3 a = (0, 0), b 6= (2, 3)

(3, 4)

 (2, 3)       a      (3, 4)     a ∨K b                         

a ∧ b :=

                       

if L 3 a 6= (0, 0), b 6= (2, 3) if a = (2, 3), b ∈ L if a 6= (2, 3), L 3 b = (0, 0) if a 6= (2, 3), L 3 b 6= (0, 0) if a, b ∈ K ∗

a ∧L b

if a, b ∈ 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)

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 6= 0

Finally, we notice that 1. x ≤V (2, 3) for every x ∈ L. This is because x ∨ (2, 3) = (2, 3) for each x ∈ L. 2. (0, 0) ≤V (0, 4) and (0, 0) ≤V (3, 0). In fact, (0, 0) ∨ (0, 4) = (0, 4) since (0, 4) 6= (2, 3), and (0, 0) ∨ (3, 0) = (3, 0) since (3, 0) 6= (2, 3). 3. x ∨ (0, 4) = (3, 4) and x ∨ (3, 0) = (3, 4) for every x ∈ L, x 6= (0, 0). In fact, let x ∈ L, x 6= (0, 0). Then Since (0, 4) 6= (2, 3), we have x ∨ (0, 4) = (3, 4). Since (3, 0) 6= (2, 3), we have x ∨ (3, 0) = (3, 4). The diagram of V is the following: 28

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University of Athens, Department of Mathematics 157 84 Panepistimiopolis, Athens, Greece e-mail: [email protected]

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