Free MV -Modules

1 Introduction

MV-algebras were defined by C.C. Chang [2] as algebras corresponding to the Lukasiewicz infinite valued propositional calculus. These algebras have appeared in the literature under different names and polynomially equivalent presentation: CN-algebras, Wajsberg algebras, bounded commutative BCK-algebras and bricks. It is discovered that MV-algebras are naturally related to the Murray-von Neumann order of projections in operator algebras on Hilbert spaces and that they play an interesting role as invariants of approximately finite-dimensional C^*-algebras. They are also naturally related to Ulams searching games with lies. MV-algebras admit a natural lattice reduct and hence a natural order structure. Many important properties can be derived from the fact, established by Chang that nontrivial MV-algebras are subdirect products of MV-chains, that is, totally ordered MV-algebras. To prove this fundamental result, Chang introduced the notion of prime ideal in an MV-algebra. The categorical equivalence between MV-algebras and lu-groups leads to the problem of defining a product operation on MV-algebras, in order to obtain structures corresponding to l-rings. A product MV-algebra (or PMV-algebra, for short) is an MV-algebra which has an associative binary operation “ .”. It satisfies an extra property which will be explained in Preliminaries. During the last years, PMV-algebras were considered and their equivalence with a certain class of l-rings with strong unit was proved. It seems quite natural to introduce modules over such algebras, generalizing the divisible MV-algebras and the MV-algebras obtained from Riesz spaces and to prove natural equivalence theorems. Hence, the notion of MV-modules was introduced as an action of a PMV-algebra over an MV-algebra by A. Di Nola [5]. Since MV-modules are in their infancy, stating and opening of any subject in this field can be useful. Hence, in this paper, we present the definition of free MV-module and we verify some properties on it. For example, in a special category of MV-modules (A _k -modules), we state a general representation for a free MV-module and obtain a method to construct a free object on a nonempty set. In fact, we open new fields to anyone that is interested to studying and development of MV-modules.

2 Preliminaries

In this section, we review related lemmas and theorems that we use in the next sections.

Definition 2.1. [3] An MV-algebra is a structure M = (M, ⊕ , ′, 0) of type (2, 1, 0) such that: (MV1) (M, ⊕ , 0) is an Abelian monoid, (MV2) (a′) ′ = a, (MV3) 0′ ⊕ a = 0′, (MV4) (a′ ⊕ b) ′ ⊕ b = (b′ ⊕ a) ′ ⊕ a, If we define the constant 1 = 0′ and operations ⊙ and ⊖ by a ⊙ b = (a′ ⊕ b′) ′, a ⊖ b = a ⊙ b′, then (MV5) (a ⊕ b) = (a′ ⊙ b′) ′, (MV6) x ⊕ 1 =1, (MV7) (a ⊖ b) ⊕ b = (b ⊖ a) ⊕ a, (MV8) a ⊕ a′ = 1, for every a, b ∈ M. It is clear that (M, ⊙ , 1) is an Abelian monoid. Now, if we define auxiliary operations ∨ and ∧ on M by a ∨ b = (a ⊙ b′) ⊕ b and a ∧ b = a ⊙ (a′ ⊕ b), for every a, b ∈ M, then (M, ∨ , ∧ , 0) is a bounded distributive lattice.

An MV-algebra M is a Boolean algebra if and only if the operation “⊕” is idempotent, that is x ⊕ x = x, for every x ∈ M. In an MV-algebra M, the following conditions are equivalent: (i) a′ ⊕ b = 1, (ii) a ⊙ b′ = 0, (iii) b = a ⊕ (b ⊖ a), (iv) ∃c ∈ M such that a ⊕ c = b, for every a, b, c ∈ M. For any two elements a, b of the MV-algebra M, a ≤ b if and only if a, b satisfy the above equivalent conditions (i) - (iv). An ideal of MV-algebra M is a subset I of M, satisfying the following conditions: (I1): 0 ∈ I, (I2): x ≤ y and y ∈ I imply x ∈ I, (I3): x ⊕ y ∈ I, for every x, y ∈ I. In an MV-algebra M, the distance functiond : M × M → M is defined by d (x, y) = (x ⊖ y) ⊕ (y ⊖ x) which satisfies (i): d (x, y) =0 if and only if x = y, (ii): d (x, y) = d (y, x), (iii): d (x, z) ≤ d (x, y) ⊕ d (y, z), (iv): d (x, y) = d (x′, y′), (v): d (x ⊕ z, y ⊕ t) ≤ d (x, y) ⊕ d (z, t), for every x, y, z, t ∈ M. Let I be an ideal of MV-algebra M. We denote x ∼ y (x ≡  _I y) if and only if d (x, y) ∈ I, for every x, y ∈ M. So ∼ is a congruence relation on M. Denote the equivalence class containing x by x I and M I = { x I : x ∈ M }. Then ( M I , ⊕ , ′ , 0 I ) is an MV-algebra, where ( x I ) ′ = x ′ I and x I ⊕ y I = x ⊕ y I, for all x, y ∈ M. Let M and K be two MV-algebras. A mapping f : M → K is called an MV-homomorphism if (H1): f (0) =0, (H2): f (x ⊕ y) = f (x) ⊕ f (y) and (H3): f (x′) = (f (x)) ′, for every x, y ∈ M. If f is one to one (onto), then f is called an MV-monomorphism (MV-epimorphism) and if f is onto and one to one, then f is called an MV-isomorphism (see [3]).

Lemma 2.2. 21 [3] In every MV-algebra M, the natural order “≤” has the following properties: (i) x ≤ y if and only if y′ ≤ x′, (ii) if x ≤ y, then x ⊕ z ≤ y ⊕ z, for every z ∈ M.

Lemma 2.3. 22 [3] Let M and N be two MV-algebras and f : M → N be an MV-homomorphism. Then the following properties hold: (i) Ker (f) is an ideal of M, (ii) if f is an MV-epimorphism, then M Kerf ≅ N.

Definition 2.4. [3] An l-group is an algebra (G, + , - , 0, ∨ , ∧), where the following properties hold: (a) (G, + , - , 0) is a group, (b) (G, ∨ , ∧) is a lattice, (c) x ≤ y implies that x + a ≤ y + a, for any x, y, a, b ∈ G. A strong unit u > 0 is a positive element with property that for any g ∈ G there exits n ∈ ω such that g ≤ nu. The Abelian l-groups with strong unit will be simply called lu-groups [9]. The category whose objects are MV-algebras and whose homomorphisms are MV-homomorphisms is denoted by textslMV. The category whose objects are pairs (G, u), where G is an Abelian l-group and u is a strong unit of G and whose homomorphisms are l-group homomorphisms is denoted by textslUg. The functor that establishes the categorial equivalence between textslMV and textslUg is Γ : Ug ⟶ MV , where Γ (G, u) = [0, u]  _G , for every lu-group (G, u) and Γ (h) = h|_[0,u], for every lu-group homomorphism h. The above results allows us to consider an MV-algebra, when necessary, as an interval in the positive cone of an l-group. Thus, many definitions and properties can be transferred from l-groups to MV-algebras. For example, the group addition becomes a partial operation when it is restricted to an interval, so we define a partial addition on an MV-algebra M as follows: x + y is defined if and only if x ≤ y′ and in this case, x + y = x ⊕ y, for every x, y ∈ M [6]. Moreover, if z + x ≤ z + y, then x ≤ y [5].

Definition 2.5. [4] An l-ring is a structure (R, + , . , 0, ≤), where (R, + , 0, ≤) is an L-group such that, for any x, y ∈ R, x ≥ 0 and y ≥ 0 implies x . y ≥ 0 . A productMV-algebra (or PMV-algebra, for short) is a structure A = (A, ⊕ , . , ′, 0), where (A, ⊕ , ′, 0) is an MV-algebra and “ .” is a binary associative operation on A such that the following property is satisfied: if x + y is defined, then x . z + y . z and z . x + z . y are defined and (x + y) . z = x . z + y . z, z . (x + y) = z . x + z . y, for every x, y, z ∈ A, where “+” is the partial addition on A. A unity for the product is an element e ∈ A such that e . x = x . e = x, for every x ∈ A. If A has a unity for product, then e = 1. A PMV-homomorphism is an MV-homomorphism which also commutes with the product operation.

An lu-ring is a pair (R, u), where (R, + , . , 0, ≤) is an l-ring and u is a strong unit of R (i.e., u is a strong unit of the underlying l-group) such that u . u ≤ u. The category whose objects are pairs (R, u), where R is an l-ring and u is a strong unit of R and whose homomorphisms are l-ring homomorphisms is denoted by textslUR. The functor that establishes the categorial equivalence between textslPMV and textslUR is Γ : UR ⟶ PMV , where Γ (R, u) = [0, u]  _R , for every lu-ring (R, u) and Γ (h) = h|_[0,u], for every lu-ring homomorphism h.

Lemma 2.6. 23 [4] Let A be a PMV-algebra. Then a ≤ b implies that a . c ≤ b . c and c . a ≤ c . b for every a, b, c ∈ A.

Definition 2.7. [5] Let A = (A, ⊕ , . , ′, 0) be a PMV-algebra, M = (M, ⊕ , ′, 0) be an MV-algebra and the operation Φ : A × M ⟶ M be defined by Φ (a, x) = ax, which satisfies the following axioms: (AM1) If x + y is defined in M, then ax + ay is defined in M and a (x + y) = ax + ay, (AM2) If a + b is defined in A, then ax + bx is defined in M and (a + b) x = ax + bx, (AM3) (a . b) x = a (bx), for every a, b ∈ A and x, y ∈ M.

Then M is called a (left) MV-module over A or briefly an A-module. We say that M is a unitaryMV-module if A has a unity for the product and (AM4) 1 _A x = x, for every x ∈ M.

Example 2.8. [5] Consider the real unit interval [0, 1]. Let x ⊕ y = min {x + y, 1} and x′ = 1 - x, for all x, y ∈ [0, 1]. Then ([0, 1] , ⊕ , ′, . , 0) is a PMV-algebra, where “+”, “-” and “ .” are the ordinary operations in ℝ. Also, [0, 1] is a [0, 1]-module, where xy = x . y. Moreover, the MV-algebras [0, u] and [0, u_{1] × ⋯ × [0, u n ] are [0, 1]-modules, for u, u1, ⋯ , u n  > 0.}

Lemma 2.9. 25 [5] Let A be a PMV-algebra and M be an A-module. Then (a) 0x = 0, (b) a0 = 0, (c) ax′ ≤ (ax) ′, (d) a′x ≤ (ax) ′, (e) (ax) ′ = a′x + (1x) ′, (f) x ≤ y implies ax ≤ ay, (g) a ≤ b implies ax ≤ bx, (h) a (x ⊕ y) ≤ ax ⊕ ay, (i) d (ax, ay) ≤ ad (x, y), (j) if x ≡  _I y, then ax ≡  _I ay, where I is an ideal of A, (k) if M is a unitary MV-module, then (ax) ′ = a′x + x′, for every a, b ∈ A and x, y ∈ M.

Definition 2.10. [5] Let A be a PMV-algebra and M₁, M₂ be two A-modules. A map f : M₁ → M₂ is called an A-module homomorphism or(A-homomorphism, for short) if f is an MV-homomorphism and (H4): f (ax) = af (x), for every x ∈ M₁ and a ∈ A.

Definition 2.11. [5] Let A be a PMV-algebra and M be an A-module. Then an ideal N ⊆ M is called an A-ideal of M if (I4): ax ∈ N, for every a ∈ A and x ∈ N.

Definition 2.12. [7] Let C be a concrete category and X be a nonempty set. Then the object F in C is called a free object on X if there exists a map i : X ⟶ F such that for any map f : X ⟶ M, where M is an object in C, there exists a unique morphism f ¯ : F ⟶ M such that f ¯ ∘ i = f.

Proposition 2.13. [5] Let F₁ and F₂ be two free objects on X₁ and X₂, respectively. If |X₁| = |X₂|, then F₁ ⋍ F₂.

Note. From now on, in this paper, we let A be a PMV-algebra, M be an MV-algebra and ∑ i = 1 n x i means x₁ ⊕ x₂ ⊕ ⋯ ⊕ x _n .

3 Free MV-modules

In this section, we present the definitions of finitely generated and free MV-modules and we give some examples and properties.

Definition 3.1. Let M be an A-module, ∅ ≠ T ⊆ M and M = { ∑ i = 1 n x i t i : x i ∈ A , t i ∈ T , for 1 ≤ i ≤ n and x₁t₁ + ⋯ + x _n t _n  is defined in M} ,

where “+” is the partial addition. (Similarly, if T is infinite set, then M can be defined) Then we say that M is generated by T and we set M =≺ T ≻. If |T|< ∞, then M is called a finitely generated A-module. Specially, if M =≺ m ≻, where m ∈ M, then M is called a cyclicA-module.

Example 3.2. (i) Let A = {0, 1, 2, 3} and the operations “⊕” and “ .” be defined on A as follows: ⊕ 0 1 2 3 0 0 1 2 3 1 1 1 3 3 2 2 3 2 3 3 3 3 3 3 . 0 1 2 3 0 0 0 0 0 1 0 1 0 1 2 0 0 2 2 3 0 1 2 3 Consider 0′ = 3,  1′ = 2,  2′ = 1 and 3′ = 0. Then it is easy to show that (A, ⊕ , ′, . , 0) is a PMV-algebra and (A, ⊕ , ′, 0) is an MV-algebra. Now, let the operation • : A × A ⟶ A be defined by a • b = a . b, for every a, b ∈ A. It is easy to show that A is an MV-module on A. Since 1 = 1.1, 2 = 2.2 and 3 = 1.1 + 2.2 = 1.1 ⊕ 2.2, A =≺ {1, 2} ≻. Furthermore, since a = a . 3 = a • 3, for every a ∈ A, A =≺3 ≻, too. (ii) Consider L₂ = {0, 1}, L 4 = { 0 , 1 3 , 2 3 , 1 }, a ⊕ b = min {1, a + b}, a′ = 1 - a and +, -, . are ordinary operations in ℝ. Then it is routine to show that (L₂, ⊕ , ′, . , 0) is a PMV-algebra and (L₄, ⊕ , ′, 0) is an MV-algebra. Let operation • : L₂ × L₄ ⟶ L₄ is defined by a • b = a . b, for every a ∈ L₂ and b ∈ L₄. Then it is easy to show that L₄ is an L₂-module. Now, since 2 3 = 1 . 2 3, 1 3 = 1 . 1 3 and 1 = 1 . 1 3 + 1 . 2 3 = 1 . 1 3 ⊕ 1 . 2 3, L 4 = ≺ { 1 3 , 2 3 } ≻. (iii) Let A = {0, 1, 2, 3} and the operations “⊕” and “ .” be defined on A as follows: ⊕ 0 1 2 3 0 0 1 2 3 1 1 1 2 3 2 2 2 2 3 3 3 3 3 3 . 0 1 2 3 0 0 0 0 0 1 0 1 1 1 2 0 1 2 2 3 0 1 2 3 Consider 0′ = 3,  1′ = 2,  2′ = 1 and 3′ = 0. Then it is easy to show that (A, ⊕ , ′, . , 0) is a PMV-algebra and (A, ⊕ , ′, 0) is an MV-algebra. Now, let the operation • : A × A ⟶ A be defined by a • b = a . b, for every a, b ∈ A. It is easy to show that A is an MV-module on A. Since a = a . 3 = a • 3, for every a ∈ A, A =≺3 ≻. (iv) Let A be a unital PMV-algebra. Then it is clear that A is an A-module, where ab = a . b, for every a, b ∈ A. Since a = a . 1, for every a ∈ A, A =≺1 ≻.

(v) Let M 2 ( ℝ ) be the ring of square matrixes of order 2 with real elements and let 0 be the matrix with all elements 0. If we define the order relation on components A = ( a ij ) i , j = 1 , 2 ≥ 0 if and only if a ij ≥ 0 for any i , j , then M 2 ( ℝ ) is an l-ring. If v = ( 1 2 1 2 1 2 1 2 ), then ( M 2 ( ℝ ) , v ) is an lu-ring and so A = Γ ( M 2 ( ℝ ) , v ) is a PMV-algebra. Let ℝ 2 be the direct product with the order relation defined on components. If u = (1, 1), then ( ℝ 2 , u ) is an lu-group and so M = Γ ( ℝ 2 , u ) is an MV-algebra. Moreover, M is an A-module, where the external operation is the usual matrix multiplication: K ( x , y ) = K ( x y ), for any K ∈ A and (x, y) ∈ M. We show that T = {(1, 0) , (0, 1)} is a generator of M. Let (a, b) ∈ M. If a , b ≤ 1 2, then ( a , b ) = ( a 0 0 0 ) ( 1 0 ) + ( 0 0 0 b ) ( 0 1 ). Let 1 2 < a ≤ 1, then a = 0 . x₁x₂⋯, where x₁ ≥ 5 and x _i  ≠ 0, for some i ≥ 2 and so a = 0 . x₁+ 0.0x₂ ⋯. If x₁ ≥ 6, then 0 . x₁ = 0 . t₁ + 0 . t₂, where t₁, t₂ ≤ 5 and t₁ + t₂ ≤ 9. It means that (a, b) can be generated by (0, 1) , (1, 0), for every 0 ≤ a, b ≤ 1. Therefore, M =≺ T ≻. The above conclusion can be generalized for any n ≥ 2. (v) Let A = Γ ( ℝ , 1 ) and M = Γ ( ℝ 2 , ( 1 , 1 ) ). Then M is an A-module, where α (x, y) = (αx, αy). It is easy to show that M =≺ {(1, 0) , (0, 1)} ≻. (vii) Let A be a nonempty set. Then it is easy to show that (P (A) , ∪ , ′, ∩ , ∅) is a PMV-algebra and (P (A) , ∪ , ′, ∅) is an MV-algebra. Let the operation • : P (A) × P (A) ⟶ P (A) be defined by A₁ • A₂ = A₁ ∩ A₂, for every A₁, A₂ ∈ P (A). Then P (A) is a P (A)-module. Now, since B = B ∩ A = B • A, for every B ∈ P (A), P (A) =≺ A ≻.

Proposition 3.3. Let X = {x₁, ⋯ , x _n } and M =≺ X ≻. Then for every x _i  ≤ x _j , where 1 ≤ i, j ≤ n, the set T _j  = {x₁, ⋯ , x_j-1, x _j  ⊖ x _i , x_j+1, ⋯ , x _n } is a generator of M, that is M =≺ T _j  ≻, too.

Proof. It is enough to show that x _j is generated by T _j . Since x _i  ≤ x _j , x _j  = x _i  ⊕ (x _j  ⊖ x _i ) and so x _j is generated by x _i and x _j  ⊖ x _i . Hence, M is generated by T _j , where x _i  ≤ x _j and 1 ≤ i, j ≤ n.□

Definition 3.4. Let M be an A-module. Then M is called a BooleanA-module if ax ⊕ ay ≤ a (x ⊕ y), for every a ∈ A and x, y ∈ M.

Note: If M is a Boolean A-module, then by Lemma 2.9 (h), a (x ⊕ y) = ax ⊕ ay.

Example 3.5. If M is a Boolean-algebra, then every A-module M is a Boolean A-module. Since x ≤ x ⊕ y and y ≤ x ⊕ y, by Lemma 2.9(f), ax ≤ a (x ⊕ y) and ay ≤ a (x ⊕ y), for every a ∈ A and x, y ∈ M and so by Lemma 2.2(ii), ax ⊕ ay ≤ a (x ⊕ y) ⊕ ay and a (x ⊕ y) ⊕ ay ≤ a (x ⊕ y) ⊕ a (x ⊕ y) = a (x ⊕ y). Hence, ax ⊕ ay ≤ a (x ⊕ y), for every a ∈ A and x, y ∈ M.

Lemma 3.6. Let M be a Boolean A-module. (i) If I is an A-ideal of M, then M I is an A-module. (ii) If N and K are two A-ideals of M such that N ⊆ K, then K N = { k N : k ∈ K } is an A-ideal of M N.

Proof. (i) Let the operation • : A × M I ⟶ M I be defined by a • m I = am I, for any a ∈ A, m ∈ M and ( a 1 , m 1 I ) = ( a 2 , m 2 I ), for any a₁, a₂ ∈ A and m₁, m₂ ∈ M. Then a₁ = a₂ and d (m₁, m₂) ∈ I. Hence, by Lemma 2.9(j), d (a₁m₁, a₂m₂) = d (a₁m₁, a₁m₂) ∈ I, for every a₁, a₂ ∈ A and m₁, m₂ ∈ M. It means that a 1 m 1 I = a 2 m 2 I and so “ • " is well-defined.

( A M I 1 ): If m I + n I is defined in M I, then m I ≤ n ′ I and so by Lemma 2.9(f, c), am I ≤ an ′ I ≤ ( an ) ′ I. Hence, am I + an I is defined in M I and so by (AM1), for every a ∈ A and m, n ∈ M, a • ( m I + n I ) = a • ( m I ⊕ n I ) = a • m ⊕ n I = a ( m ⊕ n ) I = am ⊕ an I = am I ⊕ an I = am I + an I ,( A M I 2 ): If a + b is defined in A, then am + bm is defined in M and so am ≤ (bm) ′. It results that am I ≤ ( bm ) ′ I and so am I + bm I is defined in M I. Hence, by (AM2), for every a, b ∈ A and m ∈ M, ( a + b ) • m I = ( a + b ) m I = am + bm I = am ⊕ bm I = am I ⊕ bm I = am I + bm I = a • m I + b • m I ,( A M I 3 ): By (AM3), for every a, b ∈ A and m ∈ M, ( a . b ) • m I = ( a . b ) m I = a ( bm ) I = a • bm I = a • ( b • m I ) .

Therefore, M I is an A-module.

(ii) By (i), M N is an A-module. It is clear that 0 ∈ K N and x N ⊕ y N ∈ K N, for every x N , y N ∈ K N. Let x N ≤ y N and y N ∈ K N, for every x N , y N ∈ M N. Then x ′ ⊕ y N = x ′ N ⊕ y N = 1 N and so (x′ ⊕ y) ′ = d (x′ ⊕ y, 1) ∈ N ⊆ K. It results that x ⊖ y = (x′ ⊕ y) ′ ∈ K. Since y ∈ K, by (MV6′), (y ⊖ x) ⊕ x = (x ⊖ y) ⊕ y ∈ K. Since x ≤ (y ⊖ x) ⊕ x ∈ K, x ∈ K and so x N ∈ K N. Now, let x N ∈ K N and a ∈ A. Then it is easy to show that a x N ∈ K N. Therefore, K N is an A-ideal of M N.

Theorem 3.7. Let M be a Boolean finitely generated A-module and I be an A-ideal of M. Then M I is a finitely generated A-module.

Proof. By Lemma 3.6(i), M I is an A-module. Now, let M =≺ T ≻. It is easy to see that M I = ≺ T I ≻.

Definition 3.8. Let M be an A-module and ∅ ≠ T ⊆ M. We say that T is a basis for M if M =≺ T ≻ and if ∑ i = 1 n x i t i = 0, where x _i  ∈ A and t _i  ∈ T, then x _i  = 0, for every 1 ≤ i ≤ n (that is T is a linearly independent set).

Remark. Let ∑ i = 1 n x i m i = 0, where x _i  ∈ A and m _i  ∈ M, for every 1 ≤ i ≤ n. Since x i m i ≤ ∑ i = 1 n x i m i = 0, x _i m _i  = 0, for every 1 ≤ i ≤ n.

Definition 3.9. Let M be an A-module. Then M is called a freeA-module, if M has a nonempty basis.

Example 3.10. (i) In Example 3.2 (i), if a • 3 =0, for every a ∈ A, then it is clear that a = 0 and so {3} is a linearly independent set. Hence, {3} is a basis for A and so A is a free A-module. Furthermore, {1, 2} is a basis for A, too. (ii) In Example 3.2 (ii), it is easy to show that L 4 = ≺ { 1 3 , 2 3 } ≻ is a free L₂-module. (iii) In Example 3.2(iii), it is easy to show that {3} is a linearly independent set. Hence, {3} is a basis for A and so A is a free A-module. (iv) In Example 3.2(iv), it is easy to show that A =≺1 ≻ is a free A-module. (v) In Example 3.2(v), it is easy to see that T is a linearly independent set and so Γ ( ℝ 2 , ( 1 , 1 ) ) is a free Γ ( ℝ , 1 )-module. The above conclusion can be generalized for any n ≥ 2. (vi) In Example 3.2(vii), if A₁• A = ∅, for every A₁ ∈ P (A), then it is clear that A₁ =∅ and so {A} is a linearly independent set. Hence, {A} is a basis for P (A) and so P (A) is a free P (A)-module. (vii) Let A = {a, b, c} and B = {a, b}. Then (P (A) , ∪ , ′, ∩ , ∅) is a PMV-algebra and (P (B) , ∪ , ′, ∅) is an MV-algebra. Let operation • : P (A) × P (B) ⟶ P (B) is defined by C • D = C ∩ D, for any C ∈ P (A) and D ∈ P (B). It is easy to show that P (B) is a P (A)-module. Also, it is easy to show that P (B) is not a free P (A)-module.

Example 3.11.Let B₂ be a Boolean algebra with two elements. Then any B₂-module is a free B₂-module. Because, if B₂ = {0, 1}, then it is clear that B₂ is a PMV-algebra. Now, let M be a unitary B₂-module and K = {T ⊆ M : T is a linear independent set}. Since M is a unitary B₂-module, 1 . a = a ≠ 0, for every 0 ≠ a ∈ M and so {a} is a linear independent set. It means that {a} ∈ K and so K≠ ∅. Let Y = {T _i  : i ∈ I} be a chain of elements in K. We claim that U = ⋃ _i∈IT _i is an upper bound for K, with respect to ⊆. Since we have a chain, there exists T _j  ∈ K such that U ⊆ T _j and so U ∈ K. Hence, by Zorn’s Lemma, K has a maximal element T₁. We claim that M =≺ T₁ ≻. Let M≠ ≺ T₁ ≻. Then ≺T₁ ≻ ⊊ M and so there exists m ∈ M such that m∉ ≺ T₁ ≻. We show that T₁ ∪ {m} is a linear independent set. Let x . m ⊕ x₁ . t₁ ⊕ ⋯ ⊕ x _n  . t _n  = 0, for every x, x _i  ∈ X and 1 ≤ i ≤ n. Then x . m = 0. If x ≠ 0, then x = 1 and so m = 0, which is a contradiction. Hence, x = 0 and so T₁ ∪ {m} is a linear independent set, which is again a contradiction. Therefore, M is a freeA-module.

Example 3.12. Let M be a unitary B₂-module. Then every W ⊆ M such that M =≺ W ≻, contains a basis for M. (Similar to the Example 3.11, we get this result.)

Lemma 3.13. Let M be an A-module. Then d (αm, βm) ≤ d (α, β) m, for every α, β ∈ A andm ∈ M.

Proof. Since βm ≤ (αm) ′ ⊕ βm, by Lemma 2.2(i), (αm) ⊙ (βm) ′ = ((αm) ′ ⊕ βm) ′ ≤ (βm) ′ and so (αm) ⊙ (βm) ′ + βm is defined, where “+” is the partial addition on M. Similarly, α ⊙ β′ + β is defined, too. Also, since α ⊙ β′ ≤ β′, by Lemma 2.9(d,g), (α ⊙ β′) m ≤ β′m ≤ (βm) ′ and so (α ⊙ β′) m + βm is defined. Now, α ≤ α ∨ β implies that αm ≤ (α ∨ β) m and similarly, βm ≤ (α ∨ β) m. Then αm ∨ βm ≤ (α ∨ β) m and so (αm) ⊙ (βm) ′ + βm = αm ∨ βm ≤ (α ∨ β) m = (α ⊙ β′ ⊕ β) m = (α ⊙ β′ + β) m = (α ⊙ β′) m + βm . By canceling βm, we have αm ⊖ βm = αm ⊙ (βm) ′ ≤ (α ⊙ β′) m = (α ⊖ β) m and so (αm ⊖ βm) ⊕ (βm ⊖ αm) ≤ (α ⊖ β) m ⊕ (β ⊖ α) m . Since (α′ ⊕ β) ⊕ (β′ ⊕ α) =1, α ⊖ β = (α′ ⊕ β) ′ ≤ (β′ ⊕ α) = (β ⊖ α) ′ and so (α ⊖ β) + (β ⊖ α) is defined. It is easy to show that (α ⊖ β) m + (β ⊖ α) m is defined, too. Hence, d ( α m , β m ) = ( α m ⊖ β m ) ⊕ ( β m ⊖ α m ) ≤ ( α ⊖ β ) m ⊕ ( β ⊖ α ) m = ( α ⊖ β ) m + ( β ⊖ α ) m = ( ( α ⊖ β ) + ( β ⊖ α ) ) m = ( ( α ⊖ β ) ⊕ ( β ⊖ α ) ) m = d ( α , β ) m .

Lemma 3.14. Let I be an ideal of A. (i) If A has a unity for product, then c . a ∈ I and a . c ∈ I, for every c ∈ I and a ∈ A. (ii) If x ⊕ y ≡  _I 0, for every x, y ∈ A. Then x ≡  _I 0 and y ≡  _I 0.

Proof. (i) Since a ≤ 1, by Lemma 2.6, a . c ≤ 1 . c = c ∈ I and so a . c ∈ I. Similarly, c . a ∈ I. (ii) Since x ⊕ y ≡  _I 0, (x ⊕ y) ⊕ y′ ≡  _I 0 ⊕ y′ and so 1 ≡  _I y′, for every x, y ∈ A. It results that y ≡  _I 0. Similarly, x ≡  _I 0.

Definition 3.15. Let A be an A-module. An A-ideal I of A is called a normalA-ideal, if ax ∈ I implies that a ∈ I or x ∈ I, for every a, x ∈ A.

Example 3.16. Every prime ideal in a Boolean algebra A is a normal A-ideal of A.

Theorem 3.17. Let A has a unity for product and I be an A-ideal of A. Then (i) A I is a unitary A I-module. (ii) If A be a free A-module with basis T, where T ⊆ A - I and I be a normal A-ideal, then A I is a free A I-module.

Proof. (i) If we define a 1 I . a 2 I = a 1 . a 2 I, for every a₁, a₂ ∈ A, then it is easy to show that A I is a PMV-algebra. Let the operation • : A I × A I ⟶ A I be defined by α I • a I = α . a I = α a I, for every α, a ∈ A. We must show that “•” is well-defined. Let ( α 1 I , a 1 I ) = ( α 2 I , a 2 I ), for every α₁, α₂, a₁, a₂ ∈ A. Then d (α₁, α₂) ∈ I and d (a₁, a₂) ∈ I. By Lemma 2.9(j), d (a₁α₁, a₁α₂) ∈ I. By Lemmas 3.13 and 3.14(i), d (a₁α₂, a₂α₂) ≤ d (a₁, a₂) α₂ ∈ I and so d (a₁α₂, a₂α₂) ∈ I, for every α₁, α₂, a₁, a₂ ∈ A. Since d (a₁α₁, a₂α₂) ≤ d (a₁α₁, a₁α₂) ⊕ d (a₁α₂, a₂α₂) ∈ I, then d (a₁α₁, a₂α₂) ∈ I and so a 1 α 1 I = a 2 α 2 I. Now, ( A I A I 1 ): If a 1 I + a 2 I is defined inA I, then by Lemmas 2.6 and 2.9(d), α a 1 I + α a 2 I is defined. Since A I is a PMV-algebra, then α I ( a 1 I + a 2 I ) = α a 1 I + α a 2 I , ( A I A I 2 ): if α 1 I + α 2 I is defined in A I, then α 1 a I + α 2 a I is defined and so it is clear that ( α 1 I + α 2 I ) • a I = α 1 a I + α 2 a I , ( A I A I 3 ): it is clear that ( a 1 I . a 2 I ) α I = a 1 I ( a 2 I α I ), for every α₁, α₂, a₁, a₂ ∈ A. It is clear that 1 I is a unity for A I. Therefore, A I is an A I-module. (ii) By (i), A I is an A I-module. Let a I ∈ A I. It is easy to show that a I = ∑ i = 1 n α i I • t i I, where α _i  ∈ A and t _i  ∈ T and so A I = ≺ T I ≻. Now, let ∑ i = 1 n α i I • t i I = 0 I. Then ∑ i = 1 n α i t i I = 0 I and so ∑ i = 1 n α i t i ≡ I 0. By Lemma 3.14 (ii), α _i t _i  ≡  _I 0 and so α _i t _i  ∈ I, for every 1 ≤ i ≤ n. Since t _i  ∉ I, α _i  ∈ I and so α i I = 0 I, for every 1 ≤ i ≤ n. Therefore, T I is a linear independent set and so T I is a basis for A I.□

4 Introduction of a branch of free MV-modules

In this section, we define the notions of A _k -modules and free A _k -modules and we obtain some results on them. For example, we state a general representation for a free A _k -module and we construct a free object in A _k -modules. Also, we define the notions of invariant dimension property and A _k -invariant dimension property in PMV-algebras and prove that every PMV-algebra has the A _k -invariant dimension property.

Definition 4.1. Let M be a unitary A-module and there exists k ∈ ℕ such that ∑ i = 1 n a i ′ m i ≤ ( ∑ i = 1 n a i m i ) ′, for every 1 ≤ n ≤ k, a _i  ∈ A and m _i  ∈ M. Then M is called an A _k -module. If ∑ i = 1 n a i ′ m i ≤ ( ∑ i = 1 n a i m i ) ′, for every n ∈ ℕ, then M is called an A ℕ-module. Moreover, if M is a free A-module, then M is called a free A _k -module.

Example 4.2. (i) If we define αx = 0, for every α ∈ A and x ∈ M, then M is an A ℕ-module. Because, ∑ i = 1 n a i ′ m i = 0 ≤ 1 = 0 ′ = ( ∑ i = 1 n a i m i ) ′. (ii) By Lemma 2.9(d), every A-module M is an A₁-module. (iii) If A is the Boolean algebra with two elements, then every MV-algebra M is an A _N -module. (iv) In Example 3.2(ii), L₄ is a free A _N -module, where A = L₂. (v) In Example 3.2(i), A is not an A₂-module. Because, 1′ . 2 ⊕2′ . 3 =2 ⊕ 1 =3nleq1 = 2′ = (1.2 ⊕ 2.3) ′.

Proposition 4.3.M is an A _k -module if and only if ( ∑ i = 1 n a i m i ) ′ = ( ∑ i = 1 n m i ) ′ ⊕ ∑ i = 1 n a i ′ m i, for every a _i  ∈ A, m _i  ∈ M and n ≤ k.

Proof. (⇒) Let M be an A _k -module. By Lemma 2.9(d), a i ′ m i + a i m i is defined, for every a _i  ∈ A, m _i  ∈ M and 1 ≤ i ≤ n. Since a _i  ≤ a _i , a i + a i ′ is defined. Hence, ∑ i = 1 n a i m i ⊕ ∑ i = 1 n a i ′ m i = ∑ i = 1 n ( a i m i ⊕ a i ′ m i ) = ∑ i = 1 n ( a i m i + a i ′ m i ) = ∑ i = 1 n ( a i + a i ′ ) m i = ∑ i = 1 n 1 m i = ∑ i = 1 n m i and so ( ∑ i = 1 n a i m i ) ′ = ( ∑ i = 1 n a i m i ) ′ ∨ ∑ i = 1 n a i ′ m i = ( ( ∑ i = 1 n a i m i ) ′ ⊙ ( ∑ i = 1 n a i ′ m i ) ′ ) ⊕ ∑ i = 1 n a i ′ m i = ( ∑ i = 1 n a i m i ⊕ ∑ i = 1 n a i ′ m i ) ′ ⊕ ∑ i = 1 n a i ′ m i = ( ∑ i = 1 n m i ) ′ ⊕ ∑ i = 1 n a i ′ m i .

(⇐) The proof is clear.□

Lemma 4.4. Let M be an A ℕ-module, I be an ideal of A and N be an A-ideal of M. Then IM ⊕ N = { ∑ i = 1 k r i m i ⊕ n : r i ∈ I , m i ∈ M , n ∈ N } is an A-ideal of M.

Proof. (I₁) and (I₃) are clear. (I₂): Let m ≤ b, where m ∈ M and b ∈ IM ⊕ N. Since b ∈ IM ⊕ N, there exist n ∈ N, r₁, ⋯ , r _k  ∈ I and m₁, ⋯ , m _k  ∈ M such that b = ∑ i = 1 k r i m i ⊕ n and so m ′ ⊕ ∑ i = 1 k r i m i ⊕ n = 1. By Lemma 2.9(d), r i ′ m i + r i m i is defined and it is clear that r i + r i ′ is defined, too, for every r _i  ∈ A, m _i  ∈ M and 1 ≤ i ≤ k. Hence, m ′ ⊕ ∑ i = 1 k m i ⊕ n = m ′ ⊕ ∑ i = 1 k ( r i ⊕ r i ′ ) m i ⊕ n = m ′ ⊕ ∑ i = 1 k ( r i + r i ′ ) m i ⊕ n = m ′ ⊕ ∑ i = 1 k ( r i m i + r i ′ m i ) ⊕ n = m ′ ⊕ ∑ i = 1 k ( r i m i ⊕ r i ′ m i ) ⊕ n = m ′ ⊕ ∑ i = 1 k r i m i ⊕ ∑ i = 1 k r i ′ m i ⊕ n = 1 ⊕ ∑ i = 1 k r i ′ m i = 1 and so by Proposition 4.3, m = m ∧ b = m ⊙ ( m ′ ⊕ b ) = ( m ′ ⊕ ( m ′ ⊕ ∑ i = 1 k r i m i ⊕ n ) ′ ) ′ = ( m ′ ⊕ ( 1 m ′ ⊕ ∑ i = 1 k r i m i ⊕ 1 n ) ′ ) ′ = ( m ′ ⊕ ( ∑ i = 1 k r i ′ m i ⊕ ( m ′ ⊕ n ⊕ ∑ i = 1 k m i ) ′ ) ) ′ = ( m ′ ⊕ ∑ i = 1 k r i ′ m i ⊕ 1 ′ ) ′ = ( m ′ ⊕ ∑ i = 1 k r i ′ m i ) ′ = ( 1 m ′ ⊕ ∑ i = 1 k r i ′ m i ⊕ 0 n ) ′ = ∑ i = 1 k r i m i ⊕ 1 n ⊕ ( m ′ ⊕ ∑ i = 1 k m i ⊕ n ) ′ = ∑ i = 1 k r i m i ⊕ n ∈ IM ⊕ N . (I₄): Let a ∈ A and ∑ i = 1 k r i m i ⊕ n ∈ IM ⊕ N. Since by Lemma 2.9(h),a ( ∑ i = 1 k r i m i ⊕ n ) ≤ ∑ i = 1 k a ( r i m i ) ⊕ an = ∑ i = 1 k ( a . r i ) m i ⊕ an and since by Lemma 3.14(i), a . r _i  ∈ I, for any 1 ≤ i ≤ k, ∑ i = 1 k ( a . r i ) m i ⊕ an ∈ IM ⊕ N and so by (I₂), a ( ∑ i = 1 k r i m i ⊕ n ) ∈ IM ⊕ N. Therefore, IM ⊕ N is an A-ideal of M.

Lemma 4.5. Let M be an A-module and N be an A-ideal of M. Then ( N : M ) = { a ∈ A : aM ⊆ N } is an ideal of A,.

Proof. It is clear that 0 ∈ (N : M). Let a, b ∈ (N :M). Then am, bm ∈ N, for every m ∈ N. Similarthe proof of Lemma 3.13, we have am ⊙ (bm) ′ ≤ (a ⊙ b′) m, for every a, b ∈ A. If we set a ⊕ b instead of a, then by Lemma 2.9 (g), we have(a ⊕ b) m ⊙ (bm) ′ ≤ ((a ⊕ b) ⊙ b′) m = (a ∧ b′) m≤am. Since (a ⊕ b) m = (a ⊕ b) m ∨ bm = (a ⊕ b)m ⊙ (bm) ′ ⊕ bm ≤ am ⊕ bm ∈ N, a ⊕ b ∈ (N : M). Now, let a ≤ b and b ∈ (N : M). Then by Lemma 2.9(g), am ≤ bm ∈ N and so am ∈ N, for every m ∈ M. It means that a ∈ (N : M).

Definition 4.6. A PMV-algebra A is called commutative, if x . y = y . x, for every x, y ∈ A.

Example 4.7. (i) Every Boolean algebra is a commutative PMV-algebra. (ii) In Example 3.2(i), A is a commutative PMV-algebra.

Theorem 4.8. Let A be commutative and M be a cyclic A ℕ-module. Then for every A-ideal N of M, there exists an ideal I of A such that N = IM.

Proof. Since M is a cyclic A ℕ-module, there exists m ∈ M such that M =≺ m ≻. By Lemma 4.5, (N : M) is an ideal of A and by Lemma 4.4, (N : M) M is an A-ideal of M. Now, we show that N = (N : M) M. It is clear that (N : M) M ⊆ N. Now, let n ∈ N. Then there exists a ∈ A such that n = am. Since aM = a ≺ m ≻ = { a ( a i m ) : a i ∈ A } = { ( a . a i ) m : a i ∈ A } = { ( a i . a ) m ) : a i ∈ A } = { a i ( am ) : a i ∈ A } = { a i n : a i ∈ A } ⊆ N , a ∈ (N : M) and so n ∈ (N : M) M. Hence, N ⊆ (N : M) M and so N = (N : M) M. Now, by considering I = (N : M), we get that N = IM.

Proposition 4.9. Let A be a PMV-algebra. Then ∑ i = 1 n A is a PMV-algebra.

Proof. We define { a i } i = 1 n ⊕ { b i } i = 1 n = { a i ⊕ b i } i = 1 n, { a i } i = 1 n . { b i } i = 1 n = { a i . b i } i = 1 n and ( { a i } i = 1 n ) ′ = { a i ′ } i = 1 n, for every { a i } i = 1 n , { b i } i = 1 n ∈ ∑ i = 1 n A. It is easy to show that ( ∑ i = 1 n A , ⊕ , ′ , . , { 0 } ) is a PMV-algebra.

Definition 4.10. Let M₁ and M₂ be two A-modules. Then the map f : M₁ → M₂ is called an A′-homomorphism if and only if it satisfies in (H1), (H3), (H4) and (H′2) : if x + y is defined in M₁, then h (x + y) = h (x ⊕ y) = h (x) ⊕ h (y), for every x, y ∈ M₁, where “+” is the partial addition on M₁. If h is one to one (onto), then h is called an A′-monomorphism (epimorphism). If h is onto and one to one, then h is called an A′-isomorphism and we write M₁≅′M₂.

Example 4.11. (i) Every A-homomorphism is an A′-homomorphism. (ii) In Example 3.2(ii), by Proposition 4.9, A ⊕ A is an MV-algebra. If operation • : A × (A ⊕ A) ⟶ (A ⊕ A) is defined by a • (b, c) = (a . b, a . c), for every a, b, c ∈ A, then it is easy to show that A ⊕ A is a A-module. consider φ : A ⊕ A ⟶ L₄, where φ ( 1 , 0 ) = 1 3, φ ( 0 , 1 ) = 2 3, φ (0, 0) =0 and φ (1, 1) =1. It is clear that φ is well defined. It is easy to show that φ is an A′-homomorphism but it is not an A-homomorphism. Because, φ ( ( 0 , 1 ) ⊕ ( 0 , 1 ) ) = φ ( 0 , 1 ) = 2 3 ≠ 1 = 2 3 ⊕ 2 3 = φ ( 0 , 1 ) ⊕ φ ( 0 , 1 ).

Theorem 4.12. M is a free A _k -module with basis T = {t₁, ⋯ , t _n } that ∑ j = 1 n t j = 1 and n ≤ k if and only if M ≅ ′ ∑ j = 1 n A.

Proof. (⇒) Let M be a free A _k -module with a basis T and ∑ j = 1 n t j = 1. By Proposition 4.9, ∑ j = 1 n A is a PMV-algebra. Now, let the operation • : A × ∑ j = 1 n A ⟶ ∑ j = 1 n A be defined by a • { a j } j = 1 n = { a . a j } j = 1 n = { aa j } j = 1 n, for every { a j } j = 1 n ∈ ∑ j = 1 n A and a ∈ A. It is easy to show that ∑ j = 1 n A is an A-module. Let Φ : ∑ j = 1 n A ⟶ M be defined by Φ ( { a j } j = 1 n ) = ∑ j = 1 n a j t j, for every a _j  ∈ A and t _j  ∈ T. Then it is clear that φ is well defined and φ (0) =0. If { a j } j = 1 n + { b j } j = 1 n is defined in ∑ j = 1 n A, then a _j  + b _j is defined in A and so a _j t _j  + b _j t _j is defined in M, for every 1 ≤ j ≤ n and t _j  ∈ T. Hence, φ ( { a j } j = 1 n + { b j } j = 1 n ) = φ ( { a j + b j } j = 1 n ) = ∑ j = 1 n ( a j + b j ) t j = ∑ j = 1 n ( a j t j + b j t j ) = ∑ j = 1 n ( a j t j ⊕ b j t j ) = ∑ j = 1 n a j t j ⊕ ∑ j = 1 n b j t j = φ ( { a j } j = 1 n ) ⊕ φ ( { b j } j = 1 n ) . Also, φ ( α { a j } j = 1 n ) = φ { α a j } j = 1 n ) = ∑ j = 1 n ( α . a j ) t j= ∑ j = 1 n α ( a j t j ) = α ∑ j = 1 n a j t j = α φ ( { a j } j = 1 n ) . Finally, for every { a j } j = 1 n ∈ ∑ j = 1 n A, φ ( ( { a j } j = 1 n ) ′ )= φ ( { a j ′ } j = 1 n ) = ∑ j = 1 n a j ′ t j = ∑ j = 1 n a j ′ t j ⊕ ( ∑ j = 1 n t j ) ′ = ( ∑ j = 1 n a j t j ) ′ = ( φ ( { a j } j = 1 n ) ) ′ . Hence, φ is an A′-homomorphism. Let φ ( { a j } j = 1 n ) = 0, for any { a j } j = 1 n ∈ ∑ j = 1 n A. Since T is a linearly independent set, it is easy to show that a _j  = 0, for every 1 ≤ j ≤ n and so Ker (φ) =0. It results that φ is an A′-monomorphism. It is clear that φ is an A′-epimorphism and so M ≅ ′ ∑ j = 1 n A. (⇐) Let M ≅ ′ ∑ j = 1 n A. We construct a basis for ∑ j = 1 n A. Let θ j = { u i } i = 1 n such that u i = { 0 if i ≠ j 1 if i = j We show that K = {θ _j  : 1 ≤ j ≤ n} is a basis for ∑ j = 1 n A. Let { a j } j = 1 n ∈ ∑ j = 1 n A. Since 1 . a = a . 1 = a, for every a ∈ A, { a j } j = 1 n = { a 1 , 0 , ⋯ , 0 } ⊕ { 0 , a 2 , 0 , ⋯ , 0 } ⊕ ⋯ ⊕ { 0 , ⋯ , 0 , a n } = { a 1 . 1 , 0 , ⋯ , 0 } ⊕ { 0 , a 2 . 1 , 0 , ⋯ , 0 } ⊕ ⋯ ⊕ { 0 , ⋯ , 0 , a n . 1 } = a 1 . { 1 , 0 , ⋯ , 0 } ⊕ a 2 . { 0 , 1 , ⋯ , 0 } ⊕ ⋯ ⊕ a n . { 0 , ⋯ , 0 , 1 } = a 1 . θ 1 ⊕ a 2 . θ 2 ⊕ ⋯ ⊕ a n . θ n

Hence, ∑ j = 1 n A = ≺ K ≻. Now, let ∑ j = 1 n a j . θ j = 0. Then it is easy to show that a _j  = 0, for every 1 ≤ j ≤ n. Now, if φ : ∑ j = 1 n A ⟶ M is an A′-isomorphism, then {φ (θ _j ) :1 ≤ j ≤ n} is a basis for M. Moreover, it is clear that ∑ j = 1 n θ j = 1.

Notation: Consider MV _mod  (A _k ) shows the category of all Boolean A _k -modules, which the morphisms are A′-homomorphisms.

Theorem 4.13. Let F ∈ MV _mod  (A _k ) be a free object on the set T, where |T| = n ≤ k. Then F ≅ ′∑ _n A.

Proof. Let F ∈ MV _mod  (A _k ) be a free object on T, where |T| = n ≤ k. Then similar to the proof of Theorem 4.12, K = {θ _t  : 1 ≤ t ≤ n} is a basis for ∑ t = 1 n A, as an A-module. We show that ∑ t = 1 n A is a free object on K. Let G ∈ MV _mod  (A _k ) and i : K ⟶ ∑ t = 1 n A, f : K ⟶ G be two maps. We define h : ∑ t = 1 n A ⟶ G by h ( ∑ t = 1 n a t . θ t ) = ∑ t = 1 n a t . f ( θ t ), where a _t  ∈ A, for every 1 ≤ t ≤ n. Let ∑ t = 1 n a t . θ t = ∑ t = 1 n b t . θ t, for every a _t , b _t  ∈ A. Then { a t } t = 1 n = { b t } t = 1 n and so a _t  = b _t , for every 1 ≤ t ≤ n. Hence, ∑ t = 1 n a t . f ( θ t ) = ∑ t = 1 n b t . f ( θ t ) and so h is well defined. It is clear that h (0) =0. Let ∑ t = 1 n a t . θ t + ∑ t = 1 n b t . θ t = { a t + b t } t = 1 n is defined in ∑ t = 1 n A. Then a _t  + b _t is defined in A and so by (AG1), a _t f (θ _t ) + b _t f (θ _t ) is defined in G and (a _t  + b _t ) f (θ _t ) = a _t f (θ _t ) + b _t f (θ _t ), for every 1 ≤ t ≤ n. Hence, h ( ∑ t = 1 n a t . θ t + ∑ t = 1 n b t . θ t ) = h ( { a t + b t } t = 1 n ) = ∑ t = 1 n ( a t + b t ) f ( θ t ) = ∑ t = 1 n ( a t f ( θ t ) + b t f ( θ t ) ) = ∑ t = 1 n ( a t f ( θ t ) ⊕ b t f ( θ t ) ) = ∑ t = 1 n a t f ( θ t ) ⊕ ∑ t = 1 n b t f ( θ t ) = h ( ∑ t = 1 n a t . θ t ) ⊕ h ( ∑ t = 1 n b t . θ t ) .