RADICAL CLASSES AND WEAK RADICAL

DOI: 10.2478/s12175-008-0104-1 Math. Slovaca 58 (2008), No. 6, 719–738

RADICAL CLASSES AND WEAK RADICAL MAPPINGS OF GM V -ALGEBRAS ń Jakub´ık Ja (Communicated by Anatolij Dvureˇ censkij ) ABSTRACT. We use the concept of generalized MV -algebra (GMV -algebra, in short) in the sense of Galatos and Tsinakis; the main tool in their investigation was a truncation construction. The relations between radical classes of GMV -algebras and radical classes of lattice ordered groups are investigated in the present paper. Further, we apply the truncation construction for dealing with weak retract mappings of GMV -algebras. c 2008 Mathematical Institute Slovak Academy of Sciences

1. Introduction We use the concept of generalized M V -algebra (GM V -algebra, in short) in the sense introduced and studied by G a l a t o s and T s i n a k i s [9]. They proved that GM V -algebras can be obtained by a truncation construction from lattice ordered groups. In the present paper we introduce and study the concept of radical class of GM V -algebras. We consider the relations between radical classes of GM V -algebras and radical classes of lattice ordered groups. For any GM V -algebra M = (M ; ∧, ∨, ·, /, \, e) we consider the algebra M0 = (M ; ∧, ∨, ·, e). A mapping f : M → M is defined to be a weak retract mapping of M if it is a retract mapping of the algebra M0 . We apply the truncation construction for investigating weak retract mappings of GM V -algebras. 2000 M a t h e m a t i c s S u b j e c t C l a s s i f i c a t i o n: Primary 06D35; Secondary 06F15. K e y w o r d s: generalized MV -algebra, lattice ordered group, radical class. Supported by the Slovak Research and Development Agency under the contract No APVV-0071-06. This work has been partially supported by the Slovak Academy of Sciences via the project Center of Excellence — Physics of Information (Grant I/2/2005).

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´ JAKUBÍK JAN

Direct summands and retract mappings of GM V -algebras were studied in [23]. For lattice ordered groups and M V -algebras cf. the monographs [11] and [3], respectively. Radical classes of lattice ordered groups have been dealt with in [2], [4], [5], [12]–[19], [24] and [26]. In the papers [22], [20] and [21], radical classes of M V -algebras were investigated. Retract mappings of GM V -algebras were dealt with in [23] (for further related references concerning radical mappings cf. [23]). We remark that the term ‘generalized M V -algebra’ was applied in a different sense in [25]; in the sense of [25], this notion is equivalent to the notion of pseudo M V -algebra introduced in [10]. From the series of papers dealing with this notion let us mention [1], [7] and [21]. A fundamental theorem on the representation of pseudo M V -algebras was proved in [6].

2. Preliminaries In [9], the concept of GM V -algebra was introduced in context of residuated lattices. We recall some basic notions which will be applied below. Cf. also [23]. A residuated lattice is defined to be an algebra L = (L; ∧, ∨, ·, \, /, e) of type (2, 2, 2, 2, 2, 0) such that (L; ∧, ∨) is a lattice, (L; ·, e) is a monoid and the relation x · y z ⇐⇒ x z/y ⇐⇒ y x \ z is valid for each x, y, z ∈ L. The negative cone of L is the algebra L− = (L− ; ∧, ∨, ·, \L− , /L− , e), where − L = {x ∈ L : x e} and x \L− y = (x \ y) ∧ e,

x/L− y = (x/y) ∧ e.

A generalized M V -algebra (GM V -algebra, in short) is a residuated lattice L such that x/((x ∨ y) \ x) = x ∨ y = (x/(x ∨ y)) \ x for each x, y ∈ L. A mapping γ : P → P on a partially ordered set P is a closure operator on P if (i) γ(x) γ(y) whenever x y, (ii) x γ(x) and (iii) γ(γ(x)) = x. 720


RADICAL CLASSES AND WEAK RADICAL MAPPINGS OF GMV -ALGEBRAS

We denote γ(P ) = Pγ . Assume that L is a residuated lattice and γ is a closure operator on the lattice (L; ∧, ∨) such that γ(a)γ(b) γ(a, b) for each a, b ∈ L; then γ is said to be a nucleus on L. In such a case we consider the algebra Lγ = (Lγ ; ∧, ∨γ , ◦γ , \, /, γ(e)), where γ(a) ∨γ γ(b) = γ(a ∨ b),

γ(a) ◦γ γ(b) = γ(a · b).

Then Lγ is a residuated lattice. Further, let G = (G; ∧, ∨, ·, e) be a lattice ordered group. We denote by c(G) the system of all convex -subgroups of G; this system is partially ordered by the set-theoretical inclusion. Then c(G) is a complete lattice. A nonempty class X of lattice ordered groups which is closed with respect to isomorphisms is a radical class if it satisfies the following conditions: (i) Whenever G ∈ X and G1 ∈ c(G), then G1 ∈ X. (ii) If G is a lattice ordered group and {Gi }i∈I ∈ X ∩ c(G), then

i∈I

Gi ∈ X.

If (i) is valid, then the condition (ii) is equivalent with (iii) If G is a lattice ordered group and S = {Hj }j∈J = X ∩ c(G), then the system S possesses a greatest element. For any GM V -algebra M = (M ; ∧, ∨, ·, \, /, e) we denote by (M) its underlying lattice; e.g., (M) = (M ; ∧, ∨). A subalgebra M1 of M is said to be convex if the lattice (M1 ) is a convex sublattice of the lattice (M). Let c(M) be the system of all convex subalgebras of M; the system c(M) is partially ordered by the set theoretical inclusion. Analogously as in the case of lattice ordered groups, c(M) is a complete lattice. We define a nonempty class Y of GM V -algebras to be a radical class if it is closed with respect to isomorphisms and (i1 ) each convex subalgebra of a GM V -algebra belonging to Y also belongs to Y ; (iii1 ) for each GM V -algebra M, the system of all convex subalgebras of M which belong to Y has a greatest element. Our aim is to prove that there exists a one-to-one correspondence between radical classes of lattice ordered groups and radical classes of GM V -algebras. We systematically apply the following notation: if A is an algebra, then its underlying subset is denoted by A; moreover, if A is lattice ordered, then the symbol (A) denotes the underlying lattice of A. If, e.g., we have Lemma 2.5, then instead of ‘According to Lemma 2.5’ we often write ‘According to 2.5’. 721


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3. Direct sums of GM V -algebras A residuated lattice A is a direct sum of its subalgebras B and C, in symbols A = B ⊕ C, if the map B × C → A defined by f (x, y) = xy is an isomorphism. (Cf. [9].) (We remark that we often write xy instead of x · y; thus, on the right hand of the previous equation we mean the product of elements in the residuated lattice.) Under the notation as above, put z = xy; we denote x = z(B) and y = z(C). We say that x and y is the component of z in B or in C, respectively. B and C are direct summands of A. By the obvious induction, we can define the meaning of A = B1 ⊕B2 ⊕· · ·⊕Bn , where B1 , . . . , Bn are subalgebras of A. In [23], direct sum decompositions of A with infinitely many summands have been also considered; in the present paper we will not deal with this situation. Let P be a partially ordered set and a ∈ P . For the notion of internal direct product decomposition of P with respect to the element a cf. [23, Section 3]; in accordance with [23], we apply the notation ϕa : P → P1 (a) × P2 (a) × · · · × Pn (a).

(1)

We say that Pi (a) (i = 1, 2, . . . , n) are internal direct factors of P with respect to the element a. In view of [23], we have:

3.1 Let M be a GM V -algebra. Put (M) = P and assume that (1) is valid with a = e. Then there exist A1 , . . . , An ∈ c(M) such that (A1 ) = P1 (a), . . . , (An ) = Pn (a) and M = A1 ⊕ A 2 ⊕ · · · ⊕ An .

(2)

Let G = (G; ∧, ∨, ·, e) be a lattice ordered group. Its negative cone is the algebra G− = (G− ; ∧, ∨, ·, e), where G− = {x ∈ G : x e}. The algebra G∗ = (G; ∧, ∨, ·, \, /, e) where x \ y = x−1 y and y/x = yx−1 is a GM V -algebra (cf. [9]). For the following result cf [9, Theorem (A)]; we apply a slightly modified notation (cf. also [23]).

3.2 A residuated lattice M is a GM V -algebra if and only if there are lattice ordered groups G and G1 and a nucleus γ on (G∗1 )− such that M = G∗ ⊕ Lγ , where L = (G∗1 )− . 722



Again, let (1) be valid. For x ∈ P and i ∈ {1, 2, . . . , n} let x(Pi (a)) be the component of x in Pi (a). Further, for Q ⊆ P we put Q(Pi (a)) = {x(Pi (a) : x ∈ Q}. The following result is well-known; the proof will be omitted.

3.3

Assume that P is a lattice, a ∈ P . Let (1) be valid and let Q be a convex sublattice of P , a ∈ Q. Then (i) Q(Pi (a)) = Q ∩ Pi (a) for i = 1, 2, . . . , n; (ii) we have an internal direct product decomposition ϕa0 : Q → (Q ∩ P1 (a)) × · · · × (Q ∩ Pn (a)), where

ϕa0

(3)

a

= ϕ |Q.

Applying the notation as in Proposition 3.2, we put P = (M),

P1 = (G∗ ),

P2 = (Lγ ).

For each z ∈ M we put ϕ(z) = (z(G∗ ), z(Lγ )). Then from Proposition 3.2 we immediately obtain:

3.4

The mapping ϕ : P → P1 × P 2

is an internal direct product decomposition of the lattice P with respect to the element e. It is obvious that G∗ and Lγ are elements of c(M). Assume that Q is a convex subalgebra of M having the underlying set Q. Hence there are convex subalgebras Q1 and Q2 of M (with underlying sets Q1 or Q2 , respectively) such that Q1 = Q ∩ G, Q2 = Q ∩ Lγ . Then we have:

3.5

Q = Q1 ⊕ Q2 .

P r o o f. This is a consequence of Proposition 3.2, Lemma 3.4, Lemma 3.3 and Proposition 3.1.

4. Auxiliary results Let us apply the notation as in Proposition 3.2. The operations in Lγ will be written in the standard way, i.e., Lγ = (Lγ ; ∧, ∨, ·, \, /, e).

4.1

Lγ = {z ∈ M : z ∨ t = e for each t ∈ G− }. 723


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P r o o f. Let z ∈ Lγ and t ∈ G− . We have z = ez and t = te. Also, z e. Hence in view of 3.2, we obtain z ∨ t = (ez) ∨ (te) = (e ∨ t)(z ∨ e) = ee = e. Conversely, let z ∈ M and assume that z ∨ t = e for each t ∈ G− . There exist x ∈ G and y ∈ Lγ with z = xy. We have z e, hence x e and y e. Put t = x. Then z ∨ t = (xy) ∨ (xe) = (x ∨ x)(y ∨ e) = xe = x. This yields x = e, whence z = y ∈ Lγ .

4.2 (Cf. [23].) G = {z ∈ M : there exists z1 ∈ M with zz1 = e}.

4.3 Let M be a GM V -algebra. Then the GM V -algebras G∗ and

Lγ from 3.2 are uniquely determined.

Under the notation as in 4.3 we put G∗ = α(M ) and Lγ = β(M ). It can be shown by simple examples that the lattice ordered group G1 from 3.2 need not be, in general, uniquely determined by the GM V -algebra M. Now let G1 be any lattice ordered group and let γ1 be a nucleus on the negative cone G− 1 of G1 . The meaning of Lγ1 is as above. Further, let A1 be the set of all elements a1 ∈ G1 that can be written in the form a1 = g1 g2 . . . gn with g1 , . . . , gn ∈ Lγ1 . From Riesz Theorem we can easily deduce that A1 is a convex sublattice of −1 the lattice (G− 1 ). Put A2 = {a1 : a1 ∈ A1 }. We denote by A3 the set of all elements a3 ∈ G1 such that there exist a1 ∈ A1 and a2 ∈ A2 with a1 a3 a2 . ∈ A3 . Also, A3 is a convex subIf a3 , a3 ∈ A3 , then a3 a3 ∈ A3 and a−1 3 lattice of (G1 ). Thus A3 is an underlying subset of a convex -subgroup G1 of G1 . Moreover, we have (G1 )− ∈ A1 and γ1 (a1 ) ∈ A1 for each a1 ∈ A1 . Put γ1 |A1 = γ1 . Hence γ1 is a nucleus on A1 . We obtain Lγ1 = Lγ1 . From this we conclude:

4.4 In Proposition 3.2, the triple (G1 , γ, Lγ ) can be replaced by (G1 , γ1 , Lγ1 ). In view of the above construction and according to 4.3 we obtain that G1 is the minimal convex -subgroup of G1 having the property from Lemma 4.4. For a subset H of a lattice ordered group K we denote by f (H, K) the convex -subgroup of K generated by the set H. 724



4.4.1 (i) (ii)

L1 G1

Let us apply the notation as above. Then

= f (G1 , Lγ1 );

is the set of all elements of G1 which can be expressed in the form h1 h2 . . . hm such that for each i ∈ {1, 2, . . . , m} either hi ∈ Lγ1 or h−1 i ∈ Lγ1 .

P r o o f. The assertion (i) is obvious. If a3 ∈ A3 , then a3 ∨e ∈ A2 and a3 ∧e ∈ A1 . Further, a3 = (a3 ∨ e)(a3 ∧ e). From this and from the above construction we conclude that (ii) is valid.

4.5

Let G1 be a lattice ordered group and a1 , a2 , b1 , b2 ∈ G− 1 . Suppose that a1 a2 = b1 b2 . Then there exist elements cij ∈ G− (i, j = 1, 2) such that 1 (i = 1, 2), ai = ci1 ci2 bj = c1j c2j (j = 1, 2), c12 ∨ c21 = e. P r o o f. This is a consequence of Riesz Theorem (cf., e.g., [8, Chap. V, Theorem 1]) and of the duality.

For a subset H of a lattice ordered group G1 we denote by [H] the set of all g ∈ G which can be written in the form g = h1 h2 where h1 , h2 ∈ H. − A subset H of G− 1 will be said to be a quasi-filter of (G1 ) if (i) e ∈ H, and (ii) whenever h ∈ H, then the interval [h, e] of (G− 1 ) is a subset of H. Let G1 and G2 be lattice ordered groups. Assume that for i ∈ {1, 2}, H0i is 1 2 a quasi-filter of (G− i ) and that ϕ0 : H0 → H0 is a bijection such that for each 1 1 1 a , b ∈ H0 , (i) a1 b1 ∈ H01 ⇐⇒ ϕ0 (a1 )ϕ0 (b1 ) ∈ H02 ,

(ii) if a1 b1 ∈ H01 , then ϕ0 (a1 b1 ) = ϕ0 (a1 )ϕ0 (b1 ). Under these assumptions we say that the mapping ϕ0 is regular. 2 1 If ϕ0 is regular, then the mapping ϕ−1 0 : H0 → H0 is regular as well. In what follows, we shall often apply the well-known fact that orthogonal elements commute.

4.6 Under the notation as above, let ϕ0 : H01 → H02 such that a1 a2 = b1 b2 . Then ϕ0 (a1 )ϕ0 (a2 ) = ϕ0 (b1 )ϕ0 (b2 ). P r o o f. Let us apply the notation as in 4.5. Hence we have ϕ0 (ai ) = ϕ0 (ci1 )ϕ0 (ci2 )

(i = 1, 2),

ϕ0 (bj ) = ϕ0 (c1j )ϕ0 (c2j )

(j = 1, 2),

ϕ0 (c12 ) ∨ ϕ0 (c21 ) = e.

(∗) 725


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In view of (∗) we obtain ϕ0 (c12 )ϕ0 (c21 ) = ϕ0 (c12 ) ∧ ϕ0 (c21 ) = ϕ0 (c21 )ϕ0 (c12 ). Therefore ϕ0 (a1 )ϕ0 (a2 ) = ϕ0 (c11 )ϕ0 (c12 )ϕ0 (c21 )ϕ0 (c22 ), ϕ0 (b1 )ϕ0 (b2 ) = ϕ0 (c11 )ϕ0 (c21 )ϕ0 (c12 )ϕ0 (c22 ). Thus according to (∗) we get ϕ0 (a1 )ϕ0 (a2 ) = ϕ0 (b1 )ϕ0 (b2 ).

The previous lemma and the fact that ϕ−1 is regular yield:

4.6.1

Let ϕ0 be as in 4.6. Let a1 , a2 , b1 , b2 be elements of H01 such that ϕ0 (a1 )ϕ0 (a2 ) = ϕ0 (b1 )ϕ0 (b2 ). Then a1 a2 = b1 b2 . For i ∈ {1, 2} we denote Hi1 = [H0i ]. Then H1i is a quasi-filter of (G− i ). Let t ∈ H11 . There exist h1 , h2 ∈ H01 with t = h1 h2 . Put ϕ1 (t) = ϕ0 (h1 )ϕ0 (h2 ).

4.7

ϕ1 is a bijection of H11 onto H12 .

P r o o f. In view of 4.6, ϕ1 is a correctly defined mapping of H11 into H12 . Let y ∈ H12 . There are y1 , y2 ∈ H02 with y = y1 y2 . Further, there are x1 , x2 ∈ H01 such that ϕ0 (x1 ) = y1 , ϕ0 (x2 ) = y2 . Put t = x1 x2 . Then t ∈ H11 and ϕ1 (t) = y. Hence ϕ0 is surjective. According to 4.6.1 we conclude that ϕ1 is a monomorphism.

4.8

The mapping ϕ1 is regular.

P r o o f. Let x, y, z ∈ H11 and xy = z. There exist a1 , b1 ∈ H01 with z = a1 b1 . Hence xy = a1 b1 . Thus in view of 4.5 there exist p1 , p2 , p3 , p4 ∈ G− 1 such that x = p1 p2 , y = p3 p4 , a1 = p1 p3 , b1 = p2 p4 , p2 ∨ p3 = e. We have p1 , p3 ∈ [a1 , e], p2 , p4 ∈ [b1 , e], whence p1 , p2 , p3 , p4 ∈ H01 . Also, p2 p3 = p3 p2 . Thus we obtain ϕ1 (z) = ϕ0 (a1 )ϕ0 (b1 ) = ϕ0 (p1 p3 )ϕ0 (p2 p4 ) = ϕ0 (p1 )ϕ0 (p3 )ϕ0 (p2 )ϕ0 (p4 ). In view of p2 p3 = p3 p2 we get ϕ0 (p3 )ϕ0 (p2 ) = ϕ0 (p2 )ϕ0 (p3 ), whence ϕ1 (z) = ϕ0 (p1 )ϕ0 (p2 )ϕ0 (p3 )ϕ0 (p4 ) = ϕ1 (x)ϕ1 (y). Thus ϕ1 (x)ϕ1 (y) ∈ H12 and ϕ1 (x)ϕ1 (y) = ϕ1 (xy). and ϕ−1 we get that x, y ∈ H11 and By analogous argument applying ϕ−1 0 1 2 1 ϕ1 (x)ϕ1 (y) ∈ H1 , then xy ∈ H1 . In view of 4.7, the proof is complete. 726



We have H0i ⊆ H1i for i ∈ {1, 2}. If x ∈ H01 , then ϕ(x) = ϕ1 (x). i For n ∈ N, n > 1 we define by induction Hn+1 = [Hni ] (i ∈ {1, 2, }). Further, we set i H∞ = Hni (i = 1, 2). n∈N 1 H∞ .

Let x ∈ There is n ∈ N with x ∈ Hn1 . We put ϕ(x) = ϕn (x). Then ϕ is 1 2 a correctly defined mapping of H∞ into H∞ . Moreover, ϕ is a bijection. Both 1 2 i H∞ and H∞ are closed with respect to the multiplication. For i ∈ {1, 2}, H∞ is a convex subset of (Gi ) containing the element e. By the obvious induction we obtain that ϕ is an isomorphisms with respect to the multiplication. 1 1 Let x, y ∈ H∞ , x y. Then there is z ∈ H∞ with x = yz. Thus ϕ(x) = ϕ(y)ϕ(z) and hence ϕ(x) ϕ(y). Similarly from ϕ(x) ϕ(y) we obtain x y. Thus ϕ is isomorphism with respect to the partial order. 1 1 For p, q ∈ H∞ we have p ∨ q p, whence p ∨ q ∈ H∞ . Further, p ∧ q pq, 1 1 2 thus p ∧ q ∈ H∞ . Therefore H∞ is lattice. This yields that H∞ is a lattice as well.

4.9 Let Gi (i = 1, 2) be the lattice ordered groups and let H0i (i = 1, 2) be a quasi-filter of (Gi )− . Assume that there exists a regular mapping i ϕ : H01 → H02 . Let H∞ (i = 1, 2) be as above. Then 1 2 f (G1 , H∞ ) f (G2 , H∞ ).

P r o o f. This is a consequence of the above results and the fact that i i (f (Gi , H∞ ))− = H∞ .

5. Radical classes Assume that G1 is a lattice ordered group and that γ1 is a nucleus on the negative cone G− 1 of G1 . Then (under the notation as above) Lγ is a quasi-filter of (G1 )− . We put H01 = Lγ and we can apply the notation from Section 4. Thus, if G1 is as in Lemma 4.4, then we have 1 (G1 )− = H∞ ,

f (G1 , Lγ1 ) = G1 .

We express this situation by writing Lγ1 G1 . From Proposition 4.9 we obtain:

5.1 Let Gi (i = 1, 2) be lattice ordered groups and for i ∈ {1, 2} let γi be a nucleus on (Gi )− . Assume that Lγ1 Lγ2 . Then f (G1 , Lγ1 ) f (G2 , Lγ2 ). 727


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Assume that X is a radical class of lattice ordered groups. We denote by Y the class of all GM V -algebras M such that, under the notation as in 3.2, M = G∗ ⊕ Lγ ,

(1)

L = (G∗1 )− and 1) G ∈ X; 2) there exists H1 ∈ X with Lγ H1 . Then in view of 5.1, Y is closed with respect to isomorphisms. Assume that M is a convex subalgebra of M. In view of (1) for the corresponding lattices we have (M) = (G∗ ) × (Lγ ).

(2)

Since (M ) is a convex sublattice of (M), the relation (2) yields (M ) = ((M ) ∩ (G∗ )) × ((M ) ∩ (Lγ )).

(3)

Applying [23, Theorem 6.6] we conclude that the direct product decomposition (3) of (M ) determines a direct sum decomposition of M ; hence there are GM V -algebras A and B such that (i) (A) = (M ) ∩ (G∗ ), (ii) (B) = (M ) ∩ (Lγ ), (iii) M = A ⊕ B. Then we have α(M ) = A and β(M ) = B. The underlying lattice ordered group of A will be denoted by A0 . Thus A0 is a convex -subgroup of the lattice ordered group G. Since G belongs to X, we infer that A0 belongs to X as well. The underlying subset B of B is a quasi-filter on (G− 1 ). Hence we can − construct the lattice ordered group f (G− , B); we put f (G 1 1 , B) = G2 . Then − G1 is a convex -subgroup of f (G1 , Lγ ). Since Lγ f (G− 1 , Lγ ), according to 4.9 we get H1 f (G1 , Lγ ) and hence f (G1 , Lγ ) belongs to X. This yields that G2 belongs to X. Therefore M ∈ Y . Thus Y is closed with respect to convex subalgebras. Let M0 be any GM V -algebra; in accordance with 3.2 we express it in the form M0 = G∗0 ⊕ L0γ . Further, let {Mi }i∈I be the system of all convex subalgebras of M0 which belong to Y . Applying 3.2 we write Mi = G∗i ⊕ Liγ . 728



All Gi are convex -subgroups of G0 belonging to X. Their join Gi is a i∈I Gi = G01 . Thus G∗01 is convex -subgroup of G0 and it belongs to X. Put i∈I

a convex subalgebra of G∗0 . In view of the facts mentioned by considering convex subalgebras of a GM V -algebra, for each i ∈ I there exists a convex -subgroup Hi of G1 such that Liγ Hi . Since G1 ∈ X, all Hi belong to X. Thus their join H0 = Hi also belongs i∈I

to X. Put H0 ∩ Lγ = L0γ . Then L0γ is the underlying subset of a convex subalgebra L0γ of Lγ . In view of (1) there exists a convex subalgebra M0 of M such that M0 = G∗0 ⊕ L0γ .

(4)

Then M0 ∈ Y . Thus M0 is an element of the system {Mi }i∈I . The definition of M0 yields that M0 is the greatest element of this system. Under the notation as above put ϕ(X) = Y . We proved:

5.2

Let X be a radical class of lattice ordered groups. Then ϕ(X) is a radical class of GM V -algebras. The following result is well-known.

5.3

Let {Gi }i∈I be a system of convex -subgroups of a lattice ordered group G. Put G0 = Gi . Then G0 is the set of all g ∈ G which can be i∈I Gi . expressed in the form g = g1 g2 . . . gm , where g1 , g2 , . . . , gm ∈ i∈I

Let M be a GM V -algebra; let us apply the representation (1). We consider convex subalgebras of Lγ . Analogously to 5.3, we have:

i∈I

5.4 Let {Li }i∈I be a system of convex subalgebras of Lγ . Put L = Li . Then L is the set of all elements h ∈ Lγ which can be expressed in the

form where h1 , h2 , . . . , hm ∈

i∈I

h = h1 ◦γ h2 ◦γ · · · ◦γ hm , Li .

(+)

P r o o f. Since the system of allconvex subalgebras of Lγ is a convex lattice, there exists the element L = Li in this system. Let H be the set of all i∈I

elements of Lγ which can be expressed as in (+). It is easy to verify that the set H is closed with respect to all GM V -operations. Moreover, in view of Riesz 729


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Theorem, H is a convex subset of (Lγ ). Hence there is a convex subalgebra H of Lγ whose underlying set is H. For each i ∈ I we have Li ⊆ H, hence L is a subalgebra of H. On the other hand, each element of H belongs to L. Thus H ⊆ L. Summarizing, we obtain H = L.

5.5 Let {Li }i∈I and L be as in 5.4. Let h ∈ L. Then there are elements h1 , h2 , . . . , hm ∈

i∈I

Li such that h = h1 h2 . . . hm .

P r o o f. In view of 5.4, there are h1 , h2 , . . . , hm ∈

i∈I

Li such that h = h1 ◦γ

h2 ◦γ · · · ◦γ hm . According to the definition of the operation ◦γ , h h1 h2 . . . hm . Hence Riesz Theorem yields that there are h1 , h2 , . . . , hm ∈ G1 with

Clearly, h1 , h2 , . . . , hm

(j = 1, 2, . . . , m), e hj hj h = h1 h2 . . . hm . ∈ Li .

5.6

i∈I

Let L be as in 5.4. Put f (L, G1 ) = H. Then H is the set of all elements of G1 which can be expressed in the form h1 h2 . . . hm such that for each i ∈ {1, 2, . . . , m}, either hi ∈ L or h−1 0 ∈ L. P r o o f. It suffices to apply the same method as in the construction of G1 in Section 4 (cf. Lemma 4.4 and Corollary 4.4.1). If Li is as in 5.4, then a result analogous to 5.6 is valid for f (Li , G1 ).

5.7 Let {Li } and L be as in 5.4. For i ∈ I we put Gi = f (Li , G1). Further, let H = f (L, G1 ) and G0 = Gi . Then H = G0 . i∈I

P r o o f. Since H and G0 are -subgroups of G1 , it suffices to verify that H = G0 . Assume that g ∈ G0 . Then in view of 5.3, g is a product of a finite number of elements of Gi . If i ∈ I and gi ∈ Gi , hence according to Lemma 5.6 (applied i∈I

for Li ), we have gi = gi1 gi2 . . . gim such that for j ∈ {1, 2, . . . , m}, either gij ∈ Li −1 or gij ∈ Li . Since Li ⊆ L, applying Lemma 5.6 again we get g ∈ H. i∈I

Conversely, suppose that h ∈ H. Let h be expressed as in 5.6. Let j ∈ {1, 2, . . . , m}. If hj ∈ L, then according to 5.5 there are elements h1j , h2j , . . . , hm(j)j ∈ Li such that hj = h1j h2j . . . hm(j)j . Since Li ⊆ Gi for i∈I

each i ∈ I, we get hj ∈ G0 . By a slightly modified argument we conclude that hj belongs to G0 also in the case when h−1 j ∈ L. Thus h ∈ G0 . Summarizing, we have H = G0 . 730



Now suppose that Y is a radical class of GM V algebras. We denote by ψ1 (Y ) the class of all lattice ordered groups G such that there exists M ∈ Y with M = G∗ ⊕ Gγ (under the notation as in 3.2). Further, let ψ2 (Y ) be the class of all lattice ordered groups H such that there exists M ∈ Y satisfying the relation Lγ H (under the notation as above for M). Put ψ0 (Y ) = ψ1 (Y ) ∪ ψ2 (Y ). By ψ(Y ) denote the radical class of lattice ordered groups which is generated by ψ0 (Y ). (Cf. [18].) Let G ∈ ψ1 (Y ). Let M be as in the definition of ψ1 (Y ). Then (G∗ )− is a convex subalgebra of M, hence (G∗ )− ∈ Y . We have (G∗ )− = (G− )∗ and G− G. Thus G ∈ ψ2 (Y ). Therefore ψ1 (Y ) ⊆ ψ2 (Y ) and ψ0 (Y ) = ψ2 (Y ). Now, let H ∈ ψ0 (Y ). Hence H ∈ ψ2 (Y ). Let M be as in the definition of ψ2 (Y ). Suppose that H1 ∈ c(H). Put L1γ = Lγ ∩ H1 . Then L1γ is a convex subalgebra of Lγ and Lγ is a convex subalgebra of M. Thus Lγ ⊆ M . Further, we have L1γ H1 . Considering the GM V -algebra {e} ⊕ L1γ = L1γ we conclude that H1 belongs to ψ2 (Y ). We verified that ψ0 (Y ) is closed with respect to convex -subgroups. Further, let H be any lattice ordered group. Suppose that {Hi }i∈I is a nonempty system of lattice ordered groups belonging to c(H) ∩ ψ0 (Y ). For each i ∈ I let Mi = G∗i ⊕ Liγ be as in the definition of ψ2 (Y ). Thus Liγ Hi . Since Y is closed with respect to convex subalgebras and each Liγ is a convex subalgebra i Lγ = L0γ of (H− )∗ of M = {e} ⊕ (H− )∗ , we infer that the convex subalgebra i∈I belongs to Y . In view of Lemma 5.7, we have L0γ H0 , where H0 = Hi . i∈I

Thus H0 ∈ ψ2 (Y ). We verified that ψ0 (Y ) is closed with respect to joins of convex -subgroups. It is clear that ψ2 (Y ) is closed with respect to isomorphisms. Summarizing, we obtain:

5.8 Let Y be a radical class of GM V -algebras. Then ψ0 (Y ) is a radical class of lattice ordered groups. In view of the above notation, we get ψ(Y ) = ψ0 (Y ). A simple calculation yields:

5.9

Let X be a radical class of lattice ordered groups and let Y be a radical class of GM V -algebras. Then ψ(ϕ(X)) = X,

ϕ(ψ(Y )) = Y.

We denote by R and R the collection of all radical classes of lattice ordered groups or the collection of all radical class of GM V -algebras, respectively. Both R and R are partially ordered by the class-theoretical inclusion. Hence we have mappings ϕ : R → R , ψ : R → R. 731


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From 5.9 and from the definitions of ϕ and ψ we obtain:

5.10 Both ϕ and ψ are bijections and ϕ−1 = ψ. X1 , X2 ∈ R, then X1 ⊆ X2 ⇐⇒ ϕ(X1 ) ⊆ ϕ(X2 ).

Further, if

We remark that the partial order end the class R was studied in detail in the paper [12]; it was also shown that R is a proper class. Hence R is a proper class as well.

6. Weak retract mappings A retract mapping of an algebra A is an endomorphism f of A such that f2 = f. Retract mappings of GM V -algebras were investigated in [23, Section 7]. Let M = (M ; ∧, ∨, ·, /, \, e) be a GM V -algebra; we consider the algebra M0 = (M ; ∧, ∨, ·, e). A mapping f : M → M is defined to be a weak retract mapping of M if it is a retract mapping of the algebra M0 . Each retract mapping of M is a weak retract mapping of M. If f is a weak retract mapping of M, then it need not be a retract mapping of M (cf. the example given below). We use the notation as in Proposition 3.2. Let z ∈ M . As above, we denote by z(G∗ ) the component of z in the direct summand G∗ . The meaning of z(Lγ ) is analogous. If z(G∗ ) = x and z(Lγ ) = y, then z = xy. We denote by T0 (M) the system of all triples of mappings (f1 , f2 , f3 ) such that (i0 ) f1 is a weak retract mapping of G∗ ; (ii0 ) f2 is a weak retract mapping of Lγ ; (iii0 ) f3 is a homomorphism of (Lγ )0 into the negative cone ((G∗ )− )0 of (G∗ )0 such that for each y ∈ Lγ , and each g ∈ G− , a) f1 (g) ∨ f3 (y) = e, b) f3 (f2 (y)) = f3 (y). Let R0 (M) be the set of all weak retract mappings of M. We prove that there exists a bijection ψ0 : R0 (M) → T0 (M). A related result concerning retract mappings of M was proved in [23]. First, let us assume that f is an element of R0 (M). We put f |G = f1 .

6.1

f1 is a weak retract mapping of G∗ .

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For proving this assertion it suffices to apply the same steps which were used in proving [23, Lemma 7.4]. Let y ∈ Lγ . We set f (y)(Lγ ) = f2 (y),

f (y)(G∗ ) = f3 (y).

We obtain mappings f : Lγ → Lγ ,

f3 : Lγ → G.

By the same method as in the proof of [23, Lemma 7.5] we obtain:

6.2 f2 is a weak retract mapping of Lγ . 6.3 f3 is a homomorphism of (Lγ )0 into

((G∗ )− )0 . Moreover, f3

satisfies the following conditions:

(c1 ) Let g ∈ G− and y ∈ Lγ . Then f1 (g) ∨ f3 (y) = e. (c2 ) Let y ∈ Lγ . Then f1 (f3 (y)) = e. (c3 ) Let y ∈ Lγ . Then f3 (f3 (y)) = f3 (y). P r o o f. From the definition of f3 we immediately conclude that f3 is a homomorphism of (Lγ )0 into (G∗ )0 . If y ∈ Lγ , then y e, whence f (y) f (e) = e; then f (y)(G∗ ) e, thus f3 (y) e. Therefore f3 is a homomorphism of (Lγ )0 into ((G∗ )− )0 . 1) Let g ∈ G− and y ∈ Lγ . Then g ∨ y = e. This yields f (g) ∨ f (y) = e. Since f (y) e we get f (y)(G∗ ) f (y), hence f (g) ∨ f3(y) = e. Clearly, f (y) = f1 (g). Thus (c1 ) is valid. 2) Let y ∈ Lγ . Put f3 (y) = x, f2 (y) = y1 . Thus f (y) = xy1 . Further, we set f1 (x) = x0 . In view of y e we get x e. For t1 , t2 ∈ M we write t1 ⊥t2 if t1 ∨ t2 = e. For each x ∈ G− we have x ⊥y, thus f (x )⊥f (y). Hence f (x)⊥f (y). Clearly f (x) = f1 (x) and so x0 ⊥xy1 . In view of xy1 x e we get x0 ⊥x. This yields f1 (x0 )⊥f1 (x). But f1 (x0 ) = x0 = f1 (x), thus x0 ⊥x0 , hence x0 = e. We obtain f1 (f3 (y)) = e. Therefore (c2 ) is valid. 3) As above, let z ∈ M , z = xy where x ∈ G and y ∈ Lγ . Thus f (z) = f1 (x)f3 (y)f2 (y). Put f (z) = z ,

f1 (x)f3 (y) = x ,

f2 (y) = y .

Hence z = x y and x ∈ G, y ∈ Lγ . Then f (z ) = f1 (x )f3 (y )f2 (y ) = f1 (f1 (x)f3 (y))f3 (f2 (z))f2 (f2 (y)) = f1 (x)f1 (f3 (y))f3 (f2 (y))f2 (y). 733


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In view of condition (c2 ) we get f (z ) = f1 (x)f3 (f2 (y))f2 (y). We have f1 (x)f3 (f2 (y)) ∈ G and f2 (y) ∈ Lγ . Since f (z) = f (z ) we obtain f1 (x)f3 (y) = f1 (x)f3 (f2 (y)). Therefore, f3 (y) = f3 (f2 (y)). We have verified that (c3 ) holds.

Under the notation as above we put ψ(f ) = (f1 , f2 , f3 ). From 6.1, 6.2 and 6.3 we obtain:

6.4

ψ is a mapping of the set R0 (M) into the system T0 (M).

Now let us suppose that f1 and f2 are as in the conditions (i0 ) and (ii0 ). Further, assume that f3 is a homomorphism of Lγ into (G∗ )− such that the conditions (c1 ) and (c2 ) from Lemma 6.3 are satisfied. Let z ∈ M , z = xy, x ∈ G, y ∈ Lγ . We denote h(z) = f1 (x)f3 (y).

(1)

Hence h is a mapping of M into G.

6.5

Let x0 ∈ G and y0 ∈ Ly . Then f1 (x0 )f3 (y0 ) = f3 (y0 )f1 (x0 ).

P r o o f. Both f1 (x0 ) and f3 (y0 ) are elements of G. It is well known that if p, q are elements of G, then p ∧ q = e =⇒ pq = qp. (∗) Using (∗) and the condition (c1 ) we obtain, by a simple calculation, that the assertion of the lemma is valid.

6.6

Let z1 , z2 ∈ M . Then h(z1 z2 ) = h(z1 )h(z2 ).

P r o o f. For z1 and z2 we apply analogous notation as in (1). Put z = z1 z2 . In view of Proposition 3.2 we have x = x1 x2 and y = y1 y2 . Thus according to Lemmas 6.1 and 6.3, h(z1 z2 ) = f1 (x1 x2 ) · f3 (y1 y2 ) = f1 (x1 )f1 (x2 )f3 (y1 )f3 (y2 ). Lemma 6.5 yields h(z1 z2 ) = f1 (x1 )f3 (y1 )f1 (x2 )f3 (y2 ) = h(z1 )h(z2 ).

6.7

Let p, q ∈ M , p ∨ q = e. Then pq = p ∧ q.

P r o o f. This is a consequence of [5, Lemma 2.10]. 734



6.8

Let z1 , z2 ∈ M . Then h(z1 ∧ z2 ) = h(z1 ) ∧ h(z2 ).

P r o o f. Put z = z1 ∧ z2 . According to Lemma 6.7 and condition (c1 ), h(z) = f1 (x) ∧ f3 (y). Further, in view of Proposition 3.2, x = x1 ∧ x2 and y = z ∧ y2 . Then (cf. Lemma 6.1 and Lemma 6.3) we have h(z1 ∧ z2 ) = f1 (x1 ∧ x2 ) ∧ f3 (y1 ∧ y2 ) = f1 (x1 ) ∧ f1 (x2 ) ∧ f3 (y1 ) ∧ f3 (y2 ) = (f1 (x1 ) ∧ f3 (y1 )) ∧ (f1 (x2 ) ∧ f3 (y2 )) = (f1 (x1 )f3 (y1 )) ∧ (f1 (x2 )f3 (y2 )) = h(z1 ) ∧ h(z2 ).

6.9

The mapping h is a homomorphism with respect to the oper-

ations · and ∧.

From Corollary 6.9 we also obtain that if z1 , z2 ∈ M and z1 z2 , then h(z1 ) h(z2 ).

6.10

Let z1 , z2 ∈ M . Then h(z1 ∨ z2 ) = h(z1 ) ∨ h(z2 ).

P r o o f. Put z1 ∧ z2 = u and z1 ∨ z2 = v. Using Proposition 3.2 and the wellknown properties of lattice ordered groups we obtain that there exists p ∈ M , p e such that vp = z1 , z2 p = u. Thus in view of Lemma 6.6, h(v)h(p) = f (z1 ),

h(z2 )h(p) = h(u).

Further, Lemma 6.8 yields h(u) = h(z1 ) ∧ h(z2 ). Put t = h(z1 ) ∨ h(z2 ). Since z1 v and z2 v we get h(z1 ) h(v) and h(z2 ) h(v), thus t h(v). We have to verify that t = h(v). By the way of contradiction, suppose that t < h(v) (cf. Fig. 1). We have h(z2 )h(p) = h(z1 ) ∧ h(z2 ) and hence h(p) = h(z1 ). This yields h(v)h(p) = h(p).

(2)

In view of Proposition 3.2 and according to t < h(v), the relation (2) cannot hold. Hence t = h(v). 735


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u h(v)

ut @ @ @u h(z2 )

h(z1 ) u @ @ @u

h(z1 ) ∧ h(z2 ) Figure 1

In view of Corollary 6.9 and Lemma 6.10 we get:

6.11

The mapping h : M → G is a homomorphism of M into G∗ .

Let us apply the notation as above. Let z ∈ M , z = xy, x ∈ G, y ∈ Lγ . We put f0 (z) = h(z)f2 (y) = f1 (x)f3 (y)f2 (y). For a triple (f1 , f2 , f3 ) belonging to the system T (M) we put χ((f1 , f2 , f3 )) = f0 , where f0 is as above.

6.12

χ is a mapping of the system T0 (M) into the set R0 (M).

P r o o f. Let (f1 , f2 , f3 ) ∈ T0 (M) and χ((f1 , f2 , f3 )) = f0 . In view of Lemma 6.11, h(z) is a homomorphism of M into G∗ . Further, f2 is an endomorphism of Lγ . Then according to Proposition 3.2 we conclude that f0 is an endomorphism of M. It remains to verify that f02 = f0 . To this end, if suffices to apply the same calculation as in Part 3 of the proof of Lemma 6.3. From the definitions of the mappings ψ and χ we immediately obtain that χ = ψ −1 . Hence we have:

6.13

Let M be a GM V -algebra. The mapping ψ is a bijection of the set R0 (M) onto the system T0 (M).

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The following example shows that a weak retract mapping need not be a retract mapping. Let H be lattice ordered group having more than one element. Put G = H × H, G1 = H. Further, let M be as in Proposition 2.3, where Lγ = G− 1. The elements of M will be written in the form (x, x1 , y), where (x, x1 ) ∈ G and y ∈ Lγ . For z = (x, x1 , y) ∈ M we put f (z) = (x , x1 , y ),

where

x = x, x1 = y, y = y.

Thus for (x, x1 ) ∈ G we have f1 ((x, x1 )) = (x, 0). Further, if y ∈ Lγ , then f2 (y) = y and f3 (y) = (0, y). The conditions (i0 ), (ii0 ) and (iii0 ) are satisfied, thus f is a weak retract mapping of the GM V -algebra M. There exists y ∈ Lγ with y = 0, hence f3 (y) = (0, 0); therefore in view of [23, Section 7], we infer that f fails to be retract mapping of M.

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´ JAKUBÍK JAN [16] JAKUBÍK, J.: Closure operators on the lattice of radical classes of lattice ordered groups, Czechoslovak Math. J. 38 (1988), 71–77. [17] JAKUBÍK, J.: On a radical class of lattice ordered groups, Czechoslovak Math. J. 39 (1989), 641–643. [18] JAKUBÍK, J.: Radical classes of complete lattice ordered groups, Math. Slovaca 49 (1989), 417–424. [19] JAKUBÍK, J.: Closed convex -subgroups and radical classes of convergence -groups, Math. Bohem. 122 (1997), 301–315. [20] JAKUBÍK, J.: Radical classes of MV -algebras, Czechoslovak Math. J. 49 (1999), 191–211. [21] JAKUBÍK, J.: Weak (m, n)-distributivity of lattice ordered groups and of generalized MV -algebras, Soft Comput. 10 (2006), 119–124. [22] JAKUBÍK, J.: K-radical classes and product radical classes of MV-algebras, Math. Slovaca 58 (2008), 143–154. [23] JAKUBÍK, J.: Direct summands and retract mappings of generalized MV-algebras, Czechoslovak Math. J. 58 (2008), 183–202. [24] MEDVEDEV, N. YA.: On the lattice of radicals of a finitely generated -group, Math. Slovaca 33 (1983), 185–188 (Russian). [25] RACH˚ UNEK, J.: A non-commutative generalization of MV-algebras, Czechoslovak Math. J. 52 (2002), 255–273. [26] DAO RONG TON: Product radical classes of -groups, Czechoslovak Math. J. 43 (1992), 129–142. Received 7. 9. 2007

Mathematical Institute Slovak Academy of Sciences Greˇsa ´kova 6 SK–040 01 Koˇsice SLOVAKIA E-mail : [email protected]

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