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Volume 14, Issue 1 (Jan 2016)

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On annihilators in BL-algebras

Yu Xi Zou
  • College of Mathematics, Northwest University, Xi’an, 710127, P.R. China
/ Xiao Long Xin
  • Corresponding author
  • College of Mathematics, Northwest University, Xi’an, 710127, P.R. China
  • Email:
/ Peng Fei He
  • School of Mathematics and Information Science, Shaanxi Normal University, Xi’an, 710119, P.R. China
Published Online: 2016-05-21 | DOI: https://doi.org/10.1515/math-2016-0029

Abstract

In the paper, we introduce the notion of annihilators in BL-algebras and investigate some related properties of them. We get that the ideal lattice (I(L), ⊆) is pseudo-complemented, and for any ideal I, its pseudo-complement is the annihilator I of I. Also, we define the An (L) to be the set of all annihilators of L, then we have that (An(L); ⋂,∧An(L),⊥,{0}, L) is a Boolean algebra. In addition, we introduce the annihilators of a nonempty subset X of L with respect to an ideal I and study some properties of them. As an application, we show that if I and J are ideals in a BL-algebra L, then JI is the relative pseudo-complement of J with respect to I in the ideal lattice (I(L), ⊆). Moreover, we get some properties of the homomorphism image of annihilators, and also give the necessary and sufficient condition of the homomorphism image and the homomorphism pre-image of an annihilator to be an annihilator. Finally, we introduce the notion of α-ideal and give a notation E(I ). We show that (E(I(L)), ∧E, ∨E, E(0), E(L) is a pseudo-complemented lattice, a complete Brouwerian lattice and an algebraic lattice, when L is a BL-chain or a finite product of BL-chains.

Keywords: BL-algebra; MV-algebra; Ideal; Annihilator; Homomorphism

MSC 2010: 08A72; 06B75

1 Introduction

It is well known that logic gives a technique for the artificial intelligence to make the computers simulate human being in dealing with certainty and uncertainty in information. And as uncertain information processing, nonclassical logic has become a formal and useful tool for computer science to deal with uncertain information, fuzzy information and intelligent system. Various logical algebras have been proposed and researched as the semantical systems of non-classical logical systems. Among these logical algebras, residuated lattices were introduced by Ward and Dilworth in 1939 to constitute the semantics of Höhle Monoidal Logic which are the basis for the majority of formal fuzzy logic. Apart from their logical interest, residuated lattices have interesting algebraic properties and include two important classes of algebras: BL-algebras and MV-algebras. In order to study the basic logic framework of fuzzy set system, based on continuous triangle module and under the theoretical framework of residuated lattices theory, Hájek [1] proposed a new fuzzy logic system–BL-system and the corresponding logic algebraic system–BLalgebra. MV-algebras were introduced by Chang [2] in order to give an algebraic proof of the completeness theorem of Lukasiewice system of many valued logic.

The notion of ideals has been introduced in many algebraic structures such as lattices, rings, MV-algebras. Ideals theory is a very effective tool for studying various algebraic and logical systems. In the theory of MV-algebras the notion of ideals is at the center and deductive systems and ideals are dual notions, while in BL-algebra, with the lack of a suitable algebraic addition, the focus is shifted to deductive systems also called filters. So the notion of ideals is missing in BL-algebras. To fill this gap the paper [3] introduced the notion of ideals in BL-algebras, which generalized in a natural sense the existing notion in MV-algebras and subsequently all the results about ideals in MV-algebras. The paper also constructed some examples to show that, unlike in MV-algebras, ideals and filters are dual but behave quite differently in BL-algebra. So the notion of ideal from a purely algebraic point of view has a proper meaning in BL-algebras.

A lot of work has been done with respect to the co-annihilators and the annihilators. For example, in [4], Davery studied the relationship between minimal prime ideals conditions and annihilators conditions on distributive lattices. Turunen [5] defined co-annihilator of a non-empty set X of L and proved some of its properties on BL-algebras. They got A as a prime filter if and only if A is linear and A ≠ {1}. Also, in [6] B. A.Laurentiu Leustean introduced the notion of the co-annihilator relative to F on pseudo-BL-algebras, which is a generalization of the co-annihilator, and they also extended some results obtained in [4]. Moreover, in [7], B. L. Meng et al. defined the generalized co-annihilator of BL-algebras as a generalization of co-annihilator on BL-algebras. In [8] W.H. Cornish defined the notion of α-ideals in distributive lattices, where an ideal I is an α-ideal if I¯¯=I. Since the notion of ideals in BL-algebras has been defined in paper [3], we think it is a new direction to study the ideals by the concept of annihilators, which will enrich and develop the theory of ideals in BL-algebras.

This paper is organized as follows: In Section 2, we review some basic definitions and results about BL-algebras. In Section 3, we introduce the notion of the annihilators of BL-algebras and the notion of the annihilators of a nonempty subset X with respect to an ideal I. Also, we investigate the homomorphism image of annihilators. In Section 4, we introduce the notion of α-ideals and give a notation E(I). Then we focus on the algebraic structures of the set (E(I(L)).

2 Preliminaries

([1]). An algebra structure ((L, ∧, ∨,⊙, →, 0, 1) of type (2, 2, 2, 2, 0, 0) is called a BL-algebra, if it satisfies the following conditions: for all x, y, zL

  • BL1

    (L, ∧, ∨, 0, 1) is a bounded lattice relative to the order;

  • BL2

    (L, ⊙, 1) is a commutative monoid;

  • BL3

    xyz if and only if xyz;

  • BL4

    xy = x ⊙ (xy);

  • BL5

    (xy) ∨ (yx) = 1.

In what follows, by L we denote the universe of a BL-algebra (L, ∧, ∨, ⊙, →, 0, 1). For any xL and a natural number n, we define x¯=x0,x¯¯=(x¯)¯,x0=1 and xn = xn-1x for n ≥ 1.

([1]). Let L be a BL-algebra. For all x, y; zL, the followings hold:

  1. x ⊙ (xy) ≤ y,

  2. xyxyxy,

  3. xyxy = 1,

  4. x → (yz) = (xy) → z = y → (xz),

  5. xy implies zxzy; yzxz and xzyz,

  6. yx ≤ (zy) → (zx),

  7. xy ≤ (yz) → (xz),

  8. xy = ((xy) → y) ∧ ((yx) → x),

  9. (xy) → zx → (yz),

  10. 1 → x = x, xx = 1, x → 1 = 1,

  11. 1¯=0,0¯=1,1¯¯=1,0¯¯=0

  12. xy = xy; xy = xy,

  13. xyxzyz,

  14. xy¯¯=xy¯¯,

  15. xy¯¯=x¯¯y¯¯,

  16. (L, ∧, ∨) is a distributive lattice.

For every x, yL, we adopt the notation: xy = xy.

([3]). In every BL-algebra L, the following holds:

  1. The operationis associative, that is, for every x, y, zL, (xy) ⊘ z = x ⊘ (yz);

  2. The operationis compatible with the order, that is, for every x, y, z, tL, such that xy and zt, then xzyt.

([3])If L is a BL-algebra that is not an MV-algebra, then there is an element xL such that x¯¯x. Hence x ⊘ 0 ≠ 0 ⊘ x and we conclude that the operationis not commutative in general. The associative and noncommutative operationwill be called the pseudo-addition of the BL-algebra.

([9])Let L and M be two BL-algebras. A mapping f : LM is said to be a homomorphism, if for any x, yL, we have (1) f (xy) = f (x) ⊙ f (y); (2) f (xy) = f (x) → f (y); (3) f (0) = 0. If f is an injection (a surjection), then f is said to be an injective (a surjection) homomorphism. If f is a bijection, then f is said to be an isomorphism.

Let L, M be two BL-algebras, and f : LM be a homomorphism. Then for any x, yL, (1) f (xy) = f (x) ∧ f (y); (2) f (xy) = f (x) ∨ f (y).

([9])Let (L, ∧, ∨) be a lattice and let f : LL be a closure. Then Imf is a lattice in which the lattice operations are given by inf {a, b} = ab, sup{a, b} = f (ab).

([9]). Let L be a BL-algebra and I be a nonempty subset of L. We say that I is an ideal of L if it satisfies:

  • I1:

    for every x, yL, if xy and yI, then xI ;

  • I2:

    for every x, yI, xyI.

([3]). Let L be a BL-algebra and I be an ideal of L. Then for every x, yI, we have xyI and xyI.

We recall that the smallest ideal containing A in L is called the ideal generated by the subset A in L and it is denoted by 〈A〉. It is also the intersection of all the ideals containing A.

[3]. For every subset A of a BL-algebra L, we have

  1. If A is empty, thenA〉 = {0};

  2. If A is not empty, thenA〉 = {aL | ax1x2 ⊘· · · ⊘ xn; x1, x2; · · · xnA}.

([3])Let L be a BL-algebra and P be an ideal of L. We say P is a prime ideal if it satisfies for every x, yL, xyP or yxP.

([3]). An ideal P of a BL-algebra L is prime if and only if for any x, yL, xyP implies that xP or yP.

3 Annihilators in BL-algebras

Definition 3.1. Let A be a non-void subset of a BL-algebra L, then we call the set A= {xL | ax = 0 for all aA} is an annihilator of A.

Let L ={0, a, b, c, d, 1} be a set, where 0 ≤ ab ≤ 1, 0 ≤ ad ≤ 1 and 0 ≤ cd ≤ 1. The Cayley tables are as follows:

Then (L, ∧, ∨, ⊙, →, 0, 1) is a BL-algebra. Now, we consider A = {0, a}, it is easy to check that A = {0, c}.

Let L be a BL-algebra and A be a subset of L, then A is an ideal of L. Moreover, if A ≠ {0}, then A is proper.

Proof. For every aA, we have a ∧ 0 = 0, hence 0 ∈ A, which implies A is nonempty.

I1: Assume that yA, xy, since for all aA, xaya = 0, then we have xA;

I2: Assume that xA, yA. Let aA, then a ∧ (xy)=a ⊙ (a → (xy))=a ⊙ (x → (ay)) ≤ a ⊙ (xaa ⊙ (ay))=a ⊙ ((xa) → (ay)) = a ⊙ ((xa) → 0)=a ⊙ (xa) ≤ a(a¯¯x¯¯=a¯¯x¯¯=ax¯¯=0. Hence xy = xyA, which implies A is an ideal.

If A ≠ {0}, then there is aA such that a ≠ 0, so 1 ∧ a = a ≠ 0, then we have 1 ∉ A. Therefore, A is proper. □

Let L be a BL-algebra. Then the following conclusions hold: for all x, a, bL,

  1. {1} = {0}, {0} = L;

  2. if ab, then {b} ⊆ {a};

  3. {a} ∩ {b} = {ab};

  4. {a} ∪{b} ⊆ { ab};

  5. if x ∈ {a}, then ax and xa.

Proof. (1) For all x ∈ {1}, x = x ∧ 1 = 0, so x = 0, which implies {1} ={0}. For all xL, since x ∧ 0 = 0, we have L ⊆ {0}, and evidently, {0}L, so {0} = L.

(2) For all x ∈ {b}, we have axbx = 0, so ax = 0, so we have x ∈ {a}.

(3) x ∈ {a} ∩ {b} iff x ∈ {a} and x ∈ {b} iff xa = 0 and xb = 0 iff x ∧ (ab) = 0 iff x ∈ {ab}.

(4) If x ∈ {a}[{b}, then x ∈ {a} or x ∈ {b}, so xa = 0 or xb = 0, and so xab = 0, therefore, x ∈ {ab}.

(5) If x ∈ {a}, then xa = 0, so we have axxa = 0, then ax = 0, therefore, ax and xa. □

Let L be the BL-algebra in Example 3.2. We have a = {0, c}, c = {0, a, b}, and ac = 0, so a [c = {0, a, b, c}. Hence 0 = La [c. Therefore, we do not have the equation for Proposition 3.4(4).

Let L be a BL-algebra. Then the following conclusions hold: for all x, y, a, bL,

  1. if x ∈ {a}, y ∈ {ab}, then xy ∈ {ab};

  2. if x ∈ {a}, y ∈ {ab}, then xy ∈ {ab};

  3. if x ∈ {a}, y ∈ {b}, then xy, xy, xy ∈ {ab}.

Proof. (1) Since y ∈ {ab} and bab, by Proposition 3.4 (2), we have {ab} ⊆ {b}, so y ∈ {b},

since {b} is a down set, we get xy ∈ {b} ⊆ {ab}.

(2) Since y ∈ {a → b} and bab, by Proposition 3.4 (2), we get {a → b} ⊆ {b}, so y ∈ {b}. Since {b} is a down set, we get xy ∈ {b} ⊆ {a ∧ b}.

(3) Since x ∈ {a}, y ∈ {b} and aba, b, by Proposition 3.4 (2), we have {a} ⊆ {a ∧ b}, {b} ⊆ {a ∧ b}, so x, y ∈ {a ∧ b}, and so xy ∈ {a ∧ b}. Moreover, since {a ∧ b} is a down set, we get xy, xy ∈ {a ∧ b}. □

For any ∅ ≠ XL, 〈X〉∩ X = {0}.

Proof. Assume that a ∈ 〈X〉 ∩ X, then we have ax1x2 ⊘ · · · ⊘ xn xiX, and axi = 0 i = 1, 2, · · · n.

Now we prove that a ∧(x1x2 ⊘· · · ⊘xn) ≤ (ax1) ⊘(ax2) ⊘· · · ⊘(axn). Firstly, a ∧(xy) → ((ax)⊘(ay))≤ax → ((a ∧ x) → (ay)) = ((ax) ⊙ (a ∧ x)) → (ay) = 0 → (ay) = 1, so a∧(xy) ≤ (ax) ⊘(ay). Then assume that when k = n, a ∧ (x1x2 ⊘· · · ⊘xn) ≤ (ax1) ⊘(ax2) ⊘· · · ⊘(axn) holds. When k = n + 1, a ∧ (x1x2 ⊘· · · ⊘ xnxn+1) ≤ (a ∧ (x1x2 ⊘· · · ⊘ xn) ⊘ (axn+1) ∨ (ax1) ⊘ (ax2) ⊘· · · ⊘ (axn) ⊘ (axn+1).

Considering the above we have a ∧ (x1x2 ⊘· · · ⊘ xn) ≤ (ax1) ⊘ (ax2) ⊘· · · ⊘ (axn). Then aaa ∧ (x1x2 ⊘· · · ⊘ xn) ≤ (ax1) ⊘ (ax2) ⊘· · · ⊘ (axn) = 0 ⊘ 0 ⊘· · · ⊘ 0 = 0. Therefore, 〈X〉∩ X = {0}. □

Let A be an ideal of L and A be linear(which means that A is totally ordered). Then A is a prime ideal.

Proof. Assume that A is an ideal which is linear, and abA but a, bA. Then there are x′, x″ ∈ A, such that ax′ ≠ 0, and bx″ ≠ 0. Set x = x′∨ x″. Then xA as A is an ideal. Clearly, ax = a ∧ (x′ ∨ x″) = (ax) ∨(ax″). Similarly, we have bx ≠ 0. Since axx, bxx, we conclude ax, bxA. As A is linear, we may assume that axbx. Now, 0 = (ab) ∧ x = a ∧ (bx) ≥ a ∧ (ax) = ax, which contradicts the fact ax ≠ 0, which implies that aA or bA. Therefore, A is prime. □

Let L be a BL-algebra, if XYL, then YX;

Proof. If zY , we have zy = 0. Then for any xXY, zx = 0, and so z ∈∩xX {x} = X. This means that YX. □

Let L be a BL-algebra, for any ∅ ≠ XL, the following hold:

  1. X = ∩xX {x};

  2. XX⊥⊥;

  3. X = X⊥⊥;

  4. X = 〈X〉.

Proof. (1) aX ⇔ for all xX, ax = 0 ⇔ for all xX, axa ∈∩xX {x}.

(2) By the definition of annihilator, we have X⊥⊥ = {a ∈ L | ax = 0 for all xX}. So for all xX, if bX, then bx = 0, hence bX⊥⊥.

(3) By (2) taking X = X we have XX⊥⊥⊥. Conversely, by (1) and Proposition 3.8, we have X⊥⊥⊥X, therefore, X = X⊥⊥⊥.

(4) Since X ⊆ 〈X〉, by Proposition 3.9, we have 〈XX. Now we prove X⊆ 〈X. Let yX, so for any xX, we have xy = 0. For any z ∈ 〈X〉, there are x1,x2, · · ·xnX, such that zx1x2 ⊘· · ·⊘ xn. So yzy ∧ (x1x2 ⊘· · ·⊘ xn) ≤ (yx1) ⊘ (yx2) ⊘· · ·⊘ (yxn) = 0 ⊘ 0 ⊘· · ·⊘ 0 = 0. Hence y ∈ 〈X, that is, X⊆ 〈X. Therefore, X = 〈X. □

Let L be a BL-algebra, and X, YL. Then

  1. L = {0};

  2. X⊥⊥ X = {0};

  3. (XY) = XY;

  4. XY ⊆ (XY).

Proof. (1) If there is aL such that a ≠ 0 and aL, then 1 ∧ a = a ≠ 0, this is a contradiction, so a = 0.

(2) If xX⊥⊥X, by the definition of annihilator, we have x = xx = 0, so X⊥⊥X ⊆ 0. Conversely, since for every YL there is 0 ∈ Y, we get 0 ∈ X⊥⊥ and 0 ∈ X, so 0 ∈ X⊥⊥X, finally X⊥⊥X = {0}.

(3) Since XXY and YXY, we have (XY)X and (XY)Y, so (XY)XY, conversely, for any aXY , we have aX and aY, ie. for any xX, yY, we have ax = 0 and ay = 0. So for any tXY, we always have at = 0, hence a ∈ (XY).

(4) Since XYX and XYY, we have X ⊆ (XY) and Y⊆ (XY), so XY ⊆ (XY). □

Let L be the BL-algebra in Example 3.2. Take X ={a, c}, Y = {c, b}, then X ∩ Y = {c}, so we have X ={0}, Y = {0}, and (XY) = {0, a, b}. Hence (XY)XY . Therefore, we do not have the equation for Proposition 3.11(4)

Let L be a BL-algebra, and≠ X, YL. For all a, bL, if aX, bY, then (1) ab ∈ (XY), (2) ab ∈ (XY) .

Proof. (1) If aX, bY, then abX and abY, so abXY = (XY).

(2) If aX, bY , then aX ⊆ (XY) and bY ⊆ (XY), so ab ∈ (XY). □

Let L be a BL-algebra, X, YL. Then XY ={0} if and only if XY ⊥⊥ and YX⊥⊥.

Proof. ⇒ For all aX, bY , we have abXY , since XY ={0}, we get ab = 0, by definition of annihilator, we get aY ⊥⊥ and bX⊥⊥, so XY ⊥⊥ and Y X⊥⊥.

⇐ If XY ⊥⊥ and Y X⊥⊥, then XY Y ⊥⊥Y ={0}, so XY ⊆ {0}. Clearly, {0} ⊆ XY , so XY ={0}. □

Let L be a BL-algebra, I be a non-empty subset of L. Then there is a XL, satisfied I = X if and only if I is a down set, and I ⊥⊥ = I, IX⊥⊥ ={0}, XI ={0}.

Proof. ⇒If I = X, it is clear that I is a down set, I ⊥⊥ = X⊥⊥⊥ = X = I, IX⊥⊥ = XX⊥⊥ ={0}. Then we have XI = XX⊥⊥ ={0} by Proposition 3.11 (2)

⇐ Since IX⊥⊥ ={0}, for any iI, for any xX⊥⊥, then ixIX⊥⊥ ={0}, so iX⊥⊥⊥ . We have already known X⊥⊥⊥ = X, so IX. Since XI ={0}, then similarly, we get XI ⊥⊥, since I ⊥⊥ = I, we get XI. Therefore, I = X . □

Let L be a BL-algebra, if an ideal I which is linear contains an element x ≠ 0 and xx = 1, then x is the largest element of I.

Proof. By xx = 1, we have xx¯¯ = 0. So xxxx¯¯ = 0. Let aI, then a = a ∧ 1 = a ∧ (xx) = (ax) ∨ (ax), where the last equation follows by the distributive of L. Since I is linear, by Proposition 3.8, we have I is a prime ideal. Since xx = 0 ∈ I , either xI or xI . As xx = x ≠ 0, we necessarily have xI . Hence for any aI, we have ax = 0, which implies that a = ax, thus ax. Therefore, x is the largest element of I. □

Let L be a BL-algebra. Then the ideal lattice I(L) is pseudo-complemented and for any ideal I of L, its pseudo-complement is I .

Proof. By Proposition 3.7, we have II ={0}. Let G be an ideal of L such that IG ={0}, we shall prove that GI . Let aG, for any xI, then we have xaxI, xaaG, so xaIG ={0}. Hence xa = 0 for any xI, then we have aI . So I is the largest ideal such that IG ={0}. It follows that I is the pseudo-complement of I. □

Let An(L) ={X | XL} be the set of annihilators of L. Since X = 〈X, we get that An(L) = {I | II(L)}. Hence An(L) is the set of pseudo-complements of the pseudo-complemented lattice I(L).

Let L be a BL-algebra, I, JI(L). Then

  1. {0};LAn(L);

  2. IAn(L) ⇔ I⊥⊥ = I ;

  3. ⊥⊥: XX⊥⊥ is a closure map;

  4. I ∩ (IJ) = IJ ;

  5. (IJ)⊥⊥ = I ⊥⊥J ⊥⊥;

  6. I, JAn(L), then IAn(L) J = IJ ;

  7. (IJ) = I J ;

  8. if I, JAn(L), then IAn(L) J = (I J ).

Proof. (1) By Propositions 3.4 and 3.11.

(2) Assume that IAn(L), then there exists XL such that X = I, so we get I ⊥⊥ = X⊥⊥⊥ = X = I. The converse is clear.

(3) By Propositions 3.9, we know the function f : XX⊥⊥ is isotone and by Propositions 3.10, we get that f = f 2idL. So, XX⊥⊥ is a closure map.

(4) Since (IJ) ∩ (IJ) ={0}, by Proposition 3.17, we get I ∩ (IJ)J and so I ∩ (IJ)IJ . Conversely, by IJJ, we get J ⊆ (IJ), so IJ I ∩ (IJ). Therefore, I ∩ (IJ) = IJ .

(5) Since IJI, J, we get (IJ)⊥⊥I ⊥⊥J ⊥⊥. Conversely, (IJ) ∩ (IJ) ={0}) ⇒ I ∩ (IJ)J = J ⊥⊥⊥IJ ⊥⊥ ∩ (IJ) ={0} ⇒ J ⊥⊥ ∩ (IJ)I = I ⊥⊥⊥)I ⊥⊥J ⊥⊥ ∩ (IJ) ={0}) I ⊥⊥J ⊥⊥⊆ (IJ)⊥⊥. So we get (IJ)⊥⊥ = I ⊥⊥J ⊥⊥.

(6) By (3) and Proposition 2.6, we have IAn(L) J = IJ.

(7) Since I, JIJ, we get (IJ)I J = I ⊥⊥⊥J ⊥⊥⊥ = (I J )⊥⊥. Conversely, II ⊥⊥⊆ (I J ), similarly, we have JJ ⊥⊥⊆ (I J ), so IJ ⊆ (I J ), hence (I J )⊥⊥⊆ (IJ). Therefore, (IJ) = (I J )⊥⊥ = I ⊥⊥⊥J ⊥⊥⊥ = I J .

(8) By (3) and Proposition 2.6, we have IAn(L) J = (IJ)⊥⊥, then by (7) we have IAn(L) J = (I J ) . □

Let L be a BL-algebra. Then (An(L); ∩ ; ∨An(L), ⊥, {0}; L) is a Boolean algebra.

Proof. Firstly, we show that An(L) is distributive, it suffices to prove that: for all I, J, HAn(L), H ∩ (IAn(L) J) ⊆ (HI) ∨An(L) (HJ). Now, let K = (HI) ∨An(L) (HJ), then HIK = K⊥⊥ gives HIK ={0} and so HKI . Similarly, HKJ and therefore HKI J = (I J )⊥⊥. It follows that HK ∩ (I J ) ={0} and hence H ∩ (IAn(L) J) = H ∩ (I J )K⊥⊥ = K = (HI) ∨An(L) (HJ).

Secondly, we show that An(L) is complemented, observe that L ={0}An(L) and {0}= LAn(L). Since for every IAn(L) we have II ={0} and IAn(L) I = (I I ⊥⊥) ={0} = L. So the complement of IAn(L) is I . Therefore, An(L) is a Boolean algebra. □

Let L be a BL-algebra, ∅ ≠ XL, I be an ideal of L and f be an endomorphism. We define the annihilator of X with respect to I to be the set XIf ={aL | f (a) ∧ xI, ∀xX}.

If f = idL, we denote XI:=XIidL ={aL | axI, ∀xX}.

Let L ={0, a, b, c, 1} be a set, where 0 ≤ ac ≤ 1 and 0 ≤ bc ≤ 1. The Cayley tables are as follows:

Then (L, ∧, ∨, ⊙, →, 0, 1) is a BL-algebra. Now we define a map f as follows: f (0) = 0, f (a) = b, f (b) = a, f (c) = c, f (1) = 1, then we can check that f is an endomorphism. Let X ={a, c}, and I ={0, a}, then we get XIf = {0, b}.

Let L = {0, a, b, 1} be a set, where 0 ≤ a, b ≤ 1. The Cayley tables are as follows:

Then (L, ∧, ∨, ⊙, →, 0, 1) is a BL-algebra. Let X = {0, b}, and I = {0, a}, then we get XI = {0, a}.

Let L be a BL-algebra, ∅ ≠ XL, I be an ideal of L and f be an endomorphism. Then XIf is an ideal of L.

Proof. Clearly, 0 ∈ XIf, so XIf is nonempty.

I1: Let aXIf and bL such that ba. Then f (b) ≤ f (a). It follows that f (b) ∧ xf (a) ∧ x. Since f (a) ∧ xI, we have f (b) ∧ xI, which implies bXIf;

I2: Let a, bXIf, then f (a)∧xI and f (b)∧xI for all xX. Since f (ab)∧x = (f (a)⊘f (b))∧x ≤ (f (a) ∧ x) ⊘ (f (b) ∧ x), we have f (ab) ∧ xI, that is, abXIf. □

Let L be a BL-algebra, f be an endomorphism, I, JI(L) and ∅ ≠ X, X′ ⊆ L. Then we have:

  1. IJ implies XIfXJf;

  2. XXimplies (X)IfXIf;

  3. (λΛXλ)If=λΛ(Xλ)If;

  4. XIf=xXxIf;

  5. XλΛIλf=λΛXIf;

  6. XIf=XIf;

  7. Ker(f)X{0}f,L{0}f=Ker(f).

Proof. (1) Let IJ and aXIf. Then f (a) ∧ xI for all xX. So we have f (a) ∧ xJ for all xX, that is, aXJf. Therefore, XIfXJf.

(2) Let XX′ and a(X)If. Then f (a) ∧xI for all xX. Hence f (a) ∧xI for all xX, which implies aXIf. Therefore, (X)IfXIf.

(3) By (2) we have (λΛXλ)If(Xλ)If for all λ ∈⋀, so we get (λΛXλ)IfλΛ(Xλ)If. Conversely, let aλΛ(Xλ)If, we have a(Xλ)If for all λ ∈Λ. Hence f (a) ∧ xλI for all xλXλ and

λ ∈ Λ, which implies a(λΛXλ)If. Therefore, (λΛXλ)If=λΛ(Xλ)If.

(4) It is clear by (3).

(5) We have aXλΛIλf if and only if f (a) ∧ x ∈ ∩ λΛ Iλ for all xX, and if and only if f (a) ∧ xIλ for all xX and λ ∈ Λ, which is equivalent to aXIf for all λ ∈ Λ, that is, aλΛXIλf.

(6) Since X ⊆ 〈X〉, by (2) we get XIfXIf. Conversely, let aXIf and z ∈ 〈X〉. Then f (a) ∧ xI for all xX. Since z ∈ 〈X〉, then there exist x1;x2; ··· ;xnX such that zx1, ⊘x2 ⊘···⊘ xn. It follows that f (a) ∧ zf (a) ∧ (x1x2 ⊘···⊘ xn) ≤ (f (a) ∧ x1) ⊘ (f (a) ∧ x2) ⊘···⊘(f (a) ∧ xn), since f (a) ∧ xiI for all 1 ≤ in, we get f (a) ∧ zI, which implies that a ∈ 〈XIf. Therefore, 〈XIf = XIf.

(7) Let a ∈ Ker(f), we have f (a) = 0, then f (a) ∧ x = 0 for all xX, that is, aX{0}f, hence Ker(f) ⊆ X{0}f. Let aL{0}f, then f (a) ∧ x = 0 for all xL. In particular, taking x = f (a), we have f (a) = 0, which implies a ∈ Ker(f). Hence L {0}f ⊆ Ker(f). Conversely, by Ker(f) ⊆ X{0}f, for any ∅ ≠ XL, taking X = L, so Ker(f) ⊆ L{0}f. Therefore, L{0}f =Ker(f). □

Let I be an ideal of a BL-algebra L, f be an endomorphism and a, bL. Then (1) ab implies bIfaIf ; (2) (ab)If = aIfbIf.

Proof. (1) Let xbIf, then f (x) ∧ bI and f (x) ∧ af (x) ∧ b, since ab. It follows that f (x) ∧ aI, that is, xa If;

(2) Since a, bab, by (1) we have that (ab)IfaIf, bIf, so (ab)IfaIfbIf. Conversely, let xa IfbIf, that is, af (x), bf (x) ∈ I, it follows f (x) ∧ (ab) = (f (x) ∧ a) ∨ (f (x) ∧ b) ∈ I, as I is an ideal of L, and L is a distributive lattice. Therefore, x ∈ (ab)If; □

Let L be a BL-algebra, ∅ ≠ XL, I be an ideal of L. Then

  1. IXI;

  2. XI = L if and only if XI.

Proof. (1) Let iI, then ixiI for all xX, hence iXI, that is, IXI;

(2) If XI = L, then 1 ∈ XI, so for all xX we have x = x ∧ 1 ∈ I. Conversely, if XI, then for any aL and for all xX we have axxI, so aXI. That is, LXI, which implies that XI = L. □

Let L be a BL-algebra, I, J, HI(L). Then we have:

(1) JIJI ;

(2) JHI if and only if HJI.

Proof. (1) Let xJIJ, then xJI and xJ, so we get x = xxI. That is, JIJI ;

(2) If JHI and xH, then for any yJ, we have xyJH, it follows that xyI. Hence xJI, that is HJI. Conversely, let HJI, by (1), we have JHJJII. □

Let L be a BL-algebra, I, JI(L). Then JI is the relative pseudo-complement of J with respect to I in the lattice (I(L), ⊆).

Proof. We have already known JI is an ideal, and by Proposition 3.27 (1), we have JIJI. Now we show that JI is the greatest ideal of L such that HJI, where HI(L). Assume that H is an ideal of L such that HJI, If aH, then axa, x for all xJ. Since J and H are ideals of L, we have axJH, Hence axI for all xJ, that is, aJI. Therefore, JI is the relative pseudo-complement of J with respect to I in the lattice (I(L), ⊆). □

Let L, M be two BL-algebras, f : LM be a homomorphism, ∅ ≠ AL. Then f (A) ⊆ (f (A)).

Proof. For all xf (A), there is a yA, such that x = f (y). For all zf (A), there is a tA, such that z = f (t). So we have xz = f (y) ∧ f (t) = f (yt) = f (0) = 0, therefore, x ∈ (f (A)) . □

The next example shows the following: let L, M be two BL-algebras, f : LM be a homomorphism, ∅ ≠ AL, then f (A) may not be an annihilator of a subset of M.

Let L be the BL-algebra of Example 3.22. Let M={0,12,1}, such that 0121. The Cayley tables are as follows:

Then (M, ∧, ∨, ⊙, →, 0, 1) is a BL-algebra. Let f (1) = f (a) = 1, f (0) = f (b) = 0, then f : LM is a homomorphism. Let A ={b}, then A = {0, a}, f (A) = {0, 1}, clearly {0, 1} is not a down set, so there is no BM, such that f (A) = B.

Let L, M be two BL-algebras, f : LM be a surjective homomorphism, ∅ ≠ BM. Then (f 1(B))f 1(B).

Proof. For all x ∈ (f 1(B)), and for all bB, there is a aL, such that b = f (a), so xa = 0, then f (x) ∧ b = f (x) ∧ f (a) = f (xa) = f (0) = 0. Therefore, f (x) ∈ B, which implies that xf 1(B). □

The next example shows the following: let L, M be two BL-algebras, f : LM be a surjective homomorphism, ∅ ≠ BM, then f 1(B) may not be an annihilator of a subset of L.

Let L = {0, a, b, c, d, 1}, where 0 ≤ ac ≤ 1, 0 ≤ bd ≤ 1 and 0 ≤ bc ≤ 1. The Cayley tables are as follows:

Then (L, ∧, ∨, ⊙, →, 0, 1) is a BL-algebra.

Let M = {0, 1}. The Cayley tables are as follows:

Then (M, ∧, ∨, ⊙, →, 0, 1) is a BL-algebra. Let f (0) = f (b) = f (d ) = 0, f (a) = f (c) = f (1) = 1, then f : LM is a homomorphism. Let B ={1}. Then B ={0}, f 1(B) = f 1(0) = {0; b; d }, we can check that there is no AL, such that f 1(B) = A.

Let L, M be two BL-algebras, f : LM be a homomorphism, ∅ ≠ AL. Then f (A) = (f (A)) if and only if (f (A))⊥⊥ = f (A) and (f (A)) ∩ (f (A)) ={0}.

Proof. ⇒ If f (A) = (f (A)), then (f (A))⊥⊥ = (f (A))⊥⊥⊥ = (f (A)) = f (A), (f (A)) ∩ (f (A)) = (f (A)) ∩ (f (A))⊥⊥ ={0}.

⇐ By Proposition 3.27, we have f (A) ⊆ (f (A)) . Now we prove that f (A) (f (A)) . Since (f (A)) ∩ (f (A)) ={0}, we have (f (A)) ⊆ (f (A))⊥⊥ = f (A). Therefore, f (A) = (f (A)) . □

Let L, M be two BL-algebras, f : LM be a surjective homomorphism, ∅ ≠ BM. Then (f 1(B)) = f 1(B) if and only if f 1(B) ∩ (f 1(B))⊥⊥ ={0}.

Proof. ⇒ If (f 1(B)) = f 1(B), then f 1(B) ∩ (f 1(B))⊥⊥ = (f 1(B)) ∩ (f 1(B))⊥⊥ ={0}.

⇐ Firstly, if xf 1(B), yL, such that yx, then f (y) ≤ f (x), since f (x) ∈ B, we get f (y) ∈ B, so yf 1(B), and so we have f 1(B) is a down set. Since f 1(B) ∩ (f 1(B))⊥⊥ ={0}, we get f 1(B) ⊆ (f 1(B))⊥⊥⊥ = (f 1(B)), by Proposition 3.29, we already have (f 1(B))f 1(B). Therefore, (f 1(B)) = f 1(B). □

Let L, M be two BL-algebras, f : LM be an isomorphism, ∅ ≠ AL, ∅ ≠ BM. Then f (A) = (f (A)) and (f 1(B)) = f 1(B).

Proof. (1) By Proposition 3.29, we have already known f (A) ⊆ (f (A)), we now prove f (A) ⊇ (f (A)) . For any y ∈ (f (A)), since f is surjective, there is a xL, such that f (x) = y. For any aA, we have f (a) ∈ f (A), so f (xa) = f (x) ∧ f (a) = yf (a) = 0. Since f is injective, we get xa = 0, so xA, that is yf (A). Therefore, f (A) ⊇ (f (A)).

(2) By Proposition 3.31, we have already known (f 1(B))f 1(B), we now prove (f 1(B))f 1(B). For any xf 1(B), then f (x) ∈ B. For all af 1(B), so f (a) ∈ B, f (xa) = f (x)∧f (a) = 0, since f is injective, we get xa = 0, so x ∈ (f 1(B)). Therefore, f 1(B) ∨ (f 1(B)) . □

4 α ideals in BL-algebras

An ideal I of a BL-algebra L is said to be an α-ideal if i⊥⊥I, for all iI.

Let L be the BL-algebra of Example 3.22. Consider I = {0, a}, since 0⊥⊥ = 0 ⊆ I and a⊥⊥{0, a} ⊆ I, so I is an α-ideal.

Let I be an ideal of a BL-algebra L, we define E(I ) = {xL |∃iI, ix}.

Let L = {0, a, b, c, d, 1}, where 0 ≤ ac ≤ 1, 0 ≤ bd ≤ 1 and 0 ≤ bc ≤ 1, The Cayley tables are as follows:

Then (L, ∧, ∨, ⊙, →, 0, 1) is a BL-algebra. Consider I = {0, a}, then E(I ) = L.

Let L be a BL-algebra, then E(I ) is the smallest αideal containing I, for any ideal I of L. Proof. Clearly, IE(I ).

I1: Assume that xy and yE(I ). Since yE(I ), we get there is a iI such that i y. And since xy, we get y x, so iyx, finally we get xE(I ).

I2: Assume that x, yE(I ). Since x, yE(I ), we get there are i, jI such that ix and j y. For any t ∈ (ij), we have t ∧ (ij) = 0, so ti = 0 and tj = 0, Since ix and j y we deduce that tx = 0 and ty = 0. Since t ∧ (xy) ∨ (tx) ⊘ (ty) = 0 ⊘ 0 = 0, so t ∈ (xy), so (ij) ∨ (xy). Since I is an ideal, we have ijI, therefore xyE(I ).

By I1 and I2 we get E(I ) is an ideal.

Now we prove E(I ) is an α-ideal. For any xE(I ), we get there is a iI such that ix. For any tx⊥⊥, by Proposition 3.9 we have x = x⊥⊥⊥t, so ixt, it follows that tE(I ), which implies that x⊥⊥E(I ). Therefore, E(I ) is an α-ideal.

Next we prove E(I ) is the smallest α-ideal containing I. Let K be an α-ideal such that IK. For any xE(I ), we get there is iI such that ix. Then by Propositions 3.9 and 3.10, we have x ∈ 〈x〉⊆〈x⊥⊥ = x⊥⊥i⊥⊥K. Therefore, E(I ) ⊆ K.

Let L be a BL-algebra, then the following hold:

  1. E(I ) is the intersection of all α-ideal containing I, for any ideal I of L;

  2. For any ideal I of L, I is an α-ideal if and only if E(I ) = I ;

  3. ∩ {I |I is an α · ideal of L}={0};

  4. Let I1 and I2 are ideals of L. Then E(I1) = E(I2) if and only if I1E(I2) and I2E(I1).

Proof. (1) and (2) By Theorem 4.4, they are clear.

(3) Clearly, since {0} is an α-ideal.

(4) ⇒ Let E(I1) = E(I2), so I1E(I1) = E(I2), and I2E(I2) = E(I1).

⇐ ( Since E(I ) is the smallest α-ideal containing I, by I1E(I2) we get E(I1) ⊆ E(I2) and by I2E(I1) we get E(I2) ⊆ E(I1). Therefore, E(I1) = E(I2).

Let I and J be ideals of BL-algebras L and M, respectively. Then E(I × J) = E(I ) × E(J ).

Proof. Let xL and yM, we define (x, y) = x× y. Then E(I × J) = {(x, y) |∃ (a, b) ∈ I × J : (a, b) ⊆ (x, y)} = {(x, y) |∃aI, ∃bJ : ax;by} = {(x, y) | xE(I ); yE(J )} = E(I ) × E(J ).

If Iλ are ideals of BL-algebras Lλ, for all λ ∈ Λ, then Eλ∈ΛIλ) = (Πλ∈ΛE(Iλ)

Proof. Clearly, by Proposition 4.6. □

Let L be a BL-algebra, f : LM be an isomorphism and I be an ideal of L. Then E(f (I )) = f (E(I )).

Proof. Let zE(f (I )). Then there is af (I ) such that az. Hence there exists a0I, z0L such that a = f (a0) and z = f (z0). By Proposition 3.35, we have f (a0) = (f (a0)) = az = (f (z0)) = f (z0). So a0z0. Which means z0E(I ) and so z = f (z0) ∈ f (E(I )).

Conversely, let zf (E(I )). Then z = f (z0), for some z0E(I ). Then we have there exists a0I such that a0z0, so (f (a0)) = f (a0) = f (z0) = (f (z0)) = z. It follows that zE(f (I )). □

Let L be a BL-algebra, I and J be ideals of BL-chain or a finite product of BL-chains. Then E(IJ) = E(I ) ∩ E(J ).

Proof. Firstly, we note that if I is an ideal of the BL-chain L then E(I ) = L or E(I ) ={0}. If I ≠ {0}, then there exists 0 ≠ aI. Since L is a BL-chain, we get a ={0} and so E(I ) = L. If I ={0}, then E(I ) ={x | 0 x} = {x | x = L}={0}.

Now, we show that if I and J are ideals of BL-chain L, then E(IJ) = E(I ) ∩ E(J ). If I ≠ {0} and J ≠{0}, then IJ ≠{0}. Then there exist 0 ≠ aI and 0 ≠ bJ and since L is a BL-chain we have ab or ba. Assume that ab, then 0 ≠ aIJ. So E(IJ) = L, E(I ) = L and E(J ) = L. If I ={0} or J ={0}, then IJ ={0}. Then E(I ) ={0} or E(J ) ={0} and E(IJ) ={0}. Therefore, E(IJ) = E(I ) ∩ E(J ).

Let I and J be ideals of i=1n Li, where Li are BL-chain for all 1 ≤ in, so we have I = i=1n Ii and J = i=1n Ji, where Ii and Ji are ideals of Li, for all 1 ≤ in. So E(IJ) = E(i=1n Iii=1nJi ) = E(i=1n (IiJi )) = i=1n (E(Ii ) ∩ E(Ji )) = i=1nE(Ii ) ∩ i=1nE(Ji ) = i=1nE(Ii ) ∩ i=1nE(Ji ) = E(I ) ∩ E(J ). □

Let L be a BL-algebra, for any I, JI(L), the following hold:

  1. I is an αideal;

  2. E(E(I )) = E(I );

  3. E(I ) = I ;

  4. If IJ, then E(I ) ⊆ E(J ).

Proof. (1) For any aI , we have I⊥⊥a, then we get a⊥⊥I⊥⊥⊥ = I, so I is an α-ideal.

(2) and (3) By Proposition 4.4 and (1), we can easily prove them.

(4) Clearly, by the definition of E(I). □

We denote E(I(L)) ={E(I )| II(L)}. And we know that the set of all ideals of L is a complete lattice and for every family {Fi }iI of ideals of L we have: ∧ i∈IFi = ∩ iIFi and ∨iIFi = 〈∪i∈IFi〉.

(E(I(L)), ∧ E, ∨E, E(0), E(L)) is a complete Brouwerian lattice, where E(I ) ∧ EE(J ) = E(IJ) and E(I ) ∨EE(J ) = E(IJ) and L is a BL-chain or a finite product of BL-chains.

Proof. By Theorem 4.9, we have E(IJ) = E(I ) ∩ E(J ). Hence E(I ) ∧ EE(J ) = E(I ) ∩ E(J ). Since I, J IJ, by Proposition 4.10, we have E(I ), E(J ) ⊆ E(IJ). This means that E(IJ) is an upper bound of E(I ), E(J ). Now let E(I ), E(J ) ⊆ E(K), for some KI(L). Then I, JE(K), hence IJE(K) and so E(IJ) ⊆ E(E(K)) = E(K), therefore E(IJ) is the least upper bound of E(I ) and E(J ).

Now we prove that for any family of ideals Gi, iI, we have that ∨E(E(Gi )) = E(∨(Gi )). Since E(Gi ) ∨ E(∨(Gi )), we get E(∨(Gi )) is an upper bound of E(Gi ), for all iI. Also if E(Gi ) ⊆ E(K), for all iI, then GiE(K). Then ∨(Gi ) ⊆ E(K), hence E(∨(Gi )) ∨ E(E(K)) = E(K). Therefore ∨E(E(Gi )) is the least upper bound of E(Gi ), for all iI. So (E(I(L)), ∧, ∨E, E(0), E(L))is a complete lattice. Then we have ∨E(E(I ) ∧ E(Gi )) =∨E(E(I ) ∩ E(Gi )) =∨E(E(IGi )) = E(∨(IGi )) = E(I ∩ (∨(Gi ))) = E(I ) ∧ E(∨(Gi )) = E(I ) ∧ (∨E(E(Gi ))).

Therefore, (E(I(L)), ∧E, ∨E, E(0), E(L)) is a complete Brouwerian lattice. □

The lattice (E(I(L)), ∧ E, ∨E, E(0), E(L)) is pseudo-complemented, where L is a BL-chain or a finite product of BL-chains.

Proof. Let I, JI(L), we get E(I ) ∧ EE(I ) = E(II ) = E(0). Now let E(I ) ∧ EE(K) = E(0) ={0}, by Propositions 3.17, Propositions 4.10, E(K) ⊆ (E(I ))I = E(I ), so for every ideal E(I ), its pseudo-complement is E(I ). Therefore, (E(I(L)), ∧ E, ∨E, E(0), E(L)) is pseudo-complemented. □

The lattice (E(I(L)), ∧ E, ∨E, E(0), E(L)) is an algebraic lattice, where L is a BL-chain or a finite product of BL-chains.

Proof. Firstly, we show E(〈z〉) is a compact element in the lattice E(I(L)). Assume that E(〈z〉) ⊆∨EE(Gi ), where iI and GiI(L). Then zE(〈z〉) ⊆∨EE(Gi ), so there is a ∈ ∨iI Gi, such that a⊆ z, this means that there exists xi ∈ Gi (1 ≤ in) such that ax1x2 ⊘···⊘ xn. Consider X ={G1;G2; ··· ;Gn} ⊆ ∪iIGi, so (x1x2 ⊘···⊘ xn)az, so zE(∨GiXGi ), so we get 〈z〉⊆ E(∨GiXGi ), and E(〈z〉) ⊆ E(E(∨GiXGi )) = E(∨GiXGi) = E(G1) ∨EE(G2) ∨E ··· ∨EE(Gn). Therefore E(〈z〉) is a compact element in the lattice (E(I(L)). Now consider E(I ) ∈ E(I(L)). Since I =∨aIa〉. we get E(I ) = E(∨aIa〉) =∨E{E(〈a〉) | aI}. Therefore, (E(I(L)), ∧ E, ∨E, E(0), E(L)) is an algebraic lattice. □

5 Conclusions

In this paper, motivating by the previous research on co-annihilators and ideals in BL-algebras, we introduce the concept of annihilators to BL-algebras. We conclude that the ideal lattice (I(L), ⊆) is pseudo-complemented, and for any ideal I, its pseudo-complement is I . Also, using the notion of the annihilator of a nonempty set X with respect to an ideal I, we show that JI is the relative pseudo-complement of J with respect to I in the ideal lattice (I(L), ⊆). Moreover, we give the necessary and sufficient condition under which the homomorphism image and the homomorphism preimage of annihilator become an annihilator. Finally, we introduce the notion of E(I), and we get that (E(I(L)), ∧ E, ∨E, E(0), E(L) is a pseudo-complemented lattice, a complete Brouwerian lattice and an algebraic lattice, when L is a BL-chain or a finite product of BL-chains.

Acknowledgement

The authors thank the editors and the anonymous reviewers for their valuable suggestions in improving this paper. This research is supported by a grant of National Natural Science Foundation of China (11571281) and the Fundamental Research Funds for the Central Universities (GK201603004).

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About the article

Received: 2015-10-14

Accepted: 2016-04-28

Published Online: 2016-05-21

Published in Print: 2016-01-01


Citation Information: Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0029.

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© 2016 Guliyev and Omarova, published by De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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