On implicative and maximal ideals of BL-algebras

Andrzej Walendziak

On implicative and maximal ideals of BL-algebras

Abstract. In the theory of MV-algebras, implicative ideals are studied by Hoo and Sessa. In this paper we define and characterize implicative ideals of BL-algebras. We also investigate maximal ideals of BL-algebras and prove that if an ideal is prime and implicative, then it is maximal. Moreover, we show that an ideal is maximal if and only if the quotient BL-algebra is a simple MV-algebra. Finally, we give the homomorphic properties of implicative and maximal ideals.

2000 Mathematics Subject Classification: 03G25; 06D99.

Key words and phrases: BL-algebra, MV-algebra, Implicative (maximal) ideal.

1. Introduction In 1958, C.C. Chang [2] introduced MV (Many Valued) alge- bras. In 1998, P. H´ajek [5] invented Basic Logic (BL for short) and BL-algebras, structures that correspond to this logical system. BL-algebras are a certain type of residuated lattices (see [16]). The class of BL-algebras contains MV-algebras. Apart from their logical interest, BL-algebras have important algebraic properties (see e.g.

[1], [4], [10] and [12]-[15]).

In 2013, C. Lele and J.B. Nganou [11] introduced the notion of ideal in BL- algebras as a natural generalization of that of ideal in MV-algebras. They also introduced congruences induced by ideals and proved that the quotient BL-algebra via any ideal is always an MV-algebra. Moreover, they defined prime ideals and showed that an ideal is prime if and only if the quotient BL-algebra is an MV- chain.

In this paper, we introduce the concept of implicative ideals in BL-algebras (in the theory of MV-algebras, implicative ideals are defined and studied in [6]-[8]).

Here we give some characterizations of such ideals and show that they are precisely the Boolean ideals. We also investigate maximal ideals of BL-algebras and state that if an ideal is prime and implicative, then it is maximal. Moreover we prove that an ideal is maximal if and only if the quotient BL-algebra is a simple MV-algebra.

Finally, we give the homomorphic properties of implicative and maximal ideals.

For the convenience of the reader, we give the relevant material needed in the sequel, thus making our exposition self-contained.

2. Preliminaries Let A = (A; ∧, ∨, ·, →, 0, 1) be an algebra of type (2, 2, 2, 2, 0, 0).

We say that A is a BL-algebra if the following axioms are satisfied:

(BL-1) (A; ∧, ∨, 0, 1) is a bounded lattice;

(BL-2) (A; ·, 1) is a commutative monoid;

(BL-3) x · y ¬ z ⇐⇒ x ¬ y → z;

(BL-4) x ∧ y = x · (x → y);

(BL-5) (x → y) ∨ (y → x) = 1.

Throughout this paper, A shall denote a BL-algebra. In the sequel, for brevity, we write xy instead of x · y. Set x ² = xx.

For any BL-algebra A, the reduct L(A) = (A; ∧, ∨, 0, 1) is a bounded distributive lattice ([3]), and for all x, y ∈ A, x ¬ y if and only if x → y = 1. We say that A is a BL-chain if L(A) is a chain.

A BL-algebra A is nontrivial if and only if 0 6= 1. For any x ∈ A we put x ⁻ = x → 0. We will write x ⁼ instead of (x ⁻ ) ⁻ . A BL-algebra satisfying the double negation, that is, x ⁼ = x, is called an MV-algebra. In the literature (see e.g. [9]) MV-algebras are also defined as algebras (M; ⊕, ¬, 0) of type (2, 1, 0) satisfying:

(MV-1) (M; ⊕, 0) is a commutative monoid;

(MV-2) ¬(¬x) = x;

(MV-3) x ⊕ ¬0 = ¬0;

(MV-4) ¬(¬x ⊕ y) ⊕ y = ¬(¬y ⊕ x) ⊕ x.

It is known that if A is a BL-algebra satisfying the double negation, then (A; ⊕, ¬, 0), where ¬x = x ⁻ and x ⊕ y = x ⁻ → y for all x, y ∈ A, is an MV- algebra.

For any BL-algebra A, the following properties hold for every x, y, z ∈ A (see [11]) and will be used in the sequel:

(P1) 0 ⁻ = 1 and 1 ⁻ = 0;

(P2) 1 → x = x and 0 → x = x → 1 = x → x = 1;

(P3) x → y ¬ (y → z) → (x → z);

(P4) x ¬ x ⁼ ;

(P5) xx ⁻ = 0 and x0 = 0;

(P6) x ¬ y =⇒ xz ¬ yz;

(P7) (xy) ⁼ = x ⁼ y ⁼ ;

(P8) (x ∧ y) ⁻ = x ⁻ ∨ y ⁻ and (x ∨ y) ⁻ = x ⁻ ∧ y ⁻ ; (P9) x(y ∨ z) = xy ∨ xz.

Let A be a BL-algebra. Define x + y = x ⁻ → y for all x, y ∈ A.

Lemma 2.1. ([11], Lemma 2.7) Let A be a BL-algebra.

(a) The operation + is associative, that is,

x + (y + z) = (x + y) + z for all x, y, z ∈ A.

(b) The operation + is compatible with the order, that is,

for every x, y, z, t ∈ A, if x ¬ y and z ¬ t, then x + z ¬ y + t.

Proposition 2.2. The following properties hold for any x ∈ A:

(a) 0 + x = x and x + 0 = x ⁼ ; (b) 1 + x = 1 = x + 1;

(c) x ⁼ ¬ x + x;

(d) x + x ⁻ = 1.

Proof. Let x ∈ A.

(a) By (P1) and (P2), 0 + x = 1 → x = x. The second equality is obvious.

(b) Applying (P1) and (P2) we get 1 + x = 0 → x = 1 and x + 1 = x ⁻ → 1 = 1.

(c) From (P3) and (P2) we obtain

x ⁼ = x ⁻ → 0 ¬ (0 → x) → (x ⁻ → x) = 1 → (x ⁻ → x) = x ⁻ → x = x + x.

(d) We have x + x ⁻ = x ⁻ → x ⁻ = 1. _

Proposition 2.3. A BL-algebra A satisfies the double negation if and only if (A; +, 0) is a monoid.

Proof. Let x ⁼ = x for any x ∈ A. From Lemma 2.1 (a) it follows that the operation + is associative. By Proposition 2.2 (a),

x + 0 = x ⁼ = x = 0 + x.

Therefore (A; +, 0) is a monoid.

Conversely, if x + 0 = x, then x ⁼ = x, that is, A satisfies the double negation. 

For any BL-algebra A, let B(A) denote the Boolean algebra of all complemented elements in L(A). We say that A is a Boolean algebra if B(A) = A.

Lemma 2.4. ([3], Proposition 1.9) Let A be a BL-algebra and e ∈ A. The following are equivalent:

(a) e ∈ B(A);

(b) e ² = e and e = e ⁼ ; (c) e ² = e and e ⁻ → e = e;

(d) e ∨ e ⁻ = 1.

Proposition 2.5. Let A be a BL-algebra. The following are equivalent:

(a) A is a Boolean algebra, (b) x + x = x for any x ∈ A, (c) x ∨ x ⁻ = 1 for any x ∈ A.

Proof. The implications (a) ⇒ (b) and (c) ⇒ (a) follow from Lemma 2.4.

(b) ⇒ (c). Let x ∈ A. By assumption, x+x = x. Applying (P4) and Proposition

2.2 (c) we obtain

x ¬ x ⁼ ¬ x + x = x.

Therefore x ⁼ = x for every x ∈ A. From (BL-4) and (P5) we have x ∧ x ⁻ = x(x → x ⁻ ) = x(x ⁼ → x ⁻ ) = x(x ⁻ + x ⁻ ) = xx ⁻ = 0.

Hence, by (P1) and (P8), 1 = (x ∧ x ⁻ ) ⁻ = x ⁻ ∨ x ⁼ = x ⁻ ∨ x. Thus (c) holds. 

3. Ideals An ideal of A is a subset I of A satisfying the following conditions:

(I1) 0 ∈ I;

(I2) if x, y ∈ I, then x + y ∈ I;

(I3) if x ∈ A, y ∈ I and x ¬ y, then x ∈ I.

Under this definition, {0} and A are the simplest examples of ideals.

We will denote by Id(A) the set of all ideals of A.

Remark 3.1. ([11]) Let I be an ideal. Then for every x ∈ A, x ∈ I if and only if x ⁼ ∈ I. Moreover, if x, y ∈ I, then x ∨ y ∈ I and x ∧ y ∈ I. Therefore, I is a lattice ideal (that is, I is an ideal of the lattice L(A)). It is easy to see that for all x, y ∈ A,

x, y ∈ I ⇐⇒ x ∨ y ∈ I.

Proposition 3.2. ([11], Proposition 3.2) Let I be a subset of A containing 0. Then I is an ideal of A if and only if for every x, y ∈ A, x ⁻ y ∈ I and x ∈ I imply y ∈ I.

Let A and B be BL-algebras. A function ϕ : A → B is called a homomorphism ([5]) if for all x, y ∈ A:

(H1) ϕ(0) = 0;

(H2) ϕ(xy) = ϕ(x)ϕ(y);

(H3) ϕ(x → y) = ϕ(x) → ϕ(y).

Remark 3.3. From [11] it follows that if ϕ : A → B is a homomorphism of BL- algebras, then for any x, y ∈ A we have:

(h1) ϕ(x ∧ y) = ϕ(x) ∧ ϕ(y);

(h2) ϕ(x ∨ y) = ϕ(x) ∨ ϕ(y);

(h3) ϕ(x ⁻ ) = ϕ(x) ⁻ ; (h4) x ¬ y =⇒ ϕ(x) ¬ ϕ(y);

(h5) ϕ(x + y) = ϕ(x) + ϕ(y).

If ϕ : A → B is a homomorphism of BL-algebras, then the kernel of ϕ is the set Kerϕ := {x ∈ A : ϕ(x) = 0},

that is, Kerϕ = ϕ ^← ({0}), where ϕ ^← (X) denotes the f-inverse image of X ⊆ B.

Proposition 3.4. ([11]) Let ϕ : A → B be a homomorphism and let I ∈ Id(B).

Then ϕ ^← (I) ∈ Id(A).

Proposition 3.5. Let ϕ : A → B be a surjective homomorphism and let I be an ideal of A. Then ϕ(I) ∈ Id(B).

Proof. Obviously, 0 ∈ ϕ(I). Let x, y ∈ ϕ(I). Then there are a, b ∈ I such that x = ϕ(a) and y = ϕ(b). We have x + y = ϕ(a) + ϕ(b) = ϕ(a + b) ∈ ϕ(I), because a + b ∈ I.

Now let x ∈ B, y ∈ ϕ(I) and x ¬ y. Then x = ϕ(a) and y = ϕ(b) for some a ∈ A and b ∈ I. Since ϕ(a) ¬ ϕ(b), we conclude that ϕ(a) = ϕ(a ∧ b). We obtain a ∧ b ¬ b and b ∈ I. Therefore a ∧ b ∈ I and hence x = ϕ(a) = ϕ(a ∧ b) ∈ ϕ(I).

Consequently, ϕ(I) ∈ Id(B). 

C. Lele and J.B. Nganou [11] introduced congruences induced by ideals. Let I be an ideal of A. Define the relation ∼ ^I on A by: for every x, y ∈ A, x ∼ ^I y if and only if x ⁻ y ∈ I and y ⁻ x ∈ I, that is,

(1) x ∼ ^I y ⇐⇒ x ⁻ y ∨ y ⁻ x ∈ I.

By Theorem 4.2 of [11], ∼ ^I is a congruence on A.

For any x ∈ A, we denote by x/I the equivalence class of x, that is, x/I = {y ∈ A : x ∼ ^I y }. Let A/I = {x/I : x ∈ A} and we define on the set A/I the operations: x/I ∧ y/I = (x ∧ y)/I, x/I ∨ y/I = (x ∨ y)/I, x/I · y/I = (xy)/I, x/I → y/I = (x → y)/I. The quotient algebra A/I = (A/I; ∧, ∨, ·, →, 0/I, 1/I) is a BL-algebra. Moreover we have

Proposition 3.6. ([11], Proposition 4.3) The quotient BL-algebra via any ideal is an MV-algebra.

It is clear that for all x, y ∈ A, (x/I) ⁻ = (x ⁻ )/I and x/I + y/I = (x + y)/I. By (1),

(2) x/I = y/I ⇐⇒ x ⁻ y ∨ y ⁻ x ∈ I.

Observe that

(3) x/I = 0/I ⇐⇒ x ∈ I.

Indeed,

x/I = 0/I ⇐⇒ x ∼ ^I 0 ⇐⇒ x ⁻ 0 ∨ 0 ⁻ x ∈ I ⇐⇒ x ∈ I.

Since x ⁻ 1 ∨ 1 ⁻ x = x ⁻ , we conclude from (2) that

(4) x/I = 1/I ⇐⇒ x ⁻ ∈ I.

Remark 3.7. In BL-algebras there are congruences, which are not induced by

ideals (see Example 3.8 below).

Example 3.8. ([11]) Let A = {0, a, b, 1} be such that 0 < a < b < 1. Define binary operations “·” and “→” on A by the following tables:

· 0 a b 1

0 0 0 0 0

a 0 0 a a

b 0 a b b

1 0 a b 1

→ 0 a b 1

0 1 1 1 1

a a 1 1 1

b 0 a 1 1

1 0 a b 1

Then A = (A; ∧, ∨, ·, →, 0, 1) is a BL-algebra. It is easy to see that A has two ideals:

I 1 = {0} and I ² = A. We have

∼ ^I

= {(0, 0), (a, a), (b, b), (1, 1), (1, b), (b, 1)} and ∼ ^I

= A ² .

However, the smallest congruence {(0, 0), (a, a), (b, b), (1, 1)} is not induced by ide- als.

Remark 3.9. The ideals and the congruences of any MV-algebra are in a one-to- one correspondence ([14]). In BL-algebras, it is not true. The BL-algebra A from Example 3.8 has two ideals and three congruences.

4. Implicative ideals In the theory of MV-algebras, implicative ideals are defined and studied in [6]-[8]. Now we introduce the concept of implicative ideals in BL-algebras.

Definition 4.1. An ideal I of a BL-algebra A is called implicative if it satisfies the following condition (for every x, y, z ∈ A):

(Im) (xy ⁻ z ⁻ ∈ I and yz ⁻ ∈ I) =⇒ xz ⁻ ∈ I.

Theorem 4.2. Let I ∈ Id(A). Then the following statements are equivalent:

(a) I is implicative;

(b) If x, y ∈ A and xy ⁼ y ⁼ ∈ I, then xy ∈ I;

(c) If x ² ∈ I, then x ∈ I;

(d) {x ∈ A : x ² = 0} ⊆ I;

(e) x ∧ x ⁻ ∈ I for all x ∈ A;

(f) A/I is a Boolean algebra.

Proof. (a) ⇒ (b). Let xy ⁼ y ⁼ ∈ I, that is, x(y ⁻ ) ⁻ (y ⁻ ) ⁻ ∈ I. Since y ⁻ y ⁼ = 0 ∈ I and I is implicative, we have xy ⁼ ∈ I. By (P4), y ¬ y ⁼ and therefore, using (P6) we get xy ¬ xy ⁼ . From this we deduce that xy ∈ I.

(b) ⇒ (c). Suppose that x ² ∈ I. From Remark 3.1 we have (xx) ⁼ ∈ I. By (P7), x ⁼ x ⁼ ∈ I, and hence 1x ⁼ x ⁼ ∈ I. Applying (b) we conclude that x = 1x ∈ I.

(c) ⇒ (d). Let x ² = 0. Since 0 ∈ I, we see that x ² ∈ I. From (c) it follows that

x ∈ I. Consequently, (d) holds.

(d) ⇒ (e). Let x ∈ A. Since x ∧ x ⁻ ¬ x and x ∧ x ⁻ ¬ x ⁻ , we obtain (x ∧ x ⁻ ) ² ¬ xx ⁻ = 0 by (P6) and (P5). Therefore (x ∧ x ⁻ ) ² = 0. From (d) we deduce that x ∧ x ⁻ ∈ I.

(e) ⇒ (f). Let x ∈ A. Using (P8) we have (x ∨ x ⁻ ) ⁻ = x ⁻ ∧ x ⁼ ∈ I. Then (x ∨ x ⁻ )/I = 1/I (see (4)) and hence x/I ∨ (x/I) ⁻ = 1/I. By Proposition 2.5, A/I is a Boolean algebra.

(f) ⇒ (a). Suppose that xy ⁻ z ⁻ , yz ⁻ ∈ I. Applying (3) we get x/I · y ⁻ /I · z ⁻ /I = 0/I and y/I · z ⁻ /I = 0/I.

Since A/I is a Boolean algebra, we see (by Proposition 2.5) that y ⁻ /I ∨ y/I = 1/I.

From this and from (P9) we have

xz ⁻ /I = x/I · z ⁻ /I = x/I · z ⁻ /I · 1/I

= x/I · z ⁻ /I · (y ⁻ /I ∨ y/I)

= x/I · y ⁻ /I · z ⁻ /I ∨ x/I · y/I · z ⁻ /I = 0/I ∨ 0/I = 0/I.

Therefore xz ⁻ ∈ I by (3). Consequently, I is an implicative ideal. 

Corollary 4.3. (Extension property for implicative ideals.) Let I be implicative and suppose that J is an ideal of A such that I ⊆ J. Then J is implicative.

Proof. Follows from Theorem 4.2. _

Proposition 4.4. If ϕ : A → B is a homomorphism of BL-algebras and I is an implicative ideal of B, then ϕ ^← (I) is an implicative ideal of A.

Proof. By Propostion 3.4, ϕ ^← (I) is an ideal of A. Let x ∈ A. Applying (h1) and (h3) we get ϕ(x ∧ x ⁻ ) = ϕ(x) ∧ ϕ(x) ⁻ . Since I is implicative, ϕ(x) ∧ ϕ(x) ⁻ ∈ I and hence ϕ(x ∧x ⁻ ) ∈ I. Therefore, x∧x ⁻ ∈ ϕ ^← (I). From Theorem 4.2 we deduce that

ϕ ^← (I) is implicative. _

Proposition 4.5. Let ϕ : A → B be a surjective homomorphism of BL-algebras and I an implicative ideal of A. Then ϕ(I) is an implicative ideal of B.

Proof. By Proposition 3.5, ϕ(I) is an ideal of B. Let x ∈ B. Then there exists a ∈ A such that ϕ(a) = x. We have x ∧ x ⁻ = ϕ(a) ∧ ϕ(a) ⁻ = ϕ(a ∧ a ⁻ ) ∈ ϕ(I), since a ∧ a ⁻ ∈ I. Consequently, ϕ(I) is implicative. 

Following Lele and Nganou [11], we say that an ideal I of A Boolean if x∧x ⁻ ∈ I for all x ∈ A. From Theorem 4.2 it follows that an ideal I of A is implicative if and only if it is Boolean.

5. Prime and maximal ideals We now define the concept of prime ideal in

BL-algebras, following Lele and Nganou [11]. A proper ideal I of A is called prime

if for all x, y ∈ A, (x → y) ⁻ ∈ I or (y → x) ⁻ ∈ I. It is shown in [11] that a proper

ideal I of A is prime if and only if the BL-algebra quotient A/I is an MV-chain.

Proposition 5.1. ([11], Proposition 5.11) A proper ideal I of A is prime if and only if for any x, y ∈ A, x ∧ y ∈ I implies that x ∈ I or y ∈ I.

By Proposition 5.1 we conclude that every prime ideal of A is prime as an ideal of the lattice L(A). From [11] we also have the prime ideal theorem for BL-algebras.

Proposition 5.2. ([11], Proposition 5.10) Let I be a proper ideal of A and a ∈ A−I.

Then there exists a prime ideal P of A such that I ⊆ P and a /∈ P .

Corollary 5.3. For every proper ideal I of A there exists a prime ideal P of A containing I.

By Proposition 5.9 of [11], if ϕ : A → B is a homomorphism of BL-algebras and I is a prime ideal of B, then ϕ ^← (I) is a prime ideal of A. Moreover we get the following statement:

Proposition 5.4. Let ϕ : A → B be a surjective homomorphism of BL-algebras and I a prime ideal of A. If ϕ(I) 6= B, then ϕ(I) a prime ideal of B.

Proof. Let ϕ(I) 6= B. Then, by Proposition 3.5, ϕ(I) is a proper ideal of B. Let x, y ∈ B. There are a, b ∈ A such that ϕ(a) = x and ϕ(b) = y. Then (x → y) ⁻ = (ϕ(a) → ϕ(b)) ⁻ = ϕ((a → b) ⁻ ). Similarly, (y → x) ⁻ = ϕ((b → a) ⁻ ). Since I is a prime ideal of A, (a → b) ⁻ ∈ I or (b → a) ⁻ ∈ I. Therefore (x → y) ⁻ ∈ ϕ(I) or

(y → x) ⁻ ∈ ϕ(I). Thus ϕ(I) a prime ideal of B. 

Corollary 5.5. Let ϕ : A → B be a surjective homomorphism and let I be a prime ideal of A containing Kerϕ. Then ϕ(I) a prime ideal of B.

Proof. It is sufficient to show that ϕ(I) 6= B. On the contrary, suppose that 1 ∈ ϕ(I). Hence ϕ(a) = 1 for some a ∈ I. Then ϕ(a ⁻ ) = ϕ(a) ⁻ = 0. Therefore a ⁻ ∈ Kerϕ ⊆ I. Thus a, a ⁻ ∈ I and hence, by Proposition 2.2 (d), 1 = a + a ⁻ ∈ I.

Consequently, I = A, a contradiction. _

An ideal I of A is called maximal if it proper and no proper ideal of A strictly contains I, that is, for each ideal J 6= I, if I ⊆ J, then J = A.

Next lemma is obvious and its proof will be omitted.

Lemma 5.6. Every proper ideal of A can be extended to a maximal ideal.

Lemma 5.7. If I ∈ Id(A) is maximal, then I is prime.

Proof. Let I be a maximal ideal of A. By Corollary 5.3 there is a prime ideal P of A containing I. Since I is maximal, I = P , and hence I is prime. 

Theorem 5.8. Let M be a proper ideal of A. Then the following conditions are

equivalent:

(a) M is maximal;

(b) |Id(A/M)| = 2;

(c) A/M is a simple MV-algebra (i.e., it has exactly two congruences).

Proof. (a) ⇒ (b). Let I ∈ Id(A/M). Suppose that I 6= A/M. Set J := {x ∈ A : x/M ∈ I}. We show first that J ∈ Id(A). Obviously, 0 ∈ J. Let x, y ∈ J. Then x/M, y/M ∈ I, and hence (x + y)/M = x/M + y/M ∈ I. Therefore x + y ∈ J. Now let x, y ∈ A, x ¬ y and y ∈ J. From this we deduce that x/M ¬ y/M and y/M ∈ I.

Since I ∈ Id(A/M), we conclude that x/M ∈ I. Thus x ∈ J and consequently, J is an ideal of A. Since I 6= A/M, we get J 6= A, that is, J is proper. It is easy to see that M ⊆ J. Then J = M, because M is maximal. Let x/M ∈ I. Therefore x ∈ M.

Hence x/M = 0/M. Thus I = {0/M}.

(b) ⇒ (c). From Proposition 3.6 it follows that A/M is an MV-algebra. In MV-algebras (see Remark 3.9), the ideals and the congruences are in a one-to-one correspondence. Therefore A/M has exactly two congruences. Hence A/M is simple.

(c) ⇒ (a). Let I be a proper ideal of A such that M ⊆ I. Write J := {x/M : x ∈ I }. We prove that J ∈ Id(A/M). It is clear that 0/M ∈ I and J satisfies (I2). Let x/M ∈ A/M, y/M ∈ J and x/M ¬ y/M. Then x ∈ A, y ∈ I and (x∧y)/M = x/M.

By (2), (x ∧ y) ⁻ x ∈ M and hence (x ∧ y) ⁻ x ∈ I. From Proposition 3.2 we conclude that x ∈ I. Therefore x/M ∈ J. Hence, J also satisfies (I3). Consequently, J is an ideal of A/M. Since I 6= A, we get J 6= A/M. From (c) it follows that J = {0/M}.

Let a ∈ I. Then a/M = 0/M and hence a ∈ M. Thus I ⊆ M. From this and from

M ⊆ I we get M = I. Therefore M is maximal. 

Theorem 5.9. Let I be a proper ideal of A. Then the following statements are equivalent:

(a) I is maximal and Boolean;

(b) I is prime and Boolean;

(c) ∀x ∈ A (x ∈ I or x ⁻ ∈ I).

Proof. (a) ⇒ (b). Follows from Lemma 5.7.

(b) ⇒ (c). Let x ∈ A. Since I is Boolean, x ∧ x ⁻ ∈ I. By Proposition 5.1, x ∈ I or x ⁻ ∈ I.

(c) ⇒ (a). Let I satisfy (c) and x ∈ A. By assumption, x ∈ I or x ⁻ ∈ I. Since x ∧ x ⁻ ¬ x and x ∧ x ⁻ ¬ x ⁻ , we conclude that x ∧ x ⁻ ∈ I. Therefore I is Boolean.

Observe that I is also maximal. Let J be an ideal of A such that I ⊂ J. For every z ∈ J − I, we have by (c) that z ⁻ ∈ I. Consequently, z ⁻ ∈ J. Then z + z ⁻ ∈ J.

But z + z ⁻ = 1 (see Proposition 2.2 (d)). Hence 1 ∈ J and thus J = A. Therefore

I is maximal. _

As an immediate consequence of Theorem 5.9 we obtain

Corollary 5.10. Let I be a proper Boolean ideal. Then I is prime if and only if I is maximal.

The following two theorems give the homomorphic properties of maximal ideals.

Theorem 5.11. Let ϕ : A → B be a surjective homomorphism of BL-algebras and I a maximal ideal of A. If ϕ(I) 6= B, then ϕ(I) a maximal ideal of B.

Proof. Let ϕ(I) 6= B. Then, by Proposition 3.5, ϕ(I) is a proper ideal of B. We take a proper ideal J of B such that J ⊇ ϕ(I). Hence ϕ ^← (J) ⊇ ϕ ^← (ϕ(I)) ⊇ I. From Proposition 3.4 we deduce that ϕ ^← (J) ∈ Id(A). We prove that ϕ ^← (J) is proper.

On the contrary, suppose that ϕ ^← (J) = A. Then J = ϕ(ϕ ^← (J)) = ϕ(A) = B, a contadiction. Consequently, ϕ ^← (J) is a proper ideal of A. Since I ⊆ ϕ ^← (J) and I is maximal, we conclude that ϕ ^← (J) = I. Therefore, ϕ(I) = ϕ(ϕ ^← (J)) = J. Thus

ϕ(I) is a maximal ideal of B. 

Corollary 5.12. Let ϕ : A → B be a surjective homomorphism and let I be a maximal ideal of A containing Kerϕ. Then ϕ(I) is a maximal ideal of B.

Proof. From the proof of Corollary 5.5 we see that ϕ(I) 6= B. Theorem 5.11 now

shows that ϕ(I) is a maximal ideal of B. 

Theorem 5.13. Let ϕ : A → B be a surjective homomorphism and let J be a maximal ideal of B. Then f ^← (J) is a maximal ideal of A.

Proof. From Proposition 3.4 it follows that I := f ^← (J) ∈ Id(A). It is easily seen that I 6= A. By Lemma 5.6 there is a maximal ideal M of A containing I. We have

I = ϕ ^← (J) ⊇ ϕ ^← ({0}) = Kerϕ.

Since M ⊇ I ⊇ Kerϕ, Corollary 5.12 shows that ϕ(M) is a maximal ideal of B.

Obviously, ϕ(M) ⊇ ϕ(I) = ϕ(ϕ ^← (J)) = J and hence ϕ(M) = J. Then M ⊆ ϕ ^← (ϕ(M)) = ϕ ^← (J) = I ⊆ M, that is, ϕ ^← (J) = M. Thus ϕ ^← (J) is a maximal

ideal of A. 

If X t is a set for all t ∈ T , then Q

t ∈T X t denotes the Cartesian product of X t , t ∈ T . If A ^t , t ∈ T , are BL-algebras, then Q

t ∈T A ^t is a BL-algebra (the direct product of A ^t , t ∈ T ).

Lemma 5.14. For each t ∈ T , let I ^t be an ideal of the BL-algebra A ^t . Then Q

t ∈T I t

is an ideal of Q

t ∈T A ^t .

Proof. The proof is straightforward. _

Theorem 5.15. For each t ∈ T , let I ^t be an ideal of the BL-algebra A ^t . Then Q

t ∈T I t is a maximal ideal of Q

t ∈T A ^t if and only if there is an unique index s ∈ T such that I s is a maximal ideal of A ^s and I t = A t for any t 6= s.

Proof. Let I := Q

t ∈T I t be a maximal ideal of A := Q

t ∈T A ^t . It is easily seen

that there is at least one index t such that I t is a maximal ideal of A ^t . Assume that

there are two indices t 1 and t 2 such that I t

and I t

are proper ideals of A ^t

and

A ^t

, respectively. Then J := Q

t ∈T I _t ⁰ , where I _t ⁰ = I t if t 6= t ¹ and I _t ⁰

= A t

, is a proper ideal of A containing I, which contradicts the maximality of I.

Suppose that I = Q

t ∈T I t , where I s is a maximal ideal of A ^s and I t = A t for all t 6= s. By Lemma 5.14, I ∈ Id(A). Observe that I is maximal. Indeed, let K ∈ Id(A) and K ⊃ I. Then K = Q

t∈T I _t ⁰ , where I _t ⁰ = A t if t 6= s and I s ⁰ is an ideal of A ^s such that I _s ⁰ ⊃ I ^s . Since I _s is maximal in A s , we see that I _s ⁰ = A _s , and therefore K = Q

t∈T A t = A. Consequently, I is a maximal ideal of A. 

6. Conclusion We introduced the notions of an implicative ideal and a maximal ideal in BL-algebras. We obtained the properties and characterizations of them. In particular, we showed that implicative ideals are precisely the Boolean ideals and proved that an ideal is maximal if and only if the quotient BL-algebra is a simple MV-algebra.

Some important issue for future work are studying other types of ideals and establishing some logic consequences.

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Andrzej Walendziak

Institute of Mathematics and Physics, Siedlce University Sieldlce, Poland

E-mail: walent@interia.pl

(Received: 13.06.2014)