A representation for radicals in pseudo BL-algebras

1. Introduction

In 1998, BL-algebra was introduced by P. Hajek as an algebraic structure to prove the integrity of the basic logic (BL-Logic) [11]. MV-algebras, product algebras, and Gödel algebras are the most important classes of BL-algebras. In 2000, A. Di. Nola, G. Georgescu, and A. Iorgulescu proposed pseudo BL-algebras as a non-commutative structure which is a generalization of BL-algebras and pseudo MV-algebra, and then, the relationship between this algebraic structure and other algebraic structures, including BCK-algebras, was investigated by researchers. Algebraic structures were intensively studied in [1–5].

J. K u ̈ hr proved that a pseudo BL-algebra is equivalent to a divisible residuated lattice (DRL-monoid) [14]. Additionally, A. Iorgulescu showed that the category of pseudo BL-algebra is equivalent to the category of Iseki-algebras [13].

Then some classes of pseudo BL-algebras were introduced by G. Georgescu et al. [10]. The filters theory plays an important role in studying logical algebraic structures. It is also widely used to prove the integrity of non-classical logics. Moreover, filters are specifically related to congruence relations and even have a key usage in quotient algebras.

The main aim of the present paper is to define the notion of the radical of a filter in a pseudo BL-algebra and obtain the results and properties of the radical in the pseudo BL-algebra. In the following, in addition to introducing dense, infinitesimal, nilpotent, and unity elements in a pseudo BL-algebra, we investigate the relationship between the radical of a pseudo BL-algebra and these elements, and extract important results in this regard.

This paper is organized as follows: In Section (2), we present the definition, characteristics and properties of a pseudo BL-algebra. These materials will be used in the next sections to accomplish the objective of the paper.

In Section (3), we define the notion of the radical of a filter in a pseudo BL-algebra and investigate and challenge its properties and characteristics. In this section, we define nilpotent elements in a pseudo BL-algebra, and investigate the relationship between the radical of a pseudo BL-algebra and nilpotent elements and pseudo BL-algebra in the form of several important theorems and propositions. For example, we show that nilpotent elements of a pseudo BL-algebra A are not in Rad (A), and we show that if {F _i } _i∈I is a family of filters in A, then Rad (⋂ _i∈IF _i ) = ⋂ _i∈IRad (F _i ).

In section (4), first, we define dense, infinitesimal, nilpotent, and unity elements in pseudo BL-algebra, then, express the relationship between these elements and the radical of a pseudo BL-algebra, and present the obtained results in the form of several prepositions and theorems. Additionally, we show that D _s  (A) ⊆ Rad (F) and Rad (F/D _s  (F)) = Rad (F)/D _s  (F). We also show that if a is a nonunity member of A and a∈Rad(A), then a is an infinitesimal element, and we finally conclude that Rad (A)/{1} ⊆ Inf (A) ⊆ Rad _n  (A).

2. Preliminaries

Definition 2.1. [6, 7] Let (A ; ∨ , ∧ , ⊙ , → , ⇝   , 0, 1) be an algebra of type (2, 2, 2, 2, 2, 0, 0). A is called pseudo BL-algebra if it satisfies the following axioms, for any x, y, z ∈ A: (psBL₁) (A ; ∨ , ∧ , 0, 1) is a bounded lattice; (psBL₂) (A ; ⊙ , 1) is a monoid; (psBL₃) x ⊙ y ≤ z ⇔ x ≤ y → z ⇔ y ≤ x ⇝  z; (psBL₄) x ∧ y = (x → y) ⊙ x = x ⊙ (x ⇝  y); (psBL₅) (x → y) ∨ (y → x) = (x ⇝  y) ∨   (y ⇝  x) = 1.

In the sequel, we shall agree that the operations ∨, ∧ , ⊙ have priority over the operations →, ⇝. A pseudo BL-algebra A is nontrival if and only if 0 ≠ 1 for any pseudo BL-algebra (A ; ∨ , ∧ , ⊙ , → , ⇝  , 0, 1).

The reduct L ( A ) = ( A ; ∨ , ∧ , 0 , 1 ) is a bounded distributive lattice.

A pseudo BL-chain is a pseudo BL-algebra such that its lattice order is linear. For any x ∈ A, we define x^∼ = x ⇝  0 and x^- = x → 0.

We shall write x^∼∼ instead of (x^∼) ^∼ and x^-- instead of (x^-) ^-. We define x° = 1 and x ⁿ  = x^n-1 ⊙ x for n = 1, 2, ….

The order of x ∈ A, in symbols ord (x), is the smallest n ∈ ℕ \ { 0 } such that x ⁿ  = 0. If no such n exists, then ord (x) =∞.

Example 2.2. [7] (i) Consider a pseudo MV-algebra (A ; ⊙ , ⊕ , ∼ , - , 0, 1) with binary operations → and ⇝ as follows: x → y = y ⊕ x - and x ⇝ y = x ∼ ⊕ y . Then (A ; ∨ , ∧ , ⊙ , → , ⇝  , 0, 1) is a pseudo BL-algebra.(ii) Consider l-group (G ; ∨ , ∧ , + , - , 0, 1) and let u ∈ G,  u ≤ 0. For every x, y ∈ G we put by definition: x ⊙ y = ( x + y ) ∨ u ; x ⊕ y = ( x - u + y ) ∧ 0 ; x - = u - x ; x ∼ = - x + u . Then ([u, 0] ; ⊙ , ⊕ , ∼ , - , 0 = u,  1 = 0) is a pseudo MV-algebra and we define the following operations: x → y = ( y - x ) ∧ 0 ; x ⇝ y = ( - x + y ) ∧ 0 . Then ([u, 0] ; ∨ , ∧ , ⊙ , → , ⇝  , 0 = u,  1 = 0) is a pseudo BL-algebra.Definition 2.3. [6] A pseudo BL-algebra A is called commutative if and only if x ⊙ y = y ⊙ x, for any x, y ∈ A.Proposition 2.4. [6] A pseudo BL-algebra A is commutative if and only if x → y = x ⇝  y, for all x, y ∈ A.

Any commutative pseudo BL-algebra A is a BL-algebra. Then we shall say that a pseudo BL-algebra is proper if it is not commutative, i.e. if it is not a BL-algebra.Example 2.5. (i) [17] Let a , b , c , d ∈ ℝ , where ℝ is the set of all real numbers. We define: ( a , b ) ≤ ( c , d ) ⇔ a < c or ( a = c and b ≤ d ) . For any x , y ∈ ℝ × ℝ , we define operations ∨ and ∧ as follows: x ∨ y = max { x , y } and x ∧ y = min { x , y } . Let A = { ( 1 2 , b ) ∈ ℝ 2 : b ≥ 0 } ∪ { ( a , b ) ∈ ℝ 2 : 1 2 < a < 1 , b ∈ ℝ } ∪ { ( 1 , b ) ∈ ℝ 2 : b ≤ 0 } . For any (a, b) , (c, d) ∈ A, we put by definition: ( a , b ) ⊙ ( c , d ) = ( 1 2 , 0 ) ∨ ( ac , bc + d ) ; ( a , b ) → ( c , d ) = ( 1 2 , 0 ) ∨ ( ( c a , d - b a ) ∧ ( 1 , 0 ) ) ; ( a , b ) ⇝ ( c , d ) = ( 1 2 , 0 ) ∨ ( ( c a , ad - bc a ) ∧ ( 1 , 0 ) ) . Then ( A ; ∨ , ∧ , ⊙ , → , ⇝ , ( 1 2 , 0 ) , ( 1 , 0 ) ) is a pseudo BL-algebra.

(ii)Consider the MV-algebra for x, y ∈ L _n we have: (L _n  ; ⊕ , ⊙ , - , 0, 1) where L n = { 0 , 1 n - 1 , ⋯ , n - 2 n - 1 , 1 } and for x, y ∈ L _n : x ⊕ y = min { 1 , x + y } ; x ★ = 1 - x ; x ⊙ y = max { 0 , x + y - 1 } ; x ∨ y = max { x , y } ; x ∧ y = min { x , y } ; Then L _n is an MV-sub algebra of [0, 1]. Now we define: x ⊙ y = (x^★ ⊙ y^★) ^★, x → y = x^★ ⊕ y, 1 = 0^★, x ∨ y = (x → y) → y = (y → x) → x and x ∧ y = (x^★ ⊙ y^★) ^★, then (L _n  ; ∨ , ∧ , ⊙ , →0, 1) is a BL-algebra. Now, suppose that → = ⇝ and - =∼ = ★, then (L _n  ; ∨ , ∧ , ⊙ , → , ⇝  , 0, 1) is a pseudo BL-algebra.

(iii) Consider the pseudo BL-algebra A in (ii) and putting n : =3. Then we have L 3 = { 0 , 1 2 , 1 } . Suppose L : = L 3 × L 3 = { ( 0 , 0 ) , ( 0 , 1 2 ) , ⋯ , ( 1 , 1 ) } and define: (x, y) ⩽ (a, b) ⇔ x < a or x = a,  and y ⩽ b; (x, y) ⊕ (a, b) = min {(1, 1) , (x, y) + (a, b)}; (x, y) ^★ = (1, 1) - (x, y). Then (L ; ⊕ , ★ , (0, 0)) is an MV-algebra. Also, if we define: (x, y) ⊙ (a, b) = ((x, y) ^★ ⊕ (a, b) ^★) ^★; (x, y) → (a, b) = (x, y) ^★ ⊕ (a, b); (x, y) ∨ (a, b) = ((x, y) → (a, b)) → (a, b); (x, y) ∧ (a, b) = ((x, y) ^★ ∨ (a, b) ^★) ^★. Then (L ; ∨ , ∧ , ⊙ , → , ⇝  , (0, 0) , (1, 1)) is a pseudo BL-algebra, in which ⇝   =→. (iv) Consider the set A = {0, b₁, b₂, ⋯ , 1} with the chain 0 < b₁ < b₂ < ⋯ <1, organized as a lattice by a ∨ b = max {a, b} ,  a ∧ b = min {a, b} and as the following tables: ⊙ 0 b 1 b 2 b 3 b 4 ⋯ 1 0 0 0 0 0 0 ⋯ 0 b 1 0 0 0 0 b 1 ⋯ b 1 b 2 0 0 0 0 b 2 ⋯ b 2 s b 3 0 0 0 0 b 3 ⋯ b 3 b 4 0 b 1 b 2 b 3 b 4 ⋯ b 4 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 1 0 b 1 b 2 b 3 b 4 ⋯ 1

Then (A ; ∨ , ∧ , ⊙ , → , ⇝   , 0, 1) is a linearly ordered pseudo BL-algebra.Proposition 2.6. [6, 7] In a pseudo BL-algebra A the following hold, for all x, y, z ∈ A: (1) (x ⊙ y) → z = x → (y → z);

(2) (y ⊙ x) ⇝  z = x ⇝  (y ⇝  z);

(3) x ≤ y ⇔ x → y = 1 ⇔ x ⇝  y = 1;

(4) x ≤ y implies x ⊙ z ≤ y ⊙ z and z ⊙ x ≤ z ⊙ y;

(5) if x ≤ y, then y ⇝  z ≤ x ⇝  z and

y → z ≤ x → z;

(6) x ⊙ y ≤ x, y;

(7) x ⊙ y ≤ x ∧ y;

(8) x → y  =  x → (x ∧ y) , x ⇝  y = x ⇝   (x ∧ y);

(9) x ∨ y = ((x → y) ⇝  y) ∧ ((y → x) ⇝   x);

(10) x ∨ y = ((x ⇝   y) → y) ∧ ((y ⇝   x) → x);

(11) x ⊙ y = 0 ⇔ x ≤ y^- ⇔ x ≤ y^∼;

(12) x ⊙ 0 =0 ⊙ x = 0;

(13) x ⊙ x^- = x^∼ ⊙ x = 0;

(14) 1 → x = 1 ⇝  x = x;

(15) x^- = 1 ⇔ x^∼ = 1 ⇔ x = 0;

(16) 1^- = 1^∼ = 0;

(17) x ≤ y implies y^- < x^- and y^∼ ≤ x^∼;

(18) x ≤ (x^-) ^∼,  x ≤ (x^∼) ^-;

(19) y ≤ x^∼ ⇔ x ⊙ y = 0;

(20) y ≤ x^- ⇔ y ⊙ x = 0;

(21) x → y ≤ y^- ⇝  x^-,  x → y ≤ y^∼ → x^∼;

(22) x^- = x^-∼-,  x^∼ = x^∼-∼;

(23) (x ⊙ y) ^- = x → y^-,   (x ⊙ y) ^∼ = y ⇝  x^∼;

(24) (x ∨ y) ^- = x^- ∧ y^-,   (x ∧ y) ^∼ = x^∼ ∨ y^∼;

(25) (x ∧ y) ^- = x^- ∨ y^-,   (x ∧ y) ^∼ = x^∼ ∨ y^∼;

(26) x ⊙ (y ∨ z) = (x ⊙ y) ∨ (x ⊙ z);

(27) (y ∨ z) ⊙ x = (y ⊙ x) ∨ (z ⊙ x);

(28) x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z);

(29) x^-- ⊙ y^-- ≤ (x ⊙ y) ^--;

(30) x^∼∼ ⊙ y^∼∼ ≤ (x ⊙ y) ^∼∼;

(31) x ⩽ x^∼ → y and x ⩽ x^- ⇝  y .

For any pseudo BL-algebra A, we define: G ( A ) = { x ∈ A : x ⊙ x = 0 } and M ( A ) = { x ∈ A : x = ( x ∼ ) - = ( x - ) ∼ } .

Also, let B (A) be the Boolean algebra of all complemented elements in the distributive lattice L (A) = (A ; ∨ , ∧ , 0, 1) of pseudo BL-algebra A and so B (A) = B (L (A)).

Proposition 2.7. [7] If A is pseudo BL-algebra and a ∈ G (A), b ∈ A, then

(1) a ⊙ b = a ∧ b = b ⊙ a;

(2) a ∧ a^∼ = 0 = a ∧ a^-;

(3) a → b = a ⇝   b;

(4) a^- = a^∼.

Proposition 2.8. [7] If A is a pseudo BL-algebra, then B (A) = M (A) ∩ G (A).

Proposition 2.9. [6, 7] If A is a pseudo BL-algebra, then for e ∈ A, the following are equivalent:

(1) e ∈ B (A);

(2) e ⊙ e = e and e = ((e^∼) ^-) = (e^-) ^∼);

(3) e ⊙ e = e, e^- → e = e and e ⊙ e = e, e^∼ ⇝   e = e; (4) e ∨ e^- = 1 and e ∨ e^∼ = 1.

Definition 2.10. [6] A nonempty subset F ⊆ A is called a filter of A, if the following conditions are satisfied: (F₁) if x, y ∈ F, then x ⊙ y ∈ F; (F₂) if x ∈ F, y ∈ A ,  x ≤ y, then y ∈ F. Clearly {1} and A are filters. A filter F of A is called proper if F ≠ A. Remark 2.11. [6] Any filter of pseudo BL-algebras A is a filter of the lattice L (A) and for a nonempty subset F the following are equivalent: (1) F is a filter; (2) 1 ∈ F and if x, x → y ∈ F, then y ∈ F; 1 ∈ F and if x, x ⇝   y ∈ F, then y ∈ F. Definition 2.12. [7] A filter H of A is called normal if for every x, y ∈ A, x → y ∈ H iff x ⇝  y ∈ H.

Definition 2.13. [7] A proper filter P of A is called prime if, for any x, y ∈ A, the condition x ∨ y ∈ P implies x ∈ P or y ∈ P.

Proposition 2.14. [7] Let H be a normal filter of A. H is a prime filter if and only if A/H is a pseudo BL-chain.

Theorem 2.15. [7] (Prime filter theorem) If F ∈ F ( A ) and I is an ideal of the lattice L (A) such that F∩ I = ∅, then there is a prime filter P of A such that F ⊆ P and P∩ I = ∅.

Corollary 2.16.[7] If F ∈ F ( A ) is a proper filter and a ∈ A \ F, then there is a prime filter P of A such that F ⊆ P and a ∉ P. In particular, for F = {1} we deduce that for any a ∈ A, a ≠ 1, there is a prime filter P _a such that a ∉ P _a .

Definition 2.17. [7] A filter of A is maximal (ultra filter) if it is proper and it is not contained in any other proper filter.

Theorem 2.18. [7] If H ∈ F n ( A ) , H ≠ A , then the following are equivalent: (1) H ∈ Max _n  (A); (2) for any x ∈ A, x ∉ H if and only if (x ⁿ ) ^- ∈ H, for some n ≥ 1; (3) for any x ∈ A, x ∉ H if and only if (x ⁿ ) ^∼ ∈ H, for some n ≥ 1.

Example 2.19. [17] Consider pseudo BL-algebra in Example 2.5 (i) and let H = {(1, b) :  b ≤ 0}. We show that it is a normal ultra filter of A. Obviously, H is a filter. Suppose that (a, b) , (c, d) ∈ A. Then ( a , b ) → ( c , d ) ∈ H ⇔ ( 1 2 , 0 ) ∨ ( ( c a , d - b a ) ∧ ( 1 , 0 ) ) ∈ H ⇔ c a ≥ 1 ⇔ ( 1 2 , 0 ) ∨ ( ( c a , ad - bc a ) ∧ ( 1 , 0 ) ) ∈ H ⇔ ( a , b ) ⇝ ( c , d ) ∈ H .

By definition, H is normal. We now apply Theorem 2.18 to show that H is a maximal filter. Let x = (a, b) ∉ H. Then 1 2 ≤ a < 1 , and we have x n = ( 1 2 , 0 ) for some n ∈ ℕ . Hence ( x n ) - = ( 1 2 , 0 ) - = ( 1 2 , 0 ) → ( 1 2 , 0 ) = ( 1 , 0 ) ∈ H .

Assume that x ∈ H, that is, x = (1, b) , b ≤ 0. Then x ⁿ  = (1, nb) ∈ H for all n ∈ ℕ , and so ( x n ) - = ( 1 , nb ) - = ( 1 2 , - nb ) ∉ H . Therefore, H is an ultra filter.

3. On the radicals of filters in pseudo BL-algebras

From now on, (A ; ∨ , ∧ , ⊙ , → , ⇝   , 0, 1) is a pseudo BL-algebra unless otherwise specified.

In this section, we introduce the notion of the radical of filters in a pseudo BL algebra and express its characteristics and properties.Definition 3.1. Let F be a proper filter of A. The intersection of all maximal filters of A which contain F is called radical of F and it is denoted by Rad (F). If F = A, then we put by definition: Rad (A) = A. it is clear that Rad (F) is a filter and F ⊆ Rad (F).

We denot by Rad _n  (A) the intersection of maximal and normal filters of A.Proposition 3.2. Let F be a Boolean filter of A. Then Rad (F) is too.

Proof. Assume that F is a Boolean filter of A. Then we have x ∨ x^- ∈ F(x ∨ x^∼ ∈ F), for all x ∈ A. Since F ⊆ Rad (F) , we get x ∨ x^- ∈ Rad (F) (x ∨ x^∼ ∈ Rad (F)). Therefore, Rad (F) is a Boolean filter of A. In the following example, we shows that the converse of Proposition 3.2 is not true in general.

Example 3.3. Consider the set A = {0, a, b, c, 1} and define ∗ and → as the following tables:

∗ 0 a b c 1 0 0 0 0 0 0 a 0 a a c a b 0 a b b b c 0 a a c c 1 0 a b c 1 → 0 a b c 1 0 1 1 1 1 1 a 0 1 1 1 1 b 0 c 1 c 1 c 0 b b 1 1 1 0 a b c 1 If → =⇝ and - = ∼ . Then (A ; ∨ , ∧ , ⊙ , → , ⇝   , 0, 1) is a pseudo BL-algebra. If consider F = {1, b}, then F is not a Boolean filter, because a ∨ a^- = a ∨ 0 = a ∉ F, but Rad (F) = {a, b, c, 1} is a Boolean filter. Now, if B is a proper pseudo BL-algebra, then A × B is a proper pseudo BL-algebra and if we consider F = {b, 1} × B, then Rad (F) = {a, b, c, 1} × B. Its follow that Rad (F) is a proper and Boolean filter, while F is not a Boolean filter, because for all b ∈ B we have: (a, b) ∨ (a, b) ^- = (a, b) ∨ (a^-, b^-) = (a, b) ∨ (0, b^-) = (a, b ∨ b^-) ∉ F (b ∨ b′ ∈ B, but a ∉ F).

Lemma 3.4. The following statements hold: (1) if M is a maximal filter of A, then Rad (M) = M; (2) if P is a prime filter, then there exists a unique maximal filter M of A such that P ⊆ M, and so Rad (P) = M.

Example 3.5. (i) Consider pseudo BL-algebra from Example 2.19, since H = {(1, b) ; b ⩽ 0} is a normal and maximal filter of A, then Rad _n  (H) = Rad (H) = H. (ii) Consider pseudo BL-algebra A from Example 2.5 (iv). Then M = {b₄, b₅, ⋯ , 1} is the unique maximal filter of A and we have Rad (M) = M. In the following theorem we characterize Rad (F) by elements of A, where F is an arbitrary filter of A.

Theorem 3.6. Let F be a filter of A. Then Rad ( F ) = { a ∈ A : ( a n ) - → a ∈ F , for all n ∈ ℕ } = { a ∈ A : ( a n ) ∼ ⇝ a ∈ F , for all n ∈ ℕ } . Proof. Assume that F = A. Then we have:

Rad ( A ) = { a ∈ A : ( a n ) - → a ∈ A , for all n ∈ ℕ } = { a ∈ A : ( a n ) ∼ ⇝ a ∈ A , for all n ∈ ℕ } .

Now, let F ≠ A, a ∈ Rad (F) and there exists k ∈ ℕ such that (a ^k ) ^- → a ∉ F ((a ^k ) ^∼ ⇝ a ∉ F). Then by Corollary 2.16 and Lemma 3.4 (2), there exists a prime filter P of A such that F ⊆ P and (a ^k ) ^- → a ∉ P ((a ^k ) ^∼ ⇝ a ∉ P). Since P is a prime filter of A we obtain a → (a ^k ) ^- ∈ P (a ⇝ (a ^k ) ^∼ ∈ P. Also by Lemma 3.4 (2), there exists a unique maximal filter M of A such that P ⊆ M, therefore, a → (a ^k ) ^- ∈ M (a ⇝ (a ^k ) ^∼ ∈ M). If a ∈ M, then a ⁿ  ∈ M, for all n ∈ ℕ and thus a ^k  ∈ M in particular. Also, we have a → (a ^k ) ^- ∈ M (a ⇝ (a ^k ) ^∼ ∈ M), hence (a ^k ) ^- ∈ M ((a ^k ) ^∼ ∈ M) and Thus 0 = a ^k  ⊙ (a ^k ) ^- = a ⊙ (a ^k ) ^∼ ∈ M, which is a contradiction, thus a ∉ M. Hence F ⊆ P ⊆ M and a ∉ M, hence a ∉ Rad (F), which is a contradiction, therefore (a ^k ) ^- → a ∈ F ((a ^k ) ^∼ ⇝ a ∈ F), for all k ∈ ℕ .

Conversely, let (a ⁿ ) ^- → a ∈ F ((a ⁿ ) ^∼ ⇝ a ∈ F), for all n ∈ ℕ and a ∉ Rad (F). Then there exists a maximal filter M of A such that F ⊆ M and a ∉ M, since M is a maximal filter of A, there exists n ∈ ℕ such that (a ⁿ ) ^- ∈ M ((a ⁿ ) ^∼ ∈ M). We have (a ⁿ ) ^- → a ∈ F ((a ⁿ ) ^∼ ⇝ a^∼) ∈ F) ⊆ M, hence a ∈ M. Then a ⁿ  ∈ M and so 0 = (a ⁿ ) ^- ⊙ a ⁿ  = (a ⁿ ) ^∼ ⊙ a ⁿ  ∈ M, which is a contradiction.

Theorem 3.7. Let F be a filter of A and a ∈ A. Then a ∈ Rad (F) if and only if a^- → a ⁿ  ∈ F iff a^∼ ⇝  a ⁿ  ∈ F, for all n ∈ ℕ .

Proof. Let a ∈ Rad (F). Since Rad (F) is a filter of A, we obtain a ⁿ  ∈ Rad (F), for all n ∈ ℕ . So by Theorem 3.6, ((a ^{n m} ) ^- → a ⁿ  ∈ F (a ^nm ) ^∼ ⇝ a ⁿ ) ∈ F). For all m ∈ ℕ , we have a ^nm  ≤ a. Then (a ^nm ) ^- → a ≤ a^- → a ⁿ  ((a ^nm ) ^∼ ⇝ a ≤ a^∼ ≤ a ⁿ ). Therefore, a^- → a ⁿ  ∈ F (a^∼ ⇝ a ⁿ  ∈ F), for all n ∈ ℕ .

Conversely, let a^- → a ⁿ  ∈ F (a^∼ ⇝ a ⁿ  ∈ F), for all a ∈ ℕ and a ∉ Rad (F). Then there exists a maximal filter M of A such that F ⊆ M and a ∉ M. Hence there exists m ∈ ℕ such that (a ^m ) ^- ∈ M ((a ^m ) ^∼ ∈ M). We have (a ^m ) ≤ (a ^m ) ^-- (a ^m  ≤ (a ^m ) ^∼∼), therefore, a^- → a ^m  ≤ a^- → (a ^m ) ^-- (a^∼ ⇝ a ^m  ≤ b^∼ ⇝ (b ^m ) ^∼∼), and so a^- → (a ^m ) ^-- ∈ F (a^∼ ⇝ (a ^m ) ^∼∼ ∈ F) we have: ( a m ) - → a - - = ( a m ) - → ( a - → 0 ) ( resp . ( a m ) ∼ ⇝ a ∼ ∼ = ( a m ) ∼ ⇝ ( a ∼ ⇝ 0 ) ) = a - → ( ( a m ) - → 0 ) ( resp . a ∼ ⇝ ( ( a m ) ∼ ⇝ 0 ) ) = a - → ( a m ) - - ∈ F ( resp . a ∼ ⇝ ( a m ) ∼ ∼ ∈ F ) .

Therefore, (a ^m ) ^- → a^-- ∈ F ⊆ M. (by a similar argument (a ^m ) ^∼ ⇝ a^∼∼ ∈ F ⊆ M). Since (a ^m ) ^- ∈ M ((a ^m ) ^∼ ∈ M) we get that a^-- ∈ M (a^∼∼ ∈ M) and so (a^--)  ⁿ  ∈ M ((a^∼∼)  ⁿ  ∈ M), for all n ∈ ℕ . Thus in particular (a^--)  ^m  ∈ M we have ((a^∼∼)  ^m  ∈ M) (a ^m ) ^-- = (a^--)  ^m  ∈ M (Similarly  (a ^m ) ^∼∼ = (a^∼∼)  ^m  ∈ M). Thus (a ^m ) ^-- ∈ M ((a ^m ) ^∼∼ ∈ M). Since (a ^m ) ^- ∈ M ((a ^m ) ^∼ ∈ M), hence 0 = (a ^m ) ^-- ⊙ (a ^m ) ^- ∈ M (0 = (a ^m ) ^∼∼ ⊙ (a ^m ) ^∼ ∈ M). Which is a contradiction.Let F₁ and F₂ be arbitrary filters of A. Then since 1 ∈ F₁ ∩ F₂, we get: [ F 1 ∪ F 2 ) = { x ∈ A : x ≥ ( f 1 ⊙ f 1 ′ ) ⊙ ⋯ ⊙ ( f n ⊙ f n ′ ) , for some n ≥ 1, f₁, f₂, ⋯ , f _n  ∈ F₁ and f 1 ′ , f 2 ′ , ⋯ , f n ′ ∈ F 2 } .

For F 1 , F 2 ∈ F ( A ) we put by definition: F 1 → F 2 = { a ∈ A : F 1 ∩ [ a ) ⊆ F 2 } , where [a) = {x ∈ A :  x ≥ a ⁿ ,  for some n ≥ 1}. It is clear that if F₁ ⊆ F₂ then F₁ → F₂ = A.

Proposition 3.8. Let F, G, H ∈ F (A). Then F ∩ G ⊆ H if and only if F ⊆ G → H.

Proof. Suppose that F ∩ G ⊆ H and let x ∈ G. Then [x) ⊆ G and so F ∩ [x) ⊆ F ∩ G ⊆ H. Hence [x) ⊆ G and so x ∈ F → H. Therefore, G ⊆ F → H. Conversely, suppose that G ⊆ F → H and let x ∈ F ∩ G. Then x ∈ G and so x ∈ F → H, that is [x) ∩ F ⊆ H. Now, since x ∈ [x) ∩ F ⊆ H, we get x ∈ H and so F ∩ G ⊆ H.

Recall that a ∈ A is called nilpotent element of A, if a ⁿ  = 0, for some n ∈ ℕ .Theorem 3.9. Let F and G be proper filters of A and a, b ∈ A. Then the following conditions hold: (1) if a is nilpotent element of A, then a ∉ Rad (F) and Rad (F) ⊆ {a ∈ A : ord (a) = ∞}; (2) if a, b ∈ Rad (F), then a^- → b ∈ F and a^∼ ⇝  b ∈ F; (3) if a, b ∈ Rad (F), then (a^- ⊙ b^-) ^- ∈ F and (a^∼ ⊙ b^∼) ^∼ ∈ F; (4) a^- = 0 and a^∼ = 0 for all a ∈ A \ {0}, if and only if Rad (F) = A \ {0}; (5) if A is linear pseudo BL-algebra and a ∈ Rad (F), then ((a ⁿ ) ^- ⊙ (a ⁿ ) ^-) → a = 1, for all n ∈ ℕ ; (6) if F ⊆ G, then Rad (F) ⊆ Rad (G); (7) Rad (F) = A if and only if F = A; (8) Rad (Rad (F)) = Rad (F); (9) Rad (F) ∩ Rad (G) ⊆ Rad ((F ∪ G)); (10) Rad (F) → Rad (G) ⊆ F → Rad (G); (11) Rad (F → G) ⊆ Rad (F → Rad (G)); (12) if for every a ∈ F there exists k ∈ ℕ , such that a ^k  ∈ G, then Rad (F) ⊆ Rad (G); (13) Rad (F → G) ⊆ Rad (F) → Rad (G); (14) Rad (F)→ Rad (G) = Rad (Rad (F)) → Rad (Rad (G)) .

Proof. (1) Let a be nilpotent elements of A and a ∈ Rad (F). Hence there exists m ∈ ℕ such that a ^m  = 0. By filter property of Rad (F), we get that a ^m  ∈ Rad (F). Therefore, 0 ∈ Rad (F), which is a contradiction. Hence a ∉ Rad (F). (2) Let a, b ∈ Rad (F). Then a ⊙ b ∈ Rad (F) and so (a ⊙ b) ^- → (a ⊙ b) ∈ F ((a ⊙ b) ^∼ ⇝ (a ⊙ b) ∈ F). Since a ⊙ b ≤ a, we have: (a ⊙ b) ^- → (a ⊙ b) ≤ a^- → (a ⊙ b) ((a ⊙ b) ^∼ ⇝ (a ⊙ b) ≤ a^∼ ⇝ (a ⊙ b)) . Therefore, a^- → (a ⊙ b) ∈ F (a^∼ ⇝ (a ⊙ b) ∈ F). Since a ⊙ b ≤ b; then a ⊙ b → b = 1 ∈ F (a ⊙ b ⇝ b = 1 ∈ F) and so ( a - → ( a ⊙ b ) ) ⊙ ( ( a ⊙ b ) → b ) ︸ 1 ∈ F ( a ∼ ⇝ ( a ⊙ b ) ⊙ ( a ⊙ b ⇝ b ︸ 1 ) ∈ F ) . Thus, we obtain a^- → b ∈ F (a^∼ ⇝ b ∈ F). (3) Let a, b ∈ Rad (F). Then by (2), we have a^- → b ∈ F. since b ≤ b^-- we get that a^- → b ≤ a^- → b^-- and so a^- → b^-- ∈ F. Thus we have (a^- ⊙ b^-) ^- ∈ F. By a similar argument we have (a^∼ ⊙ b^∼) ^∼ ∈ F. (4) Let a^- = 0, for all a ∈ A \ {0}. It is clear that Rad (F) ⊆ A \ {0}. We must show that A \ {0} ⊆ Rad (F). Take x ∈ A \ {0}, then by hypothesis x^- = 0 and so x^- → x ⁿ  = 0 → x ⁿ  = 1 ∈ F, for all n ∈ ℕ . Therefore, x ∈ Rad (F) by Theorem 3.7. Hence Rad (F) = A \ {0}. Conversely, let Rad (F) = A and there exists a ∈ A \ {0} such that a^- ≠ 0. Hence by hypothesis a^-, a ∈ Rad (F). So, 0 ∈ Rad (F), which is a contradiction. (5) Let A be a linear pseudo BL-algebra and a ∈ Rad (F). Then (a ⁿ ) ^- → a ≤ (a ⁿ ) ^- or (a ⁿ ) ^- ≤ (a ⁿ ) ^- → a for all n ∈ ℕ . Let (a ⁿ ) ^- → a ≤ (a ⁿ ) ^-. Since (a ⁿ ) ^- → a ∈ F Then (a ⁿ ) ^- ∈ F and so a ∈ F. Hence a ⁿ  ∈ F, for all n ∈ ℕ . So (a ⁿ ) ^- ⊙ a ⁿ  ∈ F. Therefore, 0 ∈ F which is a contradiction. Hence (a ⁿ ) ^- ≤ (a ⁿ ) ^- → a, for all n ∈ ℕ . then (a ⁿ ) ^- (→ ((a ⁿ ) ^-) →0) =1 for all n ∈ ℕ . So ((a ⁿ ) ^- (⊙ (a ⁿ ) ^-)) → a = 1, for all n ∈ ℕ . (6) The proof is clear. (7) Let Rad (F) = A. Then 0 ∈ Rad (F) and so 0 = 1 →0 = (0 ⁿ ) ^- → 0 ∈ F (0 = 1 ⇝0 = (0 ⁿ ) ^∼ ⇝ 0 ∈ F), for all n ∈ ℕ . Therefore, F = A. (8) By (6), we have Rad (F) ⊆ Rad (Rad (F)). Let x ∈ Rad (Rad (F)). Then x ∈ M, for all maximal filter M of A containing Rad (F). Let M₀ be an arbitrary maximal filter of A containing F. Then M₀ = Rad (M₀) ⊇ Rad (F) and so x ∈ M₀. Therefore, x ∈ Rad (F), that is Rad (Rad (F)) ⊆ Rad (F). Thus Rad (Rad (F)) = Rad (F). (9) The proof is clear by (6). (10) Let x ∈ Rad (F) → Rad (G). Then Rad (F) ⋂ [x] ⊆ Rad (G). Hence F ⋂ [x] ⊆ Rad (G), that is x ∈ F → Rad (G). (11) Let x ∈ Rad (F → G). Then (x ⁿ ) ^- → x ∈ F → G, for all n ∈ ℕ , that is F ⋂ ((x ⁿ ) ^- → x) ⊆ G ⊆ Rad (G), for all n ∈ ℕ . Hence (x ⁿ ) ^- → x ∈ F → Rad (G), for all n ∈ ℕ , and so x ∈ Rad (F → Rad (G)). (12) Let a ∈ F. Assume that there is a k ∈ ℕ such that a ^k  ∈ G. We have a ^k  ≤ a, thus a ∈ G. Hence F ⊆ G, so by (6), we have Rad (F) ⊆ Rad (G). (13) Since F ∩ (F → G) ⊆ G, using (6) we have: Rad (F) ∩ Rad (F → G) ⊆ Rad (G) . Thus Rad (F → G) ⊆ Rad (F) → Rad (G). (14) Since F ⊆ Rad (F), it follows that Rad (F) → Rad (G) ⊆ Rad (Rad (F) → Rad (G)). Conversely, using (13) and (8), we get Rad (Rad (F) → Rad (G)) ⊆ Rad (Rad (F) → Rad (Rad (G) = Rad (F) → Rad (G))) .

Example 3.10. (i) Consider the linear pseudo BL-algebra A from Example 2.5 (iv), and let F = {b₄, b₅, ⋯ , 1}. Then F is a filter of A and Rad (F) = F. We have:

(1) b 3 - → b 4 = ( b 3 → 0 ) → b 4 = b 3 → b 4 = 1 ∈ F and b 3 ∼ ⇝ b 4 = ( b 3 ⇝ 0 ) ⇝ b 4 = 0 ⇝ b 4 = 1 ∈ F while  b₃ ∉ Rad (F). Thus the converse of Theorem 3.9 (2), is not true in general.

(2) ( b 3 - ⊙ b 4 - ) - = 0 - = 1 ∈ F and ( b 3 ∼ ⊙ b 4 ∼ ) ∼ = 1 ∈ F , while b₃, b₁ ∉ Rad (F). Thus the converse of Theorem 3.9 (3), is not true in general. (ii) Consider the pseudo BL-algebra from Example 2.5 (iii) the set of all filters on L are: F 1 = { ( 1 , 1 ) } , F 2 = { ( 1 , 1 ) , ( 1 , 0 ) } , F 3 = { ( 1 , 1 ) , ( 0 , 1 ) } , F 4 = { ( 1 , 1 ) , ( 1 , 1 2 ) , ( 1 , 0 ) } , F 5 = { ( 1 , 1 ) , ( 1 2 , 1 ) , ( 0 , 1 ) } , F 6 = L . Then we have Rad (F₂) = F₂, Rad (F₃) = F₃, Rad (F₂ ∪ F₃) = L, but Rad (F₂) ∩ Rad (F₃) = F₁, then Rad (F₂ ∪ F₃) ≠ Rad (F₂) ∩ Rad (F₃). (iii) Consider the pseudo BL-algebra A = {0, b₁, b₂, ⋯ , 1} from Example 2.5 (iv), we know that M = {b₄, b₅, ⋯ , 1} is a unique maximal filter of A. Let F and G be proper filters of A, then we have: G ⊆ F ⊆ M ( Rad ( G ) ⊆ Rad ( F ) ) or F ⊆ G ⊆ M ( Rad ( F ) = M , Rad ( G ) = M ) and Rad ( G ) ∩ Rad ( F ) = Rad ( G ) ⊆ Rad ( F ∪ G ) = Rad ( F ) , but M = Rad ( M ) ∩ Rad ( A ) = Rad ( M ) ⊆ Rad ( M ∪ A ) = Rad ( A ) . So, (i) and (ii) in Example 3.10, shows that equality of Theorem 3.9 (9) is not true in general.

Proposition 3.11. Let A be a linear pseudo BL-algebra, F be a filter of A and a ∉ Rad (F). Then a is a nilpotent element of A.

Proof. Let a ∉ Rad (F). Then by Theorem 3.6, there exists m ∈ ℕ , such that (a ^m ) ^- → a ∉ F (Similarly (a ^m ) ^∼ ⇝  a ∉ F). Hence a < (a ^m ) ^- (a < (a ^m ) ^∼), and so a^m+1 = a ⊙ a ^m  = 0, by Proposition 2.6 (11). Therefore, a is a nilpotent element of A.Corollary 3.12. If A is a linear BL-algebra, then Rad (F) = {a : ord (a) = ∞}, for each filter F of A.Proposition 3.13. Let f : A → B be a psBL-homomorphism. Then Rad (ker(f)) = f^-1 (Rad ({1})).

Proof. By Theorem 3.7, we have: a ∈ Rad (ker(f)) if and only if (a ⁿ ) ^- → a ∈ ker(f) . By a similar argument we have (a ⁿ ) ^∼ ⇝  a ∈ ker f, for all n ∈ ℕ , ⇔ f ( ( a n ) - → a ) = 1 ( resp . f ( ( a n ) ∼ ⇝ a ) = 1 ) , ⇔ ( f ( a n ) ) - → f ( a ) = 1 ( resp . ( f ( a n ) ) ∼ ⇝ f ( a ) = 1 ) , ⇔ f ( a ) ∈ Rad ( { 1 } ) ⇔ a ∈ f - 1 ( Rad ( { 1 } ) ) .

Proposition 3.14. Let {F _i } _i∈I be a family of filters of A. Then Rad ( ∩ i ∈ I F i ) = ∩ i ∈ I Rad ( F i ) .

Proof. We have ∩ i ∈ I F i ⊆ F i ⊆ Rad ( F i ) , for all i ∈ I, then by Theorem 3.9 (6) and (8) we get that Rad ( ∩ i ∈ I F i ) ⊆ Rad ( F i ) , for all i ∈ I. Therefore, Rad ( ∩ i ∈ I F i ) ⊆ ∩ i ∈ I ( Rad ( F i ) ) . Conversely, Let x ∈ ∩ i ∈ I ( Rad ( F i ) ) . Then x ∈ Rad (F _i ), for all i ∈ I, and so (x ⁿ ) ^- → x ∈ F _i ((x ⁿ ) ^∼ ⇝  x ∈ F _i ), for all i ∈ I and n ∈ ℕ . Hence ( x n ) - → x ∈ ∩ i ∈ I F i ,   ( ( x n ) ∼ ⇝ x ∈ ∩ i ∈ I F i ) , for all n ∈ ℕ , that is x ∈ Rad ( ∩ i ∈ I F i ) . Therefore, Rad ( ∩ i ∈ I F i ) = ∩ i ∈ I Rad ( F i ) . Theorem 3.15. Let F be a filter of A. Then Rad (F) ⋂ B (A) ⊆ F.

Proof. Let x ∈ Rad (F) ⋂ B (A). Then x ∈ Rad (F) and x ∈ B (A). So, (x ⁿ ) ^- → x ∈ F (((x ⁿ ) ^∼ ⇝  x) ∈ F), for all n ∈ ℕ , and x^- → x = x (x^∼ ⇝  x = x), by Proposition 2.6 (11) and (13). Hence x ∈ F. Therefore, Rad (F) ⋂ B (A) ⊆ F.In the following example, we show that equality of Theorem 11 is not true in general.

Example 3.16. Consider pseudo BL-algebra A from Example 2.5 (iv), then F = {b₄, b₅, ⋯ , 1} is the only maximal filter of A and we have B (A) = {0, 1} hence F ≠ Rad (F) ∩ B (A) = {1}.

Corollary 3.17. Let A be a pseudo BL-algebra. Then Rad ({1}) ⋂ B (A) = {1}.

Theorem 3.18. Let F be a filter of A. Then the following hold: (1) Rad ({1/F}) = Rad (F)/F; (2) if Rad (F) ⊆ B (A), then B (A/Rad (F)) = B (A)/Rad (F); (3) if a ∈ A is nilpotent element, then a/Rad (F) ∈ A/Rad (F) is a nilpotent element.

Proof. (1) By definition of radical we have: Rad ( { 1 / F } ) = ⋂ F ⊆ N N ∈ Max ( A ) ( N / F ) = ( ⋂ F ⊆ N N ∈ Max ( A ) N ) / F = Rad ( F ) / F . ( 2 ) LetRad ( F ) ⊆ B ( A ) . Then B ( A ) / Rad ( F ) = { e / Rad ( F ) : e ∈ B ( A ) } = { e / Rad ( F ) : e ∨ e - = 1 } = { e / Rad ( F ) : e ∨ e ∼ = 1 } = { e / Rad ( F ) : e / Rad ( F ) ∨ ( e / Rad ( F ) ) - = 1 / Rad ( F ) } = { e / Rad ( F ) : e / Rad ( F ) ∨ ( e / Rad ( F ) ) ∼ = 1 / Rad ( F ) } = B ( A / Rad ( F ) ) .

(3) Let a ∈ A be a nilpotent element. Then there exists n ∈ ℕ such that a ⁿ  = 0. Hence

0/Rad (F) = a ⁿ /Rad (F) = (a/Rad (F))  ⁿ .

Therefore, a/Rad (F) ∈ A/Rad (F) is a nilpotent element.Proposition 3.19. Let A be a linear pseudo BL-algebra. a ∈ A is a nilpotent element if and only if a/Rad (F) ∈ A/Rad (F) is a nilpotent element.

Proof. Let A be a linear BL-algebra and a/Rad (F) ∈ A/Rad (F) be a nilpotent element. Then there is an n ∈ ℕ such that a ⁿ /Rad (F) = (a/Rad (F))  ⁿ  = 0/Rad (F), and so a ⁿ  ∉ Rad (F). By Proposition 3.11, there exists m ∈ ℕ such that ((a ⁿ )  ^m ) ^- → a ⁿ  ∉ F, hence ((a ⁿ )  ^m ) ^-notleqslanta ⁿ . By hypothesis we get that a ⁿ  < ((a ⁿ )  ^m ) ^-, so by Proposition 2.6 (11), we get that a ⁿ  ⊙ (a ⁿ )  ^m  = 0. Therefore, a is a nilpotent element of A. Hence by Theorem 3.18 (3), the proof is complete.Lemma 3.20. Rad ({1}) = {1} if and only if x = inf { ( x n ) - → x : for all n ∈ ℕ } = inf { ( x n ) ∼ ⇝ x : for all n ∈ ℕ } , for all x ∈ A .

Proof. Let Rad ({1}) = {1}. We have x ≤ (x ⁿ ) ^- → x (x ≤ (x ⁿ ) ^∼ ⇝ x), for all n ∈ ℕ . Hence x is a lower bound for { ( x n ) - → x : forall n ∈ ℕ } ( { ( x n ) ∼ ⇝ x : forall n ∈ ℕ } ) . We show that x is the largest lower bound for { ( x n ) - → x : forall n ∈ ℕ } ( { ( x n ) ∼ ⇝ x : forall n ∈ ℕ } ) . Let y be a lower bound for { ( x n ) - → x : forall n ∈ ℕ } ( { ( x n ) ∼ ⇝ x : forall n ∈ ℕ } ) . We must show that y ≤ x. For all n ∈ ℕ , we have y ≤ (x ⁿ ) ^- → x  (y ≤ (x ⁿ ) ^∼ ⇝ x) then y → (x ⁿ ) ^- ≤ x  (y ⇝ (x ⁿ ) ^∼ ≤ x). For all n ∈ ℕ , we know that: x ≤ y → x ⇒ x n ≤ ( y → x ) n ⇒ ( y → x n ) - ≤ ( x n ) - ⇒ y ⊙ ( ( y → x ) n ) - ≤ y ⊙ ( x n ) - ≤ x ⇒ y ⊙ ( ( y → x ) n ) - ≤ x ⇒ ( ( y → x ) n ) - ≤ y → x ⇒ ( ( y → x ) n ) - → ( y → x ) = 1 ⇒ y → x ∈ Rad ( { 1 } ) = { 1 } ⇒ y ≤ x .

Conversely, let x = inf { ( x n ) - → x : forall n ∈ ℕ } , for every x ∈ A. We must show that Rad ({1}) ⊆ {1}. Let x ∈ Rad ({1}). Then (x ⁿ ) ^- → x = 1  ((x ⁿ ) ^∼ ⇝  x = 1), for all n ∈ ℕ . So, by hypothesis we get x = 1. Therefore, Rad ({1}) = {1}.

4. Dense and infinitesimal element in pseudo BL-algebra

In this section, we first, introduce dense, infinitesimal and unity elements in a pseudo BL-algebra, and then, we investigate the relationship between the above mentioned elements and radical in the pseudo BL-algebra and obtain very important results. We put by definition: D s ( A ) = { x ∈ A : x - = 0 } = { x ∈ A : x ∼ = 0 } .

We define dense elements of F by

D s ( F ) = { x ∈ F : x - = 0 } = { x ∈ F : x ∼ = 0 } ,

where F is a filter of A.Example 4.1. {Consider linear pseudo BL-algebras A from Example 2.5 (iv). Then D s ( A ) = { b 4 , b 5 , ⋯ , 1 } . Since  x^∼ = x^- = 0, for all x ∈ D _s  (A).

Proposition 4.2. Let F be a filter of A. Then (1) D _s  (F) = D _s  (A) ∩ F;

(2) D _s  (F) is a filter of A;

(3) D _s  (F) ⊆ Rad (F).

Proof. (1) Proof is clear. (2) Let a, b ∈ D _s  (F). Then a^- = b^- = 0. On the other hand (a ⊙ b) ^- = a → b^- = a → 0 = a^- = 0, hence a ⊙ b ∈ D _s  (F). Assume that a ∈ D _s  (F) and a ≤ b. Then b^- ≤ a^- = 0 and so b ∈ D _s  (F). Hence D _s  (F) is a filter of A. (3) Assume that x is an arbitrary element of D _s  (F) such that x ∉ Rad (F). Then there exists M ∈ Max (A) such that x ∉ M. Hence A = [M ∪ {x}) and so 0 ∈ [M ∪ {x}) = {a ∈ A : x ⊙ d ≤ a}. Thus, there exists d ∈ M such that d ≤ x^- = 0 and so x^- = 0 ∈ M, which is a contradiction and so x ∈ Rad (F).Proposition 4.3. Let M ∈ Max (A) and a ∈ A. Then (1) a/Ds (A) ∈ M/Ds (A) if and only if a ∈ M; (2) Max (A/Ds (A)) = {N/Ds (A) : N ∈ Max (A)}; (3) a/Ds (A) ∈ Rad (A)/Ds (A) if and only if a ∈ Rad (A); (4) Rad (A/Ds (A)) = Rad (A)/Ds (A).Proof. It follows from [16, Pro. 3.1].Lemma 4.4. Let F be a filter of A. Then the following hold: (1) D _s  (A) ⊆ Rad (F); (2) Rad (F/D _s  (F)) = Rad (F)/D _s  (F).Proof. (1) Let x ∈ D _s  (A). Then 0 = x^- ≤ x ⁿ , (0 = x^∼ ≤ x ⁿ ), for all n ∈ ℕ , and so x^∼ → x ⁿ  = 1 ∈ F, (x^- ⇝ x ⁿ  = 1 ∈ F), for all n ∈ ℕ . Therefore, x ∈ Rad (F), by Theorem 3.9. (2) We have: Rad ( F / D s ( F ) ) = ⋂ N ∈ Max ( A ) D s ( F ) ⊆ F ⊆ N ( N / D s ( F ) ) = ( ⋂ D s ( F ) ⊆ F ⊆ N N ∈ Max ( A ) N ) / D s ( F ) = Rad ( F ) / D s ( F ) . For a Pseudo -BL algebra A we make the following notations: U ( A ) = { a ∈ A : ( a n ) ∼ ≤ a , forall n ∈ ℕ } ; V ( A ) = { a ∈ A : ( a n ) - ≤ a , forall n ∈ ℕ } . Remark 4.5. If A is a BL algebra, then Max _n  (A) = Max (A) and U (A) = V (A). Definition 4.6. An element a of A is called a unity if and only if for all n ∈ ℕ , (a ⁿ ) ^- and (a ⁿ ) ^∼ are nilpotent of A.Example 4.7. Consider pseudo BL-algebras from Example 2.5 (iv). Since (b₄ ⊙ b₄) ^- = (b₄) ^- = (b₄) ^∼ = 0, we get b₄ is a unity element. Also, b₁ is not a unity element, since ( b 1 2 ) - = ( b 1 ⊙ b 1 ) - = ( 0 ) - = 1 . In the next theorem we state the relationship between unity elements and radical of filters. Theorem 4.8. Let A be a linear pseudo BL-algebra and F be a proper filter of A. Then the following hold: (1) Rad (F) = {a∈ A : a is a unity element of A} ; (2) a is a unity element if and only if a is not a nilpotent of A; (3) let a be a unity element of A, then a^- < a and a^∼ < a.Proof. (1) Let a be a unity element of A and a ∉ Rad (F). Then there exists a maximal filter M of A such that F ⊆ M and a ∉ M. So, (a ⁿ ) ^- ∈ M ((a ) ^∼ ∈ M), for some n ∈ ℕ . By hypothesis, we have ((a ⁿ ) ^-)  ^m  = ((a ⁿ ) ^∼)  ^m  = 0, for some m ∈ ℕ , hence 0 ∈ M, which is a contradiction. So, a ∈ Rad (F). Conversely, let a ∈ Rad (F) and (a ⁿ ) ^- ((a ⁿ ) ^∼) not be a nilpotent elements of A, for some n ∈ ℕ . Therefore, by Proposition 3.11, we get (a ⁿ ) ^- ∈ Rad (F) ((a ⁿ ) ^∼ ∈ Rad (F)). Hence a ⁿ  ∈ Rad (F), for all n ∈ ℕ . Therefore, 0 = (a ⁿ ) ^- ⊙ a ⁿ  = (a ⁿ ) ^∼ ⊙ a ⁿ  ∈ Rad (F), which is a contradiction. Thus (a ⁿ ) ^- and (a ⁿ ) ^∼ are nilpotent elements of A, i. e., a is a unity element of A.

(2) The proof is clear by (1) and Corollary 3.12.

(3) Let a be a unity element of A and a^-notlessthana (a^∼notlessthana). Since A is linear pseudo BL-algebra, we get a ≤ a^- (a ≤ a^∼). So, by Proposition 2.6 (11), we have a² = a ⊙ a = 0, i. e. a is a nilpotent element of A, which is a contradiction by (2). Therefore, a^- < a and a^∼ < a.

Proposition 4.9. Rad(A) ⊑ U(A) ⊓ V(A) ⊑ U(A) ⊔ V(A) ⊑

Proof. Assume that a ∉ U(A). Then there exists n∈ N such that (an)∼ ≰ a; it follows that (an)∼ ≰ a ≠ 1. Hence, there exists a prime filter Psuch that (an)∼ ≰ a ∉ P. But Pis a prime, hence a≰ (an)∼ ∈ P. By Zorn Lemma, there exists a maximal filter Msuch that P⊑ M, hence a≰ (an)∼ ∈ M. If a∈ M, then an∈ M; it follows that (an)∼ ∈ M, too, since a, a≰ (an)∼ ∈ Mimplies (an)∼ ∈ M; we thus obtain a contradiction: 0 = an≰ (an)∼ ∈ M. It follows that a ∉ M, hence a ∉ ⊓Max(A). Thus we have proved that ⊓Max(A) ⊑ U(A). Analo-gously, ⊓Max(A) ⊑ V(A). Hence we have proved the first inclusion. Let now a ∉ ⊓Maxn(A), hence, there exists a normal maximal fil-ter Msuch that a ∉ M. Then (an)∼ ∈ M, by Theorem 2.18. If (an)∼ ≤ a, then a∈ Mcontra-diction. Hence (an)∼ ≰ a, so a ∉ U(A), hence U(A) ⊑ ⊓Maxn(A). Analogously, we have V(A) ⊑ ⊓Maxn(A).

Remark 4.10. (1) If M∈ Maxn(A), then ∹M∺ = M; (2) if M∈ Max(A) Maxn(A), then ∹M∺ = A.

Proposition 4.11 Let a, b∈ Rad(A). Then a− ⊙ b− = a∼ ⊙ b∼ = 0. Proof. Let a, b∈ Rad(A). Suppose (a− ⊙ b−)∼ ≠ 1. By Corollary 2.16, there is a prime filter Psuch that (a− ⊙ b−)∼ ∉ P. By Proposition 2.6 (23), we have (a− ⊙ b−)∼ = b− → (a−)∼ ∉ P, so, (a−)∼ → b− ∈ P, that is, ((a−)∼ ⊙ b)− ∈ P.

There is a maximal filter Msuch that P874 ⊑ M. Then (a−)∼ ⊙b∉ M. By Theorem 2.18, there is n≥ 1 such that (((a−)∼ ⊙ b)n)− ∈ M; so, if we denote c= ((a−)∼ ⊙ b)n, we have c− ∈ M. Since a, b∈ Rad(A) then we deduce that a, b∈ M, hence (a−)∼ ∈ M, so c= ((a−)∼ ⊙ b)n879 ∈ M. Hence cand c− are in M, which is a contradiction. By a similar argument we have a∼ ⊙ b∼ = 0.□

Definition 4.12. An element a of Ais infinitesimalif a ≠ 1 and an≥ a− ∨ a∼, for all n∈ N.

Example 4.13. Consider pseudo BL-algebra Afrom Example 2.5 (iv). Then all elements of {b4, b5,· · · ,1} are infinitesimal, but b3 and b2 are not infinitesimal. Since b3∼ ∨ b3− ≰ b3 ⊙ b3 = b32 = 0.

We denote by Inf (A) the set of all infinitesimals of A.

Proposition 4.14. Let a be a nonunit element a of A. Then the following statements hold:

(1) if a is infinitesimal, then a∈ Radn(A);

(2) if a∈ Rad(A), then ais infinitesimal.

Proof. (1) Assume that a≠1 is an infinitesimal and suppose that a∉ Radn(A). Thus, there is a maximal normal filter Mof Asuch that a∉ M. By Theorem 2.18, there is n≥ 1 such that (an)− ∈ M. By hypothesis an903 ≥ a− ∨ a∼ ≥ b− hence (an)− ≤ a−−, so a−− ∈ M. By Proposition 2.6 (30), we have (a−−)n905 ≤ (an)−−, hence (an)−− ∈ M. if we denote b= (an)− we conclude that b, b− ∈ M, hence 0 = b− ⊙ b∈ M, that is, M= A, which is a contradiction.

(2) Assume that 1 ≠ a∈ Rad(A) ⊑ U(A) ⊓ V(A). Then (an)− ≤ aand (an)∼ ≤ a for any n∈ N. For n= 1 we obtain that a−, a∼ ≤ a. Since for any n∈ w, an912 ∈ Rad(A) we deduce that (an)−,(an)∼ ≤ an. Since a− ⊙ an= an914 ⊙ a∼ = 0 for any n≥ 1, then by Proposition 2.6 (19) and (20) we obtain an915 ≤ (a−)∼ and an916 ≤ (a∼)− for any n≥ 1. So, for any n≥ 1, a− = ((a−)∼)− ≤ (an)−, a∼ = ((a∼)−)∼ ≤ (an)∼ and (an)−,(an)∼ ≤ an, hence a−, a∼ ≤ an, which implies an919 ≥ a− ∨ a∼, that is, ais an infinitesimal. For n= 0 the inequalities are trivial.

Corollary 4.15. Rad(A) {1} ⊑ Inf (A) ⊑ Radn(A).

Corollary 4.16. If A is a BL-algebra, then Rad(A) {1} = Inf (A).

Proposition 4.17. Let a∈ A and n∈ N, n≥ 1, if an∈ B(A) and an≥ a− ∨ a∼, then a= 1.