Radical of filters on equality algebra

1 Introduction

A new structure, called equality algebras, was introduced by Jenei in [12] and ithe study continued in [2, 22]. The equality algebra has two connectives, a meet operation and an equivalence, and a constant. Given that equality algebra can be considered as corresponding algebras with fuzzy type theory, it is important to study in this field. In [7], the author studied the relation among equality algebra and BCK-meet semilattice and he proved that every BCK(D)-meet semilattice and equality algebra equivalent. In logical algebra, a kind of subset of them which are called filters play very important role in this filed such as normal subgroup and ideal in group and ring theory, respectively.

Zebardast et al. [23] investigated the relation between equality algebras and other logical algebras and studied that under which conditions they are equivalent to others. Also, Zebardast et al. [23] studied commutative equality algebras. They considered characterizations of commutative equality algebras.

In [3, 19], the authors defined the concept of the radical of a filter in MTL-algebras and some equivalent definitions and properties are studied. Moreover, they introduced the notion of dense element and proved some properties of it and the relation between dense element and radical filters. In [18], Paad and Borzooei defined the new filters on BL-algebra and called it semi-maximal filters. Then they studied the radical of them and showed that any semi-maximal filter is a primary filter. Moreover, they investigated the relationship among semi-maximal filters and other types of filters. Specially, they showed that any fantastic filter is a semi-maximal filter and any semi-maximal filter is an (n-fold) positive implicative filter in a Gödel algebra. In [15], Motamed presented new properties for the primary filters and defined the notions of prime-like filters and BL-algebras. Then by using this notion, he/she introduced Top-like BL-algebras and by using the concept of prime-like filters, he/she studied a new topology on BL-algebras. In [21], the radical of an implicative filter in Hilbert algebras is studied and some theorems are given on its properties and the authors showed that radical form a closure operator. Also, the radical of the set of all regular elements is characterized. Also, the notion of semi-maximal filter is introduced as a closed set of that closure operator and studied it in detail. In addition, in [16], the notion of the radical of a filter in BL-algebras is defined and several characterizations of the radical of a filter are given. Also, the authors proved that A F is an MV-algebra if and only if D _s (A) ⊆ F. After that the authors defined the notion of semi-maximal filter in BL-algebras and they stated and proved some theorems which determine the relationship between this notion and the other types of filters of a BL-algebra. Moreover, they proved that A F is a semi-simple BL-algebra if and only if F is a semi-maximal filter of A. Prime filters of hyperlattices were studied in [1].

Considering the importance of the concept of filters in logical algebras and its application in creating the quotient space and creating logical algebras, one of the aims of this article is to introduce new filters on equality algebras and examine the relationship between them and the filters that are previously defined. It is shown that the introduced concepts are the expansion of previous filters, such as the prime and maximal(ultra) filters. On the other hand, we used the introduced concepts and define new open sets on equality algebra that could be used to construct a Zarisky topology. These concepts can also be used to define concepts related to algebra such as modules and minimal prime filters and results in this field. On the other hand, we used the introduced concepts and define new open sets on equality algebra that could be used to construct a Zarisky topology. These concepts can also be used to define concepts related to algebra such as modules and minimal prime filters and results in this field.

2 Preliminaries

Here, we refer to the basic concepts and features required in the field of equality algebra that use in the next sections [8, 20].

Definition 2.1. [12] An algebraic structure (L, ∧ , ∼ , 1) is said to be an equality algebra if for each e, h, u ∈ L, it satisfies in the following conditions:

(E1) (L, ∧ , 1) is a commutative idempotent integral monoid,

(E2) The operation “∼” is commutative,

(E3) e ∼ e = 1,

(E4) e ∼ 1 = e,

(E5) If e ≼ h ≼ u, then e ∼ u ≼ h ∼ u and e ∼ u ≼ e ∼ h,

(E6) e ∼ h ≼(e ∧ u) ∼(h ∧ u),

(E7) e ∼ h ≼(e ∼ u) ∼(h ∼ u),

where e ≼ h iff e ∧ h = e. The equality algebra (L, ∧ , ∼ , 1) is simply denoted by L for short.

We define an operation “ ↠ ” on L by e ↠ h : = e ∼(e ∧ h).

The equality algebra L is said to be prelinear if the unique upper bound of the set {e ↠ h, h ↠ e} is 1, and L is called commutative if (e ↠ h)  ↠ h =(h ↠ e)  ↠ e, for each e, h ∈ L. We notice that every prelinear equality algebra is a distributive lattice.

Proposition 2.2. [12] For any e, h, u ∈ L, the next assertions are valid:

(i) e ↠ h = 1 iff e ≼ h,

(ii) e ↠ (h ↠ u) = h ↠ (e ↠ u) ,

(iii) 1 ↠ e = e,  e ↠ 1 = 1,  e ↠ e = 1,

(iv) e ≼ h ↠ u iff h ≼ e ↠ u,

(v) e ≼ h ↠ e,

(vi) e ≼(e ↠ h)  ↠ h,

(vii) e ↠ h ≼(h ↠ u)  ↠ (e ↠ u) ,

(viii) If h ≼ e, then e ↠ h = e ∼ h,

(ix) e ∼ h ≼ e ↠ h,

(x) If e ≼ h, then h ↠ u ≼ e ↠ u and u ↠ e ≼ u ↠ h,

(xi) ((e ↠ h)  ↠ h)  ↠ h = e ↠ h,

(xii) e ↠ (h ∧ e) = e ↠ h.

(xiii) If L is a prelinear equality algebra, then (e ∨ h)  ↠ u =(e ↠ u) ∧(h ↠ u).}

A bounded equality algebra has the bottom element such as 0 such that 0 ≼ e, for any e ∈ L. The negation operation “¬”, in any bounded equality algebra L is defined by ¬e = e ↠ 0 = e ∼ 0, for any e ∈ L. If for each e ∈ L, ¬¬ e = e, then L is called involutive.

Let (L ; ∧  _L , ∼  _L , 1 _L ) and (D ; ∧  _D , ∼  _D , 1 _D ) be equality algebras. Then a mapping f :  L → D is said to be an equality homomorphism if for each e, h ∈ L, the following conditions hold: f ( e ∧ L h ) = f ( e ) ∧ D f ( h ) and f ( e ∼ L h ) = f ( e ) ∼ D f ( h ) . Moreover, suppose L and D are two bounded equality algebras, then an equality homomorphism f is called bounded if f(0 _L ) =0 _D .

An equality homomorphism f is said to be an equality (epimorphism) monomorphism if f is a (onto)one-to-one mapping and an equality homomorphism f is called an equality isomorphism if f is a one-to-one and onto mapping.

A subset Q of L is said to be a filter of L (see [13]) if for any e, h ∈ L:

(F₁) If e ∈ Q such that e ≼ h, then h ∈ Q.

(F₂) If e ∈ Q and e ∼ h ∈ Q, then h ∈ Q.

Denote by F ( L ) the set of all filters of L.

Lemma 2.3. [11] Suppose ∅ ≠ Q ⊆ L. Then Q ∈ F ( L ) iff 1 ∈ Q and for any e, h ∈ L, if e ∈ Q and e ↠ h ∈ Q, then h ∈ Q.

Clearly, 1 ∈ Q, for every Q ∈ F ( L ) . Also, Q ∈ F ( L ) is said to be a proper filter if Q ≠ L. Clearly, if L is bounded, then a filter of L is proper iff it is not containing 0. An equality algebra L is called simple if F ( L ) = { { 1 } , L } .

Let Q ∈ F ( L ) . Define the relation θ _Q on L as (e, h) ∈ θ _Q iff {e ↠ h, h ↠ e} ⊆ Q. Let L Q = { [ e ] ∣ e ∈ L } where [e] = {h ∈ L ∣  (e, h) ∈ θ _Q }. Then the binary relation ≼ _Q on L Q defined as follows: [ e ] ≼ Q [ h ] iff e ↠ h ∈ Q , which ( L Q , ≼ ) is a POSET. For any e, h ∈ L, define [ e ] ∼ Q [ h ] = [ e ∼ h ] and [ e ] ∧ Q [ h ] = [ e ∧ h ] . Then ( L Q , ∼ Q , ∧ Q , [ 1 ] θ Q ) is an equality algebra called quotient equality algebra and denoted by L Q .

Definition 2.4. [6] Let ∅neqQ ⊆ L such that 1 ∈ Q and e, h, u ∈ L.

(i) A prime filter is a proper filter Q of L such that e ↠ h ∈ Q or h ↠ e ∈ Q, for any e, h ∈ L. We use P ( L ) for prime filters of L.

(ii) An ultra filter(maximal) is a proper filter of L that is the greatest filter of L. We use U ( L ) for ultra filters of L.

(iii) Q is said to be a positive implicative filter of L if e ↠ (h ↠ u) ∈ Q and e ↠ h ∈ Q imply e ↠ u ∈ Q, for each e, h, u ∈ L.

(iv) Q is called an implicative filter of L if u ↠ ((e ↠ h)  ↠ e) ∈ Q and u ∈ Q imply e ∈ Q.

(v) Q is called a fantastic filter of L if u ↠ (e ↠ h) ∈ Q and u ∈ Q imply ((h ↠ e)  ↠ e)  ↠ h ∈ Q.

(vi) Suppose L is a bounded lattice equality algebra. Then Q ∈ F ( L ) is said to be a Boolean filter of L if e ∨ ¬e ∈ Q, for any e ∈ L.

Proposition 2.5. [6] (i) Every implicative filter of L is a positive implicative filter of L.

(ii) If L is bounded, then a filter Q of L is implicative iff for e ∈ L, ¬e ↠ e ∈ Q implies that e ∈ Q.

(iii) Suppose L is a bounded lattice equality algebra and Q ∈ F ( L ) . Then Q is a Boolean filter of L iff Q is an implicative filter of L.

Definition 2.6. [10] Consider X as a non-empty set. A mapping φ : P ( X ) ↠ P ( X ) is called a closure map on X, if for every Y , Z ∈ P ( X ) , we get:

(C1) Y ⊆ φ(Y),

(C2) φ²(Y) = φ(Y),

(C3) Y ⊆ Z implies φ(Y) ⊆ φ(Z).

Note. From now on, we let L be a bounded equality algebra, unless otherwise specified.

3 Radical filter of equality algebras

In the following, we introduce the concept of radical in equality algebras and study some properties of it.

Definition 3.1. Consider Q is a proper filter of L. The intersection of all ultra filters of L which contain Q is called the radical of Q and is denoted by Rad(Q).

Example 3.2. Assume L = {0, p, m, o, 1} is a POSET with the next Hasse diagram. Define the operation "∼" on L as follows:

Then (L, ∼ , ∧ , 1) is an equality algebra. Suppose Q = {p, 1} and U = {o, p, m, 1}. Thus Rad ( Q ) = ∩ U ∈ M ax ( L ) , Q ⊆ U U = { o , p , m , 1 } . Clearly Q ≠ {1} and so Rad(Q) ≠ {1}.

Remark 3.3. (i) According to Example 3.2, by our definition, Rad ( { 1 } ) = ∩ U ∈ M ax ( L ) , { 1 } ⊆ U U ≠ { 1 } .

(ii) Clearly Rad ( Q ) ∈ F ( L ) .

(iii) If Q ∈ F ( L ) , then Q ⊆ Rad(Q).

(iv) Q is a proper filter of L iff Rad(Q) is a proper filter of L.

(v) If Q ∈ U ( L ) , then Rad(Q) = Q.

Proposition 3.4. If Q is an implicative (positive implicative, fantastic, Boolean) filter of L, then Rad(Q) is, too.

Definition 3.5. Consider A ⊆ L. Then the smallest filter of L containing A is called the generated filter by A in L which is denoted by 〈A〉. Indeed, 〈 A 〉 = ∩ A ⊆ Q ∈ F ( L ) Q .

Proposition 3.6. Consider ∅ ≠ A ⊆ L. Then 〈 A 〉 = { e ∈ L ∣ ∃ n ∈ ℕ and p 1 , ... , p n ∈ A such that p 1 ↠ ( p 2 ↠ ( ... ↠ ( p n ↠ e ) ... ) ) = 1 } .

In particular, for any element p ∈ L, we have 〈 p 〉 = { e ∈ L ∣ ∃ n ∈ ℕ such that p ↠ n e = 1 } , where e^↠ 0h = h and e^↠ nh = e ↠ (e^↠ n - 1h).

If Q ∈ F ( L ) and p ∈ L \ Q, then 〈 Q ∪ { p } 〉 = { e ∈ L ∣ ∃ n ∈ ℕ such that p ↠ n e ∈ Q } .

If Q , K ∈ F ( L ) , then 〈 Q ∪ K 〉 = { e ∈ L ∣ ∃ g ∈ K such that g ↠ e ∈ Q } = { e ∈ L ∣ ∃ f ∈ Q such that f ↠ e ∈ K } .

Corollary 3.7. If L is a lattice equality algebra, then for any e, h ∈ L, we have 〈e〉∩〈h〉 = 〈e ∨ h〉.

Proposition 3.8. (i) For each proper filter Q of L there is an ultra filter of L that contains Q.

(ii) If T ∈ P ( L ) , then a unique ultra filter of L exists such that contains T.

(iii) Assume Q is a proper filter of L. Then Q ∈ U ( L ) iff 〈Q ∪ {e} 〉 = L, for any e ∈ L \ Q.

(iv) If L is bounded and U ∈ U ( L ) , then e ∈ L \ U iff there is n ∈ ℕ such that e^↠ n0 ∈ U.

(v) Q ∈ P ( L ) iff L Q is a TOSET.

(vi) Every ultra filter of L is a prime filter of L.

Theorem 3.9. Consider Q ∈ F ( L ) . Then R a d ( Q ) = { e ∈ L ∣ ( e ↠ n 0 ) ↠ e ∈ Q , for any n ∈ ℕ } ⊆ { e ∈ L ∣ ¬ e ↠ e ∈ Q } .

Proof. Suppose B = { e ∈ L ∣ ( e ↠ n 0 ) ↠ e ∈ Q , for any n ∈ ℕ } .

Assume Q is a proper filter of L. Let e ∈ Rad(Q) such that e ∉ B. Then there exists n ∈ ℕ such that (e^↠ n0)  ↠ e ∉ Q. Since Q is a proper filter of L, by Proposition 3.8 (i) and (iv), there exists T ∈ P ( L ) such that Q ⊆ T and (e^↠ n0)  ↠ e ∉ T. Since T ∈ P ( L ) , by Proposition 3.8 (i), there exists an ultra filter U of L such that T ⊆ U. Moreover, since T ∈ P ( L ) and (e^↠ n0)  ↠ e ∉ T, we have e ↠ (e^↠ n0) ∈ T, and so e ↠ (e^↠ n0) ∈ U. If e ∈ U, then e^↠ n0 ∈ U. By repeating this method, we get 0 ∈ U, a contradiction. If e ∉ U, then there exists an ultra filter of L such that Q ⊆ T ⊆ U and e ∉ U. Thus, e ∉ Rad(Q), a contradiction. Hence, for any n ∈ ℕ , (e^↠ n0)  ↠ e ∈ Q, and so e ∈ B.

Conversely, suppose e ∈ B and e ∉ Rad(Q). Since e ∈ B, for any n ∈ ℕ we have (e^↠ n0)  ↠ e ∈ Q. Moreover, e ∉ Rad(Q), then there exists an ultra filter U of L such that Q ⊆ U and e ∉ U. Then by Proposition 3.8. (iv), there exists n ∈ ℕ such that e^↠ n0inU. Since (e^↠ n0)  ↠ e ∈ Q and Q ⊆ U, we get e ∈ U, is a contradiction. Hence, e ∈ Rad(Q). Therefore, Rad ( Q ) = { e ∈ L ∣ ( e ↠ n 0 ) ↠ e ∈ Q , for any n ∈ ℕ } . Now, let e ∈ Rad(Q). Then for any n ∈ ℕ , (e^↠ n0)  ↠ e ∈ Q. By Proposition 2.2 (v), e ↠ 0 ≼ e ↠ (e ↠ 0). By repeating this method n - 1 times we have e ↠ 0 ≼ e^↠ n0. By Proposition 2.2(x), (e^↠ n0)  ↠ e ≼(e ↠ 0)  ↠ e = ¬ e ↠ e. Since Q ∈ F ( L ) and (e^↠ n0)  ↠ e ∈ Q, we get ¬e ↠ e ∈ Q. □

Next example shows Rad(Q) is not equal with {e ∈ L ∣  ¬e ↠ e ∈ Q}, in general.

Example 3.10. Consider L = {0, p, m, o, k, 1} is a POSET with the next Hasse diagram.Define the operation ∼ on L as follows:

Then (L, ∧ , ∼ , 1) is an equality algebra. Suppose Q = {1}. Clearly, Rad(Q) = {1} and ¬k ↠ k = o ↠ k = 1 ∈ Q but k ∉ Rad(Q).

Corollary 3.11. If Q is a positive implicative filter of L, then R a d ( Q ) = { e ∈ L ∣ for any n ∈ ℕ , ( e ↠ n 0 ) ↠ e ∈ Q } = { e ∈ L ∣ ¬ e ↠ e ∈ Q } .

Proposition 3.12. Assume Q ∈ F ( L ) . If for any e ∈ L, e ∧ ¬ e = 0, then Rad(Q) ⊆ {e ∈ L ∣ ¬ ¬e ∈ Q}.

Proof. Suppose e ∈ Rad(Q). Then by Theorem 3.9, ¬e ↠ e ∈ Q. Since for any e ∈ L, e ∧ ¬ e = 0, by Proposition 2.2(xii) we have, ¬e ↠ 0 = ¬e ↠ (e ∧ ¬e) = ¬e ↠ e ∈ Q, and so ¬¬e ∈ Q. Hence, Rad(Q) ⊆ {e ∈ L ∣ ¬ ¬e ∈ Q}. □

Proposition 3.13. Consider Q is a positive implicative filter of L. Then Rad(Q) = {e ∈ L ∣ ¬ ¬e ∈ Q}.

Proof. Let Q be a positive implicative filter of L and e ∈ Rad(Q). Then by Corollary 3.11, ¬e ↠ e ∈ Q. Since Q is positive implicative, we have ( ¬ e ↠ e ) ↠ ¬ ¬ e = ( ¬ e ↠ e ) ↠ ( ¬ e ↠ 0 ) = ¬ e ↠ ( e ↠ 0 ) = ¬ e ↠ ¬ e = 1 ∈ Q . Since Q ∈ F ( L ) , (¬e ↠ e)  ↠  ¬ ¬e ∈ Q and ¬e ↠ e ∈ Q, we get ¬¬e ∈ Q. Hence, Rad(Q) ⊆ {e ∈ L ∣ ¬ ¬e ∈ Q}. Conversely, suppose ¬¬ e ∈ Q. Since ¬¬e ≼ ¬e ↠ e and Q ∈ F ( L ) , we have ¬e ↠ e ∈ Q and by Corollary 3.11, e ∈ Rad(Q). Therefore, Rad(Q) = {e ∈ L ∣ ¬ ¬e ∈ Q}. □

Proposition 3.14. Assume Q and K are two proper filters of L. Then:

(i) For any e ∈ L \ {0}, ¬e = 0 iff Rad(Q) = L \ {0}.

(ii) If Q ⊆ K, then Rad(Q) ⊆ Rad(K).

(iii) Rad(Rad(Q)) = Rad(Q).

(iv) The map ψ : F ( L ) ↠ L such that for any Q ∈ F ( L ) , ψ(Q) = Rad(Q) is a closure operator.

(v) Rad(Q) ∩Rad(K) ⊆ Rad(〈Q ∪K〉).

(vi) Rad(Q ↠ K) ⊆ Rad(Q ↠ Rad(K)), where Q ↠ K = {e ∈ L ∣ Q ∩〈e〉 ⊆ K}.

(vii) Rad(Q)  ↠ Rad(K) ⊆ Q ↠ Rad(K).

(viii) If f : L ↠ L is a bounded equality homomorphism, then Rad(kerf) = f^-1(Rad({1})), where kerf = {e ∈ L ∣ f(e) =1}.

(ix) If {Q _i } _i∈I is a family of filters of L, then Rad ( ∩ i ∈ I Q i ) = ∩ i ∈ I Rad ( Q i ) .

(x) Rad ( [ 1 ] ) = Rad ( Q ) Q .

(xi) Rad(〈Q ∪ K〉) ⊆ Radpreccurlyeqft(〈Rad(Q) ∪ Rad(K) 〉).

(xii) [e] _Rad(Q∩K) = [e] _Rad(Q)∩ [e] _Rad(K).

(xiii) If K ⊆ Q, then Rad ( Q K ) = Rad ( Q ) K .

Proof. (i) Suppose for any e ∈ L \ {0}, ¬e = 0. Clearly Rad(Q) ⊆ L \ {0}. If e ∈ L \ {0}, then by assumption, ¬e = 0 and for any n ∈ ℕ , ( e ↠ n 0 ) ↠ e = = ( e ↠ ( e ↠ ( ⋯ ↠ ( e ↠ ( e ↠ 0 ) ) ⋯ ) ) ) ↠ e = ( e ↠ ( e ↠ ( ⋯ ↠ ( e ↠ 0 ) ⋯ ) ) ) ↠ e ⋮ = ( e ↠ 0 ) ↠ e = 0 ↠ e = 1 ∈ Q . Hence, by Theorem 3.9, e ∈ Rad(Q) and so L \ {0} ⊆ Rad(Q). Therefore, Rad(Q) = L \ {0}.

Conversely, suppose Rad(Q) = L \ {0}. Assume e ∈ L \ {0} such that ¬e ≠ 0. Then by assumption, e ∈ Rad(Q) and e ↠ 0 = ¬e ∈ Rad(Q). Since Rad ( Q ) ∈ F ( L ) , we have 0 ∈ Rad(Q), a contrsdiction. Hence, ¬e = 0. Therefore, for any e ∈ L \ {0}, we have ¬e = 0.

(ii) Consider e ∈ Rad(Q). Then by Theorem 3.9, for any n ∈ ℕ , (e^↠ n0)  ↠ e ∈ Q. Since Q ⊆ K, we get that for any n ∈ ℕ , (e^↠ n0)  ↠ e ∈ K, and so e ∈ Rad(K). Hence, Rad(Q) ⊆ Rad(K).

(iii) Since Rad(Q) is a proper filter of L, we get Rad(Q) ⊆ Rad(Rad(Q)). Suppose e ∈ Rad(Rad(Q)). Then for any ultra filter U of L such that Rad(Q) ⊆ U, we have e ∈ U. Suppose U 0 ∈ U ( L ) such that Q ⊆ U₀. By (ii), Rad(Q) ⊆ Rad(U₀) = U₀. Then e ∈ U₀. Thus we show that for any ultra filter U of L that Q ⊆ U, we have e ∈ U. Hence, e ∈ Rad(Q), and so Rad(Rad(Q)) ⊆ Rad(Q). Therefore, Rad(Rad(Q)) = Rad(Q).

(iv) By Remark 3.3(iii), (ii) and (iii), the proof is clear.

(v) Since Q, K ⊆ 〈Q ∪ K〉, by (ii), we have Rad(Q) , Rad(K) ⊆ Rad(〈Q ∪ K〉). Hence, Rad ( Q ) ∩ Rad ( K ) ⊆ Rad ( 〈 Q ∪ K 〉 ) . (vi) Assume e ∈ Rad(Q ↠ K). Then for any n ∈ ℕ , we have (e^↠ n0)  ↠ e ∈ Q ↠ K. Thus Q ∩〈(e^↠ n0)  ↠ e〉 ⊆ K ⊆ Rad(K), and so 〈(e^↠ n0)  ↠ e〉 ∈ Q ↠ Rad(K). Hence, e ∈ Rad(Q ↠ Rad(K)). Therefore, Rad(Q ↠ K) ⊆ Rad(Q ↠ Rad(K)).

(vii) Let e ∈ Rad(Q)  ↠ Rad(K). Then 〈e〉∩Rad(Q) ⊆ Rad(K). By Remark 3.3(iii), Q ∩〈e〉 ⊆ 〈e〉∩Rad(Q) ⊆ Rad(K), and so e ∈ Q ↠ Rad(K). Hence, Rad(Q)  ↠ Rad(K) ⊆ Q ↠ Rad(K).

(viii) For any e ∈ L, e ∈ Rad(kerf) iff for any n ∈ ℕ , (e^↠ n0)  ↠ e ∈ kerf iff f((e^↠ n0)  ↠ e) =1 iff (f(e) ^↠ nf(0))  ↠ f(e) =1 iff for any n ∈ ℕ , f(e) ∈ Rad({1}) iff e ∈ f^-1(Rad({1})).

(ix) Clearly, for any i ∈ I, ∩ i ∈ I Q i ⊆ Q i . Then by (ii), we have Rad ( ∩ i ∈ I Q i ) ⊆ ∩ i ∈ I Rad ( Q i ) . Conversely, suppose e ∈ ∩ i ∈ I Rad ( Q i ) . Then for any i ∈ I, e ∈ Rad(Q _i ). Thus for any i ∈ I and n ∈ ℕ , (e^↠ n0)  ↠ e ∈ Q _i , and so for any n ∈ ℕ , ( e ↠ n 0 ) ↠ e ∈ ∩ i ∈ I Q i . Hence, e ∈ Rad ( ∩ i ∈ I Q i ) . Therefore, Rad ( ∩ i ∈ I Q i ) = ∩ i ∈ I Rad ( Q i ) .

(x) For any e ∈ L, [e] ∈ Rad([1]) iff for any n ∈ ℕ , ([e] ^↠ n [0])  ↠  [e] ∈ [1] iff for any n ∈ ℕ , (e^↠ n0)  ↠ e ∈ Q iff e ∈ Rad(Q) iff [ e ] ∈ Rad ( Q ) Q .

(xi) By Remark 3.3(iii), we have Q ⊆ Rad(Q) and K ⊆ Rad(K). Then Q ∪ K ⊆ Rad(Q) ∪ Rad(K), and so 〈Q ∪ K〉 ⊆ langleRad(Q) ∪ Rad(K) 〉. Thus by (ii), R a d ( 〈 Q ∪ K 〉 ) ⊆ R a d ⊆ ( 〈 R a d ( Q ) ∪ R a d ( K ) 〉 ) . (xii) Since Q∩K ⊆ Q, K, by (ii) we have Rad(Q∩K) ⊆ Rad(Q) , Rad(K). Suppose h ∈ [e] _Rad(Q∩K). Then e ↠ h, h ↠ e ∈ Rad(Q∩K). Since Rad(Q∩K) ⊆ Rad(Q), we get e ↠ h, h ↠ e ∈ Rad(Q) and so h ∈ [e] _Rad(Q). By the similar way, we have h ∈ [e] _Rad(K). Hence, h ∈ [e] _Rad(Q)∩ [e] _Rad(K). Therefore, [e] _Rad(Q∩K) ⊆ [e] _Rad(Q)∩ [e] _Rad(K). Now, suppose h ∈ [e] _Rad(Q)∩ [e] _Rad(K). Then e ↠ h, h ↠ e ∈ Rad(Q) and e ↠ h, h ↠ e ∈ Rad(K). Thus e ↠ h, h ↠ e ∈ Rad(Q) ∩Rad(K). By (ix), we have e ↠ h, h ↠ e ∈ Rad(Q∩K), and so h ∈ [e] _Rad(Q∩K). Hence, [e] _Rad(Q)∩ [e] _Rad(K) ⊆ [e] _Rad(Q∩K). Therefore, [e] _Rad(Q)∩ [e] _Rad(K) = [e] _Rad(Q∩K).

(xiii) Consider Q , K ∈ F ( L ) such that K ⊆ Q. Then Rad ( Q K ) = ∩ Q K ⊆ U K U K = ∩ K ⊆ Q ⊆ U U = ∩ Q ⊆ U U K = Rad ( Q ) K . □

Next example shows that the converse of relation in Proposition 3.14 (v), (vi) and (ii) does not hold, in general.

Example 3.15. (i) Let L = {0, p, m, 1} be a POSET with the next Hasse diagram. Define the operation ∼ on L as follows:

Then (L, ∧ , ∼ , 1) is an equality algebra. Suppose Q = {p, 1} and K = {m, 1}. Clearly, Rad(Q) = Q, Rad(K) = K and 〈Q ∪ K〉 = 〈 {p, m, 1} 〉 = L. Hence, L = 〈Q ∪ K〉 ⊈ Rad(Q) ∩Rad(K) = {1}.

(ii) Assume L is the equality algebra as in Example 3.2. Suppose Q = {p, 1} and K = {m, 1}. By routine calculations, Q ↠ K = {m, 1} and so Rad(Q ↠ K) = {p, m, o, 1}. On the other side, Q ↠ Rad(K) = Q ↠  {p, m, o, 1} = L, and so L = Q ↠ Rad ( K ) ⊈ Rad ( Q ↠ K ) = { p , m , o , 1 } . (iii) Let L be the equality algebra as in Example 3.2. Obviously, K = {p, 1} and Q = {m, 1} are two proper filters of L such that { 0 , p , m , 1 } = Rad ( K ) ⊆ Rad ( Q ) = { 0 , p , m , 1 } , but {p, 1} ⊈ {m, 1}.

Proposition 3.16. Let T ∈ P ( L ) . Then there exists a uniqe ultra filter U of L such that Rad(T) = U.

Proof. By Proposition 3.8(ii), the proof is clear. □

The set of dense elements of L is denoted by D(L) = {e ∈ L ∣ ¬ e = 0}.

Example 3.17. Suppose L is the equality algebra as in Example 3.2. Clearly D(L) = {p, m, o, 1}.

Remark 3.18. (i) If Q ∈ F ( L ) , then D(Q) = D(L) ∩Q.

(ii) If Q ∈ F ( L ) , then D(Q) ⊆ Rad(Q). Since for any e ∈ D(Q), we have ¬e = 0 and for any n ∈ ℕ , (e^↠ n0)  ↠ e = 0 ↠ e = 1 ∈ Q, and so e ∈ Rad(Q).

Proposition 3.19. D ( L ) ∈ F ( L ) .

Proof. Since ¬1 =0, it is clear that 1 ∈ D(L). Suppose e, e ↠ h ∈ D(L). Then ¬e = ¬(e ↠ h) =0. By Proposition 2.2(vii), e ↠ h ≼(h ↠ 0)  ↠ (e ↠ 0). Since ¬e = 0 we get e ↠ h ≼(h ↠ 0)  ↠ 0. Moreover, by Proposition 2.2(xi), we have h ↠ 0 =((h ↠ 0)  ↠ 0)  ↠ 0 ≼(e ↠ h)  ↠ 0. From ¬(e ↠ h) =0, we obtain h ↠ 0 = 0 and so h ∈ D(L). Therefore, D ( L ) ∈ F ( L ) . □

Proposition 3.20. Assume Q ∈ F ( L ) . Then:

(i) If L Q is an involutive equality algebra, then D(L) ⊆ Q.

(ii) If L is an involutive equality algebra, then D(L) = {1}.

Proof. (i) If L Q is an involutive equality algebra, then for any e ∈ L, [¬ ¬e] = [e]. Thus ¬¬e ↠ e ∈ Q and e ↠  ¬ ¬e ∈ Q. Suppose e ∈ D(L). Then ¬e = 0 and so ¬¬e = 1. Since for any e ∈ L, e = 1 ↠ e = ¬ ¬e ↠ e ∈ Q, we get e ∈ Q. Hence D(L) ⊆ Q.

(ii) Let L be an involutive equality algebra. Since { 1 } ∈ F ( L ) , kfracL {1} is well-defined and kfracL {1} oongL. Thus kfracL {1} is an involutive equality algebra and so by (i), D(L) ⊆ {1}. On the other side, clearly ¬1 =0 and so {1} ⊆ D(L) ⊆ {1}. Hence D(L) = {1}. □

In the following example we show that the converse of Proposition 3.20(ii) does not hold:

Example 3.21. Let L be the equality algebra as in Example 3.10. Clearly D(L) = {1}, but L is not involutive, since ¬¬ m = ¬ k = o ≠ m.

Proposition 3.22. Suppose Q ∈ F ( L ) . We have,

(i) D(L) ⊆ Rad(Q).

(ii) Rad ( Q D ( Q ) ) = Rad ( Q ) D ( Q ) .

Proof. (i) Consider e ∈ D(L). Then ¬e = 0. Thus similar to Remark 3.18(ii), we can see that e ∈ Rad(Q). Hence, D(L) ⊆ Rad(Q).

(ii) For e ∈ L, [ e ] ∈ Rad ( Q D ( Q ) ) iff for any n ∈ ℕ , ( [ e ] ↠ n [ 0 ] ) ↠ [ e ] ∈ Q D ( Q ) iff [ ( e n ↠ 0 ) ↠ e ] ∈ Q D ( Q ) iff (e ⁿ  ↠ 0)  ↠ e ∈ Q iff e ∈ Rad(Q) iff [ e ] ∈ Rad ( Q ) D ( Q ) . □

Example 3.23. Let L be the equality algebra as in Example 3.15(i). Suppose K = {p, 1} and Rad(K) = K. Clearly, p ∈ Rad(K) and ¬p = m ≠ 0, and so p ∉ D(L).

Proposition 3.24. Let f : L ↠ D be an equality epimorphism such that Q ∈ F ( L ) , K ∈ F ( D ) , respectively, and kerf ⊆ Q. Then Rad(f(Q)) = f(Rad(Q)) and Rad(f^-1(K)) = f^-1(Rad(K)).

Proof. Since Q ∈ F ( L ) and kerf ⊆ Q, we get f ( Q ) ∈ F ( D ) . Suppose h ∈ Rad(f(Q)). Then h ∈ f ( R a d ( Q ) ) ⇔ there exists e ∈ R a d ( Q ) such that h = f ( e ) ⇔ for any n ∈ ℕ , ( e ↠ n 0 ) ↠ e ∈ Q such that h = f ( e ) ⇔ for any n ∈ ℕ , f ( ( e ↠ n 0 ) ↠ e ) ∈ f ( Q ) such that h = f ( e ) ⇔ for any n ∈ ℕ , ( f ( e ) ↠ n f ( 0 ) ) ↠ f ( e ) ∈ f ( Q ) such that h = f ( e ) ⇔ for any n ∈ ℕ , ( h ↠ n 0 ) ↠ h ∈ f ( Q ) ⇔ h ∈ R a d ( f ( Q ) ) . Now, we prove Rad(f^-1(K)) = f^-1(Rad(K)). For this, e ∈ f^-1(Rad(K)) iff f(e) ∈ Rad(K) iff for any n ∈ ℕ , (f(e) ^↠ nf(0))  ↠ f(e) ∈ K iff for any n ∈ ℕ , (e^↠ n0)  ↠ e ∈ f^-1(K) iff e ∈ Rad(f^-1(K)). □

Definition 3.25. If Q ∈ F ( L ) and K ∈ F ( D ) , respectively, then Q × K = { ( p , m ) ∣ p ∈ Q and m ∈ K } ∈ F ( L × D ) , where for any (p, m) , (o, k) ∈ L×D we have ( p , m ) ≈ ( o , k ) = ( p ~ L o , m ~ D k ) , ( p , m ) ⊓ ( o , k ) = ( p ∧ L o , m ∧ D k ) .

Proposition 3.26. If Q ∈ F ( L ) and K ∈ F ( D ) , respectively, then Rad(Q) × Rad(K) = Rad(Q × K).

Proof. According to Definition 3.25, we have R a d ( Q × K ) = { ( p , m ) ∈ L × D ∣ for any n ∈ ℕ , ( ( p , m ) ⇝ n ( 0 , 0 ) ) ⇝ ( p , m ) ∈ Q × K } = { ( p , m ) ∈ L × D ∣ for any n ∈ ℕ , ( ( p ↠ n 0 ) ↠ p , ( m ↠ n 0 ) ↠ m ) ) ∈ Q × K } = { ( p , m ) ∈ L × D ∣ for any n ∈ ℕ , ( p ↠ n 0 ) ↠ p ∈ Q and ( m ↠ n 0 ) ↠ m ∈ K } = { ( p , m ) ∈ L × D ∣ p ∈ R a d ( Q ) and m ∈ R a d ( K ) } = R a d ( Q ) × R a d ( K ) . Therefore, Rad(Q) × Rad(K) = Rad(Q × K). □

Corollary 3.27. If {Q _i } _i∈I is a non-empty family of filters of L, then Rad ( ∏ i ∈ I Q i ) = ∏ i ∈ I Rad ( Q i ) .

Theorem 3.28. Assume Q ∈ F ( L ) and K ∈ F ( D ) , respectively. Then L × D Rad ( Q × K ) ≃ L Rad ( Q ) × D Rad ( K ) .

Proof. Consider the natural homomorphisms π L : L → L Rad ( Q ) and π D : D → D Rad ( K ) such that for any e ∈ L and h ∈ D we have π _L (e) = [e] and π _D (h) = [h]. We define Θ : L × D → L Rad ( Q ) × D Rad ( K ) such that for any (e, h) ∈ L × D, Θ(e, h) =([e] , [h]). Then it is clear that Θ is a well-defined onto homomorphism. Now, we prove kerΘ = Rad(Q × K). For this, by Proposition 3.26 we have, k e r Θ = { ( e , h ) ∈ L × D ∣ Θ ( e , h ) = ( [ 1 ] , [ 1 ] ) } = { ( e , h ) ∈ L × D ∣ ( [ e ] , [ h ] ) = ( [ 1 ] , [ 1 ] ) } = { ( e , h ) ∈ L × D ∣ [ e ] = [ 1 ] and [ h ] = [ 1 ] } = { ( e , h ) ∈ L × D ∣ e ∈ R a d ( Q ) and h ∈ R a d ( K ) } = R a d ( Q ) × R a d ( K ) = R a d ( Q × K ) . Therefore, L × D Rad ( Q × K ) ≃ L Rad ( Q ) × D Rad ( K ) . □

Proposition 3.29. Rad({1}) = {1} iff e = inf { ( e ↠ n 0 ) ↠ e ∣ for any n ∈ ℕ } .

Proof. Suppose Rad({1}) = {1} and B = { ( e ↠ n 0 ) ↠ e ∣ for any n ∈ ℕ } . Since for any n ∈ ℕ , e ≼(e^↠ n0)  ↠ e, we get e is a lower bound of B. Now, we prove e is the greatest lower bound of B. Suppose there exists h ∈ L such that for any n ∈ ℕ , h ≼(e^↠ n0)  ↠ e. We prove h ≼ e. For this, since h ≼(e^↠ n0)  ↠ e, we have h ↠ ((e^↠ n0)  ↠ e) =1. Moreover, by Proposition 2.2(v) and (x), e ≼ h ↠ e, and so (h ↠ e)  ↠ 0 ≼ e ↠ 0. By Proposition 2.2(x), (h ↠ e)  ↠ ((h ↠ e)  ↠ 0) ≼(h ↠ e)  ↠ (e ↠ 0). Also, since (h ↠ e)  ↠ 0 ≼ e ↠ 0, we have e ↠ ((h ↠ e)  ↠ 0) ≼ e ↠ (e ↠ 0). Then ( h ↠ e ) ↠ 2 0 = ( h ↠ e ) ↠ ( ( h ↠ e ) ↠ 0 ) ≼ ( h ↠ e ) ↠ ( e ↠ 0 ) = e ↠ ( ( h ↠ e ) ↠ 0 ) ≼ e ↠ ( e ↠ 0 ) = e ↠ 2 0. By continuing this method, we get (h ↠ e) ^↠ n0 ≼ e^↠ n0, for any n ∈ ℕ . Thus (e^↠ n0)  ↠ e ≼((h ↠ e) ^↠ n0)  ↠ e, and so h ↠ ((e^↠ n0)  ↠ e) ≼ h ↠ (((h ↠ e) ^↠ n0)  ↠ e). From h ↠ ((e^↠ n0)  ↠ e) =1, we obtain h ↠ (((h ↠ e) ^↠ n0)  ↠ e) =1, and so ((h ↠ e) ^↠ n0)  ↠ (h ↠ e) =1. Thus h ↠ e ∈ Rad({1}) = {1}. Hence, h ≼ e. Therefore, e = inf { ( e ↠ n 0 ) ↠ e ∣ for any n ∈ ℕ } .

Conversely, it is clear that {1} ⊆ Rad({1}). Suppose e ∈ Rad({1}). Then for any n ∈ ℕ , (e^↠ n0)  ↠ e ∈ {1}. Thus 1 = ( e ↠ n 0 ) ↠ e ∈ inf { ( e ↠ n 0 ) ↠ e ∣ for any n ∈ ℕ } = e . Hence, e = 1. Therefore, Rad({1}) = {1}.□

4 Semi-maximal filter of equality algebras

Next, we define the concept of semi-maximal filter of equality algebra and show some properties of it.

Definition 4.1. Consider Q ∈ F ( L ) . If Rad(Q) = Q, then Q is called a semi-maximal filter of L.

All semi-maximal filters of L is denoted by S Max ( L ) .

Example 4.2. (i) Consider L = {0, p, m, o, k, 1} is a lattice with the next Hasse diagram. Define the operation "∼" on E as follows:

Then (L, ∼ , ∧ , 1) is an equality algebra. Suppose Q = {p, m, 1}. Easily Rad(Q) = Q, and so Q ∈ S Max ( L ) .

(ii) Let L be the equality algebra as in Example 3.2. If Q = {p, 1}, then easily we can see that Rad(Q) = {p, m, o, 1} ≠ Q, and so Q ∉ S Max ( L ) .

In the next example, the relation between different kinds of filter of equality algebra and semi-maximal filter of it is studied.

Example 4.3. (i) Consider L is the equality algebra of Example 4.2(i). Suppose Q = {1, o}. Since Rad(Q) = Q, we get Q ∈ S Max ( L ) which is not a positive implicative filter of L. Because p ↠ (p ↠ 0) = o ∈ Q and p ↠ p = 1 ∈ Q, but p ↠ 0 = k ∉ Q.

(ii) Assume L is the equality algebra of Example 3.10. Suppose Q = {1}. Obviously, Q ∈ S Max ( L ) which is not a fantastic filter of L. Because 0 ↠ p ∈ Q but ((p ↠ 0)  ↠ 0)  ↠ p = k ∉ Q.

(iii) Consider L is the equality algebra of Example 3.10. Suppose Q = {1}. Clearly, Q ∈ S Max ( L ) ∉ P ( L ) because m ↠ p = k ∉ Q and p ↠ m = k ∉ Q.

(iv) According to Example 3.15, suppose Q = {1}. Obviously, Q ∈ S Max ( L ) but it is not an ultra filter of L because U₁ = {p, 1} and U₂ = {m, 1} are two ultra filters of L.

(v) Suppose L = {0, p, m, 1} is a TOSET such that 0 ≼ p ≼ m ≼ 0. Define the operation ∼ on L as follows:

Then (L, ∼ , ∧ , 0, 1) is an equality algebra. Suppose Q = {m, 1}. Clearly, Q ∈ S Max ( L ) but it is not an implicative filter of L because ¬p ↠ p = 1 ∈ Q but p ∉ Q. Also, Q = {m, 1} is not a Boolean filter of L, since p ∨ ¬p = p ∉ Q.

Proposition 4.4. Assume Q ∈ F ( L ) . Then the next statements hold:

(i) If Q is an implicative filter of L, then Q ∈ S Max ( L ) .

(ii) If L is a bounded lattice equality algebra and Q is a Boolean filter of L, then Q ∈ S Max ( L ) .

(iii) If Q ∈ U ( L ) , then Q ∈ S Max ( L ) .

(iv) If Q ∈ S Max ( L ) and a positive implicative filter of L, then Q is an implicative filter of L.

(v) If Q ∈ S Max ( L ) such that e^↠ 2h = e ↠ h, for any e, h ∈ L, then Q is an implicative filter of L.

Proof. (i) Consider Q is an implicative filter of L. By Remark 3.3(iii), Q ⊆ Rad(Q). Suppose e ∈ Rad(Q). Then for any n ∈ ℕ , (e^↠ n0)  ↠ e ∈ Q. By Proposition 2.5(i), Q is a positive implicative filter of L, then e^↠ n0 = e ↠ 0, thus (e ↠ 0)  ↠ e =(e^↠ n0)  ↠ e ∈ Q, and so ¬e ↠ e ∈ Q. Since Q is an implicative filter of L, by Proposition 2.5(ii), e ∈ Q. Hence, Rad(Q) ⊆ Q. Therefore, Rad(Q) = Q and so Q ∈ S Max ( L ) .

(ii) Let Q be a Boolean filter of L. By Proposition 2.5(iii), Q is an implicative filter of L and by (i), Q ∈ S Max ( L ) .

(iii) If Q ∈ U ( L ) , clearly Rad(Q) = Q, and so Q ∈ S Max ( L ) .

(iv) Suppose Q ∈ S Max ( L ) , a positive implicative filter of L and ¬e ↠ e ∈ Q, for any e ∈ L. Since Q is a positive implicative filter of L, by Corollary 3.11, e ∈ Rad(Q) = Q. Hence, by Proposition 2.5(ii), Q is an implicative filter of L.

(v) Let e ↠ (h ↠ u) ∈ Q and e ↠ h ∈ Q. Then by Proposition 2.2(vii), we have e ↠ h ≼ ( h ↠ ( e ↠ u ) ) ↠ ( e ↠ ( e ↠ u ) ) . Since Q ∈ F ( L ) , we get e ↠ (e ↠ u) ∈ Q. Then by assumption e ↠ ( e ↠ u ) = e ↠ 2 u = e ↠ u ∈ Q . Thus Q is a positive implicative filter of L and by (iv), Q is an implicative filter of L. □

Proposition 4.5. Consider f : L ↠ D is an equality homomorphism. Then:

(i) If K ∈ S Max ( D ) , then f - 1 ( K ) ∈ S Max ( L ) .

(ii) If f is onto such that kerf ⊆ Q and Q ∈ S Max ( L ) , then f ( Q ) ∈ S Max ( D ) .

Proof. (i) Since K is a proper filter of D, it is easy to see that f^-1(K) is a proper filter of L. Moreover, K is semi-maximal, then Rad(K) = K. Thus R a d ( f − 1 ( K ) ) = { e ∈ L ∣ for any n ∈ ℕ , ( e ↠ n 0 ) ↠ e ∈ f − 1 ( K ) } = { e ∈ L ∣ for any n ∈ ℕ , ( f ( e ) ↠ n f ( 0 ) ) ↠ f ( e ) ∈ K } = { e ∈ L ∣ f ( e ) ∈ R a d ( K ) = K } = { e ∈ L ∣ e ∈ f − 1 ( K ) } = f − 1 ( K ) . Hence, f - 1 ( K ) ∈ S Max ( L ) .

(ii) Since Q ∈ S Max ( L ) , we have Rad(Q) = Q. Clearly, f ( Q ) ∈ F ( D ) and by Proposition 3.24, f(Q) = f(Rad(Q)) = Rad(f(Q)), and so f ( Q ) ∈ S Max ( D ) . □

Proposition 4.6. If Q ∈ F ( L ) , then:

(i) Rad(Q) is the smallest semi-maximal filter of L such that Q ⊆ Rad(Q).

(ii) Rad ( Q ) Q ∈ S Max ( L Q ) .

Proof. (i) By Proposition 3.14(iii), Rad(Rad(Q)) = Rad(Q). Then Rad ( Q ) ∈ S Max ( L ) such that by Remark 3.3(iii), Q ⊆ Rad(Q). Suppose K ∈ S Max ( L ) such that Q ⊆ K. Then by Proposition 3.14(ii), Rad(Q) ⊆ Rad(K). Since K ∈ S Max ( L ) , we have Rad(Q) ⊆ K, and so Rad(Q) is the smallest semi-maximal filter of L such that Q ⊆ Rad(Q).

(ii) Clearly, by Remark 3.3(iii), Rad ( Q ) Q ⊆ Rad ( Rad ( Q ) Q ) . Assume for any e ∈ L, [ e ] ∈ Rad ( Rad ( Q ) Q ) . Then for any n ∈ ℕ , ( [ e ] ↠ n [ 0 ] ) ↠ [ e ] ∈ Rad ( Q ) Q . Thus there exists h ∈ Rad(Q) such that ([e] ^↠ n [0])  ↠  [e] = [h], and so ((e^↠ n0)  ↠ e)  ↠ h ∈ Q and h ↠ ((e^↠ n0)  ↠ e) ∈ Q ⊆ Rad(Q). Since h ∈ Rad(Q) and Rad ( Q ) ∈ F ( L ) , we get (e^↠ n0)  ↠ e ∈ Rad(Q), and so by Proposition 3.14(iii), e ∈ Rad(Rad(Q)) = Rad(Q). Thus, [ e ] ∈ Rad ( Q ) Q . Hence, Rad ( Q ) Q = Rad ( Rad ( Q ) Q ) . Therefore, Rad ( Q ) Q ∈ S Max ( L Q ) . □

Corollary 4.7. (i) Let { Q i } i ∈ I ∈ S Max ( L ) . Then ∩ i ∈ I Q i ∈ S Max ( L ) .

(ii) Consider Q ∈ F ( L ) . Then Q ∈ S Max ( L ) iff [ 1 ] ∈ S Max ( L Q ) .

5 Prime-like filters of equality algebras

In this section, we introduce the notion of prime-like filter and investigate some properties of it.

Note. In this section we let L be a lattice equality algebra unless otherwise state.

Definition 5.1. Assume Q is a proper filter of L. Then Q is called a ∨-prime filter of L, if e ∨ h ∈ Q implies e ∈ Q or h ∈ Q, for any e, h ∈ L.

Example 5.2. Suppose L is the equality algebra as in Example 3.15 such that Q = {p, 1} and K = {1}. Then Q is a ∨-prime filter of L. Since p ∨ m = 1 ∈ K but p, m ∉ K, then K is not a ∨-prime filter of L.

Example 5.3. Consider L is the equality algebra as in Example 3.10 such that Q = {1}. Then Q is a ∨-prime but Q ∉ P ( L ) , because p ↠ m = m ↠ p = k ∉ Q.

Theorem 5.4. Every prime filter of L is a ∨-prime filter of L.

Proof. Suppose Q ∈ P ( L ) and Q = Q₁∩Q₂. If Q ≠ Q₁ and Q ≠ Q₂, then there exist e ∈ Q₁ \ Q and h ∈ Q₂ \ Q. Since Q ∈ P ( L ) , we have e ↠ h ∈ Q or h ↠ e ∈ Q. If e ↠ h ∈ Q ⊆ Q₁, then since e ∈ Q₁ and Q₁ is a filter of L, we get h ∈ Q₁. Thus h ∈ Q₁∩Q₂ = Q, a contradiction. By the similar way, from h ↠ e ∈ Q, we get e ∈ Q₁∩Q₂ = Q, a contradiction. Hence, Q = Q₁ or Q = Q₂. Now, suppose e ∨ h ∈ Q such that e ∉ Q and h ∉ Q. Then by Corollary 3.7, clearly, 〈Q ∪ {e} 〉∩〈Q ∪ {h} 〉 = 〈Q ∪ {eveeh} 〉 = Q. Thus 〈Q ∪ {e} 〉 ⊆ Q and 〈Q ∪ {h} 〉 ⊆ Q, and so e ∈ Q or h ∈ Q, which is a contradiction. Hence, Q is a ∨-prime of L. □

Theorem 5.5. If L is prelinear, then any prime filter of L and any ∨-prime filter of L are conside.

Proof. (⇒) Since L is a prelinear equality algebra, we get L is a lattice. Let e, h ∈ L such that e ∨ h ∈ Q. Since Q ∈ P ( L ) , we have e ↠ h ∈ Q or h ↠ e ∈ Q. If e ↠ hinQ, since e ∨ h ∈ Q, by Proposition 2.2(xiii), (e ∨ h)  ↠ h = e ↠ h ∈ Q. Since Q ∈ F ( L ) , we get h ∈ Q. By the similar way, e ∈ Q.

(⇐) Assume Q is a ∨-prime filter of L. Since L is prelinear, for any e, h ∈ L, (e ↠ h) ∨(h ↠ e) =1, and so (e ↠ h) ∨(h ↠ e) ∈ Q. Moreover, from Q is a ∨-prime filter of L, we get e ↠ h ∈ Q or h ↠ e ∈ Q. Hence, Q ∈ P ( L ) . □

Next example shows that the condition prelinear in Theorem 5.5 is necessary.

Example 5.6. Suppose L is the equality algebra as in Example 3.10. Since (p ↠ m) ∨(m ↠ p) = k ≠ 1, it is obvious that L is not prelinear. Suppose Q = {1}. Then Q is a ∨-prime but Q ∉ P ( L ) , because p ↠ m = m ↠ p = k ∉ Q.

Definition 5.7. A proper filter Q of L is called a prime-like filter of L if for any e, h ∈ L, e ∨ h ∈ Q implies e ∈ Q or h ∈ Rad(Q).

Example 5.8. (i) Assume L is the equality algebra as in Example 4.3(v). Then Q = {m, 1} is a prime-like filter of L.

(ii) Let L be the equality algebra as in Example 3.15. Then Q = {1} and Rad(Q) = {p, 1} ∩ {m, 1} = {1}. Thus p ∨ m = 1 ∈ Q and clearly p ∉ Q and m ∉ Rad(Q). Hence, Q is not a prime-like filter of L.

In the next example we show the relation between different kind of filters of L with prime-like filters of L.

Example 5.9. (i) Assume L is the equality algebra as in 4.2(i). Suppose Q = {1, o}. Then Q is a prime-like filter of L but it is not a positive implicative filter of L. Because p ↠ (p ↠ 0) = o ∈ Q and p ↠ p = 1 ∈ Q, but p ↠ 0 = k ∉ Q.

(ii) Consider L is the equality algebra as in Example 3.2. Suppose Q = {p, 1}. It is clear that Q is a prime-like filter but it is not a fantastic filter of L. Because 0 ↠ m ∈ Q but ((m ↠ 0)  ↠ 0)  ↠ m = m ∉ Q.

(iii) Let L be the equality algebra as in Example 4.3(v). Suppose Q = {m, 1}. Clearly, Q is a prime-like filter of L but it is not an implicative filter of L because ¬p ↠ p ∈ Q but p ∉ Q.

Proposition 5.10. Every ∨-prime filter of L is a prime-like filter of L.

Proof. Suppose Q is a ∨-prime filter of L and for any e, h ∈ L, e ∨ h ∈ Q. Since Q is a ∨-prime filter of L, we get e ∈ Q or h ∈ Q. If e ∈ Q, then the proof is complete. If e ∉ Q, then h ∈ Q. By Remark 3.3(iii), QsubseteqRad(Q), and so h ∈ Rad(Q). Therefore, Q is a prime-like filter of L.

In the next example, we show that the converse of above proposition may not be true, inmreak general.

Example 5.11. Let L be the equality algebra as in Example 3.3. Suppose Q = {1}. Clearly, Rad(Q) = {o, p, m, 1}, thus Q is a prime-like filter of L but it is not a ∨-prime filter of L, because p ∨ m = 1 ∈ Q but p ∉ Q and m ∉ Q.

Proposition 5.12. Assume Q is a proper filter of L such that it satisfies in at least one of the following conditions:

(i) Q ∈ U ( L ) .

(ii) Q ∈ S Max ( L ) .

(iii) Q is implicative filter.

Then Q is a prime-like filter of L iff Q is a ∨-prime filter of L.

Proof. By Propositions 4.4 and 5.8, the proof is clear. □

Corollary 5.13. Any proper filter of L can be extended to a prime-like filter of L.

Corollary 5.14. Every prime filter of L is a prime-like filter of L.

Proof. By Theorem 5.4 and Proposition 5.10, the proof is clear. □

Next example shows that the converse of Corollary 5.14 does not hold, in general.

Example 5.15. Consider L is the equality algebra as in Example 3.10. Suppose Q = {1}. Clearly, Q is a prime-like filter of L but Q ∉ P ( L ) , because m ↠ p = k ∉ Q and p ↠ m = k ∉ Q.

Corollary 5.16. Every ultra filter of L is a prime-like filter of L.

Proof. By Proposition 3.8(vi), Theorem 5.4 and Proposition 5.10, the proof is clear. □

Following example shows that the converse of Corollary 5.16 is not true, in general.

Example 5.17. Let L be the equality algebra as in Example 3.2. Suppose Q = {p, 1}. Clearly, Q is a prime-like filter of L but Q ∉ U ( L ) because U = { p , m , o , 1 } ∈ U ( L ) .

Proposition 5.18. Suppose L is prelinear and Q is a proper filter of L. Then Q is a prime-like filter of L iff for any e, h ∈ L, e ↠ h ∈ Q or h ↠ e ∈ Rad(Q).

Proof. Suppose Q is a prime-like filter of L and e, h ∈ L. Since L is prelinear, we have (e ↠ h) ∨(h ↠ e) =1 ∈ Q. Then e ↠ h ∈ Q or h ↠ e ∈ Rad(Q). Conversely, suppose Q is a proper filter of L and e, h ∈ L such that e ∨ h ∈ Q. Then by Proposition 2.2(xiii), (eveeh)  ↠ h = e ↠ h ∈ Q. Since (eveeh)  ↠ h ∈ Q, e ∨ h ∈ Q and Q ∈ F ( L ) , we get h ∈ Q. If e ↠ h ∉ Q, then h ↠ e ∈ Rad(Q). By Remark 3.3(iii), Q ⊆ Rad(Q), then eveeh ∈ Rad(Q). Again, by Proposition 2.2(xiii), (eveeh)  ↠ e = h ↠ e ∈ Rad(Q). Since (eveeh)  ↠ e ∈ Rad(Q), e ∨ h ∈ Rad(Q) and Rad ( Q ) ∈ F ( L ) , we get e ∈ Rad(Q). Therefore, Q is a prime-like filter of L. □

Theorem 5.19. Consider L is prelinear, Q is a prime-like filter of L and K is a proper filter of L such that Q ⊆ K. Then K is a prime-like filter of L.

Proof. Assume K is a proper filter of L such that for e, h ∈ L, e ∨ h ∈ K. Since Q is a prime-like filter of L, by Proposition 5.18, for any e, h ∈ L, e ↠ h ∈ Q or h ↠ e ∈ Rad(Q). If e ↠ h ∈ Q, since Q ⊆ K, then e ↠ h ∈ K. Thus by Proposition 2.2(xiii), (eveeh)  ↠ h = e ↠ h ∈ K. Since (eveeh)  ↠ h ∈ K, e ∨ h ∈ K and K ∈ F ( L ) , we get h ∈ K. If h ↠ e ∈ Rad(Q), then by Proposition 3.14(ii), since Q ⊆ K, we have Rad(Q) ⊆ Rad(K), and so h ↠ e ∈ Rad(K). Thus (eveeh)  ↠ e = h ↠ e ∈ Rad(K). Since (eveeh)  ↠ e ∈ Rad(K), e ∨ h ∈ Rad(K) and Rad ( K ) ∈ F ( L ) , we get e ∈ Rad(K). Therefore, K is a prime-like filter of L. □

Corollary 5.20. Let L be prelinear. Then radical of every prime-like filter of L is a ∨-prime filter of L.

Proof. Suppose Q is a prime-like filter of L. Since Q ⊆ Rad(Q), by Proposition 5.19, Rad(Q) is a prime-like filter of L. Moreover, by Proposition 3.14(iii), Rad(Rad(Q)) = Rad(Q), and so Rad ( Q ) ∈ S Max ( L ) . Then by Proposition 5.12, Rad(Q) is a ∨-prime filter of L. □

Proposition 5.21. Consider Q , K ∈ F ( L ) such that Q ⊆ K. If K is a prime-like filter of L, then K Q is a prime-like filter of L Q .

Proof. Let K Q be a filter of L Q such that [ e ] ∨ [ h ] ∈ K Q . Then eveeh ∈ K. Since K is a prime-like filter of L, we have e ∈ K or h ∈ Rad(K), and so [ e ] ∈ K Q or [ h ] ∈ Rad ( K ) Q . By Proposition 3.14(xiii), we have [ h ] ∈ Rad ( K Q ) . Therefore, K Q is a prime-like filter of L Q . □

Lemma 5.22. Assume L is prelinear. If Q is a prime-like filter of L, then Rad ( Q ) ∈ P ( L ) .

Proof. Suppose Q is a prime-like filter of L. Then by Proposition 5.18, for any e, h ∈ L, e ↠ h ∈ Q or h ↠ e ∈ Rad(Q). By Remark 3.3(iii), Q ⊆ Rad(Q), thus for any e, h ∈ L, e ↠ h ∈ Rad(Q) or h ↠ e ∈ Rad(Q). Hence, Rad ( Q ) ∈ P ( L ) . □

Theorem 5.23. Suppose {Q _i } _i∈I is a non-empty chain family of prime-like filters of L. We have:

(i) ∩ i ∈ I Q i is a prime-like filter of L.

(ii) If L is prelinear, then ∪ i ∈ I Q i are prime-like filters of L.

Proof. (i) Assume e, h ∈ L such that e ∨ h ∈ ∩ i ∈ I Q i . Then for any i ∈ I, e ∨ h ∈ Q _i . Since for any i ∈ I, Q _i is a prime-like filter of L, we get x ∈ Q _i or h ∈ Rad(Q _i ). If for any i ∈ I, e ∈ Q _i , then e ∈ ∩ i ∈ I Q i , and so ∩ i ∈ I Q i is a prime-like filter of L. If there exists i ∈ I such that e ∉ Q _i , then since Q _i is a prime-like filter of L, we obtain h ∈ Rad(Q _i ). From {Q _i } _i∈I is a chain, there is t ∈ I such that Q _i  ⊆ Q _t or Q _t  ⊆ Q _i . If e ∉ Q _i , then h ∈ Rad(Q _i ). If Q _i  ⊆ Q _t , by Proposition 3.14(ii), Rad(Q _i ) ⊆ Rad(Q _t ), and so h ∈ Rad(Q _t ). If Q _t  ⊆ Q _i , then e ∉ Q _t , and so h ∈ Rad(Q _t ). By the similar way, we can see that h ∈ Rad(Q _i ). Thus, for any i ∈ I, h ∈ Rad(Q _i ). Moreover, by Proposition 3.14(ix), Rad ( ∩ i ∈ I Q i ) = ∩ i ∈ I Rad ( Q i ) . Hence, h ∈ Rad ( ∩ i ∈ I Q i ) .

(ii) By Proposition 5.19, the proof is clear. □

Theorem 5.24. Consider L is a prelinear equality algebra and {Q _i } _i∈I be a non-empty chain family of prime-like filters of L. Then 〈 {Q _i } _i∈I, ∩, ∪, L〉 is a lattice.

Proof. The proof is straightforward. □

Lemma 5.25. Assume Q ∈ F ( L ) and e, h, u ∈ L. If e^↠ nu, h^↠ mu ∈ Q, for some n , m ∈ ℕ , then there exists k ∈ ℕ such that (e ∨ h) ^↠ ku ∈ Q.

Proof. The proof is straightforward. □

Theorem 5.26. Let Q be a prime-like filter of L. Then for any p ∈ LsetminusQ there exists a prime-like filter such as K such that Q ⊆ K and p ∉ K.

Proof. Define ∑ = { K ∈ F ( L ) ∣ Q ⊆ K such that p ∉ K } . Since Q ∈ ∑, we get ∑≠ ∅. Then by Zorn’ Lemma, ∑ has a maximal element such as K. Now, we prove that K is a prime-like filter of L. For this, suppose e ∨ h ∈ K such that e ∉ K and h ∉ Rad(K). It is clear that K ⊆ 〈K∪ {e} 〉 and K ⊆ 〈K∪ {h} 〉. Since K is a maximal element of ∑ we get 〈K∪ {e} 〉, 〈K∪ {h} 〉 ∉ ∑. Thus p ∈ 〈K∪ {e} 〉∩〈K∪ {h} 〉. Then there exist n , m ∈ ℕ such that e^↠ np ∈ K and h^↠ mp ∈ K. Thus by Lemma 5.25, there exists k ∈ ℕ such that (eveeh) ^↠ kp ∈ K. Since e ∨ h ∈ K and K ∈ F ( L ) , we have p ∈ K, a contradiction. Hence, K is a prime-like filter of L. □ □

Proposition 5.27. Consider T is a prime-like filter of L and Q , K ∈ F ( L ) such that Q∩K ⊆ T. Then Q ⊆ T or K ⊆ Rad(T).

Proof. Suppose Q∩K ⊆ T such that Q ⊈ T. Then there exists e ∈ Q \ T. Thus for any h ∈ K, we have e ≼ e ∨ h and h ≼ e ∨ h. Since Q , K ∈ F ( L ) , we get e ∨ h ∈ Q∩K ⊆ T. Moreover, T is a prime-like filter of L, then h ∈ Rad(T). Hence, K ⊆ Rad(T). □