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On derivations of quantales

Qimei Xiao
• Corresponding author
• School of Mathematics and Statistics, Changsha University of Science and Technology, Changsha, Hunan, 410004, P.R. China
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/ Wenjun Liu
• School of Mathematics and Statistics, Changsha University of Science and Technology, Changsha, Hunan, 410004, P.R. China
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Published Online: 2016-05-23 | DOI: https://doi.org/10.1515/math-2016-0030

Abstract

A quantale is a complete lattice equipped with an associative binary multiplication distributing over arbitrary joins. We define the notions of right (left, two) sided derivation and idempotent derivation and investigate the properties of them. It’s well known that quantic nucleus and quantic conucleus play important roles in a quantale. In this paper, the relationships between derivation and quantic nucleus (conucleus) are studied via introducing the concept of pre-derivation.

MSC 2010: 06A06; 54A10

1 Introduction

A quantale is a complete lattice equipped with associative binary multiplication distributing over arbitrary joins. Among the numerous examples of quantales are frames, various ideal lattices of rings and C* -algebras and the power set of a semigroup. The study of such partially ordered algebraic structures goes back to the work of Ward and Dilworth [1-3] on residuated lattices in the late 1930’s, motivated by ring-theoretic considerations. The notion of quantale was proposed as a combination of "quantum logic" and "locale" by Mulvey [4] in 1986, with the purpose of studying the foundations of quantum mechanics and the spectra of non-commutative C* -algebras. In 1990, Yetter [5] revealed the importance of quantales for linear logic, the logical foundation of theoretical computer science, which was proposed by Girard [6]. Since then, the theory of quantales has aroused great interest of many researchers, and a great deal of new ideas and applications of quantales have been discussed [710].

Derivation is helpful to the research of structure and property in algebraic system, which was introduced from analytic theory. There are many authors who studied derivations in various algebraic structures, such as rings, lattices, BCI-algebra and subtraction algebras, etc [11-15]. In [16], we introduced the notion of derivation for a quantale, and we discussed some related properties. It is well known that quantic nucleus and quantic conucleus play important roles in quantale theory because they determine the quotients and substructures in the category of quantales [17, 18]. The motivation of this paper is to study the properties of derivation further and the relationships between derivation and quantic nucleus (conucleus) on a quantale.

2 Preliminaries

In this section, we review some elementary notions of quantale theory [10].

A quantale is a complete lattice Q with an associative binary operation & satisfying: $a&(∨ibi)=∨i(a&bi) and (∨ibi)&a=∨i(bi&a)$

for all a ∈ Q and {bi} ⊆ Q.

Since a&— and — &a preserve arbitrary sups, they have right adjoints which we denote by a →r and al respectively. Thus a&cb if and only if c ≤ a →r b and c&ab if and only if calb.

In this paper, we denote the top element and the bottom element by ⊤ and 0 respectively. It is clear that 0&x = 0 and x&0 = 0 for all xQ.

Let Q be a quantale, aQ.

1. a is right (left) sided iff a&⊤ ≤ a (&aa).

2. a is strictly right (left) sided iff a&= a (&aa).

3. a is (strictly) two sided iff a is both (strictly) right sided and (strictly) left sided.

4. Q is two sided (right sided, left sided) iff every aQ is two sided (right sided, left sided).

5. a is idempotent iff a&a = a.

6. Q is idempotent iff every aQ is idempotent.

7. An element 1 ∈ Q is a right (left) unit iff a&1 = a (1&a = a) for all aQ.

8. 1 is a unit iff 1 is both a right and a left unit.

9. Q is (right, left) unital iff Q has a (right, left) unit.

A quantale Q is commutative iff a&b = b&a holds for all a, bQ. It is obvious that Q is commutative iff a →r c = alc for all a, cQ, and we denote by a → c. Q is right commutative iff (a&b) r c = (b&a) r c for all a, b, cQ. It is easy to observe that Q is left commutative iff a&b&c = b&a&c.

Let a, b, cQ with Q a quantale. Then

1. a&(a →r b) ≤ b;

2. (alb)&ab.

Let P be an ordered set. A map f : PP is called a closure (coclosure) operator on P if, for all a, bP,

1. af(a) (af(a);

2. abf(a) ≤ f(b);

3. ff(a) = f(a).

Let Q be a quantale. A quantic nucleus (conucleus) on Q is a closure (coclosure) operator j such that j(a)&j(b) ≤ j(a&b) for all a, bQ. We say that a quantic nucleus (conucleus) j is strict if it satisfies j(a)&j(b) = j(a&b).

Picado introduced the notion of quantic pre-nucleus in [9], which generalizes Banaschewski’s definition of a (localic) pre-nucleus [19, 20].

(Picado [9]). Let Q be a quantale. An order preserving mapping jo : QQ is called a quantic pre-nucleus iff it satisfies

1. ajo(a) for all aQ

2. a&jo(b) ≤ jo(a&b) and jo(a)&bjo(a&b) for all a, bQ.

It is easy to prove that Qjo = {aQ|jo(a) = a} is a closure system and the associated closure operator is given by

$j(a)=∧ {b∈Qjo|a≤b}.$

(Picado [9]). Let jo be a quantic pre-nucleus on a quantale Q. If j(a) = ⋀{bQjo |ab} for all aQ, then j is a quantic nucleus on Q.

Let Q be a quantale. A nonempty subset IQ. I is an ideal of Q if it satisfies the following two conditions:

1. if {ai} ⊆ Q, thenaiI;

2. for all xQ and aI, we have a&xI and x&aI.

3 Derivation on quantales

In [16], we defined the notion of derivation on quantales as follows:

Let d be a mapping on a quantale Q, then d is a derivation on Q, if it satisfies the following conditions

$d∨ibi=∨idbiandda&b=a&db∨da&b.$

for all a, bQ and {bi} ⊆ Q.

1. Let Q be a quantale. We define a mapping d by d(a) = 0 for all aQ. It’s clear that d(⋁i bi) = ⋁i d(bi) = 0 for all {bi} ⊆ Q. Since a&0 = 0&b = 0, we have (d(a)&b) ∨ (a&d(b)) = (0&b) ∨ (a&0) = 0 = d(a&b) for all a, bQ. Then d is a derivation on Q, which is called a zero derivation.

2. Let d be an identity mapping on a quantale Q. Then d is a derivation on Q, which is called an identity derivation.

According to the above definition, we obtain the following basic properties:

Let d be a derivation on a quantale Q and a, b, cQ, we have:

1. bd(a) r d(a&b), ad(b) → ld(a&b).

2. a →r bd(a) r d(b), d(a →r b) ≤ a →r d(b).

3. albd(a) → ld(b), d(alb) ≤ ald(b).

4. If ab, then d(a) ≤ d(b).

5. If a&b = b (or b&a = b) and d(a) ≥ a, then d(b) ≥ b.

6. If ca →r b, then cd(a) r d(b) and ad(c) → l d(b).

7. If a&a = a, then ad(a) r d(a) and ad(a) → ld(a).

8. If Q is right (left) unital, then a&d(⊤) ≤ d(a) (d(⊤)&ad(a)).

9. If Q is right (left) unital and d(1) = 1, then ad(a) and d(⊤) = ⊤.

10. If Q is right (left) unital and d(1) = 1, then d(a&⊤) = d(a)& =(d(⊤&a) = ⊤&d(a)).

The following corollary follows from Proposition 3.2 (5).

Let d be a derivation on a right (left) unital quantale Q. If d(1) ≥ 1, then d(a) ≥ a for all aQ.

Let d be a derivation on a quantale Q.

1. d(0) = 0.

2. d(⊤) =if Q is right (left) unital and d(1) ≥ 1.

Proof. (1) By the definition of derivation and 0&x = 0 and x&0 = 0 for all xQ, we have d(0) = d(0&0) = (d(0)&0) ∨ (0&d(0)) = 0 ∨ 0 = 0.

(2) Since Q is right (left) unital and d(1) ≥ 1, by Corollary 3.3, we have d(⊤) ≥ ⊤. Since T is the top element, we get d(⊤) = ⊤. □

Let d be a derivation on a quantale Q. If aQ is right (left, two) sided, then d(a) is right (left, two) sided.

Proof. Since a is right sided, then a&⊤ ≤ a. By the Proposition 3.2 (4), we have d(a) ≥ d(a&⊤) = (d(a)& ⊤) ∨ (a&d(⊤)) ≥ d(a)& ⊤, so d(a) is right sided. □

Let d be a derivation on a right unital quantale Q and d(1) = 1. If aQ is right (left, two) sided, then d(a) is strictly right (left, two) sided.

Proof. Since a is right sided, then a&⊤ ≤ a. By the Proposition 3.2 (4), we have d(a) ≥ d(a&⊤) = (d(a)& ⊤) ∨ (a&d(⊤)) ≥ d(a)& ⊤ and d(a) = d(a&1) ≤ d(a&⊤). That d(a&⊤) = d(a)& ⊤ follows from Proposition 3.2 (10). From the above, we get d(a) = d(a)& ⊤ which implies that d(a) is strictly right sided. □

In the following, we give some definitions about derivation which are similar to those on quantic nucleus.

Let d be a derivation on a quantale Q, aQ.

1. d is right sided iff d(⊤&a) ≤ d(a).

2. d is left sided iff d(a&⊤) ≤ d(a).

3. d is two sided iff d is both right and left sided.

4. d is idempotent iff d(a&a) = d(a).

Let Q be the complete lattice shown in Fig. 1 and the operations & on Q is shown in Table 1. It is straightforward to verify that (Q, &) is a quantale. We define d1 : QQ by d1(0) = 0, d1(a) = a, d1(⊤) = ⊤ and d2 : Q → Q by d2(0) = 0, d2(a) = d2(⊤) = ⊤. It is easy to prove that d1 and d2 are derivations on (Q, &). It is clear that d1 is left sided and d2 is two sided. Obviously, any derivation on (Q, &) is idempotent.

Table 1

Let d be a derivation on a quantale Q.If d is idempotent, right (left) sided and d(a) ≥ a for all aQ, then d(a) is an idempotent element of Q.

Proof. Since d is a right-sided derivation, we have d(a&⊤) = (d(a)& ⊤) ∨ (a&d) ∨ ≤ d(a) for all aQ. So d(a)& ⊤ ≤ d(a), and then d(a)&d(a) ≤ d(a). Making use of the idempotence of d, d(a) = d(a&a) ∨ (d(a)&a) ∨ (a&d(a)) ≤ d(a)&d(a). Therefore, d(a)&d(a) = d(a) for all aQ. □

Let d be a derivation on a quantale Q. If d(a) ≥ a for all aQ, then d is right (left) sided iff d(a) is right (left) sided.

Proof. If d(a) is right sided for all aQ, then d(a)& ⊤ ≤ d(a). So a&d(⊤) ≤ a&⊤ ≤ d(a)& ⊤ ≤ da. Thus d(a&⊤) = (d(a)& ⊤)∨ (a&d(⊤)) ≤ d(a).

Conversely, if d is right sided, we have d(a&⊤) = (d(a)& ⊤) ∨ (a&d(⊤)) ≤ d(a), then d(a)&⊤ ≤ d(a). □

Let .(Q, &) be a quantale, we denote the collection of all derivations on Q by D(Q). Define the operation "⋁" and "o" on D(Q) by (d1d2).() = d1.⋁d2.() and .d1d2 () = d1()&d2() for d1;d2D(Q).

Let Q be a quantale. Then,

1. (D(Q), ◦) is a complete lattice;

2. if Q is two sided and right commutative, then (D(Q), ◦) is a quantale.

Proof. (1) We prove that D(Q) is a complete lattice under the pointwise order.

Let {di} ⊆ D(Q) and {aj} ⊆ Q, we have ⋁idi(⋁jaj) =i(di(⋁iaj)) = ⋁i(⋁j (di(aj))) = ⋁j(⋁idi(aj)). Let a, b ∈ Q, we have ⋁i di(a&b) = \/i((di(a)&b) ∨ (a&di(b))) = (⋁i(di(a)&b)) ∨ (⋁i(a&di(b))) = ((⋁idi(a))&b) ∨ (a&(⋁idi(b))). So ⋁idiD(Q).

Since d : QQ defined by d(a) = 0 for all aQ is a derivation, then D(Q) has a bottom element. Therefore, D(Q) is a complete lattice.

(2) We need to prove "◦" is a binary operation on D(Q).

Let d1, d2D(Q), a, bQ, then (d1d2)(a&b) = d1(a&b)&d2(a&b) = ((d1(a)& b) ∨ (a&d1(b)))&((d2(a)&b) ∨ (a&d2(b)))=((d1(a)&b)&(d2(a)&b)) ∨ ((a&d1(b)) & (d2(a)&b)) ∨ ((d1(a)&b)&(a&d2(b))) ∨ ((a&d1(b))&(a&d2(b))).

Since Q is right sided, we have ((d1(a)&b)&(d2(a)&b)) ∨ ((a&d1(b))&(d2(a)&b)) = ((d1(a)&b) ∨ (a&d1(b)))&(d2(a)&b) = d1(a&b)&(d2(a)&b) ≤ d1(a&⊤)&(d2(a)&b) ≤ (d1(a)& d2(a))&b = (d1d2)(a)&b.

Since Q is left sided and right commutative, then ((d1(a)&b)&(a&d2(b))) ∨ ((a&d1(b))&(a&d2(b))) = ((d1(a)&b) ∨ (a&d1(b)))&(a&d2(b)) = d1(a&b)&(a&d2(b)) ≤ d1(⊤&b)&(a&d2(b)) ≤ d1(b)&a&d2(b) = a&d1(b)&d2(b) = a&(d1d2)(b).

From the above, we have (d1d2)(a&b) = ((d1d2)(a)&b) ∨ (a&(d1d2)(b)). So d1d2D(Q).

It is clear that the operation o is an associative binary operation satisfying the distribution over arbitrary joins, so (D(Q), ◦) is a quantale. □

Given a derivation d on a quantale (Q, &), we denote the fixed set {aQ |d(a) = a} by Qd.

Let d be a derivation on a quantale (Q, &), then (Qd, &) is a quantale.

Proof. We first prove that Qd is a complete lattice under the pointwise order of Q. Let {bi} ⊆ Qd, we have d(⋁ibi) = ⋁id(bi) =ibi, then ⋁i biQd. By Proposition 3.4, we have d(0) = 0, then 0 ∈ Qd.

Then we prove that & is a binary operation on Qd. Let a, bQd, we have d(a&b) = (d(a)&b) ∨ (a&d(b)) = a&b, so a&bQd.

From the above, we have (Qd, &) is a quantale. □

Let Q = {0, ⊤} and define a binary operation & on Q by 0&0 = 0, 0&⊤ = 0, ⊤&0 = 0, ⊤&⊤ = ⊤. The quantale (Q, &) is two sided, idempotent, unital and commutative. There are only zero derivation and identity derivation on Q, denoted by d1 and d2.It is clear that D(Q) = {d1, d2}, Qd1 = {0}and Qd2 = {0, ⊤} are quantales.

Let d be a derivation on a quantale Q. If d is two-sided, idempotent and d(a) ≥ a for all aQ, then d is a quantale homomorphism from Q onto Qd.

Proof. By Proposition 3.9 and Proposition 3.10, we have d(a) is idempotent and two-sided for all aQ. Therefore, d(d(a)) = d(d(a)&d(a)) = (d(d(a))&d(a)) ∨ (d(a)&d(d(a))) ≤ (⊤&d(a)) ∨ (d(a)&⊤) ≤ d(a). On the other hand, d(d(a)) ≥ d(a). Thus d(d(a)) = d(a) which implies that d(a) ∈ Qd for all aQ. So d is a mapping from Q onto Qd.

We need to show that d(a&b) = d(a)&d(b) for all a, bQ. Since d is idempotent, we have d(a&b) = d(d(a&b)) = d((d(a)&b) ∨ (a&d(b))) = d(d(a)&b) ∨ d(a&d(b)) ≥ d(d(a)&b) = (d(d(a))&b) ∨ (d(a)&d(b)) ≥ d(a)&d(b). On the other hand, d(a&b) = (d(a)&b)∨(a&d(b)) ≤ (d(a)&d(b))∨(d(a)&d(b)) = d(a)&d(b). So d(a&b) = d(a)&d(b). Since d preserves arbitrary sups, we know that d is a quantale homomor phism from Q onto Qd. □

By the proof of Theorem 3.14, we have the following corollary.

Let d be a derivation on a quantale Q. If d is two-sided, idempotent and d(a) ≥ a for all aQ, then Qp = {aQ|d(a) ≥ p, p&p = p} is a subquantale of Q.

In the following, we shall give a description for a derivation on general quantales.

(Paseka and Kruml [8]). Let (Q, &) be a quantale and Q[e] ={ak : aQ, k ∈ {0, e}}, where ∈ is an arbitrary element such that ∈Q. We define the supremum on Q [e]: ∈ ∨ 0 = ∈ and $⋁iQ[e](ai∨ki)=(∨i,ai)∨e,if∃ki=e,∨iai,otherwise.$

The multiplication &′ on Q[e] is as follows: $(a∨k′)&′(b∨k″)=a&b,ifk′=k″=0,(a&b)∨b,ifk′=e,k″=0,(a&b)∨a,ifk′=0,k″=e,((a&b)∨a∨b)∨e,ifk′=k″=e.$

Then (Q[e]; &′) is a unital quantale with the unit e.

Let Q be a quantale and d : Q → Q a map on Q. We define the map d : Q[e] → Q [e] such that $d¯a=da,ifa∈Q,da′,ifa=a′∨e,a′∈Q.$

It is clear that d = d|Q.

Let Q be a quantale and d : QQ a map on Q. Then d is a derivation on Q if and only if d is a derivation on Q [e].

Proof. Clearly, if d is a derivation on Q[e], then d is a derivation on Q.

Conversely, we assume that d is a derivation on Q. Let {biki}iQ[e] and biQ, ki ∈ {0, e}. We have $d¯(⋁iQ[e]bi∨ki)=d¯∨ibi∨e,if∃ki=e,d¯∨ibi,otherwise.=d(⋁ibi)=⋁idbi=⋁id¯bi∨ki$

Let ak′, bk″Q[e] and a, bQ, k′1, k″ ∈ {0, e}. Then $d¯a∨k′&′b∨k″=da&b,ifk′=k″=0,da&b∨b,ifk′=e,k″=0,da&b∨a,ifk′=0,k″=e,da&b∨a∨b,ifk′=k″=e.$

Case1:k′ = k″ = 0,[(d(ak′))&(bk″)]∨[(ak′)&′(d(bk″))] = [d(a)&′(bk″)] ∨ [(ak′)&'d(b)] = (d(a)&b) ∨ (a&d(b)) = d(a&b)i

Case 2: k′ = e, k″ = 0, [(d(ak′))&′(bk″)] ∨ [(ak′)&′( d(bk"))] = [d(a)&′(bk″)] ∨ [(ak)&′d(b)] = (d(a)&b) ∨ (a&d(b)) ∨ d(b) = d((a&b) ∨ b);

Case 3: k′ = 0, k″ = ∈, [(d(ak′))&′(bk″)] ∨ [(ak′)&′(d(bk″))] = [d(a)&′(bk″)] ∨ [(ak′)&′d(b)] = (d(a)&b) ∨ da ∨ (a&d(b)) = d((a&b) ∨ b);

Case 4: k′ = e, k″ = e, [(d(ak′))&′(bk″)] ∨ [(ak′)&′(d(bk″))] = [d(a)&′(bk″)] ∨ [(ak)&′d(b)] = (d(a)&b) ∨ da ∨ (a&d(b)) ∨ d(b) = d((a&b) ∨ ab).

Therefore, d is a derivation on Q [e]. □

4 The relation between derivation and quantic nucleus

In the following, we introduce the concept of pre-derivation which is the generalization of derivation.

Let Q be a quantale. A mapping do on Q is a pre-derivation if do satisfies: $do∨ibi≥∨idobiandda&b≥a&dob∨doa&b$

for all a, bQ and {bi} ⊆ Q.

1. If do is a pre-derivation on a right (left) unital quantale Q, then do(1) ≥ 1 if and only if do(a) ≥ a for all aQ.

2. If do is a pre-derivation on a quantale Q and do(a) ≥ a for all aQ, then do is a quanticpre-nucleus.

3. If jo is a quantic pre-nucleus on a quantale Q, then jo is a pre-derivation on Q.

Let Q be a commutative quantale and sQ, then sis a pre-derivation on Q.

Proof. Let a, bQ, by Remark 2.3, we have s&#x0026;((s →r a)&b) = (s&(s →r a))&ba&b, so (s →r a)&bs →r (a&b). Similarly, a&(slb) ≤ sl (a&b). Since Q is commutative, we have a →r c = alc for all , cQ. So s → (a&b) ≥ ((sa)&b) ∨ (a&(sb)). Since s → is order preserving, s → is a pre-derivation on Q. □

Let Q be a commutative quantale and sQ.

1. If Q is unital and s ≤ 1, then s → is a quantic pre-nucleus on Q.

2. If Q is right (left) sided, then s → is a quantic pre-nucleus on Q.

Proof. (1) Since s ≤ 1, we have 1&s = s&1 =s ≤ 1 ⇒ 1 ≤ s → 1. By Remark 4.2 and Proposition 4.3, we get the conclusion.

(2) Since Q is right (left) sided, we have s&a = a&sa&Ta for all aQ, then asa. By Remark 4.2 and Proposition 4.3, we get the conclusion. □

Let do be a pre-derivation on a quantale Q. If do(a) ≥ a and j(a) = ⋀{xQdo|ax} for all aQ, then j is a quantic nucleus on Q.

Proof. By Remark 4.2 and Theorem 2.7, we can prove it immediately. □

Let d be a derivation on a quantale Q.Ifd is idempotent, two-sided and d(a) ≥ a for all aQ, then d is a strict quantic nucleus of Q.

Proof. Let a, bQ. From the proof of Theorem 3.14, we have d(a) = d(d(a)) and d(a&b) = d(a)&d(b). Since d(a) ≥ a and d is order preserving, we have d is a strict quantic nucleus. □

A quantic nucleus j on Q is called localic iff j(a&b) = j(a) ∧ j(b) for all a, bQ.

Let d be a derivation on a quantale Q. If d(a) ≥ a for all aQ, then the following conditions are equivalent:

1. d is two sided and idempotent;

2. d(a&b) = d(a) ∧ d(b) for all a, bQ.

Proof. (1) ⇒ (2): Since d is two sided, we have d(a&b) ≤ d(a&⊤) ≤ d(a) and d(a&b) ≤ d(⊤&b) ≤ d(b). So d(a&b) ≤ d(a) ∧ d(b). By Theorem 4.6, we know that d is a strict quantic nucleus, then d(a&b) = d(d(a&b)) = d(d(a)&d(b)) ≥ d((d(a) ∧ d(b))&(d(a) ∧ d(b))) = d(d(a) ∧ d(b)) ≥ d(a) ∧ d(b). Therefore, d(a&b) = d(a) ∧ d(b).

(2) ⇒ (1): d(a&⊤) = d(a) ∧ d(⊤) ≤ d(a) and d(⊤&a) = d(⊤)∧ d(a) ≤ d(a), so d is two-sided. And d(a&a) = d(a) ∧ d(a) = d(a), then d is idempotent. □

Let d be a derivation on a quantale Q. If d is two sided, idempotent and d(a) ≥ a for all aQ, then d is localic.

Proof. By Theorem 4.6 and Lemma 4.7, we have that d is localic. □

Let do be a pre-derivation on a quantale Q. If do(a) ≤ a and g(a) = ∨{xQdo |xa} for all aQ, then g is a quantic conucleus.

Proof. For all a; bQ, we can prove that

1. g(a) ≤ a.

2. If ab, we have {xQdo |x ≤ a} ⊆ {xQdo |x ≤ b}, so g(a) ≤ g(b).

3. It is clear that g(a) = a for a ∈ Qdo. So g(g(a)) = g(⋁x|xQdo; xa) ⋁{g(x)|xQdo, xa} = ⋁(x|xQdo, xa} = g(a). On the other hand, we have g(g(a)) ≤ g(a) by (i). Therefore, g(g(a)) = g(a).

4. Since g(a) = a for aQdo and d0 is order preserving, we have g(a) = ∨{xQdo|xa} = ∨{do(x) ∈ Qdo|xa} ≤ do(a). So do(a&g(b)) ≥ a&do(g(b)) ≥ a&g(g(b)) = a&g(b). On the other hand, a&g(b) ≥ do(a&g(b)). Therefore, a&g(b) ∈ Qdo. Then g(a&g(b)) = ∨{xQdo |xa&g(b)} = a&g(b). So g(a)&g(b) = g(g(a)&g(b)) ≤ g(a&b). □

Let Q be a quantale. If I is an ideal of Q, then I = Qdo for some pre-derivation do.

Proof. Define do : QQ by do(a) = ⋁{iI|i ≤ a}. It is immediate that I = Qdo and do(⋁ bi) do(bi) for {bi} ⊆ Q. Let a, bQ. Since I is an ideal of Q, we have a&do(b) = a&⋁{iI|i ≤ b} = ⋁{a&iI|i ≤ b} ⋁{a&iI|a&i ≤ a&b} ≤ ⋁{iI|i ≤ a&b} = do(a&b): Similarly, do(a)&b ≤ do(a&b). So do(a&b) ≥ (a&do(b)) ⋁ (do(a)&b). □

From Proposition 4.10, we know that every ideal of Q can be represented by the fixed set of a pre-derivation of Q. Conversely, a fixed set of a pre-derivation of Q is not an ideal in general. In Example 3.8, we know that d2 is a derivation of Q but Qd2 = {0; ⊤} is not an ideal of Q for a&⊤ = aQd2.

Let I be an ideal of a quantale Q. If g(a) = ∨{xI|x ≤ a} for all aQ, then g is a quantic conucleus on Q.

Proof. By Proposition 4.10, there exists a pre-derivation do(a) = ⋁{iI|i ≤ a} such that I = Qdo and do(a) ≤ a. By Theorem 4.9, we have g is a quantic conucleus. □

Let a be an element of a commutative quantale Q. A mapping da : QQ defined by da(b) = a&b is a derivation on Q, we call da is a simple derivation [16]. It is obvious that zero derivation and identity derivation are special simple derivations.

(Xiao and Li [16]). Let a be an idempotent element of a commutative quantale Q. If Q is unital, then:

1. da is a strict quantic nucleus on Q if a ≥ 1,

2. da is a strict quantic conucleus on Q if a ≤ 1,

3. da is a strict quantic nucleus and a quantic conucleus on Q if a = 1.

Let a be an idempotent element of a commutative quantale Q. If Q is left (right) sided, then da is a quantic conucleus on Q.

Proof. Let b, cQ, then

1. Since Q is left sided, we have da(b) = a&b ≤&b ≤ b.

2. b ≤ c, we have da(b) ≤ da(c).

3. Since a is idempotent, then da(da(b)) = a&(da(b)) = a&(a&b) = (a&a)&b = a&b = da(b).

4. By (i) and (ii), we have da(b)&da(c) = a&b&(a&c) = a&((a&b)&(a&c)) = da(da(b)&da(c)) ≤ da(b&c).

From the above, we can obtain that da is a quantic conucleus on Q. □

5 Conclusion

In this paper, we first introduced the notions of right (left, two) sided derivation and idempotent derivation for a quantale, and discussed some interesting structural properties about these derivations on quantales. We got an important result that the collection of derivations in a quantale Q is a complete lattice, furthermore, it is a quantale if the quantale Q is right sided and right commutative. Also, the fixed set Qd of a quantale Q is a quantale and d is a quantale homomorphism from Q onto Qd if d is two sided, idempotent and expansive. Then, we introduced pre-derivation which is an extended notion of derivation and studied the relationship between derivation and quantic nucleus. We believe that these results will be useful in theoretical computer science. We will study the generalized derivations on quantales in future work.

Acknowledgement

This work was supported by the National Science Foundation of China ( No.11401052), Hunan Provincial Education Department Foundation (No.15B004) and the Science and Technology Program of Hunan of China (no. 2015RS4049).

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Accepted: 2016-05-06

Published Online: 2016-05-23

Published in Print: 2016-01-01

Citation Information: Open Mathematics, ISSN (Online) 2391-5455,

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