a note on the number of subquantales

Algebra Univers. 69 (2013) 191–200 DOI 10.1007/s00012-013-0221-8Published online February 1, 2013© Springer Basel 2013 Algebra Universalis

A note on the number of subquantales

Shengwei Han, Bin Zhao, and Shunqin Wang

Abstract. In this note, it is proved for n ≤ 5 that if Q is a finite quantale with|Q| ≥ n, then there are at least n subquantales of Q. However, the result does nothold when |Q| ≥ 6. Also, an example is given of a sequence of quantales of cardinalityp + 2 for p prime, each of which have exactly 5 subquantales.

1. Introduction

Quantales were first introduced in [4] by Mulvey in order to provide a latticetheoretic setting for studying the spectrum of non-commutative C∗-algebras,as well as a constructive foundation for quantum mechanics. As we all know,quantic nucleus and quantic conucleus are two important concepts, and playan important role in quantale theory because they determine the quotientsand subobjects in the category of quantales. So, it is worthwhile to studyquantic nuclei and quantic conuclei. In [6, 9], Paseka, Pelletier, and Rosickystudied a class of quantales without non-trivial quantic nuclei, called simplequantales. Simple quantales form an important class of quantales because anyquantale can be embedded into a simple quantale [8]. Dually, Han and Zhaoinvestigated a class of quantales without non-trivial quantic conuclei, calledtrivial quantales, and they proved that a quantale Q is trivial if and only if|Q| ≤ 2. In [6, 9], we can see that the number of quantic nuclei on a quantale Qdoes not have a simple relation to the cardinality of the quantale Q. Althoughthe cardinality of a quantale Q may be infinite, there maybe exist only twoquantic nuclei on Q. However, the number of quantic conuclei on a quantaleQ indeed has a relation to the cardinality of the quantale Q. In [1], one cansee that if Q is an infinite quantale, then the number of quantic conuclei isinfinite. For n ≤ 5, we find that the number of quantic conuclei of a finitequantale Q with |Q| ≥ n is at least n. To study the number of quantic conucleiof a quantale is interesting and very useful for getting quick counterexamples.Based on our previous work, we shall continue to investigate the number ofquantic conuclei.

Presented by J. Adamek.Received November 25, 2011; accepted in final form July 9, 2012.2010 Mathematics Subject Classification: 06F07.Key words and phrases: quantale, simple quantale, quantic conucleus, subquantale.This work is supported by the National Natural Science Foundation of China (grant nos

11001158, 11171196) and by the Basic and Advanced Research Program of He’nan ScienceCommittee (no 102300410145).

192 S.-W. Han, B. Zhao, and S.-Q. Wang Algebra Univers.

In this paper, we use 1 to denote the greatest element and 0 to denote thesmallest element in a complete lattice. First, we recall some of basic conceptsand results which we need. For notions and concepts concerned, but notexplained here, please refer to [10].

Definition 1.1 (see [10]). A quantale is an algebraic system Q = 〈Q,∨

,⊗〉such that 〈Q,

∨〉 is a complete lattice, 〈Q,⊗〉 is a semigroup, and for all a ∈ Q

and {bi}i∈I ⊆ Q,

a ⊗( ∨

i∈I

bi

)=

∨i∈I

(a ⊗ bi) and( ∨

i∈I

bi

)⊗ a =

∨i∈I

(bi ⊗ a).

For simplicity, Q denotes a quantale throughout the paper.

Definition 1.2. Let Q be a quantale and a ∈ Q.

(1) a is called right-sided (left-sided) iff a ⊗ 1 ≤ a (1 ⊗ a ≤ a). We writea ∈ R(Q), respectively, a ∈ L(Q).

(2) a is called two-sided iff a ⊗ 1 ≤ a and 1 ⊗ a ≤ a. We write a ∈ T (Q).(3) a is called prime iff x ⊗ y ≤ a =⇒ x ≤ a or y ≤ a for any x, y ∈ Q.

A subset S ⊆ Q is called a subquantale of Q if it is closed under∨

and ⊗.Clearly, if S is a subquantale of Q, then 0 ∈ S. For any quantale Q, S = {0}or S = Q is called a trivial subquantale on Q.

Remark 1.3.

(1) R(Q), L(Q), and T (Q) are subquantales of Q.(2) 0, 1 ∈ T (Q).(3) For all a ∈ R(Q) ∪ L(Q), ↓ a = {x ∈ Q : x ≤ a} is a subquantale of Q.

Definition 1.4. Let Q be a quantale. A quantic nucleus (quantic conucleus)on Q is a closure (coclosure) operator g such that g(a) ⊗ g(b) ≤ g(a ⊗ b) forall a, b ∈ Q.

A quantic conucleus g is said to be trivial if g(a) = a for all a ∈ Q, or ifg(a) = 0 for all a ∈ Q. A quantic nucleus g is called trivial if g(a) = a for alla ∈ Q, or if g(a) = 1 for all a ∈ Q.

Theorem 1.5 (see [10]). Let Q be a quantale. If g is a quantic conucleus onQ, then Qg = {a ∈ Q : g(a) = a} is a subquantale of Q. Moreover, if S is anysubquantale of Q, then S = Qg for some quantic conucleus g.

The following proposition is part of the folklore.

Proposition 1.6. Let Q be a quantale, S be a subquantale of Q and H ⊆ S.If H is a subquantale of S, then H is also a subquantale of Q.

2. The number of subquantales

In this section, we shall continue to discuss the number of quantic conu-clei of a quantale. By Theorem 1.5, it suffices to investigate the number of

A note on the number of subquantales 193

subquantales of quantale. For the number of subquantales of a quantale, thefollowing 4 results can be found in [1].

Proposition 2.1. Let Q be an infinite quantale. Then the number of sub-quantales of Q is infinite.

Proposition 2.2. Let Q be a finite quantale with |Q| ≥ 3. Then there are atleast 3 subquantales of Q.

Corollary 2.3. Let Q be a quantale with only one non-trivial subquantale.Then |Q| = 3.


In a complete lattice Q, we denote by M(Q) the set of all maximal elementsof Q below 1. In what follows we shall use the notation a < b to mean a ≤ b

and a �= b. An element a ∈ Q is called comparable if either x ≤ a or a ≤ x forall x ∈ Q.

Lemma 2.5. Let Q be a finite quantale with |Q| ≥ 5. If there are onlytwo non-trivial subquantales S1 and S2 with |S1| ≤ |S2|, then we have thefollowing:

(1) S1 = {0, 1}, S2 = {0, a, 1} and a ⊗ a = 1.(2) R(Q) = L(Q) = T (Q) = {0, 1}.(3) a is a maximal element below 1.(4) 0 is a prime element.(5) |M(Q)| �= 1.

Proof. (1): Firstly, we prove that R(Q), L(Q), and T (Q) are non-trivial sub-quantales of Q. Suppose that R(Q) = Q. Since |Q| ≥ 5, by Remark 1.3(3),we have that there are at least three non-trivial subquantales of Q, a con-tradiction. Therefore, R(Q) �= Q. Clearly, R(Q) contains both 0 and 1. Itfollows that R(Q) is a non-trivial subquantale of Q. Similarly, we can provethat L(Q) and T (Q) are also non-trivial subquantales of Q.

Secondly, we show that S1 ∩ S2 �= {0}. Assume that S1 ∩ S2 = {0}. Sincethere are only two non-trivial subquantales S1 and S2, by Propositions 1.6and 2.2, we have |S1| = |S2| = 2. Without loss of generality, we let S1 ={0, x}, S2 = {0, y} where x �= y. Since R(Q), L(Q), and T (Q) are non-trivialsubquantales of Q, we have that |T (Q)| = |R(Q)| = |L(Q)| = 2, which impliesthat S1 = {0, 1} or S2 = {0, 1}, i.e., x = 1 or y = 1. If x = 1, then y⊗1 ∈ R(Q)and 1 ⊗ y ∈ L(Q). Since |R(Q)| = |L(Q)| = 2, we have 1 ⊗ y, y ⊗ 1 ∈ {0, 1},which implies that {0, y, 1} is a non-trivial subquantale of Q, a contradiction.If y = 1, then {0, x, 1} is a non-trivial subquantale of Q, a contradiction.Therefore, we have S1 ∩ S2 �= {0}.

Finally, we prove that S1 = {0, 1}, S2 = {0, a, 1}, and a ⊗ a = 1. ByProposition 1.6, we have that S1 ∩ S2 is a non-trivial subquantale of Q. Since


there are only two non-trivial subquantales and |S1| ≤ |S2|, by Propositions2.2 and 2.4, we have that S1 ∩ S2 = S1, |S1| = 2, and |S2| = 3. By the abovedescription, we have {0, 1} ⊆ S2. We let S2 = {0, a, 1}. In what follows, weshall show that S1 = {0, 1} and a ⊗ a = 1.

Suppose that S1 �= {0, 1}. Then we have S1 = {0, a} and a ⊗ a ≤ a. Sincethere are only two non-trivial subquantales of Q, by Propositions 1.6 and 2.2,we have that a is a minimal element above 0 and 1 ⊗ 1 = a. Otherwise, wewould have three non-trivial subquantales {0, a}, {0, 1}, and {0, a, 1}. Thus,a ∈ T (Q). Let d ∈ Q with a < d; then {0, a, d} is a non-trivial subquantaleand {0, a, d} = S2, which implies that a is a maximal element below 1. Since|Q| ≥ 5, we have |M(Q)| �= 1. Thus, there exists an element b ∈ M(Q)\{a}such that {0, a, b, 1} is a non-trivial subquantale of Q, a contradiction. So, wehave S1 = {0, 1}.

If a ⊗ a ≤ a, then {x ∈ Q : x ≤ a} is a non-trivial subquantale of Qdifferent from both S1 and S2, a contradiction. Since S2 is a subquantale, wehave a ⊗ a ∈ S2, which implies that a ⊗ a = 1.

(2): By (1), we have a⊗a = a⊗1 = 1⊗a = 1, which implies that a �∈ R(Q)and a �∈ L(Q). Since R(Q), L(Q), and T (Q) are non-trivial subquantales ofQ, we have R(Q) = L(Q) = T (Q) = {0, 1}.

(3): Assume that there exists an element c ∈ Q\{1} with a < c. It is easyto verify that {0, a, c, 1} is a non-trivial subquantale of Q, a contradiction.

(4): Let x, y ∈ Q and x ⊗ y = 0. If y �= 0, then S = {t ∈ Q : t ⊗ y = 0}is a subquantale of Q. If 1 ∈ S, then 0 = 1 ⊗ y = y ⊗ y. Hence, {0, y} isa non-trivial subquantale of Q, i.e., y = 1, a contradiction with 0 = y ⊗ y.Hence, we have 1 �∈ S, and it follows that S = {0}, i.e., x = 0, which impliesthat 0 is prime.

(5): Assume that |M(Q)| = 1. By (3), we have M(Q) = {a} and a is acomparable element.

Claim (i): For all x ∈ Q\{0}, a ⊗ x = 1 = x ⊗ a.Suppose that there exists an element x ∈ Q\{0} such that a ⊗ x �= 1. By(1), we have x �= a and x �= 1. Since M(Q) = {a}, we have a ⊗ x ≤ a. It iseasy to verify that S = {t ∈ Q : a ⊗ t ≤ a} is a subquantale of Q and x ∈ S

but 1 �∈ S; this implies that S is a non-trivial subquantale, a contradiction.Similarly, we can prove that x ⊗ a = 1. We denote by Ma(Q) the set of allmaximal elements of Q\{1, a}.

Claim (ii): There are no comparable elements in Q\{0, a, 1}.Assume that there exists an element b ∈ Q\{0, a, 1} such that b is a comparableelement. Thus, b < a. Since b is a comparable element, we have b ⊗ b > b.Otherwise, S = {t ∈ Q : t ⊗ b ≤ b} is a non-trivial subquantale of Q andS �= S1, S �= S2.

It is easy to prove that S′ = {t ∈ Q : t ⊗ b > b} ∪ {0} is a subquantale ofQ and b ∈ S′. So we have that S′ = Q, i.e., for all x ∈ Q\{0}, x ⊗ b > b.Thus, S′′ = {x ⊗ b : x ∈ Q} is a non-trivial subquantale of Q. Since there are


only two non-trivial subquantales, we have that S′′ = {0, 1} or S′′ = {0, a, 1},which implies that b ⊗ b = 1 or b ⊗ b = a. If b ⊗ b = 1, then {0, b, 1} is anon-trivial subquantale, a contradiction. If b ⊗ b = a, then {0, a, b, 1} is anon-trivial subquantale, a contradiction. Therefore, there are no comparableelements in Q\{0, a, 1}. This implies that |Ma(Q)| �= 1.

Claim (iii): For all x, y ∈ Ma(Q), x ⊗ y �= 1.Assume that there exist x, y ∈ Ma(Q) such that x⊗ y = 1. It is easy to checkthat S = {t ∈ Q : t ⊗ y = 1} ∪ {0} is a subquantale and x ∈ S, which impliesthat S = Q. Thus, {0, y, 1} is a non-trivial subquantale of Q, a contradiction.Therefore, for all x, y ∈ Ma(Q), x ⊗ y �= 1.

By Claims (ii) and (iii), we have that there exist x1, x2, y ∈ Ma(Q) withx1 �= x2 such that x1 ⊗ y ≤ a and x2 ⊗ y ≤ a. By (i), we have a ≥ (x1 ⊗ y) ∨(x2 ⊗ y) = (x1 ∨ x2)⊗ y = a⊗ y = 1, a contradiction. Therefore, |M(Q)| �= 1.

�

Lemma 2.6. Let Q be a finite quantale with |Q| ≥ 5. If there are onlytwo non-trivial subquantales, S1 and S2 with |S1| ≤ |S2|, then we have thefollowing:

(1) S2 = {0, a, 1}, a ⊗ a = 1, and a is a maximal element below 1.(2) For all b, c ∈ M(Q)\{a}, b ⊗ c �= 1.(3) For all x, y ∈ Q\{0} with y �= 1, x ⊗ y �≤ y and y ⊗ x �≤ y.(4) For all x ∈ Q\{0}, a ⊗ x = x ⊗ a = 1.(5) For all x ∈ Q\{a, 1}, x⊗ y = 1 or y ⊗ x = 1 implies that y = a or y = 1.

Proof. (1): The proof follows from Lemma 2.5(1) and (3).(2): By Lemma 2.5(5), we have |M(Q)\{a}| ≥ 1. Assume that there exist

b, c ∈ M(Q)\{a} such that b ⊗ c = 1. By Lemma 2.5(2) and (4), we canverify that S = {t ∈ Q : t ⊗ c = 1} ∪ {0} is a subquantale of Q and b ∈ S,which implies that S = Q. Thus, {0, c, 1} is a non-trivial subquantale of Q, acontradiction.

(3): Assume that there exist x, y ∈ Q\{0} with y �= 1 such that x ⊗ y ≤ y.It is easy to show that S = {t ∈ Q : t ⊗ y ≤ y} is a subquantale of Q. ByLemma 2.5(2), we have 1 �∈ S, which implies that S = {0}. Thus, x = 0, acontradiction. Similarly, we can prove that y ⊗ x �≤ y.

(4): We first show that for all b ∈ M(Q)\{a}, a⊗b = 1 = b⊗a. Assume thatthere exists an element b ∈ M(Q)\{a} such that a⊗b �= 1. By (3) and Lemma2.5(5), we have that |M(Q)| ≥ 3 and there exists an element c ∈ M(Q)\{a, b}with a ⊗ b ≤ c. Since Q is a finite quantale, by (2) and (3) we have thereexist ai, aj ∈ M(Q) and k ∈ M(Q)\{b} such that ai �= aj , ai ⊗ b ≤ k, andaj ⊗b ≤ k, which implies that k ≥ (ai⊗b)∨ (aj ⊗b) = (ai∨aj)⊗b = 1⊗b = 1,a contradiction. Thus, for all b ∈ M(Q)\{a}, a ⊗ b = 1. Similarly, we canprove b ⊗ a = 1.

By Lemma 2.5(2) and (4), we can verify that S = {t ∈ Q : a⊗ t = 1} ∪ {0}is a subquantale of Q. By Lemma 2.5(5), we have that there exists an element


b ∈ M(Q)\{a} such that a ⊗ b = 1, which implies that S = Q, i.e., for allx ∈ Q\{0}, a ⊗ x = 1. Similarly, we can prove that x ⊗ a = 1.

(5): Let x ∈ Q\{a, 1} with x ⊗ y = 1. Then y �= 0. Assume that y �= a andy �= 1. By Lemma 2.5(2) and (4), we see that S = {t ∈ Q : x⊗ t = 1}∪{0} is asubquantale of Q. Since y ∈ S, we have S = Q, which implies that {0, x, 1} isa non-trivial subquantale of Q, a contradiction. Similarly, we can prove thaty ⊗ x = 1 implies that y = a or y = 1. �

Lemma 2.7. Let Q be a finite quantale with |Q| ≥ 5. If there are onlytwo non-trivial subquantales, S1 and S2 with |S1| ≤ |S2|, then we have thefollowing:

(1) S2 = {0, a, 1} and a is a maximal element below 1.(2) |M(Q)\{a}| ≥ 2.(3) For each element y ∈ M(Q)\{a}, there exists a unique element x ∈

M(Q)\{a} such that x ⊗ y < a.(4) Let y ∈ M(Q)\{a}. Then for each element xi ∈ M(Q)\{a}, there exists

a unique element ki ∈ M(Q)\{y} such that xi ⊗ y ≤ ki.(5) There exists x0 ∈ Q\{0} with x0 < a such that for all xi ∈ M(Q)\{a},

x0 �≤ xi.

Proof. (1): The proof follows from Lemma 2.5(1) and (3).(2): By (1), we have that a is a maximal element below 1. Assume that

|M(Q)\{a}| ≤ 1. By Lemma 2.5(5), we have that |M(Q)\{a}| = 1. Forx ∈ M(Q)\{a}, by Lemma 2.6(2) and (3), we can see that x ⊗ x ≤ a. Ifx ⊗ x = a, then {0, x, a, 1} is a non-trivial subquantale of Q, a contradiction.Thus, x ⊗ x < a. We let x0 = x ⊗ x. By Lemma 2.6(3) and (5), we havex ⊗ x0 ≤ a and x0 ⊗ x0 ≤ a. So, we have a ≥ (x ⊗ x0) ∨ (x0 ⊗ x0) =(x ∨ x0) ⊗ x0 = 1 ⊗ x0 = 1, a contradiction. Therefore, |M(Q)\{a}| ≥ 2.

(3): Firstly, we prove that for all x, y ∈ M(Q)\{a}, x ⊗ y �= a. Assumethere exist x, y ∈ M(Q)\{a} such that x⊗ y = a. Then by Lemma 2.6(4) and(5), we have 1 = a⊗y = (x⊗y)⊗y = x⊗ (y⊗y), which implies that y⊗y = a

or y ⊗ y = 1. It is easy to verify that {0, a, y, 1} is a non-trivial subquantaleof Q, a contradiction.

Secondly, we prove that for each element y ∈ M(Q)\{a}, there exists anelement x ∈ M(Q)\{a} such that x ⊗ y < a. Suppose there exists an elementy ∈ M(Q)\{a} such that for all x ∈ M(Q)\{a}, x⊗ y �< a. By Lemma 2.6(2)and (3), we have that for all xi ∈ M(Q)\{a}, there exists ki ∈ M(Q)\{a, y}such that xi ⊗ y ≤ ki. Since Q is a finite quantale, we have that there existxi, xj ∈ M(Q)\{a} and k ∈ M(Q)\{a, y} such that xi �= xj , xi ⊗ y ≤ k, andxj⊗y ≤ k, which implies that k ≥ (xi⊗y)∨(xj⊗y) = (xi∨xj)⊗y = 1⊗y = 1,a contradiction.

Finally, we prove uniqueness. Assume that for y ∈ M(Q)\{a} that thereexist two different elements x1, x2 ∈ M(Q)\{a} such that x1 ⊗ y < a andx2⊗y < a; this implies that 1 = 1⊗y = (x1∨x2)⊗y = (x1⊗y)∨ (x2⊗y) < a,a contradiction.


(4): Let y ∈ M(Q)\{a}. By Lemma 2.6(2) and (3), we have that for allxi ∈ M(Q)\{a}, there exists an element ki ∈ M(Q)\{y} such that xi⊗y ≤ ki.Assume that there exists xi ∈ M(Q)\{a} such that there are two differentelements ki, k

′i ∈ M(Q)\{y} with xi ⊗ y ≤ ki and xi ⊗ y ≤ k′

i. Since Q

is a finite quantale, we have that there exist xm, xn ∈ M(Q)\{a} and k ∈M(Q)\{y} such that xm �= xn, xm ⊗ y ≤ k, and xn ⊗ y ≤ k; this implies thatk ≥ (xm ⊗ y) ∨ (xn ⊗ y) = (xm ∨ xn)⊗ y = 1⊗ y = 1, a contradiction. So, forall xi ∈ M(Q)\{a}, there exists a unique element ki ∈ M(Q)\{y} such thatxi ⊗ y ≤ ki.

(5): By (2), we have that there exists an element y ∈ M(Q)\{a}. From(3), we can see that there exists a unique element x ∈ M(Q)\{a} such thatx ⊗ y < a. Let x0 = x ⊗ y. Then by Lemma 2.5(4), we have that x0 �= 0. By(4), we have that for all xi ∈ M(Q)\{a}, x0 �≤ xi. �

The following proposition was introduced in [1], but its proof has not beengiven because of the complexity of the proof. Now, we will give the proof.


Proof. Assume that there are not five subquantales of Q. By Proposition 2.4,Q has at least 4 subquantales, two of them trivial. Hence, there are only twonon-trivial subquantales, S1 and S2, of Q. Without loss of generality, we let|S1| ≤ |S2|. By Lemma 2.5(1) and(3), we have S1 = {0, 1}, S2 = {0, a, 1}, anda is a maximal element below 1.

From Lemma 2.7(2), we have that there exists an element c ∈ M(Q)\{a}.By Lemma 2.7(5), there exists x0 ∈ Q\{0} with x0 < a such that for allxi ∈ M(Q)\{a}, x0 �≤ xi. From Lemma 2.6(3), (4), and (5), we have thatthere exists an element k0 ∈ M(Q)\{c} such that x0 ⊗ c ≤ k0. Since Q

is a finite quantale, by Lemma 2.7(4), we have that there exist xm, xn ∈(M(Q)\{a}) ∪ {x0} and k ∈ M(Q)\{c} such that xm �= xn, xm ⊗ c ≤ k, andxn ⊗ c ≤ k. So, we have k ≥ (xm ⊗ c) ∨ (xn ⊗ c) = (xm ∨ xn) ⊗ c = 1 ⊗ c = 1,a contradiction.

Therefore, there are at least five subquantales of Q. �

One may ask whether the above result will hold for every natural numbern ∈ N, i.e., for a finite quantale Q with |Q| ≥ n, there are at least n subquan-tales of Q. Unfortunately, the above result does not hold when |Q| ≥ 6.

Example 2.9. Let Q = {0, a, b, c, d, e, 1} be a complete lattice, with the order“≤” on Q determined by Figure 1. Let ⊗ be the binary operation on Q definedby Table 1. It is easy to verify that Q = 〈Q,

∨,⊗〉 is a quantale, and there

are only five subquantales S1 = {0}, S2 = {0, a}, S3 = {0, 1}, S4 = {0, a, 1}and S5 = Q.

Having seen the result of Proposition 2.1, one might think that the numberof subquantales increases with the increasing cardinality of the quantale. In


•• • • • •

•

1

a b c d e

0

��

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�

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Figure 1

⊗ 0 a b c d e 10 0 0 0 0 0 0 0a 0 a b c d e 1b 0 b c d e a 1c 0 c d e a b 1d 0 d e a b c 1e 0 e a b c d 11 0 1 1 1 1 1 1

Table 1

fact, the number of subquantales does not have a necessary relation to thecardinality of the quantale. The following example will indicate this case.

Example 2.10. Let Q = {0, a0, a1, a2, . . . , ap−1, 1} be a complete lattice andp a prime number; the order “≤” on Q is determined by Figure 2.

•• • • •

•

1

a0 a1 a2 ap−1

0

��

��

. . . . . .��

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��

Figure 2

Define a binary operation � on Q\{0, 1} as follows: for all ai, aj ∈ Q\{0, 1},ai � aj = ak, where 0 ≤ k ≤ p − 1 equals i + j modulo p. It is easy to verifythat 〈Q\{0, 1},�〉 is a group and for all x ∈ Q\{0, a0, 1}, x is a generator of〈Q\{0, 1},�〉.

Define a binary operation ⊗ on Q as follows

∀x, y ∈ Q, x ⊗ y =

⎧⎪⎪⎨⎪⎪⎩

1 if x = 1 or y = 1,

ai � aj if x = ai, y = aj ,

0 if x = 0 or y = 0.

One can show that Q = 〈Q,∨

,⊗〉 is a quantale, and there are only five sub-quantales S1 = {0}, S2 = {0, a0}, S3 = {0, 1}, S4 = {0, a0, 1} and S5 = Q.


3. The number of ideals

The concept of an ideal is a useful tool to study the structure of quantales(see [1, 5, 12]), and an ideal can be regarded as a special Q-module. Forresults on Q-modules, see [2, 3, 7, 11]. Here, we want to know whether thenumber of ideals of an infinite quantale must be infinite just as is the numberof subquantales of an infinite quantale.

Definition 3.1 (see [10]). A subset I of a quantale Q is called an ideal of Q

if it is closed under∨

and if x ∈ I and y ∈ Q imply x ⊗ y, y ⊗ x ∈ I.

Remark 3.2. An ideal is a subquantale.

Definition 3.3 (see [1]). Let Q be a quantale. An ideal conucleus on Q is acoclosure operator g such that g(a) ⊗ b ≤ g(a ⊗ b) and a ⊗ g(b) ≤ g(a ⊗ b) forall a, b ∈ Q.

Theorem 3.4 (see [1]). Let Q be a quantale. If g is an ideal conucleus on Q,then Qg = {a ∈ Q : g(a) = a} is an ideal of Q. Moreover, if I is any ideal ofQ, then I = Qg for some ideal conucleus g.

Theorem 3.5 (see [1]). Let Q be a Girard quantale. Then the correspondingrelationship between quantic nuclei and ideal conuclei is one-to-one on Q.

Let S be a complete lattice and let Q(S) denote the complete lattice of allsup-preserving endomorphisms on S with the pointwise order. In [6], Pasekaproved that 〈Q(S),

∨,⊗〉 is a simple quantale in which ⊗ is the composition of

mappings. In [3], Kruml and Paseka prove that 〈Q(S),⊗〉 is a Girard quantaleif and only if S is a completely distributive lattice. So, by Theorem 3.4 andTheorem 3.5, we have the following proposition.

Proposition 3.6. For a completely distributive lattice S, there are only twoideals on quantale 〈Q(S),

∨,⊗〉.

Now, we answer the problem which we proposed. Let S be an infinite,completely distributive lattice; then 〈Q(S),

∨,⊗〉 is an infinite quantale and

there are only two ideals on 〈Q(S),∨

,⊗〉.

Acknowledgment. We wish to express our sincere thanks to the referees forcareful reading, helpful comments, and suggestions.

References

[1] Han, S.-W., Zhao, B.: The quantic conuclei on quantales. Algebra Universalis 61,97–114 (2009)

[2] Kruml, D.: Points of Quantales. PhD thesis, Masaryk University (2002)[3] Kruml, D., Paseka, J.: Algebraic and categorical aspects of quantales. In: Handbook

of Algebra, vol. 5, pp. 323–362. Elsevier/North-Holland, Amsterdam (2008)[4] Mulvey, C.J.: &. Rend. Circ. Mat. Palermo (2) Suppl. 12, 99–104 (1986)[5] Pan, F.-F., Han, S.-W.: Ideal and congruence on quantale. Journal of Chongqing

University (Natural Science Edition) 30(12), 89–94 (2007) (Chinese)


[6] Paseka, J.: Simple quantales. In: Proceedings of the Eighth Prague TopologicalSymposium (Prague, 1996), pp. 314–328 (electronic). Topology Atlas, North Bay(1997)

[7] Paseka, J.: Quantale Modules. Habilitation thesis, Masaryk University Brno (1999)[8] Paseka, J., Kruml, D.: Embeddings of quantales into simple quantales. J. Pure Appl.

Algebra 148, 209–216 (2000)[9] Pelletier, J.W., Rosicky, J.: Simple involutive quantales. J. Algebra 195, 367–386

(1997)[10] Rosenthal, K.I.: Quantales and Their Applications. Longman, New York (1990)[11] Solovyov, S.A.: On the category Q-Mod. Algebra Universalis 58, 35–58 (2008)[12] Wang, S.-Q, Zhao, B.: Ideal of quantale. Journal of Shaanxi Normal University

(Natural Science Edition) 31(4), 7–10 (2003)

Shengwei Han and Bin Zhao

Department of Mathematics, Shaanxi Normal University, Xi’an 710062, P. R. Chinae-mail, S.-W. Han: [email protected]

e-mail, B. Zhao: [email protected]

Shunqin Wang

School of Mathematics and Statistics, Nanyang Normal University, Nanyang 473061,P. R. Chinae-mail : [email protected]

a note on the number of subquantales

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