Bulletin of the Section of Logic Volume 29/1 (2000), pp. 1–4
Tomasz Kowalski
A REMARK ON QUASIVARIETIES OF MODAL ALGEBRAS
A class K of algebras will be said to have the congruence extension property (CEP) iff for any algebras A, B from K and any congruence Θ ∈ ConA, we have: if A ⊆ B, then there is a Φ ∈ ConB with Θ = Φ|A . This property is sometimes called the absolute CEP. A congruence Θ on a member A of K is called a K-congruence iff A/Θ belongs to K as well. ConK A will stand for the set of all K-congruences of A. K will be said to have the relative CEP (RCEP) iff for any A, B from K and any congruence Θ ∈ ConK A, we have: if A ⊆ B, then there is a Φ ∈ ConK B with Θ = Φ|A . Clearly, CEP and RCEP coincide on varieties. On quasivarieties, however, the two notions are independent, as for instance BCK-algebras have RCEP but not CEP, whereas for some quasivarieties of modal algebras the converse is true. Recall, that modal algebras are structures A = hA; ∧, ∨, ¬, l, 0, 1i, such that hA; ∧, ∨, ¬, 0, 1i is a boolean algebra and the unary operation l satisfies: (i) l1 = 1, (ii) l(x ∧ y) = lx ∧ ly. If A satisfies moreover: (iii) lx ≤ x, (iv) lx ≤ llx, then it is an interior algebra.
2
T. Kowalski
Recall also that congruences on modal algebras are 1-regular and therefore completely determined by cosets of 1, also called congruencefilters as they are indeed boolean filters closed under l. We will denote the congruence-filter associated with Θ by FΘ . Czelakowski and Dziobiak in [1] prove, among other things, the following theorem (Theorem 5.5.): Theorem. Let Q be a quasivariety of interior algebras. Then, Q is a variety iff Q has RCEP. Then, they pose this: Problem. Let K be a quasivariety of modal algebras. Is it true that K has RCEP iff K is a variety? Let’s note here that the terminology in Czelakowski and Dziobiak’s paper differs from ours; they call the relevant property simply CEP–they do not use the notion of (the absolute) CEP defined above. We shall, however, stick to our terminology mainly because of the following part of folklore: Proposition 1. Let Q be a quasivariety of modal algebras. Then, Q has CEP (in the absolute sense). Moreover, for A ⊆ B ∈ Q, and a congruence Θ ∈ ConA, the congruence-filter FΦ determining the smallest Φ ∈ ConB with Φ|A = Θ, is defined by {b ∈ B | ∃a∈FΘ : a ≤ b}. Let now A be the quasivariety of modal algebras defined by the following single quasiidentity: (v) lx = 1 ⇒ x = 1. Fact 0. A congruence Θ on A is an A-congruence iff FΘ is closed under l−1 . Proof. Suppose FΘ is not closed under l−1 . Then for some a 6∈ FΘ we have la ∈ FΘ . In the quotient algebra A/Θ we get then: a/Θ 6= 1 and la/Θ = 1. This violates (v). For the other direction simply revert the reasoning. Fact 1. A is not a variety.
Implicative logics, sequential deductive systems...
3
Proof. Take the modal algebra A = h{0, a, ¬a, 1}; ∧, ∨, ¬, l, 0, 1i, with l0 = la = l¬a = ¬a. It is easily seen that it satisfies (v). The set {¬a, 1} is a congruence filter. Let φ be the corresponding congruence. Take the quotient algebra A/φ. It has two elements and verifies l0 = 1. Thus, it falsifies (v). Fact 2. A has the relative congruence extension property. Proof. Let A ⊆ B be algebras from A, and Θ ∈ ConA A. We proceed to construct an A-congruence Ψ with Ψ|A = Θ. Observe that by Proposition 1 above, there is the smallest congruence Φ ∈ ConB (not necessarily in ConA B) with Φ|A = Θ. Take FΦ , and define inductively: • G0 = FΦ , • Gn+1S= Gn ∪ {b ∈ B : lb ∈ Gn }, • G = n∈ω Gn . We will show that G is the desired congruence-filter. In fact, we will first prove that Gn is a congruence-filter for any n ∈ ω. To start the induction, G0 is a congruence-filter by definition. Suppose Gn is a congruence-filter. Let g, f ∈ Gn+1 , and h ≥ g. Then, lg, lf ∈ Gn ⊆ Gn+1 . This proves closure under l. Moreover, lh ≥ lg, and by the inductive hypothesis we get that lh ∈ Gn . Thus, h ∈ Gn+1 , which proves upward closedness. Further, lg ∧ lf = l(g ∧ f ) ∈ Gn , whence, g ∧ f ∈ Gn+1 , which proves downward directedness, and ends this part of the proof. That G is also a congruence-filter follows immediately. Now, suppose lb ∈ G. Then, lb ∈ Gn , for some n, and thus b ∈ Gn+1 ; hence b ∈ G. This proves that G is closed under l−1 , and it follows by Fact 0 that the congruence determined by G, call it Ψ, is an A-congruence. All that remains is to show that Ψ restricted to A equals Θ. It is enough to prove that no element of A belongs to G1 \G0 . Suppose otherwise. Then some a ∈ A has: a 6∈ FΦ , but la ∈ FΦ . As FΦ |A = FΘ , this yields: a 6∈ FΘ , but la ∈ FΘ . This contradicts the assumption that Θ is an Acongruence. Thus, we have seen that the answer to Czelakowski and Dziobiak’s question is negative in general setting. The ease with which one can produce a counterexample suggests that this setting was indeed something of
4
T. Kowalski
an overshot. In the same paper, however, they formulated the following property, that may or may not hold in a variety V: (0) For any Q, subquasivariety of V, Q has RCEP iff Q is a variety. This leads to the following cautious reformulation of their problem: Question. Which varieties of modal algebras have property (0)?
Reference [1] J. Czelakowski, W. Dziobiak, The parameterized local deduction theorem for quasivarieties of algebras and its application, Algebra Universalis 35 (1996), pp. 373–419.
Japan Advanced Institute of Science and Technology Tatsunokuchi, Ishikawa, 923-1292, Japan email:
[email protected] 