T.S. Blyth Jie Fang Congruence Coherent Symmetric Extended de Morgan Algebras Abstract. An algebra A is said to be congruence coherent if every subalgebra of A that contains a class of some congruence ϑ on A is a union of ϑ-classes. This property has been investigated in several varieties of lattice-based algebras. These include, for example, de Morgan algebras, p-algebras, double p-algebras, and double MS-algebras. Here we determine precisely when the property holds in the class of symmetric extended de Morgan algebras. Keywords: congruence coherent, extended de Morgan algebra If A is an algebra and ϑ is a congruence on A then A is said to be ϑ-coherent every subalgebra that contains a ϑ-class is a union of ϑ-classes. An algebra A is said to be congruence coherent if it is ϑ-coherent for every ϑ Con A. This notion has been investigated in several varieties of latticebased algebras. For example, Beazer [2] considered this in the context of de Morgan algebras (bounded distributive lattices with an involutory dual automorphism) and proved that a de Morgan algebra is congruence coherent if and only if it is boolean, or is simple, or is the 4-element de Morgan chain; and that a p-algebra is congruence coherent if and only if it is boolean. Subsequently it was shown by Adams, Atallah and Beazer [1] that examples of congruence coherent double p-algebras are those that are congruence regular (if two congruences have a class in common then they coincide). More general types of distributive lattice with additional unary operations provide a rich area for further investigations of congruence coherence. In this connection we mention the wide class of Ockham algebras, as introduced by Urquhart [7]. An Ockham algebra is a bounded distributive lattice together with a dual endomorphism (see also [6]). In particular, an MS-algebra (or de Morgan Stone algebra) is an Ockham algebra (L;f) in which the dual endomorphism f is such that id L f 2. The special case where f 2 = id L gives a de Morgan algebra. In [3] we investigated congruence coherence in the context of double MSalgebras, and showed that a double MS-algebra is congruence coherent if and only if it is a trivalent Lukasiewicz algebra, or is simple, or is boolean, or is the 4-element de Morgan chain. Presented by M.E. Adams; Received January, 2006.
2 T.S. Blyth and Jie Fang In this paper our purpose is to add to the above collection the following class of symmetric extended de Morgan algebras. Definition. By an extended Ockham algebra L (L;,,f,k,0,1) we mean an algebra of type (2,2,1,1,0,0) consisting of a bounded distributive lattice (L;,, 0, 1) together with a pair of unary operations f, k such that (1) f is a dual lattice endomorphism with f(1) = 0 and f(0) = 1; (2) k is a lattice endomorphism with k(1) = 1 and k(0) = 0; (3) f and k commute. By an extended de Morgan algebra we mean an extended Ockham algebra L (L;,,f,k,0,1) in which f 2 = id L. An extended de Morgan algebra in which k 2 = id L is said to be symmetric. For the basic properties of extended Ockham algebras and those of symmetric extended de Morgan algebras we refer the reader to [4],[5]. We note that when k = id L we can regard an extended de Morgan algebra L simply as a de Morgan algebra, in which case we shall say that L is primitive. Example 1. Consider the algebra (L;f,k) depicted as follows: 1 s t p q r i j m n c d e a b 0 0 a b c d e i j m n p q r s t 1 f 1 s t p q r i j m n c d e a b 0 k 0 b a e d c n m j i r q p t s 1 Roughly speaking, f acts vertically and k acts horizontally. It is readily seen that (L;f,k) is a symmetric extended de Morgan algebra. Example 2. Consider the algebra (L;f,k) depicted as follows:
Congruence Coherent Symmetric Extended de Morgan Algebras 3 1. c2 a 2 b2 Let f : L L be described by c1 a 1 f(0) = 1, f(1) = 0, b1 f(z i ) = z i for z i {a i,b i,c i }; c0 a 1 and let k : L L be described by b 1 k(0) = 0, k(1) = 1, c 1 a 2 k(a i ) = b i, k(b i ) = a i, k(c i ) = c i. b 2 c 2. 0 Clearly, f is a dual lattice endomorphism with f 2 = id L, and k is a lattice endomorphism with k 2 = id L. By observing that fk(a i ) = b i = kf(a i ), that fk(b i ) = a i = kf(b i ), and that fk(c i ) = c i = kf(c i ), we see that fk = kf. Thus L is a symmetric extended de Morgan algebra. Example 3. Let A and B be finite boolean lattices each having n+1 atoms. Let a 0,a 1,,a n be those of A, and let b 0,b 1,,b n be those of B. Consider the lattice vertical sum L = A B. With 0 = 0 A, 1 = 1 B and α = 1 A = 0 B, consider the algebra L (L;,,f,k,0,1) where f,k : L L are given by f(0) = 1, f( a i ) = b i, f( b i ) = a i ; k(0) = 0, k( a i ) = a n i, k( b i ) = b n i. Note that these identities give f(1) = f( n b i ) = n a i = 0. Likewise, k(1) = 1, f(α) = α, and k(α) = α. To see that f is a dual lattice endomorphism on L there are three cases to consider, namely x,y A, x,y B, and x A,y B. For the first of these let x = a i and y = a i. Then j J f(x) f(y) = ( b i b j ) = ( j J t I J f(x) f(y) = ( b i b j ) = ( j J t I J i=0 b t ) = b t ) = i=0 b t t I J t I J = f(x y); b t = f(x y). The case where x,y B is similar. As for x A and y B, this is trivial since in this case x y.
4 T.S. Blyth and Jie Fang Since A is boolean it follows from these equalities that [f(x)] = f(x ) for every x A; and similarly for x B. Consequently we have that f 2 ( a i ) = f( b i ) = [f( b i )] = ( a i ) = a i, and similarly f 2 ( b i ) = b i. Hence we see that f 2 = id L. Likewise we can see that k is a lattice endomorphism on L with k 2 = id L. Finally, we observe that kf( a i ) = k( b i ) = [k( b i )] = b n i = fk( a i ), and similarly, kf( b i ) = a n i = fk( b i ). Thus fk = kf. We conclude that L is a symmetric extended de Morgan algebra. For convenience in what follows, for a symmetric extended de Morgan algebra L (L;,,f,k,0,1), we shall write x for f(x) and x + for k(x). We note that it is of course possible for a subalgebra of L to be primitive. Clearly, the smallest subalgebra of L that is primitive is {0,1}, and the biggest is Fix + L = {x L x+ = x}. In order to avoid unnecessary repetition, we shall assume throughout what follows that L is not primitive. Concerning congruences, we shall use throughout the standard notation ϑ(a,b) for the principal congruence on L that identifies a and b with a b, and ϑ lat (a,b) for the corresponding principal lattice congruence. We begin with the following observation. Theorem 1. Let the symmetric extended de Morgan algebra L be congruence coherent. If ϕ Con L is such that Ker ϕ {0} then ( a L) a a a + Ker ϕ. Proof. By way of obtaining a contradiction, suppose that there exists a L such that a a a + / Ker ϕ, and let b = a a +. Then b b / Ker ϕ; for otherwise a a + (a a + ) Ker ϕ whence we have the contradiction a a + a Ker ϕ. Let A be the sublattice of L that is generated by {b,b } and consider the subset K = {0,1} [b] ϕ [b ] ϕ [b b ] ϕ [b b ] ϕ. Clearly, K is a subalgebra of L. Now for every x A we have 0 / [x] ϕ, and so [x] ϕ Ker ϕ = for every x A. But since L is congruence coherent we have Ker ϕ K. There follows the contradiction Ker ϕ = {0}.
Congruence Coherent Symmetric Extended de Morgan Algebras 5 The above result leads in a natural way to a consideration of elements of L of the form x x + x and, dually, of the form x x + x. In particular, we shall focus attention on the subsets K(L) = {x L x x x + = 0}, K d (L) = {x L x x x + = 1}. A basic property of these sets is the following. Theorem 2. The following properties are equivalent : (1) K(L) = L; (2) ( x L) x x + x + = 1; (3) ( x Fix + L) x x = 1; (4) K d (L) = L; (5) ( x L) x x + x + = 0; (6) ( x Fix + L) x x = 0; (7) the primitive subalgebra (Fix + L; ) is boolean ; (8) ( x L) ϑ(0,x) = ϑ lat (0,x x + ). Proof. (1) (2) Applying + to the identity x x x + = 0 we obtain (2). (2) (3) For every x L we have x x + Fix + L. Writing x x + for x in (2), we obtain (3). (3) (4) By (3) we have x x + (x x + ) = 1 whence (4) follows since (x x + ) x. (4) (5) This is dual to (1) (2). (5) (6) (1) Apply the same procedure as in (2) (3) (4). The equivalence of (7) to the above follows from that of (3) and (6). (7) (8) For every x L we have ϑ(0,x) = ϑ(0,x x + ) ϑ(0,x x + ) by [5, Theorem 1] = ϑ lat (0,x x + ) ϑ lat ( (x x+ ),1 ) by [4, Theorem 2.1] = ϑ lat (0,x x + ), the last equality following from the fact that, by (7), (x x + ) is the complement of x x +. (8) (4) If (8) holds then since (0,x) ϑ(0,x) we have (x,1) ϑ(0,x) = ϑ lat (0,x x + ) whence x x x + = 1.
6 T.S. Blyth and Jie Fang Definition. If the symmetric extended de Morgan algebra L satisfies any of the equivalent properties in Theorem 2 then we shall say that L is special. Example 4. Every finite boolean lattice B can be made into a special symmetric extended de Morgan algebra. In fact, as shown in [4, Example 1.4], if a 0,a 1,...,a n are the atoms of B then B = (B;,,f,k,0,1) is an extended de Morgan algebra where f,k are given, for x = a i, by f(x) = a n i, f(0) = 1; k(x) = [f(x)]. Since f commutes with we see that k 2 = id B and so B is symmetric. Here we clearly have x f(x) k(x) = 0 and so B is special. We now obtain an important further structural characterisation of when L is special. For this purpose, we require the following simple observation, in which Z(L) denotes the centre of L. Theorem 3. If a, x a, x a Z(L) then x Z(L). Proof. x a = (x a) a Z(L) whence x = (x a) (x a ) Z(L). As the following characterisation shows, the choice of a boolean lattice reduct for the algebra B in the above example is no accident. Theorem 4. The following statements are equivalent : (1) L is special ; (2) L has a boolean lattice reduct in which x is the relative complement of x + in the interval [(x x + ) x +, (x x + ) x + ]. Proof. (1) (2) Suppose that (1) holds and let x L. Then we observe that (a) x x is complemented with complement (x x ) +. In fact, replacing x by x in the identity x x x + = 0 we obtain x x x + = 0 whence x x (x + x + ) = 0. Likewise, replacing x by x + in the identity x x x + = 1 we obtain x + x + x = 1, and replacing x by x in this gives x + x + x = 1. Together these produce x + x + (x x ) = 1. It follows that x x has complement (x x ) +. (b) x x + is complemented with complement x x +. In fact, replacing x by x + in the identity x x x + = 0 we obtain x + x + x = 0. Consequently, using the above, we see that x x + (x + x ) = 0.
Congruence Coherent Symmetric Extended de Morgan Algebras 7 Likewise, x x + (x x + ) = 1. Hence x x + is complemented with complement x x +. By (a) and (b) we have, for every x L, x (x x + ) Z(L) where a = x x + Fix + L. Also, by Theorem 2(7), we have Fix + L Z(L). Now by applying + to Theorem 2(2), we see that x a = x x x + = 1 Z(L). It therefore follows by Theorem 3 that x Z(L). Hence L has a boolean lattice reduct. Now by Theorem 2 and the above it is clear that (x x + ) x + x (x x + ) x +. Moreover, in the interval [(x x + ) x +, (x x + ) x + ] the relative complement of x is [(x x + ) x + ] {[(x x + ) x + ] x} = [(x x + ) x + ] (x + x) = (x x + x) x + = x +. (2) (1) For every x Fix + L we have (x x + ) x + = x and likewise (x x + ) x + = x, whence by (2) we have x = x. Thus (Fix + L; ) is boolean and so L is special. Concerning the problem at hand, namely that of determining precisely which symmetric extended de Morgan algebras are congruence coherent, the following result provides the first half of the answer. Theorem 5. Every special symmetric extended de Morgan algebra is congruence coherent. Proof. Suppose that L is special. Let A be a subalgebra of L and let ϕ Con L. Suppose that [z] ϕ A for some z L. Then clearly [z ] ϕ A and [z + ] ϕ A. It follows from this that [0] ϕ A and [1] ϕ A. In fact, if x [0] ϕ then x z [z] ϕ, x z [z ] ϕ, x z + [z + ] ϕ whence x z, x z, x z + A. Consequently, x = x 0 = x (z z z + ) = (x z) (x z ) (x z + ) A and therefore [0] ϕ A. Similarly, we see that [1] ϕ A. Now suppose that a A. By Theorem 4, a is complemented. Let a be its complement. Then if y [a] ϕ we have y a [0] ϕ A and y a [1] ϕ A. Thus y a = (y a ) a A whence y = (y a) (y a ) A and consequently [a] ϕ A. It follows from this that L is congruence coherent.
8 T.S. Blyth and Jie Fang In order to investigate the situation when L is not special we require some further properties of congruences. For this purpose, consider for each a L the relation ϕ a given by (x,y) ϕ a (x a) a = (y a) a. If (x,y) ϕ a then (x a,y a) ϑ lat (a,1). Since (x,x a) ϑ lat (0,a) and (y,y a) ϑ lat (0,a) we see that ϕ a ϑ lat (0,a) ϑ lat (a,1). On the other hand, (0,a) ϕ a and (a,1) ϕ a whence we have the reverse inequality. Consequently, ϕ a = ϑ lat (0,a) ϑ lat (a,1). Theorem 6. ( x L) ϑ(0,x) = ϕ x x +. Proof. We recall from the proof of (7) (8) in Theorem 2 that, for every x L, ϑ(0,x) = ϑ lat (0,x x + ) ϑ lat ( (x x+ ),1 ). The result therefore follows from the above. Corollary. ( x Fix + L) Ker ϑ(0,x) = {a L a x x}. Proof. For every x L we have Ker ϕ x = {a L (a x) x = x x } = {a L a x x} and, by the above, if x Fix + L then ϑ(0,x) = ϕ x. We deduce from the above the following. Theorem 7. If L is congruence coherent and is not special then (1) K(L) Fix + L = {0,1}; (2) K(L) = {x L x x+ = 0} {1}; (3) K d (L) = {x L x x+ = 1} {0}. Proof. (1) By way of obtaining a contradiction, suppose that there exists x K(L) Fix + L with x / {0,1}. Then x x = 0 and x x = 1. Now for any a / K(L) we have, by Theorem 1, a a a + Ker ϑ(0,x). The Corollary to Theorem 6 then gives a a a + x x whence a a a + x = 0. Likewise we see that a a a + x = 0. From this we obtain the contradiction that a a a + = a a a + (x x) = 0, whence (1) holds.
Congruence Coherent Symmetric Extended de Morgan Algebras 9 (2) Clearly, {x L x x+ = 0} {1} K(L). To establish the reverse inclusion, suppose that x K(L) with x x + 0. Then x x x + = 0 whence x x + x + = 0. Writing y = x x + we see that y K(L) Fix + L with y 0. It follows by (1) that y = 1 whence x = 1 as required. (3) This is the dual of (2). For each a L we consider also the relation ϑ a given by (x,y) ϑ a { x (a a+ ) = y (a a + ); x (a a + ) = y (a a + ). It is clear that ϑ a Con L. This congruence has the following property. Theorem 8. Let L be congruence coherent and not special. If a / K(L) then Ker ϑ a = {0} = Kerϑ a +. Proof. Suppose, by way of obtaining a contradiction, that Kerϑ a {0}. Then by Theorem 1 we have x x x + Ker ϑ a for every x L. In particular, for x = a we have a a a + = a a a + (a a + ) = 0 whence the contradiction a K(L). As for the second equality, it suffices to observe that a / K(L) implies that a + / K(L). Theorem 8 is instrumental in the following sequence of results that will lead us to our goal. We recall the notation x y which means x y or y x, and the covering notation x y which means that if x z y then z = x or z = y. In what follows we shall use the notation x y to mean x y or y x. Theorem 9. Let L be congruence coherent and not special. If x Fix + L and x / K(L) then we have x x. Proof. By Theorem 8 we have Ker ϑ x = {0}. Consider the primitive subalgebra J = {0,1,x x,x x }. Since x Fix + L we have ϑ x = ϑ lat (x x,x ) and consequently, since L is congruence coherent, x [x x ] ϑx J. It follows from this that x x. Suppose first that x x. If x y x then we have (x,y) ϑ x = ϑ lat (x x,x ) and J = {0,1,x,x }. Since L is congruence coherent we have y [x] ϑx J whence necessarily y = x or y = x. A similar conclusion is reached if x x. Thus we see that x x. Corollary. If a / Fix + L and x = a a + 1 then x x.
10 T.S. Blyth and Jie Fang Proof. Since a a + 1, it follows by Theorem 7(3) that a / K d (L), whence a a a + 1; equivalently, a a a + 0. Then x x x + = x x = (a a + ) (a a + ) = (a a a + ) (a + a a + ) 0 and consequently x / K(L). Since x Fix + L it now follows by Theorem 9 that x x. In what follows we shall let Fix L = {x L x = x } and define the set of fixed points of L to be Fix L = Fix + L Fix L. Theorem 10. Let L be congruence coherent and not special. If a / Fix + L then (1) when a a + 0 we have a a + Fix L and a a + = 1; (2) when a a + 1 we have a a + Fix L and a a + = 0. Proof. (1) Let x = a a +. Since x 0 it follows by Theorem 7(2) that x / K(L). Then Ker ϑ x = {0} by Theorem 8; and x x by Theorem 9. We show as follows that x = x. Suppose, by way of obtaining a contradiction, that x x. There is no loss in generality in supposing henceforth that x x. Then x a x x gives either x = a x or x = a x. In the former case we have a x whence a + x + = x whence the contradiction x = a a + x. As for the latter, this gives a + = a + x = a + (a x ) = (a + a) (a + x ). Now if we assume that a a + = 1 then we obtain a + = a + x, so that a + x whence a = a ++ x + = x and again we have the contradiction x = a a + x. We must therefore have a a + 1. Writing z = a a + we then have z z by the Corollary to Theorem 9. Since x x and x < z we must have z z. Now x z x x gives either x = z x or x = z x. The former gives z x = z z x = z x = z whence z x = z and there follows the contradiction x = z x = (z x ) x = x. As for the latter, this gives x z whence z x x z and consequently x = z. Now since x a z = x we have that (a,x) ϑ x = ϑ lat (x,x ) and consequently a [x] ϑx J = {0,1,x,x } which produces the contradiction a Fix + L. From the above sequence of contradictions we conclude that x = x and therefore a a + Fix L. Moreover, we have a a + = 1, for if y = a a + 1 then as in the above we have y y. Since x = x and x < y we must have y y, which contradicts y < x = x < y. (2) If a a + 1 then a a + 0 whence, by (1), a a + Fix L and a a + = 1. It follows that a a + Fix L and a a + = 0.
Congruence Coherent Symmetric Extended de Morgan Algebras 11 Theorem 11. If L is congruence coherent and not special then Fix + L = Fix L {0,1}. Proof. It is clear that Fix L {0,1} Fix + L. To obtain the reverse inclusion suppose that a Fix + L\{0,1}. Since L is not primitive there exists b L such that b / Fix + L. There are then three cases to consider. (1) b b + 0. In this case Theorem 10(1) gives b b + FixL and b b + = 1. Let β = b b + and suppose, by way of obtaining a contradiction, that a a = 0. Then β a. In fact, if β a then β a = 0 whence, since β Fix L, we have the contradiction a = β a = 1; and if β a then β a a a = 1 whence we have the contradiction 0 = (β a ) = β a = a. Moreover, β a 0. For otherwise β = β 1 = β (a a ) = β a whence β a which gives a β and the contradiction a = β a = 0. Now let y = β a. We have y = β a = β a β β a = y Fix + L\{0,1}. It follows from Theorem 7(1) that y / K(L) and so, by Theorem 9, we see that y y. Then since β Fix L we have β = y = β a which gives the contradiction β a. We conclude from the above that a a 0. Now since a Fix + L this means that a / K(L). Then by Theorem 9 we have a a. We may assume that a a. Then a (a b) a gives (a b) a = a or (a b) a = a. The former gives a b = a b whence a b + = a b + and therefore a = a 1 = a (b b + ) = a (b b + ) = a. As for the latter, this gives a b a b whence a b = a b. Then a b + = a b + and so a β = a β. Since β Fix L we have a β = a β and so, by distributivity, a = a. In this case therefore we have a Fix L. (2) b b + = 0, b b + 1. In this case, by Theorem 10(2), b b + FixL. Thus b b + = (b b + ) FixL and b b + = 1. Then, as in case (1) with b replaced by b, we have a FixL. (3) b b + = 0, b b + = 1. In this case let x = a b. Then x / Fix + L since otherwise a b = a b + whence we have the contradiction x = a b = a b b + = 0. Now x x + = (a b) (a b) + = (a b) (a b + ) = a (b b + ) = a 1 = a 1.
12 T.S. Blyth and Jie Fang It then follows by Theorem 10(2) that a = x x + Fix L. In conclusion, in all cases we have a Fix L whence the result follows. The above results can be used to provide the second half of the answer to the problem of which (non-primitive) symmetric extended de Morgan algebras are congruence coherent. Theorem 12. If L is not special then L is congruence coherent if and only if it is simple. Proof. Suppose that L is congruence coherent and let ϕ Con L be such that ϕ ω. Then there exist a,b L with a < b and (a,b) ϕ. We first have by Theorem 11 that {a a +,b b +,a a +,b b + } Fix L {0,1}. Now, since (a,b) ϕ, we have (a a +,b b + ) ϕ and (a a +,b b + ) ϕ. We cannot have both a a + = b b + and a a + = b b + since otherwise, it follows that a a + = a b + = b a + = b b + and a a + = a b + = b a + = b b +, from which there follows by distributivity the contradiction that a = b. Thus we have either a a + b b + or a a + b b +, whence there are two possibilities to consider: (1) a a + b b +. In this case, we have a a + Fix L {0}. If, on the one hand, a a + = 0 then, we have either b b + Fix L or b b + = 1. The latter clearly gives (0,1) ϕ; and the former gives 0 = a a + ϕ b b + = (b b + ) = b b + ϕ a a + = (a a + ) = 0 = 1 whence again (0,1) ϕ. If, on the other hand, a a + FixL. Then, since fixed points are incomparable, we must have that b b + = 1, whence 1 = b b + ϕ a a + = (a a + ) = a a + ϕ b b + = (b b + ) = 1 = 0, and consequently, (0,1) ϕ. Thus we see that ϕ = ι. (2) a a + b b +. In this case, a similar argument to that in (1) also gives ϕ = ι. We conclude that L is simple. The converse is clear. As was established in [4, Theorem 5.1], there are precisely nine nonisomorphic subdirectly irreducible symmetric extended de Morgan algebras, all of which are simple. Of these the (non-primitive) algebras that are congruence coherent and not special are four in number, namely the algebra
Congruence Coherent Symmetric Extended de Morgan Algebras 13 1 j s q t e c i n r p a d b 0 m 0 a b c d e i j m n p q r s t 1 f 1 q j c t p i b s n e a r m d 0 k 0 b a c m p n q d i e j r t s 1 and the subalgebras {0,a,b,c,j,q,1}, {0,a,b,c,e,j,p,q,1}, {0,d,e,m,p,r,s,t,1}. These four algebras therefore comprise the second half of the answer. References [1] M. E. Adams, M. Atallah and R. Beazer, Congruence distributive double p-algebras, Proc. Edinburgh Math. Soc. (2) 39, 1996, 71-80. [2] R. Beazer, Coherent de Morgan algebras, Algebras Universalis, 24, 1987, 128-136. [3] T. S. Blyth and Jie Fang, Congruence coherent double MS-algebras, Glasgow Math. J., 41, 1999, 289-295. [4] T. S. Blyth and Jie Fang, Extended Ockham algebras, Communications in Algebra, 28(3), 2000, 1271-1284. [5] T. S. Blyth and Jie Fang, Symmetric extended Ockham algebras, Algebra Colloquium, 10(4), 2003, 479-489. [6] T. S. Blyth and J. C. Varlet, Ockham Algebras, Oxford University Press, 1994. [7] A. Urquhart, Lattices with a dual homomorphic operation, Studia Logica, 38, 1979, 201 209; ibid. 40, 1981, 391 404. T.S. Blyth Mathematical Institute, University of St Andrews, St Andrews KY16 9SS, Scotland tsb@st-and.ac.uk Jie Fang Department of Mathematics, Shantou University, Shantou, 515063 Guangdong, P.R. China jfang@stu.edu.cn