Page 85 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 85
II: Skew Lattices
Proof. Repeated applications of Lemma 2.6.10 enable one to pass through the prime filters of T
and successfully strip primary factors off of S to obtain the decomposition. To see uniqueness, let
X be a D-class corresponding to a join-irreducible element p of T and let P = p∨T be the induced
prime filter. If P = T, X was just the minimal D-class of S and XP = X. Otherwise there is a
maximal D-class of S lying beneath X, call it Y. Applying the Lemma to the subalgebra X∪Y, X
must factor as XP × Y where to within isomorphism XP is y∨X∨y for any y in Y. Thus XP is
indeed unique to within isomorphism. Finally, if any XP for some P ≠ T is trivial, then for Y < X
as just given, K(X, Y) is an isomorphism and S is not reduced. But if no XP is trivial, except
possibly for P = T, then no K(A, B) with A > B can be an isomorphism since then A as an element
of T belongs to a prime filter P that excludes B and thus XP is not trivial so that B is to within
isomorphism a proper factor of A. £
Theorem 2.6.5 follows immediately from Theorems 2.6.9 and 2.6.11.
What happens when the maximal distributive lattice image D of S is not finite? S still
factors as the fibered product of its maximal lattice image T and its maximal reduced image Sred:
S ≅ T ×D Sred. Since Sred is necessarily distributive, the question reduces to what can be said
about arbitrary distributive, symmetric, normal skew lattices. Being a variety of algebras, every
such algebra decomposes into a subdirect product of subdirectly irreducible algebras. Which
distributive, symmetric, normal skew lattices are the subdirectly irreducible?
To answer this, let 2 denote the lattice 1 > 0, let R2 [L2] denote the right [left]
rectangular skew lattice on {1, 2} and let 3R and 3L be the results of adjoining a zero element 0 to
R2 and L2, respectively (Thus 1, 2 > 0 in both cases.) Finally, let 5 denote the fibered product
algebra 3L ×2 3R of order 5. Equivalently, 5 = (L2 × R2)0.
Theorem 2.6.12. The only nontrivial subdirectly irreducible strongly distributive skew
lattices are copies of 2, R2, L2, 3R or 3L. Every strongly distributive skew lattice is thus a
subdirect product of some of these algebras and in particular can be embedded in a power of 5.
Proof. We show that a nontrivial subdirectly irreducible distributive, symmetric, normal skew
lattice S must be a copy of one of the stated algebras. If S is such an algebra, a pair of elements a
≠ b in S must exist that are congruent under all nonidentity congruences on S. If a and b are not
D-equivalent, then since D separates them, D is the identity Δ and S is a lattice that must be a
copy of 2. So assume that a ≠ b in some D-class. We define two congruences on S as follows:
aa
x y if a∧x = a∧y and x * y if x∧a = y∧a
Each is clearly an equivalence on S that, thanks to normality and Theorem 2.3.4, is indeed a
aa
congruence. Since a D b but a ≠ b, either or * must separate a and b and thus equal Δ. In
either case A must be the maximal D-class of S. (Otherwise some aʹ > a exists in a properly
83
Proof. Repeated applications of Lemma 2.6.10 enable one to pass through the prime filters of T
and successfully strip primary factors off of S to obtain the decomposition. To see uniqueness, let
X be a D-class corresponding to a join-irreducible element p of T and let P = p∨T be the induced
prime filter. If P = T, X was just the minimal D-class of S and XP = X. Otherwise there is a
maximal D-class of S lying beneath X, call it Y. Applying the Lemma to the subalgebra X∪Y, X
must factor as XP × Y where to within isomorphism XP is y∨X∨y for any y in Y. Thus XP is
indeed unique to within isomorphism. Finally, if any XP for some P ≠ T is trivial, then for Y < X
as just given, K(X, Y) is an isomorphism and S is not reduced. But if no XP is trivial, except
possibly for P = T, then no K(A, B) with A > B can be an isomorphism since then A as an element
of T belongs to a prime filter P that excludes B and thus XP is not trivial so that B is to within
isomorphism a proper factor of A. £
Theorem 2.6.5 follows immediately from Theorems 2.6.9 and 2.6.11.
What happens when the maximal distributive lattice image D of S is not finite? S still
factors as the fibered product of its maximal lattice image T and its maximal reduced image Sred:
S ≅ T ×D Sred. Since Sred is necessarily distributive, the question reduces to what can be said
about arbitrary distributive, symmetric, normal skew lattices. Being a variety of algebras, every
such algebra decomposes into a subdirect product of subdirectly irreducible algebras. Which
distributive, symmetric, normal skew lattices are the subdirectly irreducible?
To answer this, let 2 denote the lattice 1 > 0, let R2 [L2] denote the right [left]
rectangular skew lattice on {1, 2} and let 3R and 3L be the results of adjoining a zero element 0 to
R2 and L2, respectively (Thus 1, 2 > 0 in both cases.) Finally, let 5 denote the fibered product
algebra 3L ×2 3R of order 5. Equivalently, 5 = (L2 × R2)0.
Theorem 2.6.12. The only nontrivial subdirectly irreducible strongly distributive skew
lattices are copies of 2, R2, L2, 3R or 3L. Every strongly distributive skew lattice is thus a
subdirect product of some of these algebras and in particular can be embedded in a power of 5.
Proof. We show that a nontrivial subdirectly irreducible distributive, symmetric, normal skew
lattice S must be a copy of one of the stated algebras. If S is such an algebra, a pair of elements a
≠ b in S must exist that are congruent under all nonidentity congruences on S. If a and b are not
D-equivalent, then since D separates them, D is the identity Δ and S is a lattice that must be a
copy of 2. So assume that a ≠ b in some D-class. We define two congruences on S as follows:
aa
x y if a∧x = a∧y and x * y if x∧a = y∧a
Each is clearly an equivalence on S that, thanks to normality and Theorem 2.3.4, is indeed a
aa
congruence. Since a D b but a ≠ b, either or * must separate a and b and thus equal Δ. In
either case A must be the maximal D-class of S. (Otherwise some aʹ > a exists in a properly
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