Page 83 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 83
II: Skew Lattices
Theorem 2.6.9 (The Reduction Theorem) Let S be a symmetric, normal skew lattice
with maximal lattice image T and reduced skew lattice Srd. If D is the maximal lattice image of
Srd. then the canonical epimorphisms S → T and S → Srd induce an isomorphism of S with the
fibered product of T and Srd over D, S ≅ T ×D Srd; both Sred and D, moreover, are distributive.
S ⎯⎯→ Srd
↓ pullback ↓ T= S/D and D = Srd /D.
T ⎯⎯→
D
Proof. We need only show that when S is reduced, it is distributive. We do so by showing that
none of the following types of subalgebras can arise in S where in what follows A, B. C, J and M
represent distinct D-classes of S.
J or J
A
A B C C
B
M
M
Suppose first that M is trivial so that in both diagrams each element of C commutes with all
elements in A and B. In the left diagram C must also commute with all elements of J so that C
must trivial. Similarly in the left diagram A and B must also be trivial and thus the left diagram
reduces to a lattice. Dropping the added assumption that M be trivial, we see that the natural
rectangular functor K of S when restricted to the left diagram is a functor of isomorphisms
between distinct classes. Thus K cannot be the rectangular functor of a reduced algebra. In the
right subalgebra, each n ∈ J has commuting factorizations as a∨c and b∨c with a ∈ A, b ∈ B and
c ∈ C where clearly a > b. As a consequence, A ≅ B under ≥. Even if M is not trivial, the skew
lattice on the right must factor as M × the M-trivial case. Hence still A ≅ B under ≥, which again
cannot happen in a reduced symmetric, normal skew lattice. Thus the right diagram also cannot
occur in S. Hence lattice D is distributive and the normal skew lattice Srd must also be
distributive by Theorems 2.3.2 and/or 2.3.4. £
The structure of symmetric, normal skew lattices has been reduced to that of lattices and
symmetric, normal, distributive skew lattices that is, to that of lattices and strongly distributive
skew lattices. We now examine the latter, first when S/D is finite where we show that S is
decomposable, thus proving a significant special case of Theorem 2.5.4.
So let S be strongly distributive with S/D finite and denote the latter by T. Being a finite
distributive lattice, T has a finite set π of join-irreducible elements, including its minimal element
0. The class of prime filters of T is thus given by
Pr(T) = {p∨T⎮p ∈ π}.
81
Theorem 2.6.9 (The Reduction Theorem) Let S be a symmetric, normal skew lattice
with maximal lattice image T and reduced skew lattice Srd. If D is the maximal lattice image of
Srd. then the canonical epimorphisms S → T and S → Srd induce an isomorphism of S with the
fibered product of T and Srd over D, S ≅ T ×D Srd; both Sred and D, moreover, are distributive.
S ⎯⎯→ Srd
↓ pullback ↓ T= S/D and D = Srd /D.
T ⎯⎯→
D
Proof. We need only show that when S is reduced, it is distributive. We do so by showing that
none of the following types of subalgebras can arise in S where in what follows A, B. C, J and M
represent distinct D-classes of S.
J or J
A
A B C C
B
M
M
Suppose first that M is trivial so that in both diagrams each element of C commutes with all
elements in A and B. In the left diagram C must also commute with all elements of J so that C
must trivial. Similarly in the left diagram A and B must also be trivial and thus the left diagram
reduces to a lattice. Dropping the added assumption that M be trivial, we see that the natural
rectangular functor K of S when restricted to the left diagram is a functor of isomorphisms
between distinct classes. Thus K cannot be the rectangular functor of a reduced algebra. In the
right subalgebra, each n ∈ J has commuting factorizations as a∨c and b∨c with a ∈ A, b ∈ B and
c ∈ C where clearly a > b. As a consequence, A ≅ B under ≥. Even if M is not trivial, the skew
lattice on the right must factor as M × the M-trivial case. Hence still A ≅ B under ≥, which again
cannot happen in a reduced symmetric, normal skew lattice. Thus the right diagram also cannot
occur in S. Hence lattice D is distributive and the normal skew lattice Srd must also be
distributive by Theorems 2.3.2 and/or 2.3.4. £
The structure of symmetric, normal skew lattices has been reduced to that of lattices and
symmetric, normal, distributive skew lattices that is, to that of lattices and strongly distributive
skew lattices. We now examine the latter, first when S/D is finite where we show that S is
decomposable, thus proving a significant special case of Theorem 2.5.4.
So let S be strongly distributive with S/D finite and denote the latter by T. Being a finite
distributive lattice, T has a finite set π of join-irreducible elements, including its minimal element
0. The class of prime filters of T is thus given by
Pr(T) = {p∨T⎮p ∈ π}.
81
