Page 206 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
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athan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
insures that S is cancellative. Conversely, given a cancellative skew lattice, we must show that
all pointed skew diamonds in it factor as stated. Our task quickly reduces to showing that a
cancellative pointed skew lattice S factors as stated. So let S = {J > A, B > M} be such a skew
diamond with say M = {0}. A pair of primitive subalgebras are A0 = A ∪ {0} and B0 = B ∪ {0}.
Claim: An isomorphism σ: A0 × B0 ≅ S is given by σ[(x, y)] = x∨y for all x ∈ A0 and y ∈ B0.
Thanks to Theorem 2.2.1, σ is easily seen to be surjective. It is clearly bijective from
{0}×{0} to {0}, from A×{0} to A and from {0}×B to B. Thus σ is bijective overall if it is
bijective from A×B to J. Since S is symmetric, a∨b = b∨a for each pair (a, b) in A × B since a∧b
= 0 = b∧a. Thus the bijectivity of A×B with J is given by Theorem 5.3.10.
Finally for all x1, x2 ∈A0 and y1, y2 ∈ B0,
σ[(x1, y1)∨(x2, y2)] = σ[(x1∨x2, y1∨y2)] = x1∨x2 ∨ y1∨y2
= x1∨y1 ∨ x2∨y2 = σ[(x1, y1)] ∨ σ[(x2, y2)],
since elements from A0 commute with elements from B0. Expanding σ[(x1, y1)∧(x2, y2)] and
σ[(x1, y1)] ∧ σ[(x2, y2)], we get respectively: (x1∧x2) ∨ (y1∧y2) and (x1∨y1) ∧ (x2∨y2).
Case 1) One of the xi and one of the yj is 0. Here both expressions above reduce to 0.
Case 2) Neither xi is 0 but one of the yj is 0. We get x1∧x2 on the left and on the right either
(x1∨y1) ∧ x2 or x1 ∧ (x2∨y2). Since the xi and yj commute, normality plus absorption
gives,
(x1∨y1)∧x2 = (y1∨x1)∧x2∧x1∧x2 = (y1∨x1)∧x1∧x2 = x1 ∧ x2
and
x1∧(x2∨y2) = x1∧x2∧x1∧(x2∨y2) = x1∧x2∧(x2∨y2) = x1 ∧ x2.
Case 3) One of the xi is 0, but neither of the yj is 0. This is similar to Case 2.
Case 4) None of the xi or yj are 0. By x-y commutation plus u∧v = v∨u on D-classes, both
(x1∧x2)∨(y1∧y2) and (x1∨y1)∧(x2∨y2) are easily seen to reduce to x2∨y2∨x1∨y1.
Thus σ is an isomorphism. The case where J = {1} follows by the dual argument. £
There is more. In general, every skew diamond S (cancellative or not) is a union of
maximal pointed subalgebras in two ways. S = ∪m∈M m∨S∨m = ∪j∈J j∧S∧j where
m∨S∨m = {m∨x∨m⎪x ∈S} = {x ∈S⎪x ≥ m} is pointed below with zero m, and
j∧S∧j = { j∧x∧j⎪x ∈S} = {x ∈S⎪j ≥ x} is pointed above with identity 1,
Given m, mʹ ∈M, define f: m∨S∨m → mʹ∨S∨mʹ by f(x) = mʹ∨x∨mʹ and g: mʹ∨S∨mʹ → m∨S∨m
by g(x) = m∨x∨m. Regularity and the fact x∨y∨x = x holds on M imply that f and g are a
reciprocal bijections. Again regularity gives mʹ∨(x∨y)∨mʹ = (mʹ∨x∨mʹ)∨(mʹ∨y∨mʹ). Thus f and
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insures that S is cancellative. Conversely, given a cancellative skew lattice, we must show that
all pointed skew diamonds in it factor as stated. Our task quickly reduces to showing that a
cancellative pointed skew lattice S factors as stated. So let S = {J > A, B > M} be such a skew
diamond with say M = {0}. A pair of primitive subalgebras are A0 = A ∪ {0} and B0 = B ∪ {0}.
Claim: An isomorphism σ: A0 × B0 ≅ S is given by σ[(x, y)] = x∨y for all x ∈ A0 and y ∈ B0.
Thanks to Theorem 2.2.1, σ is easily seen to be surjective. It is clearly bijective from
{0}×{0} to {0}, from A×{0} to A and from {0}×B to B. Thus σ is bijective overall if it is
bijective from A×B to J. Since S is symmetric, a∨b = b∨a for each pair (a, b) in A × B since a∧b
= 0 = b∧a. Thus the bijectivity of A×B with J is given by Theorem 5.3.10.
Finally for all x1, x2 ∈A0 and y1, y2 ∈ B0,
σ[(x1, y1)∨(x2, y2)] = σ[(x1∨x2, y1∨y2)] = x1∨x2 ∨ y1∨y2
= x1∨y1 ∨ x2∨y2 = σ[(x1, y1)] ∨ σ[(x2, y2)],
since elements from A0 commute with elements from B0. Expanding σ[(x1, y1)∧(x2, y2)] and
σ[(x1, y1)] ∧ σ[(x2, y2)], we get respectively: (x1∧x2) ∨ (y1∧y2) and (x1∨y1) ∧ (x2∨y2).
Case 1) One of the xi and one of the yj is 0. Here both expressions above reduce to 0.
Case 2) Neither xi is 0 but one of the yj is 0. We get x1∧x2 on the left and on the right either
(x1∨y1) ∧ x2 or x1 ∧ (x2∨y2). Since the xi and yj commute, normality plus absorption
gives,
(x1∨y1)∧x2 = (y1∨x1)∧x2∧x1∧x2 = (y1∨x1)∧x1∧x2 = x1 ∧ x2
and
x1∧(x2∨y2) = x1∧x2∧x1∧(x2∨y2) = x1∧x2∧(x2∨y2) = x1 ∧ x2.
Case 3) One of the xi is 0, but neither of the yj is 0. This is similar to Case 2.
Case 4) None of the xi or yj are 0. By x-y commutation plus u∧v = v∨u on D-classes, both
(x1∧x2)∨(y1∧y2) and (x1∨y1)∧(x2∨y2) are easily seen to reduce to x2∨y2∨x1∨y1.
Thus σ is an isomorphism. The case where J = {1} follows by the dual argument. £
There is more. In general, every skew diamond S (cancellative or not) is a union of
maximal pointed subalgebras in two ways. S = ∪m∈M m∨S∨m = ∪j∈J j∧S∧j where
m∨S∨m = {m∨x∨m⎪x ∈S} = {x ∈S⎪x ≥ m} is pointed below with zero m, and
j∧S∧j = { j∧x∧j⎪x ∈S} = {x ∈S⎪j ≥ x} is pointed above with identity 1,
Given m, mʹ ∈M, define f: m∨S∨m → mʹ∨S∨mʹ by f(x) = mʹ∨x∨mʹ and g: mʹ∨S∨mʹ → m∨S∨m
by g(x) = m∨x∨m. Regularity and the fact x∨y∨x = x holds on M imply that f and g are a
reciprocal bijections. Again regularity gives mʹ∨(x∨y)∨mʹ = (mʹ∨x∨mʹ)∨(mʹ∨y∨mʹ). Thus f and
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