Page 216 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 216
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Theorem 6.1.6. Let S be a distributive, symmetric skew lattice with both maximal and
minimal elements, 1 and 0. Let B be a finite Boolean lattice such that {1, 0} ⊆ B ⊆ Z(S). Then:
1) For each atom α of B, α∧S = {α∧x⎮x ∈ S} = {y ∈ S⎮0 ≤ y ≤ α} is a skew
lattice with maximal element α.
2) For atoms α ≠ β, and all x ≤ α and y ≤ β, x∧y = 0 = y∧x and x∨y = y∨x.
3) Given atoms α1, … ,αn of B, an isomorphism ϕ: S ≅ ∏1n (αi ∧ S) is defined by
ϕ(x) = (α1∧x, … , αn∧x) with ϕ–1(x1, … , xn) = x1 ∨ … ∨ xn.
Proof. Since B ⊆ Z(S), α∧x∧α∧y = α∧x∧y and
(α∧x)∨(α∧y) = (α∧x∧α)∨(α∧y∧α) = α∧(x∨y)∧α = α∧(x∨y).
Thus assertion (1) follows. Next, since α∧β = 0 for x and y as stated in (2),
x∧y = x∧α∧y∧β = x∧y∧α∧β x∧y∧0 = 0.
Similarly y∧x = 0, so that x∨y = y∨x by symmetry and (2) follows. Moreover, ϕ is tat least a
homomorphism. For all x ∈ S,
x = x ∧ 1 ∧ x = x ∧ (α1 ∨ … ∨ αn) ∧ x
= (x ∧ α1 ∧ x) ∨…∨ (x ∧ αn ∧ x) = (α1 ∧ x) ∨…∨ (αn ∧ x),
which guarantees that ϕ is also one-to-one. Finally, given any (x1, …, xn) in ∏1n (αi ∧ S) , each xi
= αi∧xi so that
αj∧(x1 ∨ … ∨ xn) = (αj∧x1) ∨ … ∨ (αj∧xn) = (αj∧xj) = xj.
Hence ϕ(x1 ∨ … ∨ xn) = (x1, …, xn), making the map ϕ surjective and hence an isomorphism. £
Our next goal is to refine our description of the right extension S of a lattice T is a ring.
In this regard, given commuting idempotents e and f in a ring R, S(e⎮f) will denote the D-class of
e in the skew diamond that is the right extension S of the lattice T = {e, f, e∨f, e∧f} in E(R).
Theorem 6.1.7. Let T be a lattice in a ring R with identity 1 and let S be the right
extension of T. If e ∈ T, then Se = ∩{S(e | f)⎮f ∈T}. For e, f ∈T, moreover,
S(e⎮f) = e + efRf(1– e) + eR(1– e)(1– f).
In particular, given e and f in T:
i) if e and f are comparable, then S(e⎮f) = Re.
ii) if e and f are disjoint (ef = 0), then S(e⎮f) = e + eR(1– e)(1– f).
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Theorem 6.1.6. Let S be a distributive, symmetric skew lattice with both maximal and
minimal elements, 1 and 0. Let B be a finite Boolean lattice such that {1, 0} ⊆ B ⊆ Z(S). Then:
1) For each atom α of B, α∧S = {α∧x⎮x ∈ S} = {y ∈ S⎮0 ≤ y ≤ α} is a skew
lattice with maximal element α.
2) For atoms α ≠ β, and all x ≤ α and y ≤ β, x∧y = 0 = y∧x and x∨y = y∨x.
3) Given atoms α1, … ,αn of B, an isomorphism ϕ: S ≅ ∏1n (αi ∧ S) is defined by
ϕ(x) = (α1∧x, … , αn∧x) with ϕ–1(x1, … , xn) = x1 ∨ … ∨ xn.
Proof. Since B ⊆ Z(S), α∧x∧α∧y = α∧x∧y and
(α∧x)∨(α∧y) = (α∧x∧α)∨(α∧y∧α) = α∧(x∨y)∧α = α∧(x∨y).
Thus assertion (1) follows. Next, since α∧β = 0 for x and y as stated in (2),
x∧y = x∧α∧y∧β = x∧y∧α∧β x∧y∧0 = 0.
Similarly y∧x = 0, so that x∨y = y∨x by symmetry and (2) follows. Moreover, ϕ is tat least a
homomorphism. For all x ∈ S,
x = x ∧ 1 ∧ x = x ∧ (α1 ∨ … ∨ αn) ∧ x
= (x ∧ α1 ∧ x) ∨…∨ (x ∧ αn ∧ x) = (α1 ∧ x) ∨…∨ (αn ∧ x),
which guarantees that ϕ is also one-to-one. Finally, given any (x1, …, xn) in ∏1n (αi ∧ S) , each xi
= αi∧xi so that
αj∧(x1 ∨ … ∨ xn) = (αj∧x1) ∨ … ∨ (αj∧xn) = (αj∧xj) = xj.
Hence ϕ(x1 ∨ … ∨ xn) = (x1, …, xn), making the map ϕ surjective and hence an isomorphism. £
Our next goal is to refine our description of the right extension S of a lattice T is a ring.
In this regard, given commuting idempotents e and f in a ring R, S(e⎮f) will denote the D-class of
e in the skew diamond that is the right extension S of the lattice T = {e, f, e∨f, e∧f} in E(R).
Theorem 6.1.7. Let T be a lattice in a ring R with identity 1 and let S be the right
extension of T. If e ∈ T, then Se = ∩{S(e | f)⎮f ∈T}. For e, f ∈T, moreover,
S(e⎮f) = e + efRf(1– e) + eR(1– e)(1– f).
In particular, given e and f in T:
i) if e and f are comparable, then S(e⎮f) = Re.
ii) if e and f are disjoint (ef = 0), then S(e⎮f) = e + eR(1– e)(1– f).
214
