Page 222 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 222
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Proof. (i) Suppose A and B are D-classes in S with A > B. Let a ∈ A and b ∈ B so that a ≻ b.
Consider bsasb = a + b – aba. From
(a + b – aba)a(a + b – aba) = a + b – aba and a(a + b – aba)a = a,
bsasb ∈A follows. On the other hand, clearly bsasb ≥ b. Thus given D-classes A > B in
S, for all b ∈ B there exists a ∈ A such that a > b. Hence S is regular by Theorem 1.2.18.
(ii) Let a ≻ b and a ≻ c. Since S is regular, bab = b, bac = bc, cab = cb and cac = c.
Hence (bsc)a(bsc) = bsc so that a ≻ bsc. Thus the D-class of bsc is the join-class of Db
and Dc and D is a s-congruence.
(iii) follows from routine calculations such as as (ab) = a + ab + aba – aaba – abaab
wich immediately reduces to a + ab + aba – aba – ab = a.
(iv) Just observe that asb and bsa differ only by their third terms, ba and ab.
(v) For the first part, regularity gives
a(bsc)a = a(b + c +bc – bcb – cbc)a
= aba + aca +abaca – abacaba – acabaca = aba s aca.
For the other identity, first observe that [b(csa)]2 = (bc + ba +bac – bcac – baca)2 expands as
(bc + bcba – bca) + (babc + ba – baca) + (babc + bacba – baca)
– (bcabc + bcaba – bca) – (babc + bacba – baca)
= bc + ba – baca + babc – bcabc
which in turn must equal b(csa) = bc + ba +bac – bcac – baca. Equating and canceling
common terms gives babc – bcabc = bac – bcac, that is, the identity
babc + bcac = bac + bcabc. (6.2.2)
Thus
(asbsa)(ascsa) = (a + b – bab)( a + c – cac) = a + bc – bcac – babc + bac
= a + bc – bac – bcabc + bac = a + bc – bcabc = as(bc) sa.
(vi) Given aca = bcb, the regularity of • implies cac = c(aca)c = c(bcb)c = cbc.
Cancelling in a + c – cac = b + c – cbc (a∇c∇a = b∇c∇b) in turn gives a = b. £
We turn to other properties observed in skew lattices. The next result demonstrates the
important role played by instances of commutativity, especially on the algebraic reducts (S, •)
and (S, s) of any s-band S.
220
Proof. (i) Suppose A and B are D-classes in S with A > B. Let a ∈ A and b ∈ B so that a ≻ b.
Consider bsasb = a + b – aba. From
(a + b – aba)a(a + b – aba) = a + b – aba and a(a + b – aba)a = a,
bsasb ∈A follows. On the other hand, clearly bsasb ≥ b. Thus given D-classes A > B in
S, for all b ∈ B there exists a ∈ A such that a > b. Hence S is regular by Theorem 1.2.18.
(ii) Let a ≻ b and a ≻ c. Since S is regular, bab = b, bac = bc, cab = cb and cac = c.
Hence (bsc)a(bsc) = bsc so that a ≻ bsc. Thus the D-class of bsc is the join-class of Db
and Dc and D is a s-congruence.
(iii) follows from routine calculations such as as (ab) = a + ab + aba – aaba – abaab
wich immediately reduces to a + ab + aba – aba – ab = a.
(iv) Just observe that asb and bsa differ only by their third terms, ba and ab.
(v) For the first part, regularity gives
a(bsc)a = a(b + c +bc – bcb – cbc)a
= aba + aca +abaca – abacaba – acabaca = aba s aca.
For the other identity, first observe that [b(csa)]2 = (bc + ba +bac – bcac – baca)2 expands as
(bc + bcba – bca) + (babc + ba – baca) + (babc + bacba – baca)
– (bcabc + bcaba – bca) – (babc + bacba – baca)
= bc + ba – baca + babc – bcabc
which in turn must equal b(csa) = bc + ba +bac – bcac – baca. Equating and canceling
common terms gives babc – bcabc = bac – bcac, that is, the identity
babc + bcac = bac + bcabc. (6.2.2)
Thus
(asbsa)(ascsa) = (a + b – bab)( a + c – cac) = a + bc – bcac – babc + bac
= a + bc – bac – bcabc + bac = a + bc – bcabc = as(bc) sa.
(vi) Given aca = bcb, the regularity of • implies cac = c(aca)c = c(bcb)c = cbc.
Cancelling in a + c – cac = b + c – cbc (a∇c∇a = b∇c∇b) in turn gives a = b. £
We turn to other properties observed in skew lattices. The next result demonstrates the
important role played by instances of commutativity, especially on the algebraic reducts (S, •)
and (S, s) of any s-band S.
220
