Page 48 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 48
athan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Proof. (1) Given e, f ∈ B, since B is a right-regular band,
(e + f – ef)2 = e + ef – ef + fe + f – fef – efe – ef + ef = e + fe + f – ef – fe = e + f – ef.
To see (2), observe first that for all g, h ∈B,
(e + f – ef)g(e + f – ef) = ege + egf – egef + fge + fgf – fgef – efge – efgf + efgef
= ge + egf – gef + fge + gf – gef – fge – egf + gef
= ge – gef + gf
= g(e + f – ef)
Thus any product g0(e + f – ef)g1(e + f – ef)g2 … gn–1(e + f – ef)gn generated from
B {e + f – ef } reduces to g0g1g2… gn–1(e + f – ef)gn. From this observation, (2) follows. £
Theorem 2.1.9. Every maximal right [left] regular band in a ring R is also closed under
the circle operation e ○ f = e + f – ef and thus forms a skew lattice in R. £
In general, maximal regular bands in rings need not be closed under ○. We give an
example of this in Section 2.3. Nonetheless, what we have seen thus far suffices to prove:
Theorem 2.1.10. A band can be ∧-embedded in a skew lattice if and only if it is a
regular band.
Proof. The condition is clearly necessary. Suppose that B is a regular band. By the Kimura
Decomposition B can be embedded in B/R × B/L. In the semigroup ring &[B/R], B/R generates
a regular band C1 that is closed under the circle operation. Likewise B/L generates a regular band
C2 in &[B/L] that is closed under the circle operation. Thus B/R × B/L is a sub-band of C1 × C2
in &[B/R] × &[B/L]. Since C1 × C2 is closed under the circle operation it is a skew lattice and B
is embedded in C1 × C2 . £
Thus, even if a maximal regular band in a ring does not form a skew lattice in that ring,
some copy of it will generate a skew lattice under ○ and multiplication in another ring. We
complete this section with an important elementary fact about skew lattice in rings. Recall that a
skew lattice is distributive if it satisfies identities D3 and D4 in Section 1.3.
Theorem 2.1.11. Skew lattices in rings (using multiplication and ○) are distributive.
Proof. a ∧ (b ∨ c) ∧ a = a(b + c – bc)a = aba + aca – abca
= aba + aca – abaca (regularity)
= (a ∧ b∧ a) ∨ (a ∧ c ∧ a) .
For the dual identity, D4, observe first that a∨b∨a = a + b – ab – ba + aba. Thus
a ∨ (b∧c) ∨ a = a + bc – abc – bca + abca.
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Proof. (1) Given e, f ∈ B, since B is a right-regular band,
(e + f – ef)2 = e + ef – ef + fe + f – fef – efe – ef + ef = e + fe + f – ef – fe = e + f – ef.
To see (2), observe first that for all g, h ∈B,
(e + f – ef)g(e + f – ef) = ege + egf – egef + fge + fgf – fgef – efge – efgf + efgef
= ge + egf – gef + fge + gf – gef – fge – egf + gef
= ge – gef + gf
= g(e + f – ef)
Thus any product g0(e + f – ef)g1(e + f – ef)g2 … gn–1(e + f – ef)gn generated from
B {e + f – ef } reduces to g0g1g2… gn–1(e + f – ef)gn. From this observation, (2) follows. £
Theorem 2.1.9. Every maximal right [left] regular band in a ring R is also closed under
the circle operation e ○ f = e + f – ef and thus forms a skew lattice in R. £
In general, maximal regular bands in rings need not be closed under ○. We give an
example of this in Section 2.3. Nonetheless, what we have seen thus far suffices to prove:
Theorem 2.1.10. A band can be ∧-embedded in a skew lattice if and only if it is a
regular band.
Proof. The condition is clearly necessary. Suppose that B is a regular band. By the Kimura
Decomposition B can be embedded in B/R × B/L. In the semigroup ring &[B/R], B/R generates
a regular band C1 that is closed under the circle operation. Likewise B/L generates a regular band
C2 in &[B/L] that is closed under the circle operation. Thus B/R × B/L is a sub-band of C1 × C2
in &[B/R] × &[B/L]. Since C1 × C2 is closed under the circle operation it is a skew lattice and B
is embedded in C1 × C2 . £
Thus, even if a maximal regular band in a ring does not form a skew lattice in that ring,
some copy of it will generate a skew lattice under ○ and multiplication in another ring. We
complete this section with an important elementary fact about skew lattice in rings. Recall that a
skew lattice is distributive if it satisfies identities D3 and D4 in Section 1.3.
Theorem 2.1.11. Skew lattices in rings (using multiplication and ○) are distributive.
Proof. a ∧ (b ∨ c) ∧ a = a(b + c – bc)a = aba + aca – abca
= aba + aca – abaca (regularity)
= (a ∧ b∧ a) ∨ (a ∧ c ∧ a) .
For the dual identity, D4, observe first that a∨b∨a = a + b – ab – ba + aba. Thus
a ∨ (b∧c) ∨ a = a + bc – abc – bca + abca.
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