Page 215 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 215
VI: Skew Lattices in Rings
Second: these complementary pairs are trivial D-classes and thus form part of the center
of the skew lattice.
Third: this case is centrally complemented in that each central (trivial) D-class has a
complementary class.
Finally, we seemingly have an internal direct product. Indeed, setting
⎡1 0 0 0⎤ ⎡0 0 0 0⎤ ⎡0 0 0 0⎤ ⎡0 0 0 0⎤
⎢0 0⎥ ⎢0 ⎥ ⎢0 ∗⎥ ⎢0 ⎥
S1 = ⎢⎢0 0 0 0⎥⎥ {0} and S2 = ⎢⎢0 1 0 0 ⎥ ⎢⎢0 1 0 ∗ ⎥⎥ ⎢⎢0 1 ∗ ∗ ⎥ {0}
0 0 0 1 0 ⎥ 0 1 0 0 0 ⎥
⎣⎢0 0 0 0⎥⎦ ⎢⎣0 0 0 1⎦⎥ ⎢⎣0 0 0 0⎦⎥ ⎢⎣0 0 0 0⎥⎦
an isomorphism of µ: S1 × S2 ≅ S is given by setting µ(x, y) = x ∨ y = x + y (since xy = 0). This
leads us to the following three results:
Theorem 6.1.4. If S is a cancellative skew lattice with a unique maximum 1 and a
unique minimum 0, then all pairs of complementary D-classes in S are trivial and thus lie in the
center of S. In particular, this occurs for all skew lattices in rings with unity 1 that contain both
1 and 0.
Proof. Let A, B be a complementary D-classes with say a, aʹ ∈ A and b ∈ B. From
a ∨ b = aʹ ∨ b = 1 and a ∧ b = aʹ ∧ b = 0, cancellation implies a = aʹ. £
Theorem 6.1.5. Every maximal [right-handed] skew lattice (under ○ and •) in a ring
with identity 1 is (a) centrally complemented and (b) contains all central idempotents of the ring.
Proof. (a) If e ∈ Z(S) but not 1 – e, then 1 – e ∉ S so that Sʹ = eS + (1 – e)S is a larger [right-
handed] skew lattice in R. Indeed given x + xʹ, y + yʹ ∈ Sʹ, we get
(x + xʹ)(y + yʹ) = xy + xʹyʹ and (x + xʹ) ○ (y + yʹ) = x○y + xʹ○yʹ.
Thus Sʹ is a skew lattice in R containing both S and 1 – e. Given the maximal status of S,
1 – e ∈ S and hence 1 – e ∈ Z(S).
(b) If e is a central idempotent that is not in S, then clearly Sʹ = eS + (1 – e)S contains
both e and S. As in (a), Sʹ must be a skew lattice, which by the maximal status of S equals the
latter, and e ∈ S follows. £
213
Second: these complementary pairs are trivial D-classes and thus form part of the center
of the skew lattice.
Third: this case is centrally complemented in that each central (trivial) D-class has a
complementary class.
Finally, we seemingly have an internal direct product. Indeed, setting
⎡1 0 0 0⎤ ⎡0 0 0 0⎤ ⎡0 0 0 0⎤ ⎡0 0 0 0⎤
⎢0 0⎥ ⎢0 ⎥ ⎢0 ∗⎥ ⎢0 ⎥
S1 = ⎢⎢0 0 0 0⎥⎥ {0} and S2 = ⎢⎢0 1 0 0 ⎥ ⎢⎢0 1 0 ∗ ⎥⎥ ⎢⎢0 1 ∗ ∗ ⎥ {0}
0 0 0 1 0 ⎥ 0 1 0 0 0 ⎥
⎣⎢0 0 0 0⎥⎦ ⎢⎣0 0 0 1⎦⎥ ⎢⎣0 0 0 0⎦⎥ ⎢⎣0 0 0 0⎥⎦
an isomorphism of µ: S1 × S2 ≅ S is given by setting µ(x, y) = x ∨ y = x + y (since xy = 0). This
leads us to the following three results:
Theorem 6.1.4. If S is a cancellative skew lattice with a unique maximum 1 and a
unique minimum 0, then all pairs of complementary D-classes in S are trivial and thus lie in the
center of S. In particular, this occurs for all skew lattices in rings with unity 1 that contain both
1 and 0.
Proof. Let A, B be a complementary D-classes with say a, aʹ ∈ A and b ∈ B. From
a ∨ b = aʹ ∨ b = 1 and a ∧ b = aʹ ∧ b = 0, cancellation implies a = aʹ. £
Theorem 6.1.5. Every maximal [right-handed] skew lattice (under ○ and •) in a ring
with identity 1 is (a) centrally complemented and (b) contains all central idempotents of the ring.
Proof. (a) If e ∈ Z(S) but not 1 – e, then 1 – e ∉ S so that Sʹ = eS + (1 – e)S is a larger [right-
handed] skew lattice in R. Indeed given x + xʹ, y + yʹ ∈ Sʹ, we get
(x + xʹ)(y + yʹ) = xy + xʹyʹ and (x + xʹ) ○ (y + yʹ) = x○y + xʹ○yʹ.
Thus Sʹ is a skew lattice in R containing both S and 1 – e. Given the maximal status of S,
1 – e ∈ S and hence 1 – e ∈ Z(S).
(b) If e is a central idempotent that is not in S, then clearly Sʹ = eS + (1 – e)S contains
both e and S. As in (a), Sʹ must be a skew lattice, which by the maximal status of S equals the
latter, and e ∈ S follows. £
213
