Page 74 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 74
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
IV. Set x∧y = (x∧n∧x) ∧ (y∧m∧y) for any m, n in the join class of x and y such that m ≤ x
and n ≤ y.
Since the equivalence classes of x and y are orthogonal in both their join and meet classes, any p,
q, m and n satisfying the stated conditions must produce the same values for x∨q∨x, y∨p∨y,
x∧n∧x and y∧m∧y so that the extended binary operations are uniquely determined. The full
algebra (S; ∨, ∧) is called the algebraic closure of (S; ≻, ∨, ∧).
Example 2.5.1. Given a ring R, two partial skew lattice structures exist for the entire set
of idempotents: (E(R); ≻L, ∨, ∧) and (E(R); ≻R, ∨, ∧). Here e ≻L f iff fe = f and e ≻R f iff ef = f.
In both cases e∨f = e○f and e∧f = ef, given that e and f are ≻-related. While Condition I above is
met in many classical rings (e.g., full matrix rings over fields), Condition II is rarely met. £
Theorem 2.5.2. A partial skew lattice (S; ≻, ∨, ∧) is the canonical partial algebra of a
full skew lattice (S; ∨, ∧) if and only if Conditions I and II above hold, in which case (S; ∨, ∧) is
precisely the algebraic closure of (S; ≻, ∨, ∧).
Proof. Necessity is clear. Moreover, if such a lattice extension exists, it must be the algebraic
closure by Theorem 2.4.9. We show that the algebraic closure must be a skew lattice. So let x
and y be given. Then x∨y is calculated as xʹ∨yʹ for some xʹ and yʹ in the join class of x and y such
that x ≤ xʹ and y ≤ yʹ. Thus in the partial skew lattice we have
x ∧ (x∨y) = x ∧ (xʹ∨yʹ) = x ∧ (x∨xʹ∨yʹ) = x.
The other instances of absorption are similarly seen. Suppose next that both x, y ≺ s in S. By
extremal associativity, x∨y = x∨q∨p∨y for any p ≥ x and q ≥ y in the join class of x and y. Choose
q such that y ≤ q ≤ y∨s∨y. (If need be, replace the given q by (y∨s∨y)∧q∧(y∨s∨y)). Applying
linear associativity we get
(x ∨ y) ∨ s = (x∨q∨p∨y) ∨ s = (x∨q) ∨ p ∨ (y ∨ s ∨ y) ∨ s
= (x∨q) ∨ (y ∨ s ∨ y) ∨ s = x ∨ (q ∨ y ∨ s ∨ y) ∨ s = x ∨ (y ∨ s).
We similarly obtain the three other cases of outer associativity: if r ≤ x, y ≤ s, then also
s ∨ (x ∨ y) = (s ∨ x) ∨ y, r ∧ (x ∧ y) = (r ∧ x) ∧ y and (x ∧ y) ∧ r = x ∧ (y ∧ r).
With the available conditional associative identities we obtain the unconditional identities. For
instance, let s lie in the join class of x, y and z. Then x∧(y∧(z∧s)) must equal both ((x∧y)∧z)∧s
and (x∧(y∧z))∧s. Likewise ((s∧x)∧y)∧z must equal both s∧((x∧y)∧z) and s∧(x∧(y∧z)). It follows
that x∧(y∧(z∧s))∧((s∧x)∧y)∧z must equal both ((x∧y)∧z)∧s∧s∧((x∧y)∧z) which reduces to
(x∧y)∧z and (x∧(y∧z))∧s∧s∧(x∧(y∧z)) which reduces to x∧(y∧z). Thus (x∧y)∧z = x∧(y∧z). The
dual form of associativity is likewise seen. The algebraic closure is thus indeed a skew lattice and
the theorem follows. £
72
IV. Set x∧y = (x∧n∧x) ∧ (y∧m∧y) for any m, n in the join class of x and y such that m ≤ x
and n ≤ y.
Since the equivalence classes of x and y are orthogonal in both their join and meet classes, any p,
q, m and n satisfying the stated conditions must produce the same values for x∨q∨x, y∨p∨y,
x∧n∧x and y∧m∧y so that the extended binary operations are uniquely determined. The full
algebra (S; ∨, ∧) is called the algebraic closure of (S; ≻, ∨, ∧).
Example 2.5.1. Given a ring R, two partial skew lattice structures exist for the entire set
of idempotents: (E(R); ≻L, ∨, ∧) and (E(R); ≻R, ∨, ∧). Here e ≻L f iff fe = f and e ≻R f iff ef = f.
In both cases e∨f = e○f and e∧f = ef, given that e and f are ≻-related. While Condition I above is
met in many classical rings (e.g., full matrix rings over fields), Condition II is rarely met. £
Theorem 2.5.2. A partial skew lattice (S; ≻, ∨, ∧) is the canonical partial algebra of a
full skew lattice (S; ∨, ∧) if and only if Conditions I and II above hold, in which case (S; ∨, ∧) is
precisely the algebraic closure of (S; ≻, ∨, ∧).
Proof. Necessity is clear. Moreover, if such a lattice extension exists, it must be the algebraic
closure by Theorem 2.4.9. We show that the algebraic closure must be a skew lattice. So let x
and y be given. Then x∨y is calculated as xʹ∨yʹ for some xʹ and yʹ in the join class of x and y such
that x ≤ xʹ and y ≤ yʹ. Thus in the partial skew lattice we have
x ∧ (x∨y) = x ∧ (xʹ∨yʹ) = x ∧ (x∨xʹ∨yʹ) = x.
The other instances of absorption are similarly seen. Suppose next that both x, y ≺ s in S. By
extremal associativity, x∨y = x∨q∨p∨y for any p ≥ x and q ≥ y in the join class of x and y. Choose
q such that y ≤ q ≤ y∨s∨y. (If need be, replace the given q by (y∨s∨y)∧q∧(y∨s∨y)). Applying
linear associativity we get
(x ∨ y) ∨ s = (x∨q∨p∨y) ∨ s = (x∨q) ∨ p ∨ (y ∨ s ∨ y) ∨ s
= (x∨q) ∨ (y ∨ s ∨ y) ∨ s = x ∨ (q ∨ y ∨ s ∨ y) ∨ s = x ∨ (y ∨ s).
We similarly obtain the three other cases of outer associativity: if r ≤ x, y ≤ s, then also
s ∨ (x ∨ y) = (s ∨ x) ∨ y, r ∧ (x ∧ y) = (r ∧ x) ∧ y and (x ∧ y) ∧ r = x ∧ (y ∧ r).
With the available conditional associative identities we obtain the unconditional identities. For
instance, let s lie in the join class of x, y and z. Then x∧(y∧(z∧s)) must equal both ((x∧y)∧z)∧s
and (x∧(y∧z))∧s. Likewise ((s∧x)∧y)∧z must equal both s∧((x∧y)∧z) and s∧(x∧(y∧z)). It follows
that x∧(y∧(z∧s))∧((s∧x)∧y)∧z must equal both ((x∧y)∧z)∧s∧s∧((x∧y)∧z) which reduces to
(x∧y)∧z and (x∧(y∧z))∧s∧s∧(x∧(y∧z)) which reduces to x∧(y∧z). Thus (x∧y)∧z = x∧(y∧z). The
dual form of associativity is likewise seen. The algebraic closure is thus indeed a skew lattice and
the theorem follows. £
72
