Page 20 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
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Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
The classic example of a Boolean algebra is the power set algebra 2X of all subsets of a
given set X with ∨ and ∧ being ∪ and ∩ respectively, and complementation being ordinary set
complementation. More generally recall that a ring of sets is any family R of subsets of a given
set X that is closed under finite unions and finite intersections. R is a field of sets if it is also
closed under complementation. Before stating the next lemma, recall that an ideal P in a lattice is
a prime ideal if x∧y ∈ P implies that either x ∈ P or y ∈ P.
Lemma 1.1.11. Let I be an ideal and F be a filter that are disjoint in a distributive
lattice. Then a prime ideal P exists such that I ⊆ P but F ∩ P = ∅.
Proof. Let P be an ideal that is maximal subject to the stated conditions. Suppose that a∧b ∈ P
for some a ∉ P and b ∉ P. Let P1 and P2 be the ideals generated respectively from P ∪ {a} and
P ∪ {b}. Since they are properly larger than I, P1 contains an element p1∨a ∈ F and P2 contains
an element p2∨b ∈ F where p1, p2 ∈ P. But then F contains
(p1∨a)∧(p2∨b) = (p1∧p2)∨(p1∧b)∨(p2∧a)∨(a∧b)
which is also in P, a contradiction. Thus a and b do not exist and P is indeed prime. £
This leads us to the first of several fundamental results about Boolean lattices:
Theorem 1.1.12. (M. H. Stone) A lattice is (distributive) Boolean if and only if it is
isomorphic to a (ring) field of sets.
Proof. The “if” direction is clear. So suppose that a lattice L is distributive. Let P denote the set
of all nonempty prime ideals of L. To each x ∈ L, set π(x) = {P ∈ P ⎢x ∉ P}. It is easily seen that
π(x∨y) = π(x) ∪ π(y) and π(x∧y) = π(x) ∩ π(y). Moreover, if x ≠ y then either x↓ ∩ y↑ = ∅ or else
y↓ ∩ x↑ = ∅. In either case, by the lemma, a prime ideal P exists containing exactly one of x and
y. Hence x ≠ y implies π(x) ≠ π(y). Thus π: L → 2P is an embedding of distributive lattices.
If L is Boolean lattice, then first π(0) = ∅. Next, in the Boolean case we consider
only proper prime ideals. We still have π(x∨y) = π(x) ∪ π(y) and π(x∧y) = π(x) ∩ π(y),
but now π(1) = P. Moreover, for all x, π(x) ∩ π(xʹ) = π(x ∧ xʹ) = π(0) = ∅. Likewise,
π(x) ∪ π(xʹ) = π(x ∨ xʹ) = π(1) = P. Thus π(x) and π(xʹ) are complements in 2P and so
π(xʹ) = P \ π(x). Thus π is an embedding of Boolean algebras. £
Recall that an atom in a lattice with 0 is an element a > 0 such that no element x exists
properly between 0 and a. A lattice is atomic if every element is a supremum of atomic elements.
In particular, each x is the supremum of the set α(x) of all atoms lying beneath x. Again, the
proofs of the following three results are easily accessible. We give the proof of the third.
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The classic example of a Boolean algebra is the power set algebra 2X of all subsets of a
given set X with ∨ and ∧ being ∪ and ∩ respectively, and complementation being ordinary set
complementation. More generally recall that a ring of sets is any family R of subsets of a given
set X that is closed under finite unions and finite intersections. R is a field of sets if it is also
closed under complementation. Before stating the next lemma, recall that an ideal P in a lattice is
a prime ideal if x∧y ∈ P implies that either x ∈ P or y ∈ P.
Lemma 1.1.11. Let I be an ideal and F be a filter that are disjoint in a distributive
lattice. Then a prime ideal P exists such that I ⊆ P but F ∩ P = ∅.
Proof. Let P be an ideal that is maximal subject to the stated conditions. Suppose that a∧b ∈ P
for some a ∉ P and b ∉ P. Let P1 and P2 be the ideals generated respectively from P ∪ {a} and
P ∪ {b}. Since they are properly larger than I, P1 contains an element p1∨a ∈ F and P2 contains
an element p2∨b ∈ F where p1, p2 ∈ P. But then F contains
(p1∨a)∧(p2∨b) = (p1∧p2)∨(p1∧b)∨(p2∧a)∨(a∧b)
which is also in P, a contradiction. Thus a and b do not exist and P is indeed prime. £
This leads us to the first of several fundamental results about Boolean lattices:
Theorem 1.1.12. (M. H. Stone) A lattice is (distributive) Boolean if and only if it is
isomorphic to a (ring) field of sets.
Proof. The “if” direction is clear. So suppose that a lattice L is distributive. Let P denote the set
of all nonempty prime ideals of L. To each x ∈ L, set π(x) = {P ∈ P ⎢x ∉ P}. It is easily seen that
π(x∨y) = π(x) ∪ π(y) and π(x∧y) = π(x) ∩ π(y). Moreover, if x ≠ y then either x↓ ∩ y↑ = ∅ or else
y↓ ∩ x↑ = ∅. In either case, by the lemma, a prime ideal P exists containing exactly one of x and
y. Hence x ≠ y implies π(x) ≠ π(y). Thus π: L → 2P is an embedding of distributive lattices.
If L is Boolean lattice, then first π(0) = ∅. Next, in the Boolean case we consider
only proper prime ideals. We still have π(x∨y) = π(x) ∪ π(y) and π(x∧y) = π(x) ∩ π(y),
but now π(1) = P. Moreover, for all x, π(x) ∩ π(xʹ) = π(x ∧ xʹ) = π(0) = ∅. Likewise,
π(x) ∪ π(xʹ) = π(x ∨ xʹ) = π(1) = P. Thus π(x) and π(xʹ) are complements in 2P and so
π(xʹ) = P \ π(x). Thus π is an embedding of Boolean algebras. £
Recall that an atom in a lattice with 0 is an element a > 0 such that no element x exists
properly between 0 and a. A lattice is atomic if every element is a supremum of atomic elements.
In particular, each x is the supremum of the set α(x) of all atoms lying beneath x. Again, the
proofs of the following three results are easily accessible. We give the proof of the third.
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