Page 238 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 238
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Lemma 6.4.5 When R is idempotent-closed and dominated, then:
i) For all x, y ∈ R and all e, f ∈ E(R), xefy = xfey.
ii) In particular, if e D f then xey = xfy.
Consequently, α(e, f) ∈ ann(R) for all e, f ∈ E(R) so that e∇f = e ○ f when ann(R) = {0}.
Proof. Given that R is idempotent-dominated and E(R) is normal, xefy = xfey must hold as
stated. If e D f then xey = xeffey = xfeefy = xfy. Thus
xα(e, f ) = x(ef + fe − efe − fef) = x(ef + fe − fe − ef) = 0
for all x ∈ R, and similarly, α(e, f)x = 0. £
Given the above assumptions it is easy to see that ann(R) = {0} when E(R) is either left-
handed (efe = ef) or right-handed (efe = fe). While ann(R) need not vanish, this is not the full
story.
Lemma 6.4.6 If R is idempotent-dominated, then
i) ann(R) = {x ∈ R⎪xe = 0 = ex for all e ∈ E(R)}
ii) ann[R/ann(R)] = {0}.
Proof. In general, ann(R) = {x ∈ R⎪xy = 0 = yx for all y ∈ R} and if π: R → R/ann(R) is the
induced homomorphism, then ann(R) ⊆ π−1{ann[R/ann(R)]} with
π−1{ann[R/ann(R)]} = {x ∈ R⎪xyz = 0 = yzx = yxz for all y, z ∈ R}.
But if R is idempotent-dominated, then ann(R) = {x ∈ R⎪xe = 0 = ex for all e ∈ E(R)}. Hence
π−1{ann[R/ann(R)]} ⊆ ann(R) and equality follows and ann[R/ann(R)] vanishes in R/ann(R).
Thus, if R is idempotent-dominated, the description of ann(R) in the lemma insures that
π−1{ann[R/ann(R)]} ⊆ ann(R). Equality follows, as does the lemma. £
Lemma 6.4.7 If R is a ring with ideal I ⊆ ann(R), then the induced epimorphism
π: R → R/I restricts to a bijection πE: E(R) → E(R/I). If E(R) is also multiplicative, so is E(R/I)
and πE is an isomorphism of skew Boolean algebras.
Proof. πE is a well-defined map between the stated sets. If π(e) = π(f) for e, f ∈ E(R), then
e = f + a for some a ∈ ann(R) and squaring gives e = f. Thus πE is at least injective. Given
x + I ∈ E(R/I), x2 = x+a for some a ∈ ann(R) and hence x4 = x2 so that x2 ∈ E(R) and πE is
bijective. Since π is a ring homomorphism, the rest of the lemma follows. £
Since eRe ∩ ann(R) = {0} for all idempotents e in any ring R, we have:
236
Lemma 6.4.5 When R is idempotent-closed and dominated, then:
i) For all x, y ∈ R and all e, f ∈ E(R), xefy = xfey.
ii) In particular, if e D f then xey = xfy.
Consequently, α(e, f) ∈ ann(R) for all e, f ∈ E(R) so that e∇f = e ○ f when ann(R) = {0}.
Proof. Given that R is idempotent-dominated and E(R) is normal, xefy = xfey must hold as
stated. If e D f then xey = xeffey = xfeefy = xfy. Thus
xα(e, f ) = x(ef + fe − efe − fef) = x(ef + fe − fe − ef) = 0
for all x ∈ R, and similarly, α(e, f)x = 0. £
Given the above assumptions it is easy to see that ann(R) = {0} when E(R) is either left-
handed (efe = ef) or right-handed (efe = fe). While ann(R) need not vanish, this is not the full
story.
Lemma 6.4.6 If R is idempotent-dominated, then
i) ann(R) = {x ∈ R⎪xe = 0 = ex for all e ∈ E(R)}
ii) ann[R/ann(R)] = {0}.
Proof. In general, ann(R) = {x ∈ R⎪xy = 0 = yx for all y ∈ R} and if π: R → R/ann(R) is the
induced homomorphism, then ann(R) ⊆ π−1{ann[R/ann(R)]} with
π−1{ann[R/ann(R)]} = {x ∈ R⎪xyz = 0 = yzx = yxz for all y, z ∈ R}.
But if R is idempotent-dominated, then ann(R) = {x ∈ R⎪xe = 0 = ex for all e ∈ E(R)}. Hence
π−1{ann[R/ann(R)]} ⊆ ann(R) and equality follows and ann[R/ann(R)] vanishes in R/ann(R).
Thus, if R is idempotent-dominated, the description of ann(R) in the lemma insures that
π−1{ann[R/ann(R)]} ⊆ ann(R). Equality follows, as does the lemma. £
Lemma 6.4.7 If R is a ring with ideal I ⊆ ann(R), then the induced epimorphism
π: R → R/I restricts to a bijection πE: E(R) → E(R/I). If E(R) is also multiplicative, so is E(R/I)
and πE is an isomorphism of skew Boolean algebras.
Proof. πE is a well-defined map between the stated sets. If π(e) = π(f) for e, f ∈ E(R), then
e = f + a for some a ∈ ann(R) and squaring gives e = f. Thus πE is at least injective. Given
x + I ∈ E(R/I), x2 = x+a for some a ∈ ann(R) and hence x4 = x2 so that x2 ∈ E(R) and πE is
bijective. Since π is a ring homomorphism, the rest of the lemma follows. £
Since eRe ∩ ann(R) = {0} for all idempotents e in any ring R, we have:
236
