Page 240 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 240
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Given a ring R, KR denotes the ideal {x ∈ R⎪uxv = 0 for all u, v ∈ R}. Clearly
KRKRKR = {0} so that KR is a nilpotent ideal of index 3.
Lemma 6.4.9 If R is idempotent-dominated, then
KR = {x ∈ R⎪exf = 0 for all e, f ∈ E(R)}.
If R is also idempotent-closed, then KR = {x ∈ R | exe = 0 for all e ∈ E(R)}.
Proof. The first statement is clear. If R is also idempotent-closed and exe = 0 for all e ∈ E(R),
then for all e, f ∈ E(R), exf = e(e∇f)x(e∇f)f = e0f = 0 and the second statement follows. £
Theorem 6.4.10 If R is an idempotent-dominated and idempotent-closed ring, then the
ring R/KR is the maximal abelian image of R. It is also idempotent-covered with
E(R/KR) ≅ E(R)/D. Conversely, if R is an idempotent-dominated ring for which R/KR is abelian,
then R is idempotent-closed.
(In general, if R is any ring with an ideal K such that RKR = {0}, then R is idempotent-closed if
and only if R/K is idempotent-closed, in which case E(R/K) is a homomorphic image of E(R) with
both sharing a common maximal lattice image.)
Proof. The quotient ring R/K is automatically idempotent-dominated. Suppose x + K ∈ E(R/K)
so that x + K = xn + K for all n ≥ 1. Then x2 = x + k for some k ∈ K. From this we get
x4 = x3 + xkx = x3, and hence x6 = x3 so that x + K = e + K for some e ∈ E(R). Thus all
idempotents in R/K come from idempotents in R, making E(R/K) multiplicative. By Lemma
6.4.5, ef − fe ∈ K for all e, f ∈ E(R), making E(R/K) commutative. Being idempotent-dominated,
this forces R/K to be idempotent-covered. Indeed, given any idempotent-dominated abelian ring
S, for any x = e1x1e1 + · · · + enxnen ∈ S, upon setting f = e1∇e2∇ . . . ∇en we have f ≥ all ei so
that fxf = f.
Next, let α: R → A be a homomorphism onto a ring A having only central idempotents. A
is automatically idempotent-dominated and hence idempotent-covered so that KA = {0}. Given
k ∈ K, since α is surjective, α(k) is in KA so that α(k) = 0. Thus K ⊆ ker(α) and the maximality of
the abelian image R/K follows.
The converse follows from the remark following the theorem. So let R be a ring with an
ideal K such that RKR = {0}, and let e, f ∈ E(R) be given. If R/K is idempotent-closed, then at
least efef = ef + k for some k ∈ K. But then
efef = e(efef)f = e(ef + k)f = ef + ekf = ef,
so that R is idempotent-closed. Likewise, if R is idempotent-closed, then so is R/K by the above
argument for R/K with E(R/K) again a homomorphic image of E(R). Given idempotent closure,
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Given a ring R, KR denotes the ideal {x ∈ R⎪uxv = 0 for all u, v ∈ R}. Clearly
KRKRKR = {0} so that KR is a nilpotent ideal of index 3.
Lemma 6.4.9 If R is idempotent-dominated, then
KR = {x ∈ R⎪exf = 0 for all e, f ∈ E(R)}.
If R is also idempotent-closed, then KR = {x ∈ R | exe = 0 for all e ∈ E(R)}.
Proof. The first statement is clear. If R is also idempotent-closed and exe = 0 for all e ∈ E(R),
then for all e, f ∈ E(R), exf = e(e∇f)x(e∇f)f = e0f = 0 and the second statement follows. £
Theorem 6.4.10 If R is an idempotent-dominated and idempotent-closed ring, then the
ring R/KR is the maximal abelian image of R. It is also idempotent-covered with
E(R/KR) ≅ E(R)/D. Conversely, if R is an idempotent-dominated ring for which R/KR is abelian,
then R is idempotent-closed.
(In general, if R is any ring with an ideal K such that RKR = {0}, then R is idempotent-closed if
and only if R/K is idempotent-closed, in which case E(R/K) is a homomorphic image of E(R) with
both sharing a common maximal lattice image.)
Proof. The quotient ring R/K is automatically idempotent-dominated. Suppose x + K ∈ E(R/K)
so that x + K = xn + K for all n ≥ 1. Then x2 = x + k for some k ∈ K. From this we get
x4 = x3 + xkx = x3, and hence x6 = x3 so that x + K = e + K for some e ∈ E(R). Thus all
idempotents in R/K come from idempotents in R, making E(R/K) multiplicative. By Lemma
6.4.5, ef − fe ∈ K for all e, f ∈ E(R), making E(R/K) commutative. Being idempotent-dominated,
this forces R/K to be idempotent-covered. Indeed, given any idempotent-dominated abelian ring
S, for any x = e1x1e1 + · · · + enxnen ∈ S, upon setting f = e1∇e2∇ . . . ∇en we have f ≥ all ei so
that fxf = f.
Next, let α: R → A be a homomorphism onto a ring A having only central idempotents. A
is automatically idempotent-dominated and hence idempotent-covered so that KA = {0}. Given
k ∈ K, since α is surjective, α(k) is in KA so that α(k) = 0. Thus K ⊆ ker(α) and the maximality of
the abelian image R/K follows.
The converse follows from the remark following the theorem. So let R be a ring with an
ideal K such that RKR = {0}, and let e, f ∈ E(R) be given. If R/K is idempotent-closed, then at
least efef = ef + k for some k ∈ K. But then
efef = e(efef)f = e(ef + k)f = ef + ekf = ef,
so that R is idempotent-closed. Likewise, if R is idempotent-closed, then so is R/K by the above
argument for R/K with E(R/K) again a homomorphic image of E(R). Given idempotent closure,
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