Page 242 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 242
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Here a(a + k)a = a + aka = a and (a + k)a(a + k) = (a + ka)(a + ak) = a + k. Thus a D (a + k)
making E(A) is a lattice section of E(S). It is easily seen that any element (a + ka)(a + ak) is
idempotent whenever a ∈ E(A) and k ∈ K. £
Thus the observations of Example 1 generalize to all idempotent-closed and dominated
rings R such that E(R) has a lattice section. For rings satisfying the chain conditions of the next
section, this is the case. (The latter include all finite-dimension matrix ring examples.) E(R) also
has a lattice section if it has a maximal D-class M, since for any e ∈ M, {f ∈ E(R)⎪f ≤ e} is a
lattice section.
While KR generally exceeds ann(R), this is not the whole story. KR also contains two
related canonical ideals, the left and right annihilator ideals of R:
annL(R) = {x ∈ R⎪xy = 0 for all y ∈ R} and annR(R) = {y ∈ R⎪xy = 0 for all x ∈ R}.
It happens that annL(R) + annR(R) = KR for all idempotent-dominated and idempotent-closed
rings. We prove this when E(R) is bounded with a maximal D-class. The general case follows
from the next proposition and Theorems 6.5.1 and 6.5.3 below.
An idempotent-dominated and closed ring R is bounded if E(R) has a maximal D-class,
M, consisting of all m in E(R) such that m∇e∇m = m and eme = e for all e ∈E(R). Such an m is
called maximal in E(R) and xmy = xy for all x, y in R. Every element x ∈R is a sum of elements
miximi where each mi is in M
Proposition 6.4.12 Let R be a bounded idempotent-dominated and closed ring and let m
be maximal in E(R). Then as an additive group, KR = KRm ⊕ ann(R) ⊕ mKR where
KRm = {km⎪k ∈ KR} and mKR = {mk⎪k ∈ KR}. In particular,
annR(R) = KRm ⊕ ann(R) and annL(R) = ann(R) ⊕ mKR.
Finally, µ: R → mRm defined by µ(x) = mxm is a ring homomorphism onto the abelian subring
mRm, with kernel KR.
Proof. Setting K = KR, the identity k = km + (k – km − mk) + mk gives K = Km + ann(R) +
mK. Let that k, k′ ∈ K and a ∈ ann(R) be such that km + a + mk′ = 0. Then
mk′ = m(km + a + mk′) = m0 = 0
and similarly km = 0, leaving a = 0 also. Thus the sum is direct. Next let x ∈annL(R). Being in
K, x has the form km + a + mk′. Thus 0 = xm = km, so that x = a + mk′, giving
annL(R) ⊆ ann(R) ⊕ mK. The reverse inclusion is trivial. That annR(R) = Km ⊕ ann(R) is seen
in similar fashion. Since xmy = xy in R, the final statement is clear. Thus as additive subgroups,
R = mRm ⊕ Km ⊕ ann(R) ⊕ mK. £
240
Here a(a + k)a = a + aka = a and (a + k)a(a + k) = (a + ka)(a + ak) = a + k. Thus a D (a + k)
making E(A) is a lattice section of E(S). It is easily seen that any element (a + ka)(a + ak) is
idempotent whenever a ∈ E(A) and k ∈ K. £
Thus the observations of Example 1 generalize to all idempotent-closed and dominated
rings R such that E(R) has a lattice section. For rings satisfying the chain conditions of the next
section, this is the case. (The latter include all finite-dimension matrix ring examples.) E(R) also
has a lattice section if it has a maximal D-class M, since for any e ∈ M, {f ∈ E(R)⎪f ≤ e} is a
lattice section.
While KR generally exceeds ann(R), this is not the whole story. KR also contains two
related canonical ideals, the left and right annihilator ideals of R:
annL(R) = {x ∈ R⎪xy = 0 for all y ∈ R} and annR(R) = {y ∈ R⎪xy = 0 for all x ∈ R}.
It happens that annL(R) + annR(R) = KR for all idempotent-dominated and idempotent-closed
rings. We prove this when E(R) is bounded with a maximal D-class. The general case follows
from the next proposition and Theorems 6.5.1 and 6.5.3 below.
An idempotent-dominated and closed ring R is bounded if E(R) has a maximal D-class,
M, consisting of all m in E(R) such that m∇e∇m = m and eme = e for all e ∈E(R). Such an m is
called maximal in E(R) and xmy = xy for all x, y in R. Every element x ∈R is a sum of elements
miximi where each mi is in M
Proposition 6.4.12 Let R be a bounded idempotent-dominated and closed ring and let m
be maximal in E(R). Then as an additive group, KR = KRm ⊕ ann(R) ⊕ mKR where
KRm = {km⎪k ∈ KR} and mKR = {mk⎪k ∈ KR}. In particular,
annR(R) = KRm ⊕ ann(R) and annL(R) = ann(R) ⊕ mKR.
Finally, µ: R → mRm defined by µ(x) = mxm is a ring homomorphism onto the abelian subring
mRm, with kernel KR.
Proof. Setting K = KR, the identity k = km + (k – km − mk) + mk gives K = Km + ann(R) +
mK. Let that k, k′ ∈ K and a ∈ ann(R) be such that km + a + mk′ = 0. Then
mk′ = m(km + a + mk′) = m0 = 0
and similarly km = 0, leaving a = 0 also. Thus the sum is direct. Next let x ∈annL(R). Being in
K, x has the form km + a + mk′. Thus 0 = xm = km, so that x = a + mk′, giving
annL(R) ⊆ ann(R) ⊕ mK. The reverse inclusion is trivial. That annR(R) = Km ⊕ ann(R) is seen
in similar fashion. Since xmy = xy in R, the final statement is clear. Thus as additive subgroups,
R = mRm ⊕ Km ⊕ ann(R) ⊕ mK. £
240
