Page 241 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
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VI: Skew Lattices in Rings
if e > f in E(R), then e + K ≠ f + K in E(R/K), for otherwise e = f + k for some k ∈ K so that
e = e(f + k)e = efe = f in R. This forces images of distinct D-classes in E(R) to remain distinct in
E(R/K), so that E(R)/D ≅ E(R/K)/D. £
In general, ann(R) ⊆ KR with the inclusion often proper; but thanks to Theorems 6.4.8
and 6.4.10, for an idempotent-closed and dominated ring R, ann(R) = KR precisely when both
ideals vanish and R idempotent-covered with central idempotents.
When E(R) has a lattice section, the ideal KR has a natural near-complement in the
ring R.
Theorem 6.4.11 Let R be idempotent-dominated and idempotent-closed. If E(R) has a
lattice section E0, then setting A = {x ∈ S⎪∃e ∈ E0, x = exe}, we have the following:
i) A is an idempotent-covered abelian subring of R.
ii) As additive groups, R = A ⊕ KR.
iii) The natural epimorphism R → R/KR induces a ring isomorphism A ≅ R/ KR.
Conversely, let S be any ring with an abelian subring A and an ideal K such that SKS = {0} and
as additive groups, S = A ⊕ K so that as rings, S/K ≅ A. Then S is idempotent-closed and E(A) is
a lattice section of E(S); moreover, for all a ∈ E(A), the D-class Da in E(S) consists of all
elements of the form (a + ka)(a + ak) with k ∈K.
Proof. Suppose exe = x and fyf = y for e, f ∈ E0. Setting g = e∇f = eof in E0, g ≥ e, f so that
gxg = x and gyg = y also. This gives g(x ± y)g = gxg ± gyg = x ± y and g(xy)g = (gx)(yg) = xy.
Thus A must be an idempotent-covered subring for which E(A) is multiplicative. By Theorem
6.4.4, (i) follows. Suppose that we are given x = e1x1e1 + · · · + enxnen ∈ R. For each ei in E(R)
let fi ∈E0 be such that ei D fi and set y = f1x1f1 + · · · + fnxnfn ∈ A. We claim that x − y is in KR.
We need only show that each eixiei − fixifi ∈ KR. But due to Lemma 6.4.5(ii), for all u, v in R:
u(eixiei − fixifi)v = ueixieiv − ufixifiv = ueixieiv − ueixieiv = 0.
Hence x − y ∈ KR so that x = y + (x − y) ∈ A + KR. Thus R = A + KR and clearly A ∩ KR = {0}
so that (ii) follows. From (i) and (ii), (iii) follows.
Conversely, that S is idempotent-closed follows from the remark after Theorem 6.4.10.
Given a ∈ A and k ∈ K, (a + k)2 = a2 + (ka + ak + k2) ∈ A⊕K. Thus a + k is idempotent if and
only if a2 = a in A and k = ka + ak + k2 in K. The latter gives k2 = kak so that when idempotent,
a + k = (a + k)2 = (a + ka)(a + ak).
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if e > f in E(R), then e + K ≠ f + K in E(R/K), for otherwise e = f + k for some k ∈ K so that
e = e(f + k)e = efe = f in R. This forces images of distinct D-classes in E(R) to remain distinct in
E(R/K), so that E(R)/D ≅ E(R/K)/D. £
In general, ann(R) ⊆ KR with the inclusion often proper; but thanks to Theorems 6.4.8
and 6.4.10, for an idempotent-closed and dominated ring R, ann(R) = KR precisely when both
ideals vanish and R idempotent-covered with central idempotents.
When E(R) has a lattice section, the ideal KR has a natural near-complement in the
ring R.
Theorem 6.4.11 Let R be idempotent-dominated and idempotent-closed. If E(R) has a
lattice section E0, then setting A = {x ∈ S⎪∃e ∈ E0, x = exe}, we have the following:
i) A is an idempotent-covered abelian subring of R.
ii) As additive groups, R = A ⊕ KR.
iii) The natural epimorphism R → R/KR induces a ring isomorphism A ≅ R/ KR.
Conversely, let S be any ring with an abelian subring A and an ideal K such that SKS = {0} and
as additive groups, S = A ⊕ K so that as rings, S/K ≅ A. Then S is idempotent-closed and E(A) is
a lattice section of E(S); moreover, for all a ∈ E(A), the D-class Da in E(S) consists of all
elements of the form (a + ka)(a + ak) with k ∈K.
Proof. Suppose exe = x and fyf = y for e, f ∈ E0. Setting g = e∇f = eof in E0, g ≥ e, f so that
gxg = x and gyg = y also. This gives g(x ± y)g = gxg ± gyg = x ± y and g(xy)g = (gx)(yg) = xy.
Thus A must be an idempotent-covered subring for which E(A) is multiplicative. By Theorem
6.4.4, (i) follows. Suppose that we are given x = e1x1e1 + · · · + enxnen ∈ R. For each ei in E(R)
let fi ∈E0 be such that ei D fi and set y = f1x1f1 + · · · + fnxnfn ∈ A. We claim that x − y is in KR.
We need only show that each eixiei − fixifi ∈ KR. But due to Lemma 6.4.5(ii), for all u, v in R:
u(eixiei − fixifi)v = ueixieiv − ufixifiv = ueixieiv − ueixieiv = 0.
Hence x − y ∈ KR so that x = y + (x − y) ∈ A + KR. Thus R = A + KR and clearly A ∩ KR = {0}
so that (ii) follows. From (i) and (ii), (iii) follows.
Conversely, that S is idempotent-closed follows from the remark after Theorem 6.4.10.
Given a ∈ A and k ∈ K, (a + k)2 = a2 + (ka + ak + k2) ∈ A⊕K. Thus a + k is idempotent if and
only if a2 = a in A and k = ka + ak + k2 in K. The latter gives k2 = kak so that when idempotent,
a + k = (a + k)2 = (a + ka)(a + ak).
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