Page 250 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 250
athan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Lemma 6.5.10 Given A and S as above and s ∈ S, then in A[S]
LS = {∑aisi ⎪∑ai = 1 in A & ai ≠ 0 ⇒ si L s in S}
and
RS = {∑bjtj⎪∑bj = 1 in A & bj ≠ 0 ⇒ tj R s in S}
are the sets of idempotents respectively L-related or R-related to s in A[S]. Moreover,
MS = LSRS = {∑(aibj)(sitj)⎪∑ai = 1 = ∑bj in A with aibj ≠ 0 ⇒ si L s R tj in S}
is the maximal rectangular band in A[S] containing s and hence all of S.
Proof. Indeed, given x = ∑aisi where ∑ai = 1 in A and si L s in S if ai ≠ 0, one easily sees that
xs = x, sx = s and thus x2 = xsx = x. Conversely, if ∑aisi is an idempotent that is L-related to s,
then
s = s(∑aisi)s = ∑ai(ssis) = ∑ais = (∑ai)s so that ∑ai = 1.
Moreover,
∑aisi = (∑aisi)s = ∑ai(sis).
Since all sis are L-related to s in S, by the uniqueness of the representation, all si with nonzero
coefficients in ∑aisi are L-related to s in S. Thus LS and likewise RS are indeed as described.
Finally, for any rectangular band M, given e ∈ M one has Re = eM, Le = Me in M so that
M = MeM = LeRe. Thus we need only show that under multiplication MS is a rectangular band.
This follows from the easily verified identity in A[S]:
(∑aisi)(∑bjtj)(∑ckuk)(∑dmvm) = (∑aisi)(∑dmvm)
given ∑ai = ∑bj = 1 = ∑ck = ∑dm in A and si, tj uk and vm are in S. £
One can use the above “Inflation Lemma” to show that for nontrivial A and S, MS will
properly include S except in three cases: &2[S] for |S| = 2 (two cases) and &2[S] where S is the 4-
element rectangular band of Example 3 below. On the other hand, if S has an L- or R-class with
≥ 3 distinct elements a, b, c, then a new element in the inflated class is given by a − b + c. Or if
α ∈ A \ {0, 1}, then given a ≠ b in an L- or R-class of S, αa + (1 − α)b is a new element in the
inflated class. Now, that the status of S and MS in A[S] is decided, we turn to all of E(A[S]).
Theorem 6.5.11 A[S] is idempotent-dominated and E(A[S]) is the 0-rectangular band
MS ∪ {0} and KA[S] = {(∑aisi⎪∑ai = 0 in A}.
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Lemma 6.5.10 Given A and S as above and s ∈ S, then in A[S]
LS = {∑aisi ⎪∑ai = 1 in A & ai ≠ 0 ⇒ si L s in S}
and
RS = {∑bjtj⎪∑bj = 1 in A & bj ≠ 0 ⇒ tj R s in S}
are the sets of idempotents respectively L-related or R-related to s in A[S]. Moreover,
MS = LSRS = {∑(aibj)(sitj)⎪∑ai = 1 = ∑bj in A with aibj ≠ 0 ⇒ si L s R tj in S}
is the maximal rectangular band in A[S] containing s and hence all of S.
Proof. Indeed, given x = ∑aisi where ∑ai = 1 in A and si L s in S if ai ≠ 0, one easily sees that
xs = x, sx = s and thus x2 = xsx = x. Conversely, if ∑aisi is an idempotent that is L-related to s,
then
s = s(∑aisi)s = ∑ai(ssis) = ∑ais = (∑ai)s so that ∑ai = 1.
Moreover,
∑aisi = (∑aisi)s = ∑ai(sis).
Since all sis are L-related to s in S, by the uniqueness of the representation, all si with nonzero
coefficients in ∑aisi are L-related to s in S. Thus LS and likewise RS are indeed as described.
Finally, for any rectangular band M, given e ∈ M one has Re = eM, Le = Me in M so that
M = MeM = LeRe. Thus we need only show that under multiplication MS is a rectangular band.
This follows from the easily verified identity in A[S]:
(∑aisi)(∑bjtj)(∑ckuk)(∑dmvm) = (∑aisi)(∑dmvm)
given ∑ai = ∑bj = 1 = ∑ck = ∑dm in A and si, tj uk and vm are in S. £
One can use the above “Inflation Lemma” to show that for nontrivial A and S, MS will
properly include S except in three cases: &2[S] for |S| = 2 (two cases) and &2[S] where S is the 4-
element rectangular band of Example 3 below. On the other hand, if S has an L- or R-class with
≥ 3 distinct elements a, b, c, then a new element in the inflated class is given by a − b + c. Or if
α ∈ A \ {0, 1}, then given a ≠ b in an L- or R-class of S, αa + (1 − α)b is a new element in the
inflated class. Now, that the status of S and MS in A[S] is decided, we turn to all of E(A[S]).
Theorem 6.5.11 A[S] is idempotent-dominated and E(A[S]) is the 0-rectangular band
MS ∪ {0} and KA[S] = {(∑aisi⎪∑ai = 0 in A}.
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