Page 251 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 251
VI: Skew Lattices in Rings
Proof. Since A[S] = ∑s∈SAs and As = sA[S]s, the first assertion holds. We show that any f = f2 ≠ 0
in A[S] lies in MS. Let f = ∑aisi. Given s in S, with i and j representing the same index values we
have
fsf = (∑aisi)s(∑ajsj) = ∑aiajsissj = ∑aiajsisj = (∑aisi)(∑ajsj) = ff = f
and
sfs = s(∑aisi)s = ∑aissis = ∑ais = (∑ai)s.
Since (sfs)2 = sfsfs = sfs, ∑ai ∈ E(A). E(A) = {0, 1} by assumption. If ∑ai = 0, so that sfs = 0,
then also f = fsfsf = 0, contradicting f ≠ 0. This leaves ∑ai = 1, so that
f = fsf = (∑aisi)s(∑aisi) = (∑aisis)(∑aissi) ∈LSRS = MS.
Next, since for all a, b ∈ A and all s, t ∈ S, (as)(∑aisi)(bt) = [a(∑ai)b](st), the condition ∑ai = 0 is
necessary for ∑aisi to be in KA[S]. Since A[S] is additively generated from all a•s terms, it is also
sufficient. £
In the general rectangular case, letting M(R) denote E(R) \ {0}, we have:
Theorem 6.5.12 If ring R is a rectangular ring, then
i) Γ(R) = ∪e ∈M(R) eRe with (eRe)(fRf) = efRef for all e, f ∈ M(R).
ii) Given e, f ∈ M(R), the map χ: x → fxf defines a ring isomorphism of eRe with
fRf. Thus every element y in fRf is uniquely expressed as fxf for some x in eRe.
iii) Given y = fxf ∈ fRf and z = gx′g ∈ gRg, yz = (fg)(xx′)(fg) ∈ fgRfg.
iv) R = ∑ e ∈M(R) eRe with all summands being isomorphic subrings.
Proof. Lemma 6.4.7 gives the first equality in (i). The second equality in (i) follows from
(eRe)(fRf) = efeRefRfef ⊆ efRef = (efRe)f ⊆ (eRe)(fRf). To see (ii), note that χ is at least an
additive homomorphism from eRe to fRf. Let x, y ∈ eRe be given. Then
f(xy)f = f(xey)f = f(xefey)f = (fxef)(feyf) = (fxf)(fyf)
so that χ is a homomorphism of rings. Indeed it is an isomorphism with inverse isomorphism
given by y → eye from fRf back to eRe. (iii) follows from
yz = (fxf)(gx′g) = (fexef)(gex′eg) = fexex′eg = fgexx′efg = fgxx′fg.
The final assertion is now clear. £
249
Proof. Since A[S] = ∑s∈SAs and As = sA[S]s, the first assertion holds. We show that any f = f2 ≠ 0
in A[S] lies in MS. Let f = ∑aisi. Given s in S, with i and j representing the same index values we
have
fsf = (∑aisi)s(∑ajsj) = ∑aiajsissj = ∑aiajsisj = (∑aisi)(∑ajsj) = ff = f
and
sfs = s(∑aisi)s = ∑aissis = ∑ais = (∑ai)s.
Since (sfs)2 = sfsfs = sfs, ∑ai ∈ E(A). E(A) = {0, 1} by assumption. If ∑ai = 0, so that sfs = 0,
then also f = fsfsf = 0, contradicting f ≠ 0. This leaves ∑ai = 1, so that
f = fsf = (∑aisi)s(∑aisi) = (∑aisis)(∑aissi) ∈LSRS = MS.
Next, since for all a, b ∈ A and all s, t ∈ S, (as)(∑aisi)(bt) = [a(∑ai)b](st), the condition ∑ai = 0 is
necessary for ∑aisi to be in KA[S]. Since A[S] is additively generated from all a•s terms, it is also
sufficient. £
In the general rectangular case, letting M(R) denote E(R) \ {0}, we have:
Theorem 6.5.12 If ring R is a rectangular ring, then
i) Γ(R) = ∪e ∈M(R) eRe with (eRe)(fRf) = efRef for all e, f ∈ M(R).
ii) Given e, f ∈ M(R), the map χ: x → fxf defines a ring isomorphism of eRe with
fRf. Thus every element y in fRf is uniquely expressed as fxf for some x in eRe.
iii) Given y = fxf ∈ fRf and z = gx′g ∈ gRg, yz = (fg)(xx′)(fg) ∈ fgRfg.
iv) R = ∑ e ∈M(R) eRe with all summands being isomorphic subrings.
Proof. Lemma 6.4.7 gives the first equality in (i). The second equality in (i) follows from
(eRe)(fRf) = efeRefRfef ⊆ efRef = (efRe)f ⊆ (eRe)(fRf). To see (ii), note that χ is at least an
additive homomorphism from eRe to fRf. Let x, y ∈ eRe be given. Then
f(xy)f = f(xey)f = f(xefey)f = (fxef)(feyf) = (fxf)(fyf)
so that χ is a homomorphism of rings. Indeed it is an isomorphism with inverse isomorphism
given by y → eye from fRf back to eRe. (iii) follows from
yz = (fxf)(gx′g) = (fexef)(gex′eg) = fexex′eg = fgexx′efg = fgxx′fg.
The final assertion is now clear. £
249
