Page 234 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 234
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
(i) any regular band S in a ring that satisfies C must satisfy (I) and (II), and
(ii) for all e, f ∈ S, the regular band Sʹ generated from S ∪ {esf} also satisfies C.
Clearly:
Theorem 6.3.2. Any regular band of idempotents in a ring that satisfies a s-inductive
condition C will generate a s-band (its s-closure) in that ring. £
We consider three conditions that are s-inductive (in this parlance). We give new proofs
based on the concept of s-induction. Since the first two cases are already known, we omit their
elementary proofs based on s-induction.
Corollary 6.3.3. Left [right] regularity is a s-inductive condition. £
Corollary 6.3.4. Normality is a s-inductive condition. £
Recall that a band S is totally pre-ordered if for all e, f ∈S, either e ≺ f or f ≺ e, that is,
either efe = e or fef = f. The following nice result is due to Karin Cvetko-Vah [2005a].
Theorem 6.3.5. Being totally pre-ordered is a s-inductive condition. Thus every
maximal totally pre-ordered regular band in a ring is a s-band.
Proof. In verifying (I), we use the fact that ea(ef)bf multiplies out to eaebf + eafbf – eaefbf.
There are two cases to consider.
Case 1: the ≺-maximum is e or f. Say e. Then eaebf + eafbf – eaefbf reduces to
eabf + eafbf – eafbf = eabf. The subcase for f is similar.
Case 2: the ≺-maximum is a or b. Say a. Here eaebf + eafbf – eaefbf reduces to
ebf + efbf – efbf = ebf = eabf. The subcase for b is similar.
For (II), first suppose that e ≺ f, so that e = efe. Then both sides of (II) reduce to 0:
a(e – ef)abc(f – ef)c = a(e – ef)fabc(f – ef)c = a0abc(f – ef)c = 0
and
a(e – ef)b(f – ef)c = a(e – ef)fb(f – ef)c = a0b(f – ef)c = 0.
Likewise, this occurs when f ≺ e, so that (II) follows.
Hence S ∪ {esf} generates a regular band Sʹ in R. Is Sʹ also totally pre-ordered?
Suppose that e ≺ f in S. (The f ≺ e case is similar.) Here esf = f + fe – fef. Thus f(esf)f = f
while (esf) f (esf) = f (esf) = f (f + fe – fef) = esf. Since esf D f in Sʹ, every element uʹ in
Sʹ must be D-equivalent to some element u in S. Since S is totally pre-ordered, so is Sʹ. £
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(i) any regular band S in a ring that satisfies C must satisfy (I) and (II), and
(ii) for all e, f ∈ S, the regular band Sʹ generated from S ∪ {esf} also satisfies C.
Clearly:
Theorem 6.3.2. Any regular band of idempotents in a ring that satisfies a s-inductive
condition C will generate a s-band (its s-closure) in that ring. £
We consider three conditions that are s-inductive (in this parlance). We give new proofs
based on the concept of s-induction. Since the first two cases are already known, we omit their
elementary proofs based on s-induction.
Corollary 6.3.3. Left [right] regularity is a s-inductive condition. £
Corollary 6.3.4. Normality is a s-inductive condition. £
Recall that a band S is totally pre-ordered if for all e, f ∈S, either e ≺ f or f ≺ e, that is,
either efe = e or fef = f. The following nice result is due to Karin Cvetko-Vah [2005a].
Theorem 6.3.5. Being totally pre-ordered is a s-inductive condition. Thus every
maximal totally pre-ordered regular band in a ring is a s-band.
Proof. In verifying (I), we use the fact that ea(ef)bf multiplies out to eaebf + eafbf – eaefbf.
There are two cases to consider.
Case 1: the ≺-maximum is e or f. Say e. Then eaebf + eafbf – eaefbf reduces to
eabf + eafbf – eafbf = eabf. The subcase for f is similar.
Case 2: the ≺-maximum is a or b. Say a. Here eaebf + eafbf – eaefbf reduces to
ebf + efbf – efbf = ebf = eabf. The subcase for b is similar.
For (II), first suppose that e ≺ f, so that e = efe. Then both sides of (II) reduce to 0:
a(e – ef)abc(f – ef)c = a(e – ef)fabc(f – ef)c = a0abc(f – ef)c = 0
and
a(e – ef)b(f – ef)c = a(e – ef)fb(f – ef)c = a0b(f – ef)c = 0.
Likewise, this occurs when f ≺ e, so that (II) follows.
Hence S ∪ {esf} generates a regular band Sʹ in R. Is Sʹ also totally pre-ordered?
Suppose that e ≺ f in S. (The f ≺ e case is similar.) Here esf = f + fe – fef. Thus f(esf)f = f
while (esf) f (esf) = f (esf) = f (f + fe – fef) = esf. Since esf D f in Sʹ, every element uʹ in
Sʹ must be D-equivalent to some element u in S. Since S is totally pre-ordered, so is Sʹ. £
232
