Page 252 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
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athan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
⎡0 0 0⎤ ⎡ 0 a ab ⎤
Example 6.5.1 continued for n =3, i = j = k = 1, e = ⎢ 0 1 0 ⎥ and f = ⎢ 0 1 b ⎥ .
⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎦⎥
Here eRe = ⎪⎩⎪⎨⎧⎡⎣⎢⎢⎢ 0 0 0 ⎤ d ∈ F ⎬⎫⎪ , fRf = ⎧⎨⎪⎩⎪⎢⎣⎡⎢⎢ 0 ad abb ⎤ d ∈ F ⎬⎪⎫ and the isomorphism x → fxf
0 d 0 ⎥ ⎭⎪ 0 d db ⎥ ⎭⎪
0 0 0 ⎥ 0 0 0 ⎥
⎦⎥ ⎦⎥
⎡0 0 0 ⎤ ⎡ 0 ad abb ⎤
of Theorem 6.5.12 (ii), sends ⎢ 0 d 0 ⎥ in eRe to ⎢ 0 d ⎥ in fRf. £
0 0 ⎥⎦ ⎢ 0 0 db ⎥
⎣⎢ 0 ⎢⎣
0 ⎥⎦
How close is this class of A[S]-rings to the class of all rectangular rings? We begin our
answer with the following canonical co-representation:
Proposition 6.5.13 Let R be a rectangular ring. If A = fRf for a fixed f ∈ M = M(R), then
A is a ring with identity 1 = f such that E(A) = {0, f}; moreover the map β: A[M] → R defined by
β(∑aiei) = ∑eiaiei in R is a homomorphism onto R that is bijective between the copy {1e⎪e ∈ M}
of M in A[M] and M in R, and also between subrings Ae in A[M] and their images eAe in R.
Proof. By Theorem 6.5.12 (ii) and (iv), R = ∑e ∈M eRe with all summands being isomorphic to A.
Thus β is an additive epimorphism that is bijective where stated. That it preserves multiplication
follows from Theorem 6.5.12 (iii) and distribution. £
Thus, all rectangular rings essentially arise as rings of the form A[S] as above and certain
homomorphic images of these rings. The entire situation is put more precisely as follows.
Theorem 6.5.14 Given a ring A with identity such that E(A) = {0, 1} and a rectangular
band S, then A[S] is a rectangular ring. Moreover given any ideal K of A[S] such that K ⊆ KA[S],
the quotient ring A[S]/K is also rectangular. Conversely, to within isomorphism every
rectangular ring is obtained in this fashion.
Proof. The first assertion is Theorem 6.5.11. The second comes from Theorem 6.4.14. To see
the converse note that for the map β above, if β(∑ aiei) = ∑ eiaiei = 0 in R, then in fRf = A,
∑ ai = ∑ faif = f(∑ ai)f = f0f = 0. Thus ker(β) ⊆ KA[M] by Theorem 6.5.11. The converse
now follows by Theorem 6.4.14. £
Example 6.5.2 Let A = &2 and let S be the rectangular band determined by the array:
a Rb
L L e.g. a∧d = b and a∧d = c.
c Rd
Setting s = a+b+c+d, the sixteen elements of &2[S] are arrayed in the following diagram.
x 0 a b c d a+b a+c a+d .
s−x s b+c+d a+c+d a+b+d a+b+c c+d b+d b+c
250
⎡0 0 0⎤ ⎡ 0 a ab ⎤
Example 6.5.1 continued for n =3, i = j = k = 1, e = ⎢ 0 1 0 ⎥ and f = ⎢ 0 1 b ⎥ .
⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎦⎥
Here eRe = ⎪⎩⎪⎨⎧⎡⎣⎢⎢⎢ 0 0 0 ⎤ d ∈ F ⎬⎫⎪ , fRf = ⎧⎨⎪⎩⎪⎢⎣⎡⎢⎢ 0 ad abb ⎤ d ∈ F ⎬⎪⎫ and the isomorphism x → fxf
0 d 0 ⎥ ⎭⎪ 0 d db ⎥ ⎭⎪
0 0 0 ⎥ 0 0 0 ⎥
⎦⎥ ⎦⎥
⎡0 0 0 ⎤ ⎡ 0 ad abb ⎤
of Theorem 6.5.12 (ii), sends ⎢ 0 d 0 ⎥ in eRe to ⎢ 0 d ⎥ in fRf. £
0 0 ⎥⎦ ⎢ 0 0 db ⎥
⎣⎢ 0 ⎢⎣
0 ⎥⎦
How close is this class of A[S]-rings to the class of all rectangular rings? We begin our
answer with the following canonical co-representation:
Proposition 6.5.13 Let R be a rectangular ring. If A = fRf for a fixed f ∈ M = M(R), then
A is a ring with identity 1 = f such that E(A) = {0, f}; moreover the map β: A[M] → R defined by
β(∑aiei) = ∑eiaiei in R is a homomorphism onto R that is bijective between the copy {1e⎪e ∈ M}
of M in A[M] and M in R, and also between subrings Ae in A[M] and their images eAe in R.
Proof. By Theorem 6.5.12 (ii) and (iv), R = ∑e ∈M eRe with all summands being isomorphic to A.
Thus β is an additive epimorphism that is bijective where stated. That it preserves multiplication
follows from Theorem 6.5.12 (iii) and distribution. £
Thus, all rectangular rings essentially arise as rings of the form A[S] as above and certain
homomorphic images of these rings. The entire situation is put more precisely as follows.
Theorem 6.5.14 Given a ring A with identity such that E(A) = {0, 1} and a rectangular
band S, then A[S] is a rectangular ring. Moreover given any ideal K of A[S] such that K ⊆ KA[S],
the quotient ring A[S]/K is also rectangular. Conversely, to within isomorphism every
rectangular ring is obtained in this fashion.
Proof. The first assertion is Theorem 6.5.11. The second comes from Theorem 6.4.14. To see
the converse note that for the map β above, if β(∑ aiei) = ∑ eiaiei = 0 in R, then in fRf = A,
∑ ai = ∑ faif = f(∑ ai)f = f0f = 0. Thus ker(β) ⊆ KA[M] by Theorem 6.5.11. The converse
now follows by Theorem 6.4.14. £
Example 6.5.2 Let A = &2 and let S be the rectangular band determined by the array:
a Rb
L L e.g. a∧d = b and a∧d = c.
c Rd
Setting s = a+b+c+d, the sixteen elements of &2[S] are arrayed in the following diagram.
x 0 a b c d a+b a+c a+d .
s−x s b+c+d a+c+d a+b+d a+b+c c+d b+d b+c
250
