Research Article
Subdirect Representations in A*-Algebras
Department of Mathematics,Mekelle University, Mekelle, Ethiopia
LiveDNA: 91.513
P. Koteswara Rao
Department of Commerce, Nagarjuna University, Nagarjuna Nagar-522 510, A.P., India
In a drafted study Manes (1989) introduced the concept of Ada based on extending the If-Then-Else concept more on the basis of Boolean algebras and later Manes (1993) introduced an Ada (A, ∧, ∨, (-)1, (-)π, 0, 1, 2), based on C-algebras introduced by Guzman and Squier (1990). Koteswara Rao (1994) introduced the concept of A*-algebra, analogous to the Manes (1993) Adas. Venkateswa Rao (2000) studied extensively the concept of subdirect representations in A*-algebras based on the subdirect representations in rings (Lambek, 1966). Koteswara Rao and Venkateswa Rao (2003) studied about algebraic structures of Boolean Algebras and A*-algebras and obtained the methods of generating A*-algebras from Boolean algebras and vice-versa. Koteswara Rao and Venkateswa Rao (2004) introduced the concepts of Prime Ideals and Congruences in A*-Algebras and developed the general ideal theory. Koteswara Rao and Venkateswa Rao (2005) obtained a Cayley theorem for A*-Algebras. Koteswara Rao and Venkateswara Rao (2008) introduced the concepts of A*-Modules and If-Then-Else Algebras over A*-algebras. Koteswara Rao and Venkateswa Rao (2010) introduced Polynomials over A*-Algebras.
In this research, we investigate the subdirect representations in A*-algebras as 3 = {0, 1, 2} is the only subdirectly irreducible A*-algebra and every A*-algebra is a sub A*-algebra of product of copies of 3.
PRELIMINARIES
Definition 1
An algebra (A, ∧, *, (-)~, (-)π, 1) is an A*-algebra if it satisfies: For a, b, c ε A, (I) aπ ∨ (aπ)~ = 1, (aπ)π = aπ where a ∨ b = (a~ ∧ b~)~; (ii) aπ ∨ bπ = bπ ∨ aπ; (iii) (aπ ∨ bπ ) ∨ cπ = aπ ∨ (bπ ∨ cπ); (iv) (aπ ∧ bπ) ∨ (aπ ∧ (bπ)~) = aπ; (v) (a ∧ b)π = aπ ∧ bπ, (a ∧ b)# = a# ∨ b# where, a# = (aπ ∨ a~π)~ ; (vi) a~π = (aπ ∨ a#)~, a~# = a#; (vii) (a * b)π = aπ, (a * b)# = (aπ)~ ∧ (b~π)~; (viii) a = b if and only if aπ = bπ, a # = b#.
We write, 0 for 1~, 2 for 0 * 1.
Example
3 = {0, 1, 2} with the operations defined below is an A*-algebra.
Note
From 1.1 (I) to 1.1 (iv) and by Huntington's theorem (Birkhoff, 1948)) we see that, B(A) = {aπ/a ε A} is a Boolean algebra with ∧, ∨, (-)~, 0 and a ε B(A)⇒aπ = a. Since, 1, 0, (aπ)~ ε B(A), we have 1π = 1, 0π = 0, (aπ)~π = (aπ)~ and aπ ∧ a# = 0, a * 0 = aπ.
Sub Direct Representations
Lemma 1
In every A*-algebra A (i) x ∧ x = x, (ii) x ∧ (x~ ∨ y) = x ∧ y, (iii) x ∨ y = 0⇒x = 0, (iv) x ∨ y = 1⇒x ∨ x~ = 1.
Proof
From 1.1 (viii) we have (i), (ii). (iii): x = x ∨ 0 = x ∨ x ∨ y = x ∨ y = 0. Therefore x = 0. (iv): 1 = x ∨ y = x ∨ (x~ ∧ y) (from dual of (ii)) = (x ∨ x~) ∧ (x ∨ y) = (x ∨ x~) ∧ 1 = x ∨ x~. Therefore x ∨ x~ = 1.
Definition 2
For a ε A, define θa = {(x, y)/ aπ xπ = aπ yπ, aπ x# = aπ y#}
Lemma 2
θa is a congruence relation on A.
Proof
Clearly, θa is an equivalence relation on A.
(x, y) ε θa⇒aπ xπ = aπ yπ, aπ x# = aπ y#⇒aπ~ ∨ xπ~ = aπ~ ∨ yπ~ ⇒ aπ ∧ (aπ~ ∨ xπ~) = aπ ∧ (aπ~ ∨ yπ~)⇒aπ xπ~ = aπ yπ~. Similarly, aπ x#~ = aπ y#~⇒aπ xπ~ x#~ = aπ yπ~ y#~⇒aπ x~π = aπ y~π Clearly, aπ x~# = aπ y~#⇒(x~, y~) ε θa Clearly, xθa y⇒xπ θa yπ, x# θa y#. Suppose, x θay, bθac⇒aπ xπ = aπ yπ, aπ x# = aπ y#, aπ bπ = aπ cπ, aπ b# = aπ c# Clearly, (x ∧ b)θa (y ∧ c), aπ(x * b)π = aπ (y * c)π Now, we will show that, aπ(x * b)# = aπ(y * c)# aπ xπ = aπ yπ⇒aπ xπ~ = aπ yπ~ xθay⇒x~θa y~⇒aπ x~π = aπ y~π⇒aπ x~π~ = aπ y~π~.
Similarly, bθac⇒aπ b~π = aπ c~π, aπ b~π~ = aπ c~π~ aπ xπ~ b~π~ = aπ xπ~ aπ b~π~ = aπ y~π aπ c~π~ = aπ y~π c~π~ Therefore, aπ(x * b)# = aπ(y * c)# Hence, (x * b)θa (y * c). Therefore, θa is a congruence relation on A.
Lemma 3
θa ∩ θα~⊆ θa ∨ a~
Notation
ΔA = {(x, x) / x ε A}
Lemma 4
θa = ΔA ⇔ a = 1.
Proof
aπ(a ∧ x)π = aπ ∧ xπ. aπ(a ∧ x)# = aπ(a# ∨ x#) = aπ x# Therefore, (a ∧ x, x) ε θa⇒(a ∧ x, x) ε ΔA⇒a ∧ x = x Therefore, a ∧ x = x for every x ε A. So, a ∧ x = 1⇒a = 1.
Lemma 5
a θa~ b, b θb~ a⇒a = b
Proof
a = a ∨ (a~ ∧ a) (from 1.1 (viii)) = a ∨ (a~ ∧ b) = a ∨ b (from 1.1 (viii)). Similarly, b = a ∨ b. Therefore, a = b.
Lemma 6
Let A be an A*-algebra with 0, 1 (0 ≠ 1). Suppose that for any x ε A {0, 1}, x ∨ x~ ≠ 1. Define f: A→3 = {0, 1, 2} by f(1) = 1, f(0) = 0 and f(x) = 2 for all x ≠ 0, 1. Then f is an A*-homomorphism.
Proof
To show that, f (x~) = (f(x))~ for all x ε A For x = 0, 1, f(x~) = (f(x))~ Suppose, x ≠ 0, 1⇒x~ ≠ 0⇒f(x~) = 2.
Since, x ≠ 0, 1, f(x) = 2⇒(f(x))~ = 2 Therefore, f(x~) = (f(x))~.
Claim
f(x ∨ y) = f(x) ∨ f(y), for all x ε A
Case (i)
For x = 0, f(x ∨ y) = f(x) ∨ f(y), for all y ε A is clear
Case (ii)
For x = 1, f(x ∨ y) = f(x) ∨ f(y), for all y ε A
For, y = 0, 1 the result is clear. Suppose, y ≠ 0, 1⇒x ∨ y ≠ 0, 1. If y ≠ 0, 1⇒x ∨ y ≠ 0, 1. (Since, (x ∨ y) = 0⇒x = 0, y = 0 ; (x ∨ y) = 1⇒y ∨ y~ = 1) So, 1 ∨ y = 0, 1. So, f(1 ∨ y) = 2 = 1 ∨ 2 = f(1) ∨ f(y) Therefore, f(1 ∨ y) = f(1) ∨ f(y)
Case (iii)
For x ≠ 0, 1⇒x ∨ y ≠ 0, 1 f(x ∨ y) = 2 = 2 ∨ f(y) = f(x) ∨ f(y).
Therefore, f(x ∨ y) = f(x) ∨ f(y), for all x ε A. To prove that, f(xπ) = (f(x)) π for all x ε A. If, x ≠ 0, 1, it is clear. Suppose, x ≠ 0, 1. Claim: xπ = 0 Since, x ≠ 0, 1, x ∨ x~ ≠ 1. xπ ∨ x~π = 1⇒xπ~ = x~π⇒x# = 0⇒x = xπ x ∨ x~ = xπ ∨ xπ~ = 1, a contradiction. Therefore, xπ ∨ x~π ≠ 1. Suppose, xπ ∨ x~π ≠ 0⇒xπ ∨ x~π ≠ 0, 1. (xπ ∨ x~π) ∨ (xπ ∨ x~π)~ ≠ 1, a contradiction. Therefore, xπ ∨ x~π = 0⇒xπ = 0 and x~π = 0 Since, xπ = 0, f(xπ) = f(0) = 0 = 2π = (f(x)) π Therefore, f(xπ) = (f(x))π, for all x ε A.
Claim 1
f(x#) = (f(x))#, for all x ε A f(x#) = f(xπ ∨ x~π)~ = [f(xπ ∨ x~π)]~ = [f(xπ) ∨ f(x~π)]~ = [(f(x))π ∨ (f(x~))π]~ = [(f(x)) π ∨ f(x)~π]~ = (f(x))# Therefore, f(x#) = (f(x))# for all x ε A.
Claim 2
For all x, y ε A, f(x * y) = f(x) * f(y). [f(x * y)]π = f((x * y)π) = f(xπ) = f(x)π But, [f(x) * f(y)]π = [f(x)]π Therefore, f(x * y)π = [f(x) * f(y)]π, for all x, y ε A. f(x * y)# = f((x * y)#) = f(xπ~ ∧ y~π~) = f(xπ~) ∧ f(y~π~) = f(x)π~ ∧ f(y)~ π~ = (f(x) * f(y))#. Therefore, f(x * y)# = (f(x) * f(y))# for all x, y ε A. Therefore, f(x * y) = f(x) * f(y) for all x, y ε A. Therefore, f is an A*-homomorphism.
Definition 3
An algebra A is called subdirectly irreducible if intersection of non-zero congruences is non-zero i.e.) ∩ θ≠ ΔA
Theorem 1
3 = {0, 1, 2} is the only subdirectly irreducible A*-algebra.
Proof
Suppose, A is a subdirectly irreducible A*-algebra. Let, a ε A {0, 1} Suppose, a ∨ a~ = 1 θa ∩ θa~ ⊆ θ1. θa ∩ θa~ ⊆ ΔA⇒θa ∩ θa~ = ΔA, a contradiction. (Because θa ≠ ΔA, θa~ ≠ ΔA from 2.6). a ε A-{0, 1}⇒a ∨ a~ ≠ 1 Let, θ = θa. Define, f: A →{0, 1, 2} by f(0) = 0, f(1) = 1, f(x) = 2 if x ≠ 0, 1. Therefore, f is a homomorphism by 2.8. Define, φ on A by x φ y ⇔ f(x) = f(y).
Then, φ is a congruence relation on A. Consider, φ ∩ θ Let, (x, y) ε φ ∩ θ⇒(x, y) ε φ and (x, y) ε θ⇒(x, y) ε φ, x = y (by 2.7) Therefore, φ ∩ θ = ΔA⇒φ = ΔA (since θ ≠ ΔA, A is subdirectly irreducible).
Let x, y ε A, f(x) = f(y)⇒x φ y⇒x = y (since φ = ΔA) Therefore, f(x) = f(y)⇒x = y Therefore, f: A →3 is injective.
Therefore, ker f = {0}⇒A ≅ f(A). But, f(A) is an A*-subalgebra of 3. So, f(A) = 3. Therefore, A ≅ 3.
Corollary 1
Every A*-algebra is a sub A*-algebra of a product of copies of 3
Proof
By 2.10 and a theorem of Birkhoff [1] every A*-algebra is a sub-direct product of copies of 3.
Corollary 2
In every A*-algebra x ∧ 0 = x ∧ x~.