INTRODUCTION
The study of lattice theory had been made by Birkhoff (1948).
In a draft paper, the equational theory of disjoint alternatives by Manes
(1989) introduced the concept of Ada (Algebra of disjoint alternatives)
(A, ∧, ∨, ()', ()_{π}, 0,1,2) which is however differ
from the definition of the Ada of E.G. Manes (1993)
later study Adas and the equational theory of ifthenelse. While the Ada of
the earlier draft seems to be based on extending the IfThenElse concept more
on the basis of Boolean Algebra and the later concept is based on Calgebra
(A, ∧, ∨,') introduced by Guzman and Squier (1990).
Koteswara (1994) first introduced the concept A*Algebra
(A, ∧, ∨,*, ()^{∼},()_{π}, 0,1,2) not only
studied the equivalence with Ada, Calgebra, Ada’s connection with 3 Ring,
Stone type representation but also introduced the concept of A*clone. Venkateswara
Rao and Koteswara Rao (2003) studied about Boolean algebras
and A*algebras and the methods of generating A*algebras from Boolean algebras
and viceversa. Venkateswara Rao and Koteswara Rao (2008)
studied about IfThenElse structure over A*algebra and also Venkateswara Rao
and Koteswara Rao (2004) studied about Prime ideals
of A*algebra. Venkateswara Rao and Koteswara Rao (2005)
obtained Cayley theorem for A*Algebras. Venkateswara Rao
(2000) introduced the concept of Pre A*algebra (A, ∧, ∨,*, ()^{∼})
as the variety generated by the 3element algebra A={0,1,2} which is an algebraic
form of three valued conditional logic. It was proved that the only sub directly
irreducible Pre A*algebra are either A or two element Boolean algebra B={0,1}.
Venkateswara Rao et al. (2009) generated Pre
A*algebras from Boolean algebras and defined congruence relation and Ternary
operation on Venkateswara Rao and Srinivasa Rao (2009)
defined a partial ordering on a Pre A*algebra A and the properties of A as
a poset are studied.
In this study we formulate the definitions of an ideal, prime ideal, maximal ideal of Pre A*algebra A and discuss certain examples. We also show that for any ideals I and J of a Pre A*algebra A, the ideal (I: J) is precisely equal to the intersection of all prime ideals containing J and not containing I. We also show that every maximal ideal is necessarily prime, while the converse is true for special cases only.
PRELIMINARIES
Definition 1: Jocobson (1994) Boolean algebra
is an algebra (B, ∨, ∧, ()' 0,1) with two binary operations, one unary
operation (called complementation) and two nullary operations which satisfies:
• 
(B, ∨, ∧) is a distributive lattice 

We can prove that x" = x,(x ∨ y)' = x' ∧ y', (x ∧
y)' = x' ∨ y', for all x, y ∈B. 

Here, we concentrate on the algebraic structure of Pre A*algebra
and state some results which will be used in the later text. 
Definition 2: Venkateswara Rao (2000): An algebra
(A, ∧, ∨, ()^{~}) where A is nonempty set with 1, ∧, ∨
are binary operations and ()^{∼} is a unary operation satisfying:
• 
x ∧ y = y ∧ x, ∀ x, y∈A 
• 
(x ∧ y)^{~} = x^{~} ∨ y^{~},
∀ x, y ∈A 
• 
x ∧ (y ∧ z) = (x ∧ y) ∧ z, ∀x, y, z
∈A 
• 
x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z), ∀x,
y, z ∈A 
• 
x ∧ y = x ∧ (x^{~} ∨ y), ∀x, y,
z ∈A is called a Pre A*algebra 
Example (Venkateswara Rao, 2000): 3 = {0, 1, 2}
with operations ∧, ∨, ()^{~} defined below is a Pre A*algebra.
Note: The elements 0, 1, 2 in the above example satisfy the following laws:
• 
(b) 1 ∧ x = x for all x ∈ 3 
• 
0 ∨ x = x for all x ∈ 3 
• 
2 ∧ x = 2 ∨ x = 2 for all x ∈ 3 
Example: 2 = {0, 1} with operations ∧, ∨, ()^{~} defined
below is a Pre A*algebra.
Note
• 
(2, ∨, ∧, ()^{~}) is a Boolean algebra. So,
every Boolean algebra is a Pre A* algebra 
• 
The identities 1.1(a) and 1.1(d) imply that the varieties
of Pre A*algebras satisfies all the dual statements of 1.1(b) to 1.1(g) 
Note: Let, A be a Pre A*algebra then A is Boolean algebra iff x∨(x∧y)
= x, x∧ (x∨ y) = x (absorption laws holds).
Lemma 1: Venkateswara Rao and Srinivasa Rao (2009).
Every Pre A*algebra satisfies the following laws.
• 
(x∨x^{~})∧y = (x∧y) ∨ (x^{~}∧y) 
• 
(x∨y)∧z = (x∧z)∨(x^{~}∧y∧z) 
Definition 3: (Venkateswara Rao and Srinivasa Rao, 2009):
Let, A be a Pre A*algebra. An element x ∈A is called central element of
A if x∨x^{~} = 1 and the set {x ∈A/x∨x^{~} = 1} of
all central elements of A is called the centre of A and it is denoted by B (A).
Note that if A is a Pre A*algebra with 1 then 1, 0∈B (A). If the centre
of Pre A*algebra coincides with {0, 1} then we say that A has trivial centre.
Theorem 1: Venkateswara Rao and Srinivasa Rao (2009).
Let, A be a Pre A*algebra with 1, then B (A) is a Boolean algebra with the
induced operations ∧, ∨, ()^{~}.
Lemma 2: Venkateswara Rao and Srinivasa Rao (2009).
Let, A be a Pre A*algebra with 1:
• 
If y ∈B(A) then x∧x^{~}∧y = x∧x^{~},
∀ x∈A 
• 
x∧(x∨ y) = x∨(x∧y) = x if and only if x, yB∈(A) 
IDEALS OF PRE A*ALGEBRA
Definition 4: A nonempty subset U of a Pre A*algebra A is said to be an ideal of A if the following hold:
• 
a ∈U ⇒ x∧a∈U for each x∈A 
Example: All the ideals of Pre A*algebra A={0, 1, 2} are I_{1 }={2}
I_{2}={0, 2} and A itself.
Now we give some examples of Pre A*algebras and collect all the ideals of these.
Example: Let, G={a_{1}, a_{2}, a_{3}, a_{4}, a_{5}} where a_{1 }=(1, 2), a_{2 }= (0, 2), a_{3 }= (2, 1), a_{4} = (2, 0), a_{5 } = (2, 2). Then G is a Pre A*algebra (a sub algebra of AxA) under the point wise operation. This algebra (G, ∧, ∨, ()^{~}) is a Pre A*algebra without 1.All the ideals of G are”
I_{1}={a_{5}}
I_{2} 
= 
{a_{2}, a_{5}} 
I_{3} 
= 
{a_{4}, a_{5}} 
I_{4} 
= 
{a_{1}, a_{2}, a_{5}} 
I_{5} 
= 
{a_{2}, a_{4}, a_{5}} 
I_{6} 
= 
{a_{3}, a_{4}, a_{5}} 
I_{7} 
= 
{a_{1}, a_{2}, a_{4}, a_{5}} 
I_{8} 
= 
{a_{2}, a_{3}, a_{4}, a_{5}} 
I_{9} 
= 
G 
Example: Let, H = {b_{1}, b_{2}, b_{3}, b_{4},
b_{5}, b_{6}, b_{7}} where b_{1 }=(1, 2), b_{2
} = (0, 2), b_{3 }= (2, 1), b_{4} = (2, 0), b_{5 }
= (2, 2), b_{6 }=(1, 1), b_{7}=(0, 0). Then H is a Pre A*algebra
(a sub algebra of AxA) under the point wise operation. This algebra (H, ∧,
∨, ()^{~}) is a Pre A*algebra with 1 and 1= b_{6}. All
the ideals of H are:
I_{1} 
= 
{b_{5}} 
I_{2} 
= 
{b_{2}, b_{5}} 
I_{3} 
= 
{b_{4}, b_{5}} 
I_{4} 
= 
{b_{1}, b_{2}, b_{5}} 
I_{5} 
= 
{b_{3}, b_{4}, b_{5}} 
I_{6} 
= 
{b_{1}, b_{2}, b_{4}, b_{5}} 
I_{7} 
= 
{b_{2}, b_{3}, b_{4}, b_{5}} 
I_{8} 
= 
{b_{2},b_{4},b_{5}, b_{7}} 
I_{9} 
= 
{b_{1},b_{2}, b_{3}, b_{4},b_{5}} 
I_{10} 
= 
{b_{1},b_{2,} b_{4},b_{5}, b_{7}} 
I_{11} 
= 
{b_{1}, b_{2},b_{3}, b_{4}, b_{5},
b_{7}} 
I_{12} 
= 
{b_{2},b_{3}, b_{4}, b_{5}, b_{7}} 
I_{13} 
= 
H 
Example: Let, F= AxA = {c_{1}, c_{2}, c_{3},
c_{4}, c_{5}, c_{6}, c_{7}, c_{8}, c_{9}}
where c_{1 }=(1, 2), c_{2 }= (0, 2), c_{3 } = (2, 1),
c_{4} = (2, 0), c_{5 }= (2, 2), c_{6 }=(1, 1), c_{7}=(0,
0), c_{8}=(1, 0), c_{9}=(0, 1). Then F is a Pre A*algebra under
the point wise operation.
This algebra (F, ∧, ∨, ()^{~}) is a Pre A*algebra with 1 and 1= c_{6}. The ideals of AxA need not be of the form I_{1}xI_{2 }where, I_{1}, I_{2 }are ideals of A for such ideals of AxA are 9 in number (since, C has only 3 ideals namely {2}, {0, 2} and A) where, we are exhibiting 14 ideals in AxA.
All the ideals of F are:
I_{1} 
= 
{c_{5}} 
I_{2} 
= 
{c_{2}, c_{5}} 
I_{3} 
= 
{c_{4}, c_{5}} 
I_{4} 
= 
{c_{1}, c_{2}, c_{5}} 
I_{5} 
= 
{c_{3}, c_{4}, c_{5}} 
I_{6} 
= 
{c_{1}, c_{2}, c_{4}, c_{5}} 
I_{7} 
= 
{c_{2}, c_{3}, c_{4}, c_{5}} 
I_{8} 
= 
{c_{2}, c_{4}, c_{5}, c_{7}} 
I_{9} 
= 
{c_{1}, c_{2}, c_{3}, c_{4}, c_{5}} 
I_{10} 
= 
{c_{1}, c_{2}, c_{4}, c_{5}, c_{7}} 
I_{11} 
= 
{c_{1}, c_{2}, c_{3}, c_{4}, c_{5},
c_{7}} 
I_{12} 
= 
{c_{1}, c_{2}, c_{4}, c_{5}, c_{7},
c_{8}} 
I_{13} 
= 
{c_{2}, c_{3}, c_{4}, c_{5}, c_{7},
c_{9}} 
I_{14} 
= 
F 
Lemma 3: Let, A be a Pre A*algebra and p∈A. Then {x∧p /x∈A}
is the smallest ideal containing p.
Proof: Let, U={x∧p /x∈A}. If s, t∈U then s= x∧p and t= y∧p for some x, y∈A s∨t=(x∧p) ∨ (y∧p)= (x∧p∧p) ∨ (y∧p∧p) = [(x∧p)∨ (y∧p)]∧p =(s∨t) ∧ p∈U. Therefore, s∨t∈U. Let, x∧p∈U and z∈A. Then, z∧(x∧p) = (z∧x)∧p∈U (since, z, x∧A∈z⇒x∈A). Therefore, U is an ideal of A. Also, U is an ideal containing p (since, p=p∧p∈U). Let, I be the ideal containing p. Then, xp∧I for any x∈A and hence U⊆I. Thus, U is the smallest ideal containing p.
Definition 5: Let, A be a Pre A*algebra and p ∈A then {x∧p /x∈A} is called the Principal ideal generated by p and is denoted by < p >.
Now we give some examples of principal ideals of certain Pre A*algebras.
Example: Principal ideals of the Pre A*algebra A={0, 1, 2} are:
< 0 > = {0, 2}
< 1 > = {0, 1, 2}
< 2 > = {2}
Example: Principal ideals of the Pre A*algebra G={a_{1}, a_{2}, a_{3}, a_{4}, a_{5}} earlier are:
< a_{1 }> = {a_{1,}a_{2}, a_{5}}
< a_{2 }> ={a_{2}, a_{5}}
< a_{3 }> = {a_{3, }a_{4}, a_{5}}
< a_{4 }> = {a_{4}, a_{5}}
< a_{5 }> = {a_{5}}
Example: Principal ideals of the Pre A*algebra H ={b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}} earlier are:
< b_{1 }> = {b_{1}, b_{2}, b_{5}}
< b_{2 }> = {b_{2}, b_{5}}
< b_{3 }> = {b_{3}, b_{4}, b_{5}}
< b_{4 }> = {b_{4}, b_{5}}
< b_{5 }> = {b_{5}}
< b_{6 }> = H
< b_{7 }> = {b_{2}, b_{4}, b_{5}, b_{7}}
Example: Principal ideals of the Pre A*algebra F= {c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}, c_{7}, c_{8}, c_{9}} earlier are:
< c_{1 }> = {c_{1}, c_{2}, c_{5}}
< c_{2}> = {c_{2}, c_{5}}
< c_{3 }> = {c_{3}, c_{4}, c_{5}}
< c_{4}> ={c_{4}, c_{5}}
< c_{5 }> = {c_{5}}
< c_{6 }> = F
< c_{7}> ={c_{2}, c_{4}, c_{5}, c_{7}}
< c_{8 }> = {c_{1}, c_{2}, c_{4}, c_{5},
c_{7}, c_{8}}
< c_{9 }> = {c_{2}, c_{3}, c_{4}, c_{5},
c_{7}, c_{9}}
Lemma 4: Let, A be a Pre A*algebra and p, y∈A then y∈< p >⇔y = y∧p.
Proof: Suppose, y∈< p > ⇒y= x∧p for some x∈A. Now y∧p= (x∧p)∧p = x∧p = y.
Conversely suppose that y = y∧p. Then, y∧p∈< p >.Therefore y∈< p >.
Lemma 5: Let, A be a Pre A*algebra and p, q∈A then < p > = < q > if and only if p=q.
Proof: Suppose < p > = < q >. Clearly p∈< p >
= < q > and q∈< q > =< p >:
⇒p= p∧q and q=q∧p (by lemma 5)
⇒p = q
On the other hand, p=q ⇒ p∧q=p and q∧p=q:
⇒p ∈< q > and q ∈< p >
⇒< p > ⊆ < q > and < q > ⊆ < p >
⇒< p > = < q >
Lemma 6: Let A be a Pre A*algebra and p, q∈A. Then < p > ∩ < q > = < p∧q >.
Proof: Let, p, q∈A. Let, x∈< p∧q >⇒x=x∧ p∧q:
⇒x∧ p=x∧ p∧q∧ p=x∧ p∧q=x
⇒x∈< p >
Also, x∧q=x∧ p∧q∧q=x∧ p∧q=x:
⇒x∈< q >
⇒x∈< p > ∩ < q >
Therefore, <p∧q> ⊆ < p > ∩ < q >. Let, x∈< p >∩ < q >. Then, x∈< p > and x∈< q >:
⇒x=x∧ p and x=x∧q
⇒x=x∧q=(x∧ p)∧q= x∧ (p∧q)
⇒x∈< p∧q >

Therefore, < p > ∩ < q >⊆< p∧q
>. Thus, < p > ∩ < q > = < p∧q >. 
Note: We do not have < p > ∪ < q > = < p∧q >,
in general for example in the Pre A*algebra G={a_{1}, a_{2},
a_{3}, a_{4}, a_{5}} < a_{1 }> ∪ <
a_{3 }>={a_{1,}a_{2},a_{5}} {a_{3},
a_{4}, a_{5}}= G. But a_{1} ∨ a_{3=} a_{5
}and < a_{5 }> ≠ G.
Definition 6: For any subset S of a Pre A*algebra A and an ideal U, we define (S : U)={a∈A\ s∧a∈U for each s∈S}. If S={c} then we write it as (c : U)
Theorem 2: (S : U) is an ideal of A.
Proof: Let, a, b ∈ (S : U). Then s∧a, s∧b ∈ U for all s∈S. s∧(a∨b) = (s∧a) ∨ (s∧b) ∈ U (since, s∧a, s∧b∈U and U is ideal). Hence, a∨b∈ (S : U). Let, a ∈ (S : U), x∈A ⇒ a∧ s∈U for some s∈S. Then, (x∧a) ∧s=x∧(a ∧ s) ∈ U (since, a∧ s∈U, x∈A and U is ideal). Therefore, x∧a∈ (S : U). Thus, (S : U) is an Ideal of A.
Theorem 3: Let, S be a subset and U and W are the two ideals of Pre A*algebra A then, (S : U∩W)= (S : U) ∩ (S : W).
Proof: Let, x∈ A. Then, x∈ (S : U∩W) ⇔ x ∧s ∈
U∩W, for all s∈S.
Therefore, (S : U∩W) = (S : U) ∩(S : W).
Theorem 4: Let, S and T are be any two subsets of Pre A*algebra A and S ⊆ T then, (T : U) ⊆ (S :U).
Proof: Let, x∈(T : U) then x ∧t ∈ U for all t ∈ T and hence, x ∧s ∈ U for all s∈S (since S ⊆ T). So, that x ∈ (S : U). Therefore, (T : U) ⊆ (S : U).
Theorem 5: Let, S and T are be any two subsets of Pre A*algebra A then, (S ∪ T: U) = (S : U)∩ (T: U).
Proof: We know that S and T ⊆ S ∩T and hence by lemma 6. (S ∪ T: U) ⊆ (S : U) ∩ (T : U). Now let, x∧ (S : U) ∩ (T : U). Then, x ∧ s ∈ U and x ∧ t ∈ U for all s ∈ S and t ∈T. Therefore, x ∧a∈U for all a ∈ S ∪T.
⇒x∈(S ∪T: U)

Therefore, (S : U) ∩ (T : U) ⊆ (S ∪ T: U). Thus,
(S ∪ T: U) = (S : U) ∩ (T : U). 
Theorem 6: Let A be a Pre A*algebra S⊆A for any ideals U and W
of A < S >∩W⊆U ⇔W⊆ (S : U).
Proof: Suppose that W⊆ (S : U). Let, x∈< S >∩(S : U). Then:
and x∈(S : U) where, s_{i}∈S, y_{i}∈ A:
and x∧s_{i}∩ U for each i. Now x = x ∧ x:
Therefore, x ∈ U (since, x∧s_{i}∈ U). Therefore, <
S >∩ (S : U) ∩ U and hence < S >∩W⊆ U (since, W⊆
(S : U)). On the other hand W is a ideal of A such that < S >∩W⊆U.
Let, x∈W. Then, for any s ∈ S, x∧s∈< S >∩W (since
W and < S > are ideals).
Yx∧s∈U (by supposition)
Therefore, x ∈ (S : U). Thus, W⊆ (S : U).
Lemma 7: Let, S be any subset of a Pre A*algebra A. Then (< S > : U) = (S : U).
Proof: We know that S ⊆ < S > and therefore by theorem 4 (< S > : U) ⊆ (S : U). Let, x ∈(S:U).Then, x = x ∧s ∈ U for all s∈S. Let, t∈< S >. Then,
for some s_{i} ∈ S, y_{i} ∈ A. Now,

Then, x∧t∈ U. Thus, x∈(< S > : U). Therefore,
(S : U) ⊆(< S > : U) and hence (< S > : U) = (S : U) 
PRIME IDEALS OF PRE A*ALGEBRA
Definition 7: Let, A be a Pre A*algebra. A proper ideal of P of A is called a prime ideal if for any a, b∈A, a∧b∈P ⇒ either a ∈ P or b ∈ P.
Example: Consider the Pre A*algebra A={0, 1, 2}. In this {0, 2} is the only prime ideal.
Example: The prime ideals of the Pre A*algebra G={a_{1}, a_{2},
a_{3}, a_{4}, a_{5}} (see 2.3 example) are I_{4}={a_{1,}a_{2},a_{5}},
I_{6}={a_{3}, a_{4}, a_{5}},I_{7}={a_{1},
a_{2}, a_{4}, a_{5}},I_{8}={a_{2}, a_{3,}
a_{4}, a_{5}} where as I_{1}={a_{5}}, I_{2}={a_{2},
a_{5}}, I_{3}={a_{4}, a_{5}}, I_{5}={a_{2,
}a_{4}, a_{5}} are not prime ideals.
Example: Some of the prime ideals of the Pre A*algebra H= {b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}} (see 2.4 example) are, I_{10}={b_{1},b_{2,} b_{4},b_{5}, b_{7}}, I_{12}={b_{2},b_{3}, b_{4}, b_{5}, b_{7}} where as I_{8}={b_{2},b_{4},b_{5}, b_{7}}is an ideal but not a prime ideal.
Example: Some of the prime ideals of the Pre A*algebra F= {c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}, c_{7}, c_{8}, c_{9}} (see 2.5 example) are I_{12}={c_{1}, c_{2}, c_{4}, c_{5}, c_{7}, c_{8}}, I_{13}= {c_{2}, c_{3}, c_{4}, c_{5}, c_{7}, c_{9}} where as I_{8}={c_{2}, c_{4}, c_{5}, c_{7}} is an ideal but not a prime ideal.
Theorem 7: The following are equivalent for any proper ideal P of Pre A*algebra A
• 
For any ideal I and J of A, I ∩ J ⊆ P⇒ I ⊆
P or J ⊆ P 
• 
For any ideal I and J of A, P= I ∩ J ⇒ P=I or P=J 
Proof: (1) ⇒(2): Let, P be prime ideal and I and J are ideals of
A such that I ∩J ⊆P. Assume that I é P. Then there exist a∈
I such that a ∉P. Let, b ∈ J. Then, a∧b ∈ I ∩ J ⊆
P (since, I ∈J ⊆P). Since, P is prime, a∈P or b∈P. But a∉P
we have b∈P. Thus, J ⊆ P.
Suppose that I ∩ J=P ⇒ P=I or P=J for any ideals I and J of A. Let,
a and b ∈A such that a∧b∈P. Then, by lemma 7 < a > ∩
< b > = < a ∧ b > ⊆ P and hence (< a >∨ P) ∩
(< b >∨P) = (< a > ∩ < b >) ∨ P = P so that <
a > ∨ P =P or < b >∨P =P or equivalently < a >⊆P or
< b >⊆P and therefore a ∈ P or b ∈ P. Thus, P is prime
ideal.
Theorem 8: Let, I be an ideal of a Pre A*algebra A and a∈A\I. Then there exist a prime ideal P containing I and not containing a.
Proof: Let, a∉ I. Let = {J/J is an ideal of A, I ⊆ J and a ∉ J}. Clearly I∉. Therefore, is a nonempty and partially ordered set under set inclusion. By zorn’s lemma there exist a maximal member M in . Then clearly M is a proper ideal of A and I⊆ M.
We shall prove that M is a prime ideal of A. Let x, y∈A such that x ∉M and y ∉M. Let, M is properly contained in M ∨< x > and M ∨< y > and hence by maximality of M, we have a∈ M ∨< x > ∩ M ∨< y > =M ∨(< x > ∧ < y >) = M ∨(< x∧y >).
Therefore, x∧y∉M (for if x∧y∈M then M ∨ (< x > ∧ < y >) = M which is a contradiction, since a ∉ M). Therefore, M is prime ideal containing I and not containing a.
Corollary: For any ideal I of a Pre A*algebra A, I =∩{P / P is a prime ideal of A and I⊆P}.
Proof: If a∉ P then by theorem 8, there exist a prime ideal P containing I and not containing a and therefore, a:
Thus, I =∩ {P / P is a prime ideal of A and I⊆P}
Theorem 9: Let, I be an ideal of a Pre A*algebra A and S a nonempty subset of A which is closed under the operation ∧ and is disjoint with I. Then there exist a prime ideal P of A containing I and disjoint with S.
Proof: Let, I be an ideal of a Pre A*algebra A, S⊆A and S∩I=φ. Let, = {J/J is an ideal of A, I⊆J and J∩S=φ}.
Clearly I∈. Therefore, is a nonempty and partially ordered set under set inclusion. is clearly closed under unions of chains. Therefore, by zorn’s lemma there exist a maximal member M in . Then I ⊆ M and M ∩ S = φ.
We shall prove that M is prime ideal of A. Since, S is nonempty and M∩S = φ, M is proper ideal of A. Let, x, y∈A such that x∉M and y∉M.
Then M ∨< x > and M ∨< y > are ideals containing M properly. By the maximality of M, we get (M ∨ < x >) ∩ S ≠φ, (M ∨< y >) ∩ S ≠ φ.
Choose s ∈ (M ∨ < x >) ∩ S and t∈(M ∨< y >) ∩ S. Then, s ∧ t∈ S (since, S is closed under ∧ and s, t ∈ S) and s ∈ M ∨ < x > and t ∈ M ∨< y >.
Then, s ∧ t ∈ (M ∨< x >) ∩ (M∨< y >) = M ∨(< x > ∩ < y >)= M ∨(< x∧y >) and hence, s∧t∈ M ∨ (< x ∧ y >) ∩ S.
Then it follows that x∧y∉M (for, if x∧y∈M then M ∨ (< x∧y >) = M and hence s∧t∈M∩S, which is contradiction to the fact that M∩ S = φ)
Thus, M is the prime ideal of A, I⊆M and M∩S = φ.
Theorem 10: For any ideals I and J of Pre A*algebra A, (I: J) = ∩ {P / P is a prime ideal of A, J ⊆P and I éP}.
Proof: We have I∩(I: J) ⊆ J. If P is prime ideal containing J then I ∩ (I: J) ⊆ J ⊆ P and if in addition I é P then (I: J) ⊆ P. Therefore, (I: J) ⊆ ∩ {P/P is a prime ideal of A, J ⊆ P and I é P}. On the other hand, suppose x ∉ (I: J).
Then there exist a ∈ I such that x ∧ a ∉ J. Therefore, there exists prime ideal P such that J ⊆ P and x ∧ a ∉ P. Then, x ∉ P and a ∉ P (since, P is prime) and hence, I é P (since, a ∈ I and a ∉ P). Therefore, x ∉1 {P / P is a prime ideal of A, J⊆P and IéP} and hence, ∩{P / P is a prime ideal of A, J⊆P and I é P ⊆ (I: J). Thus, (I: J) = ∩ {P / P is a prime ideal of A, J ⊆ P and I éP}.
MAXIMAL IDEALS OF PRE A*ALGEBRA
Definition 8: Let A be a Pre A*algebra. A proper ideal of M of A is called a Maximal ideal of A if M is maximal among the proper ideals of A.
Lemma 8: Let A be a Pre A*algebra. Every maximal ideal in A is a prime ideal.
Proof: Let M is a maximal ideal of A. Then clearly M is a proper ideal of A. Suppose a, b∈A such that a ∉ M and b ∉M.
Then, M <
M ∪ {a} > = A; M <
M ∪ {b} > =A (since M is maximal). Then:
since, b ∈A we have b ∈ < M ∪{a}.
That is:
for some y_{i}∈A, x_{i} ∈ M ∪{a}. Now b=b ∧ b:
If x_{i} ∈ M then clearly b ∧ y_{i} ∧ x_{i}
∈ M. Since, b ∉ M and
it follows that b ∧ y_{i}∧a∉M for some y_{i }(since,
x_{i}∈ M ∪ {a} ⇒ x_{i} =M or x_{i} =a).
Therefore, b∧a∉M⇒a∧b∉M (since M is ideal).
Thus, M is prime ideal of A. The validity of the converse of the theorem 10 is not known. That is, are there prime ideals which are not maximal. In certain special cases, we know that every prime ideal is maximal, for example Boolean algebras. In the following we discuss another class of Pre A*algebra when every prime ideal is maximal.
Theorem 11: Let, A be a Pre A*algebra with 1.Suppose that for any x ∈A, there exists a smallest x_{0}∈B(A) such that x∧ x_{0 }= x and x_{0}^{~}∧x = 1. Then, every prime ideal of A is a maximal ideal.
Proof: Let, P is a prime ideal of A and Q be any ideal such that P Q.
Since, P Q
there exits x∈Q such that x ∉P.
Then there exists a smallest x_{0} ∈ B(A) such that x∧ x_{0 }= x and x_{0}^{~}∨x = 1. Therefore, x_{0}∉P (for if x_{0} ∈ P then x ∈ P, a contradiction).
Since, P is prime x_{0}^{~}∈ P⊆Q. Therefore both x and x_{0}^{~} belongs to Q and hence, 1= x_{0}^{~}∨x∈Q (since, Q is ideal).
Then a=1∧ a∈Q for all a∈A (since, 1∈Q, a∈ A and Q is ideal, 1∧ a ∈Q). Therefore, Q=A. Thus P is maximal ideal of A.
Definition 9: A prime ideal P of a Pre A*algebra A is said to be a minimal prime ideal of A if there is no prime ideal Q of A such that Q ⊂P.
Lemma 9: Let, A be a Pre A*algebra. Every prime ideal contains a minimal prime ideal.
Proof: Let, P be a minimal prime ideal of A. Let, = {Q / Q is a prime ideal of A, Q⊆P}. Then, is a nonempty (P∈) and T is a partially ordered set with respect to inclusion. If {Q_{α}}_{αεΔ} is a chain in then:
in .
Therefore every chain in T has a lower bound in . Thus, by zorn’s lemma has a minimal member say M. Then M is a prime ideal of A and M ⊆ P. Suppose there is a prime ideal R of A such that R ⊆ M. Then, R ⊆ M ⊆ P. Therefore, R ⊆ . By the minimality of M in we get that R=M. Hence, M is the minimal prime ideal contained in P.