Asian Journal of Current Engineering and Maths 2: 2 March – April (2013) 115 - 117. Contents lists available at www.innovativejournal.in ASIAN JOURNAL OF CURRENT ENGINEERING AND MATHS Journal homepage: http://www.innovativejournal.in/index.php/ajcem 115 EXTERNAL APPROACH OF IDEALS IN SUBTRACTION ALGEBRAS Rabah Kellil* College of Science at Zulfi Majmaah University; KSA ARTICLE INFO ABSTRACT Corresponding Author Rabah Kellil College of Science at Zulfi Majmaah University; KSA Key Words: subtraction algebra, prime ideal, irreducible ideal, subtraction algebra Morphism. In this paper we are inspired by the work of Young Bae Jun and Kyung Ho Kim. We introduce another definition of the notion of ideal of a subtraction algebra, which infers then a different notion of prime ideal. This notion leaves the principle that if an ideal contains an element it contains all the elements which are lower than it, while that proposed in the paper quoted previously if an element y is in the ideal and if the element x – y is in the ideal then x is also in the ideal. Certain results steel valid in our case. We think that the definitions which we introduced are less restrictive than those of Young Bae Jun and Kyung Ho Kim. We introduce besides the notion of morphism of a subtraction algebra and establish some relative characterizations of the kernels of such morphisms. 1991 Mathematics Subject Classification. 06A99, 06A06, 03E20,03E15. ©2013, AJCEM, All Right Reserved. INTRODUCTION Definition 1. An algebra (X;−) is a set endowed X with a single binary operation ” − ” is called a subtraction algebra if for all x; y; z ∈ X the followingconditions hold: 1) x − (y − x) = x, 2) x − (x − y) = y − (y − x), 3) (x − y) − z = (x − z) − y. The subtraction determines a partial order ” ≤ ” on X by a ≤ b ⇔a−b = 0; where 0 = a − a is an element that doesn’t depend on the choice of a ∈ X. The ordered set (X ; ≤ ) is a semi-Boolean algebra in the sense of [1], that is, it is a meet semilattice with zero 0 in which every interval [0; a] is a Boolean algebra with respect to induced order. Here a ∧ b = a − (a − b) and the complement of an element b ∈ [0; a] is a − b. Proposition 1.1. In a subtraction algebra we have the following 1) (x − y) − y = x − y; 2) x −0 = x and 0 − x = 0; 3) (x − y) − x = 0; 4) x − (x − y) ≤ y; 5) (x − y) − (y − x) = x − y; 6) x − (x − (x − y)) = x − y; 7) (x − y) − (z − y) ≤ x − z; 8) x ≤ y if and only if x = y − w for some w ∈ X; 9) x ≤ y implies x − z ≤ y − z and z − y ≤ z − x for all z ∈ X; 10) x; y ≤ z implies x − y = x ∧ (z − y); 11) (x ∧ y) − (x ∧ z) ≤ x ∧ (y − z); 12) (x − y) − z = (x − z) − (y − z). 2. Ideals of a subtraction algebra Definition 2. Let (X ; −) be a subtraction algebra. A nonempty subset I of X is called an ideal if x − y ∈ I; for every x ∈ I; y ∈ X and denoted by I  X. Remark. If I is an ideal then it is not empty so there exists at least an elementx ∈ I so 0= x – x is also in I. Proposition 2.1. If x ≤ y and y ∈ I, where I is an ideal, then 1) x = y − (y − x) 2) x ∈ I. Proof. One has ∀ x, y ∈ X; y − (y − x) = x − (x − y) but x − y = 0 ; so y − (y − x) = x −0 = x. But y ∈ I; then x ∈ I. Proposition 2.2. Let X be a subtraction algebra and I ⊂ X. Then the followingare equivalent: 1) ∀ x ∈ I; ∀ y ∈ X; x − y ∈ I. 2) If x ≤ y and y ∈ I then x ∈ I: Proof. 1) (1) ⇒(2) Let y ∈ I and x ∈ X such that x ≤ y. Hence x − y = 0. Then: x = x − (x − y) = y − (y − x) ∈ I by (1). 2) (2) ⇒(1) Let x ∈ I and y ∈ X. Since (x − y) − x = 0, (x − y) ≤ x. Hence by (2) x − y ∈ I. Theorem 2.3. Let I be an ideal of a subtraction algebra X. For any elementa ∈ X the set Ia = {x ∈X / a∧ x ∈ I} is an ideal of X containing I. Proof. 1) Let x ∈ Ia and y ∈ X, we have to prove that a ∧ (x − y) ∈ I. We have by the property 1.1 of the proposition 2.2,a ∧ (x−y)−a ∧ x ≤ a ∧ ((x−y)−x) and by the third property of the same proposition; a ∧ ((x − y) − x) = 0. But in this case a ∧ (x − y) − a ∧ x = 0 and a ∧ (x − y) ≤ a ∧ x. In the other hand a ∧ x ∈ I so by the proposition 2.5; a ∧ (x − y) ∈ I and Ia is an ideal of X. 2) Let now x ∈ I then a ∧ x ≤ x ∈ I so by the proposition 2.5; a ∧ x ∈ I and then x ∈ Ia.