DHARMALGEBRAIC STRUCTURES 95
For example, ring (I, +, •) is an integral domain but ring (2I, +, •) is not an integral
domain because it is not the ring with unity.
Also from the previous example ring (Y, + 6 , × 6 ) is not an integral domain because, (Y, + 6 ,
× 6 ) is not a ring with unity. Although this a commutative ring and not a ring with zero divisors.
Invertible element
Let (X, +, •) be ring and an element x ∈ X, then x is said to be invertible element if there exists
an element y ∈ X i.e., x • y = 1 (multiplicative identity).
For example, consider a ring (I, +, •) of integers. Take an element 2 from I, now find an
element y in I, i.e., 2. y = 1. It is only possible when y = ½ ∉ I. Hence, element 2 is not invertible
in this ring. Consider another element 1 ∈ I, then 1 • y = 1 • y = 1 ⇒ I so 1 is invertible.
Similarly, – 1 is also invertible. Therefore, [– 1, 1] is the set of invertible elements for the ring
(I, +, •).
Consider another example of ring of reals i.e., (R, +, •). In the set R all elements are
invertible because, let a ∈ R then there exists an element b ∈ R i.e., a • b = 1 ⇒ b = 1/a ∈ R.
Boolean ring
A ring (X, +, •) is called a Boolean ring if, for all x ∈ X, we have x^2 = x (Law of Idempotent]. Form
the definition of Boolean ring if every element is idempotent then following conditions hold,
(i)x + x = 0 for all x ∈ X
(ii)x + y = 0 ⇒ x = y for x, y ∈ X
(iii) Ring X is commutative
(iv) Ring X with more than two elements contains divisors of zero.
Proof
(i) Since, x^2 = x ⇒ (x + x)^2 = x + x [by replacing x by (x + x)]
LHS ⇒ (x + x) (x + x)
⇒ x(x + x) + x(x + x) [Distributive law]
⇒ x^2 + x^2 + x^2 + x^2
⇒ x + x + x + x= x + x [∴ x^2 = x]
Hence by cancellation ⇒ x + x = 0
[Such that every element of X is additive inverse to its own (– x = x)]
(ii) Since, x + y = 0 ⇒ x + y = x + x [∴ x + x = 0 from condition (i)]
⇒ y = x [left cancellation law]
(iii) Let x, y ∈ X so we have
x + y = (x + y)^2 = (x + y) (x + y)
⇒ x(x + y) + y(x + y) [Distributive law]
⇒ x^2 + x y + y x + y^2
So, x + y = x + x y + y x + y [∴ x^2 = x & y^2 = y]
By cancellation, we obtain
x y + y x = 0
⇒ x y = – y x = y x [∴ – y = y self inverse]
⇒ x y = y x
Therefore, ring is a commutative ring.