Some Remars on regular near-ring

Boolean Near-Ring

A near-ring N is termed Boolean if every element a in N satisfies a 2 = a.

Regular Element

An element a in a near-ring is called regular if there exists x in N such that a = axa.

Center of a Near-Ring

The center of a near-ring N, denoted Z(N), consists of all elements that commute with every element in N under both addition and multiplication.

Main Theorems and Proof

3.1 Theorem: If 𝑁 is a regular near-ring then 𝑁 is not a division near-ring. Proof: (i) A Regular near-ring is one where for every element, there exist an element 𝑦 such that 𝑥 = 𝑥𝑦𝑥. (ii) A division near-ring is a near -ring with a multiplicative identity where every nonzero element has a multiplicative inverse. Above two properties are mutually exclusive, that means no near-ring possess both. Therefore a regular near-ring 𝑁 is not a division near-ring.

Theorem:

If 𝑁 is Boolean near-rings then 𝑁 is Regular near-ring. Proof: Let 𝑁 denote Boolean near-ring. Since 𝑁 is Boolean near-ring, then 𝑎. 𝑎 = 𝑎 ∀a∈ 𝑁, so we have 𝑎𝑎𝑎 = 𝑎𝑎 = 𝑎. Since every element 𝑁 is a Regular element, then 𝑁 is a Regular near-ring.

Theorem:

If 𝑁 is a division near-ring, then (𝑁, +) is abelian. Proof: The addition operation (+) within the division near-ring 𝑁 is commutative i. e. For any 𝑎, 𝑏 ∈ 𝑁, 𝑎 + 𝑏 = 𝑏 + 𝑎. This is a key property of division near-ring, while they don't necessarily have a commutative multiplication; their additive structure is always abelian.

Theorem:

If every element 𝑥 of a regular near-ring 𝑁 has multiplicative inverse, then the center of a regular near-ring is also regular.

Proof: Let 𝑍(𝑁) be a center of regular near-ring, then 𝑍(𝑁) is non empty set and 𝑍(𝑁) ⊆ 𝑁. To prove that 𝑍(𝑁) is a regular near-ring, we will first prove that (𝑍(𝑁), +) is an abelian group. Furthermore, 𝑥𝑟 + 𝑦𝑟 + 𝑧𝑟 = 𝑟𝑥 + 𝑟𝑦 + 𝑟𝑧, so that((𝑥 + 𝑦) + 𝑧)𝑟 = 𝑟((𝑥 + 𝑦) + 𝑧), with (𝑥 + 𝑦) + 𝑧 ∈ 𝑍(𝑁) and 𝑥 + (𝑦 + 𝑧) ∈ 𝑍(𝑁).

Hence 𝑥 + (𝑦 + 𝑧)=(𝑥 + 𝑦) + 𝑧 .  To show that 0 is an identity element of (𝑁), we need to prove 0 ∈ 𝑍(𝑁). Firstly let, 0 = 𝑟0 = 0𝑟, ∀ 𝑟 ∈ 𝑁, so that 0 ∈ 𝑍(𝑁) . Since 0 ∈ 𝑍(𝑁) and 𝑍(𝑁) ⊆ 𝑁, then the addition is also commutative. So we have 𝑥 + 0 and 0 + 𝑥, 𝑥 ∈ 𝑍(𝑁). Next we will prove that 0 is identity element in (𝑁) . Suppose 𝑥 ∈ 𝑍(𝑁), then 𝑥𝑟 = 𝑟𝑥, ∀ 𝑟 ∈ 𝑁. This proves that -𝑥 is an inverse of 𝑥 ∈ 𝑍(𝑁), and this means that every 𝑥 ∈ 𝑍(𝑁) has inverse -𝑥 ∈ 𝑍(𝑁).

(5 ) Suppose 𝑥, 𝑦 ∈ 𝑍(𝑁), so that 𝑥𝑟 = 𝑟𝑥 andyr = 𝑟𝑦, for every 𝑟 ∈ 𝑁, we get

It means that +𝑦 = 𝑦 + 𝑥, that is the addition in the Z(𝑁) is commutative.

From (1) to ( 5) we conclude that 𝑍(𝑁) is Abelian group. Furthermore, to show that (𝑍(𝑁), +, . ) is a Near-ring. We need to prove (𝑍(𝑁), . ) is a semi-group. The associative property is inherited from the associative property in the Near-ring. Let 𝑥, 𝑦 ∈ 𝑍(𝑁) and 𝑟 ∈ 𝑁. Since 𝑍(𝑁) 𝑁, then 𝑥, 𝑦 ∈ 𝑁. Since the multiplication is also associative in the Near-ring𝑁, we get (𝑥𝑦)𝑟 = 𝑥(𝑦𝑟) = 𝑥(𝑟𝑦) = (𝑥𝑟)𝑦 = (𝑟𝑥)𝑦 = 𝑟(𝑥𝑦).

Because of (𝑥𝑦)𝑟 = 𝑟 (𝑥𝑦), we get 𝑥, 𝑦 ∈ 𝑍(𝑁). Hence the multiplication is a binary operation in the (𝑁) . So we get a prove that (𝑍(𝑁), .) is a semigroup. As Left or Right Distributive property in 𝑍(𝑁) is given by Distributive property in Near-ring 𝑁, then we find that (𝑍(𝑁), . ) is a Near-ring. Now, it is remaining for us to show that every element 𝑥 ∈ 𝑍(𝑁)is a regular element. Assume that 𝑥 is an element of regular nearring 𝑁which have multiplicative inverse. If 𝑁 is a regular near-ring with center 𝑍(𝑁), then 𝑍(𝑁) 𝑁. For every element of 𝑥 ∈ 𝑍(𝑁) result in 𝑥 is also an element of 𝑁. Since 𝑁 is a regular near-ring, then there exist 𝑤 ∈ 𝑁 such that 𝑥𝑤𝑥 = 𝑥.

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