condition on a near ring to be a ring

Every ring is a near-ring. The converse is true only when additional conditions are imposed on the near-ring.

Theorem 1.

Let $(R,+,\cdot)$ be a near ring with a multiplicative identity $1$ such that the $\cdot$ also left distributes over $+$ ; that is, $c\cdot(a+b)=c\cdot a+c\cdot b$ . Then $R$ is a ring.

In short, a distributive near-ring with $1$ is a ring.

Before proving this, let us list and prove some general facts about a near ring:

1.

Every near ring has a unique additive identity: if both $0$ and $0^{\prime}$ are additive identities, then $0=0+0^{\prime}=0^{\prime}$ .
2.

Every element in a near ring has a unique additive inverse. The additive inverse of $a$ is denoted by $-a$ .

Proof.

If $b$ and $c$ are additive inverses of $a$ , then $b+a=0=a+c$ and $b=b+0=b+(a+c)=(b+a)+c=0+c=c$ . ∎
3.

$-(-a)=a$ , since $a$ is the (unique) additive inverse of $-a$ .
4.

There is no ambiguity in defining “subtraction” $-$ on a near ring $R$ by $a-b:=a+(-b)$ .
5.

$a-b=0$ iff $a=b$ , which is just the combination of the above three facts.
6.

If a near ring has a multiplicative identity, then it is unique. The proof is identical to the one given for the first Fact.
7.

If a near ring has a multiplicative identity $1$ , then $(-1)a=-a$ .

Proof.

$a+(-1)a=1a+(-1)a=(1+(-1))a=0a=0$ . Therefore $(-1)a=-a$ since $a$ has a unique additive inverse. ∎

We are now in the position to prove the theorem.

Proof.

Set $r=a+b$ and $s=b+a$ . Then

$\displaystyle r-s$	$\displaystyle=r-(b+a)$	$\displaystyle\quad\qquad\text{substitution}$
	$\displaystyle=r+(-1)(b+a)$	$\displaystyle\quad\qquad\text{by Fact \ref{w} above}$
	$\displaystyle=r+((-1)b+(-1)a)$	$\displaystyle\quad\qquad\text{by left distributivity}$
	$\displaystyle=r+(-b+(-a))$	$\displaystyle\quad\qquad\text{by Fact \ref{w} above}$
	$\displaystyle=(a+b)+(-b+(-a))$	$\displaystyle\quad\qquad\text{substitution}$
	$\displaystyle=((a+b)+(-b))+(-a)$	$\displaystyle\quad\qquad\text{additive associativity}$
	$\displaystyle=(a+(b+(-b))+(-a)$	$\displaystyle\quad\qquad\text{additive associativity}$
	$\displaystyle=(a+0)+(-a)$	$\displaystyle\quad\qquad-b\text{ is the additive inverse of }b$
	$\displaystyle=a+(-a)$	$\displaystyle\quad\qquad 0\text{ is the additive identity}$
	$\displaystyle=0$	$\displaystyle\quad\qquad\text{same reason as above}$

Therefore, $a+b=r=s=b+a$ by Fact 5 above. ∎

Title	condition on a near ring to be a ring
Canonical name	ConditionOnANearRingToBeARing
Date of creation	2013-03-22 17:19:54
Last modified on	2013-03-22 17:19:54
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	14
Author	CWoo (3771)
Entry type	Theorem
Classification	msc 20-00
Classification	msc 16-00
Classification	msc 13-00
Related topic	UnitalRing