# 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. 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. 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. 3.

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

4. 4.

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

5. 5.

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

6. 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. 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 ConditionOnANearRingToBeARing 2013-03-22 17:19:54 2013-03-22 17:19:54 CWoo (3771) CWoo (3771) 14 CWoo (3771) Theorem msc 20-00 msc 16-00 msc 13-00 UnitalRing