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

$r-s$$=r-(b+a)$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{substitution}$

$=r+(-1)(b+a)$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{by Fact }\text{<span class="ltx\_text ltx\_ref\_tag">7</span>}\text{ above}$

$=r+((-1)b+(-1)a)$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{by left distributivity}$

$=r+(-b+(-a))$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{by Fact }\text{<span class="ltx\_text ltx\_ref\_tag">7</span>}\text{ above}$

$=(a+b)+(-b+(-a))$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{substitution}$

$=((a+b)+(-b))+(-a)$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{additive associativity}$

$=(a+(b+(-b))+(-a)$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{additive associativity}$

$=(a+0)+(-a)$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}-b\text{ is the additive inverse of }b$

$=a+(-a)$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}0\text{ is the additive identity}$

$=0$$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{same reason as above}$

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