condition on a near ring to be a ring
Every ring is a nearring. The converse^{} is true only when additional conditions are imposed on the nearring.
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 nearring 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 $ab:=a+(b)$.

5.
$ab=0$ 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
$rs$  $=r(b+a)$  $\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{substitution}$  
$=r+(1)(b+a)$  $\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{by Fact}\text{7}\text{above}$  
$=r+((1)b+(1)a)$  $\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{by left distributivity}$  
$=r+(b+(a))$  $\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{by Fact}\text{7}\text{above}$  
$=(a+b)+(b+(a))$  $\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{substitution}$  
$=((a+b)+(b))+(a)$  $\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{additive associativity}$  
$=(a+(b+(b))+(a)$  $\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{additive associativity}$  
$=(a+0)+(a)$  $\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}b\text{is the additive inverse of}b$  
$=a+(a)$  $\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}0\text{is the additive identity}$  
$=0$  $\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}}\text{same reason as above}$ 
Therefore, $a+b=r=s=b+a$ by Fact 5 above. ∎
Title  condition on a near ring to be a ring 

Canonical name  ConditionOnANearRingToBeARing 
Date of creation  20130322 17:19:54 
Last modified on  20130322 17:19:54 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  14 
Author  CWoo (3771) 
Entry type  Theorem 
Classification  msc 2000 
Classification  msc 1600 
Classification  msc 1300 
Related topic  UnitalRing 