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# 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. 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. ∎

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## Comments

## additive identity, additive inverse

The terms "additive identity" and "additive inverse" currently do not link properly. This should be fixed, and there are several ways that this can be accomplished:

1. Declare that some entry defines these terms and file an addendum to that entry if necessary. The candidates are "ring" and "near ring".

2. Declare that "uniqueness of additive identity in a ring" and "uniqueness of additive inverse in a ring" define these terms respectively and file addenda to these entries if necessary. (In light of the fact that an additive identity and additive inverses exist in near rings, perhaps these two entries can be generalized.)

3. Define these terms in a separate entry (either together or one for each).

Or maybe there is an even better option that has not come to my mind.

Any thoughts?

Warren

## Re: additive identity, additive inverse

I like the first option, since an additive identity and an additive inverse of an element are defined without requiring that they be unique.

Better yet, you can define left and right (additive) identities and inverses. In this more general setting, you may end up with multiple identities and inverses.

## Re: additive identity, additive inverse

> Better yet, you can define left and right (additive) identities and inverses. In this more general setting, you may end up with multiple identities and inverses.

Good point. A set does not need to be a ring (or even a near ring) to have an addition operation defined on it. On the other hand, I am unsure about applications of having a noncommutative addition operation, especially if the set also has a multiplication operation defined on it. Do people study such sets?

## Re: additive identity, additive inverse

Wkbj79 writes:

> I am unsure about applications of having a

> noncommutative addition operation, especially

> if the set also has a multiplication operation

> defined on it. Do people study such sets?

The obvious example (other than near-rings) would be the ordinal numbers. (This is a proper class rather than a set, but there are subsets closed under addition and multiplication.)