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[parent] condition on a near ring to be a ring (Theorem)

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'$ are additive identities, then $ 0=0+0'=0'$.
  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$. $ \qedsymbol$
  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. $ \qedsymbol$
We are now in the position to prove the theorem.
Proof. Set $ r=a+b$ and $ s=b+a$. Then
$\displaystyle r-s$ $\displaystyle =r-(b+a)$    substitution    
  $\displaystyle =r+(-1)(b+a)$    by Fact 7 above    
  $\displaystyle = r+((-1)b+(-1)a)$    by left distributivity    
  $\displaystyle = r+(-b+(-a))$    by Fact 7 above    
  $\displaystyle = (a+b)+(-b+(-a))$    substitution    
  $\displaystyle = ((a+b)+(-b))+(-a)$    additive associativity    
  $\displaystyle = (a+(b+(-b))+(-a)$    additive associativity    
  $\displaystyle = (a+0)+(-a)$ $\displaystyle \quad \qquad -b$    is the additive inverse of $\displaystyle b$    
  $\displaystyle = a+(-a)$ $\displaystyle \quad \qquad 0$    is the additive identity    
  $\displaystyle = 0$    same reason as above    

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



"condition on a near ring to be a ring" is owned by CWoo. [ full author list (2) ]
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See Also: unital ring


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alternative proof of condition on a near ring to be a ring (Proof) by Wkbj79
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Cross-references: associativity, left distributivity, proof, iff, inverse, identity, additive, distributive, left distributes over, multiplicative identity, converse, near-ring, ring
There is 1 reference to this entry.

This is version 11 of condition on a near ring to be a ring, born on 2007-06-28, modified 2007-12-15.
Object id is 9684, canonical name is ConditionOnANearRingToBeARing.
Accessed 659 times total.

Classification:
AMS MSC: 13-00 (Commutative rings and algebras :: General reference works )
 16-00 (Associative rings and algebras :: General reference works )
 20-00 (Group theory and generalizations :: General reference works )
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additive identity, additive inverse by Wkbj79 on 2007-06-30 05:17:39
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
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