Near-Ring

A near ring is a non-abelian analog of a ring. Explicitly, a H-Set $R$ equipped with an addition $+:R\to R\to R$ and multiplication $\cdot:R\to R\to R$ is a right near ring when

• $(R,+)$ is a (not necessarily Abelian) Group
• $(R,\cdot)$ is a Monoid
• $(x+y)\cdot z=x\cdot z+y\cdot z$

Likewise, a $(R,+,\cdot)$ is a left near ring if multiplication distributes over addition on the left.

Note that some authors only require $(R,\cdot)$ to be a Semigroup, but this restriction seems unnatural, as the canonical example is unital.

Examples

• The set of all functions $G\to G$ for a (potentially non-abelian) group $G$ forms a right near-ring under pointwise multiplication and composition.

  	$\displaystyle(f+g)\circ h$	$\displaystyle=\lambda x.\;f(h(x))+g(h(x))$
  		$\displaystyle=(f\circ h)+(g\circ h)$

  Note that if $f$ is a Group Homomorphism, then we have a left distributivity law.

  	$\displaystyle f\circ(g+h)$	$\displaystyle=\lambda x.\;f(g(x)+h(x))$	$\displaystyle(\text{by defn.})$
  		$\displaystyle=\lambda x.\;f(g(x))+f(h(x))$	$\displaystyle(\text{$f$ is a group homomorphism})$
  		$\displaystyle=f\circ g+f\circ h$	$\displaystyle(\text{by defn.})$

  However, if $G$ is not abelian, then the set of Group Homomorphisms $G\to G$ does not form a group. Notably, the pointwise sum of two group homomorphisms need not be a group homomorphism.

• The Polynomial Ring $R[X]$ over a Commutative Ring $R$ forms a left near ring under addition and Substitution. This suggests that near rings are related to Duoidal Categories.

Properties

• In a right near ring, $0\cdot x=x$.

  	$\displaystyle 0\cdot x$	$\displaystyle=(x-x)\cdot x$
  		$\displaystyle=x^{2}-x^{2}$
  		$\displaystyle=0$

  Likewise, in a left near ring, $x\cdot 0=0$.

• In a right near ring, $(-x)\cdot y=-(x\cdot y)$

  As $(A,+)$ is a Monoid, Invertible Elements are unique, so it suffices to prove $\alpha+(-x)\cdot y=0$ and $-(x\cdot y)+\alpha=0$ for some $\alpha$.

  In particular, let $\alpha=(x\cdot y)$.

  We then have

  	$\displaystyle(x\cdot y)+(-x)\cdot y$	$\displaystyle=(x+-x)\cdot y$
  		$\displaystyle=0\cdot y$
  		$\displaystyle=0$

  and

  $\displaystyle-(x\cdot y)+x\cdot y=0$

  A dual argument lets us deduce that $x\cdot(-y)=-(x\cdot y)$ in a left near ring.