Near-Rings with Noncommutative Addition and Two-Sided Distributivity

[lpetrich]

TL;DR

All that I have found on near-rings features one-sided distributivity. Why not two-sided? I post some results that I have derived for the two-sided case, like lack of a multiplicative identity for nontrivial multiplication.

As an introduction, an abstract-algebra ring generalizes integers with addition and multiplication. The multiplication operation is distributive over the addition operation, this multiplication forms a semigroup over set R of its elements, and this addition forms an abelian group over R. But can addition be made noncommutative? Near-ring - Wikipedia describes what one finds when one does so and makes the distributive property one-sided. But what happens if one makes it two-sided? I have derived some results.

From two-sided distributivity,
Additive identity 0: a*0 = 0*a = 0 for all a in R -- multiplicative zero / annihilator / absorber
Additive inverse -(): (-a)*b = a*(-b) = -(a*b) for all a,b in R

(a+b)*(c+d) - take left, then right distributivity and subtract right, then left distributivity. One finds:

a*c + b*d = b*d + a*c for all a,b,c,d in R

In short, R*R is commutative. It includes the additive identity and it is closed under additive inversion. What happens when multiplication has an identity? Set a = b = 1. Then c + d = d + c and the additive group is commutative. So for that group to be noncommutative, multiplication must not have an identity.

Consider fixing one multiplicand and varying the other one over R. Thus,
a*R = (some abelian subgroup of the additive group)
R*a = (some abelian subgroup of the additive group, not necessarily the same one)
These subgroups may differ for different a's; a = 0 gives a*R = R*a = {0}.

Both multiplications are group homomorphisms, and for nonabelian +, the only trivial one is to {0}. Thus, for nonzero products, the additive group must have a nontrivial composition series with an abelian quotient group in it. Thus, the additive group must not be simple, though it need not be solvable.

The quaternionic icosahedral group is not possible, because its center is {I,-I} with quotient group (icosahedral group), a simple group.

However, the reflection icosahedral group is possible, at least if one ignores multiplicative associativity, because it is {I,-I} * (icosahedral group) and that can have quotient group Z2, from the {I,-I} part.

 

[lpetrich]

This suggests a simple special case that we can examine. The additive group has a subgroup with index 2, with a quotient group with size 2. It thus has one coset. Call the subgroup members "even" and the coset members "odd".

Let R*R be some order-2 additive subgroup of R: {0,a}, where a+a = 0. Here, a can be either even or odd.

Then, using the previous result of (result subgroup) ~ (quotient group), and also using the same decomposition for multiplication on both sides,

(even)*(even) = (even)*(odd) = (odd)*(even) = 0 and (odd)*(odd) = a

Let us check multiplicative associativity: (a*b)*c ?= a*(b*c) for all a,b,c in R.

If any of the multiplicands is even, then multiplying by it gives 0, and we do not learn anything further. So we try all odd.

((odd)*(odd))*(odd) = a*(odd) = (a even) 0, (a odd) a
(odd)*((odd)*(odd)) = (odd)*a = (a even) 0, (a odd) a

Thus satisfying associativity.

 

[fresh_42]

TL;DR Summary: All that I have found on near-rings features one-sided distributivity. Why not two-sided?

We need one-sided distributivity in case there is a $1$ as proven on nLab:
https://ncatlab.org/nlab/show/near-ring

So, two-sided distributivity makes only sense if there is no unit which strongly constrains the set of possible examples. So either allow a two-sided distributivity or allow units. People have chosen units.

 

[lpetrich]

So, two-sided distributivity makes only sense if there is no unit which strongly constrains the set of possible examples. So either allow a two-sided distributivity or allow units. People have chosen units.

That's what I showed earlier. A ring without unity is sometimes called a "rng" ("rung"): Rng (algebra) - Wikipedia -- "ring" without "i" (identity).

Let's now consider some further features. Consider multiplication x*a for all x in R for some a. That is a homomorphism of (R,+) onto some abelian subgroup of it, M(a). The kernel of that homomorphism, K(a) is

K(a) . a = 0

K(a) is a normal subgroup of (R,+), and M(a) is isomorphic to (R,+)/K(a).

Taking the intersection over all a in R, we find a kernel K with a quotient group that is the union of all (R,+)/K(a) for a in R. This implies that R*R is a subgroup of R.

Thus, for x tagged with quotient-group element i, x(i)*a = b(i,a) with i in (R,+)/K

This kernel is most properly the left kernel, and repeating this operation on a*x instead of x*a gives a right kernel and a right quotient group. These may be different from their corresponding left ones.

Thus, x(i)*x(j) = b(i,j)

where i is in the left quotient group and j in the right quotient group. Strictly speaking, the elements must be tagged for both groups:

x(i1,j1) * x(i2,j2) = b(i1,j2)

b in turn will be tagged by their quotient-group elements: ix = ix(i1,j2), jx = jx(i1,j2)

Associativity becomes
(x(i1,j1) * x(i2,j2)) * x(i3,j3) = b(i1,j2) * x(i3,j3) = b(ix(i1,j2),j3)
x(i1,j1) * (x(i2,j2) * x(i3,j3)) = x(i1,j1) * b(i2,j3) = b(i1,jx(i2,j3))

So ix( ix(i1,j2), j3) = ix( i1, jx(i2,j3) ) and iy( ix(i1,j2), j3) = iy( i1, jx(i2,j3) ) .

For the same kernel on both sides, we get some simplification: x(i,j) = x(i) from j = i.

x(i1) * x(i2) = b(i1,i2) = b(ix(i1,i2))
ix( ix(i1,i2), i3 ) = ix( i1, ix(i2,i3) )
So ix must be associative.

 

[lpetrich]

The combined tagging may be expressed as x = {i,j} where i is the left-side quotient-group element and j is the right-side quotient-group element.

For the product of x(xtag) and y(ytag), the result's tag is B(xtag,ytag) = Bval(xtag(1),ytag(2))

From the associativity of the product, B(B(x,y),z) = B(x,B(y,z)) -- B is thus associative, and with the tag values, it forms a semigroup.

The tag values themselves need not span the entire range of combinations of left-side and right-side quotient-group elements. Left-side = right-side is a possible solution, for instance.

I must note that this tagging does not fix the actual product values, only specifying what cosets that they are in.

In my even-odd example, the near-ring's elements have tags either {e,e} or {o,o}, which we may shorten to e and o. B(e,e) = B(e,o) = B(o,e) = e from products (even)*(even) = (even)*(odd) = (odd)*(even) = 0, and B(o,o) = e or o, from (odd)*(odd) = b where b is either 0 or any element with additive order 2, either even or odd.