# Types of rings satisfying certain properties

by ford jones   Last Updated October 09, 2019 14:20 PM

I was just wondering if there are rings that satisfy exactly five, but not all of the six ring axioms and a finite ring other than the trivial ring {0}. Just to make sure we’re on the same page, the axioms are described below: Let $$R$$ be a ring. Then $$R$$ satisfies the following:

1) Commutativity of addition: $$\forall a,b\in R, a+b \in R \Leftrightarrow b+a\in R$$.

3) Associativity of multiplication

4) Distributive property

5) Additive Identity ($$\forall a\in R, \exists “0”\in R (a+0)= a$$).

6) Additive Inverse ($$\forall a\in R, \exists “-a”\in R (a+(-a) = 0$$)

Note: the additive identity and additive inverse do not have to be $$0$$ and $$-a$$ respectively.

If we define addition and multiplication in the integers by the usual operations, then the set of integers is a ring.

I feel like it’s easy to find rings that don’t satisfy the distributive property:

If we define addition and multiplication as $$ab$$ and $$a+b$$, then the ring satisfies all the axioms except for distributivity.

As for a finite ring, I know that the integers $$modulo\space n$$, where $$n\in\mathbb{Z}$$, is a finite, unital ring.

Tags :

If we define addition and multiplication as 𝑎𝑏 and 𝑎+𝑏, then the ring satisfies all the axioms except for distributivity.

Good try, but not quite. If addition is defined as $$a\dotplus b=ab$$ where the thing on the right is the old product in a ring, then it won't be possible to have an $$\dotplus$$ inverse for $$0$$. The problem is that $$0$$ is absorbing with respect to $$\dotplus$$ (that is, $$0\dotplus b=0$$ for all $$b$$, so $$f(b)=0\dotplus b$$ is not 1-1) but in any abelian group the function $$f(b)=e+b$$ is 1-1, because of the existence of inverses. So it does not satisfy as many axioms as you thought.

## Examples

If you take a nontrivial finite abelian group $$G$$, the set of functions $$G\to G$$ satisfies all requirements except 4), because it doesn't distribute on one side. (I'm pretty sure nontrivial abelian suffices, but maybe there are some edge cases.) This is an example of a near-ring.

If you just take the nonnegative integers $$\mathbb N$$ with regular addition and multiplication, you get something that is only missing 6). This is an example of a semi-ring.

You can't have 6) without 5) because 6) is defined in terms of 5), so that one is not possible.

The cross product on $$\mathbb R^3$$, along with standard vector addition, provides an example of a nonassociative algebra, one which is only missing 3).

It seems you can demonstrate there is a near-ring that is distributive on both sides, which has a nonabelian group underlying the $$+$$ operation. That satisfies everything except 1).

I am not immediately sure about an example only lacking 2)... I'd have to get back to you on that one.

rschwieb
October 09, 2019 14:19 PM

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