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: $\forall a,b\in R, a+b \in R \Leftrightarrow b+a\in R$.

2) Associativity of addition

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 : ring-theory


Answers 1


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
rschwieb
October 09, 2019 14:19 PM

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