1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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. I found this earlier question which seems to be asking for confirmation they found such an example, but I have not proofread it.

$\endgroup$
  • $\begingroup$ Also, just an unrelated, simple other question: An additive inverse or identity of an element $a$ in a ring $R$ does not need to be “-a” or “0” respectively. But then, wouldn’t the set of $(\mathbb{N}\bigcup \{0\})$ modulo $3$ be such a ring? $\endgroup$ – user706791 Oct 11 at 20:01
  • $\begingroup$ Q1: I'm pretty sure you can find an example of this for yourself if you just select a small abelian group and try its set of functions. Q2: presumably you meant distributivity holds on both sides when you said 4. Q3: yes. Q4: the only cross product on $\mathbb R^2$ that I know is not a binary operation... so I don't understand your question. $\endgroup$ – rschwieb Oct 11 at 20:09
  • $\begingroup$ @rschweib, oh yeah, I forgot that cross products only occur on $\mathbb{R}^3$ so please ignore that stupid question. Also, you still haven’t answered my question of whether the additive inverse has to be “-a” and the additive identity has to be “0”. Also, do you have any tips as to how you came up with such examples and how to come up with such rings in the first place? I know there probably isn’t a general formula for finding such rings, but I often find that difficult to do. Is it true that an in-depth understanding of the material can help in finding them? $\endgroup$ – user706791 Oct 11 at 21:04
  • $\begingroup$ @rschweib, do you have any tips as to how you’ve managed to understand ring theory to such depth? Like, have you read a lot of ring theory books and understood them? Do you rewrite proofs you don’t understand the first time? Do you write proofs for similar problems? $\endgroup$ – user706791 Oct 11 at 21:10
  • $\begingroup$ Sorry for spelling your nickname wrong @rschwieb $\endgroup$ – user706791 Oct 11 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy