Can a ring have no zero divisors while being non-commutative and having no unity?

Question

I was wondering if, in a ring, the property of having no zero-divisors (except for zero itself) is independent from the ring being commutative or from having a unity (i.e.multiplicative identity) so I started looking for a ring with the following properties:

1. non-commutative
2. no unity (i.e. no multiplicative identity: a so-called "rng")
3. no zero-divisors

I came up with the set of 2 x 2 matrices with even entries: $M_2(2\Bbb Z)$ endowed with the usual matrix addition and matrix multiplication. It is:

1. non-commutative: $$\begin{pmatrix}2&2\\2&0\end{pmatrix}\begin{pmatrix}0&2\\2&2\end{pmatrix}\neq\begin{pmatrix}2&2\\2&0\end{pmatrix}\begin{pmatrix}0&2\\2&2\end{pmatrix}$$
2. no unity: $$\begin{pmatrix}1&0\\0&1\end{pmatrix}\notin M_2(2\Bbb Z)$$

But unfortunately it does have zero divisors: $$\begin{pmatrix}2&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&2\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$

So, can you come up with a ring having those three properties? Or a proof that such a group cannot exist?

Answer 1

Well, each semigroup $(S,\circ)$ with no unit element can be extended to a monoid $(S\cup\{e\},\circ,e)$ which has a unit element $e$ not belonging to $S$. This is a general construction.

So the question whether the ring (i.e., multiplicative semigroup) has a unit element or not is more or less pointless.

Answer 2

Let $R$ be the ring of polynomials with integer coefficients in two non-commuting variables $x,y$, and let $I$ be the ideal generated by $\{x,y\}$.

Then $I$, regarded as a ring, satisfies all your specified conditions.

Answer 3

One example of such a ring is the free algebra on two elements over the rng $2\Bbb{Z}$.

Answer 4

I think the example of a nontrivial ideal of the free algebra on two noncommuting variables is already an optimally simple answer, but I'd like to offer a few other noncommutative, nonfield domains that would also work.

Any nontrivial ideal of the Hurwitz quaternions or Lipschitz quaternions would do.

Answer 5

Take the ring $A = \{ a + bi +cj+ dk \, : \, a,b,c,d \in 2\mathbb{Z}\}$ in the quaternions.