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


non-commutative
no unity (i.e. no multiplicative identity: a so-called "rng")
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:


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


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

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