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?

  • $\begingroup$ A 'rng' is not a noncommutative ring; it is a ring without unity. $\endgroup$ – Servaes Jul 22 at 12:29
  • $\begingroup$ Also; such a ring cannot be finite because in a finite ring, every element is either a unit or a zero divisor. $\endgroup$ – Servaes Jul 22 at 12:32
  • $\begingroup$ @Servaes ah you're right! Just edited my post. $\endgroup$ – UndefinedBehavior Jul 22 at 13:26

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.

  • $\begingroup$ So basically 𝐼 contains all the polynomials in 𝑅 with degree >=1 plus the 0 polynomial (additive identity). Agree? $\endgroup$ – UndefinedBehavior Jul 25 at 13:46
  • 1
    $\begingroup$ Not exactly. The ideal $I$ consists of all polynomials in $R$ with constant term equal to zero. $\endgroup$ – quasi Jul 25 at 13:50
  • $\begingroup$ That's what I meant. Sorry for my misleading wording. I meant polynomials not containing any monomial of degree 0. Also... if i'm not mistaken, the quotient ring is (isomorphic to) ℤ $\endgroup$ – UndefinedBehavior Jul 25 at 14:22
  • $\begingroup$ Yes, that's right. $\endgroup$ – quasi Jul 25 at 14:27

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


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.


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.

  • 3
    $\begingroup$ -1 This is a good comment but not a good answer. $\endgroup$ – Servaes Jul 22 at 12:28
  • $\begingroup$ Well wrt my post it'd be more interesting if the opposite were true: i.e. if from any given monoid you could just remove the identity and get a bona-fide semi-group. Unfortunately this is not the case, you need to add the constraint that the monoid you start with must not be a group. $\endgroup$ – UndefinedBehavior Jul 22 at 13:24
  • $\begingroup$ Moreover, while it's true that you ca easily extend a (multiplicative) semigroup with an identity element, if it is a ring you need to also define addition for the new element. $\endgroup$ – UndefinedBehavior Jul 22 at 14:24

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.