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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?

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  • $\begingroup$ A 'rng' is not a noncommutative ring; it is a ring without unity. $\endgroup$
    – Servaes
    Jul 22, 2019 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, 2019 at 12:32
  • $\begingroup$ @Servaes ah you're right! Just edited my post. $\endgroup$ Jul 22, 2019 at 13:26

5 Answers 5

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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.

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  • $\begingroup$ So basically 𝐼 contains all the polynomials in 𝑅 with degree >=1 plus the 0 polynomial (additive identity). Agree? $\endgroup$ Jul 25, 2019 at 13:46
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    $\begingroup$ Not exactly. The ideal $I$ consists of all polynomials in $R$ with constant term equal to zero. $\endgroup$
    – quasi
    Jul 25, 2019 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 $\endgroup$ Jul 25, 2019 at 14:22
  • $\begingroup$ Yes, that's right. $\endgroup$
    – quasi
    Jul 25, 2019 at 14:27
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One example of such a ring is the free algebra on two elements over the rng $2\Bbb{Z}$.

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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.

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

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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.

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    $\begingroup$ -1 This is a good comment but not a good answer. $\endgroup$
    – Servaes
    Jul 22, 2019 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. $\endgroup$ Jul 22, 2019 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. $\endgroup$ Jul 22, 2019 at 14:24

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