A (unital) ring $R$ with the property that every element other than the identity $1_R$ is a (two-sided) zero divisor, seems to be commonly called a "$0$-ring" or "$\mathcal O$-ring". These rings were first studied by P.M. Cohn (though only in the commutative setting) in

  • Rings of zero divisors, Proc. Amer. Math. Soc. 9 (1958), 914-919.

Moreover, every right (or left) artinian $\mathcal O$-ring is, in fact, a boolean ring (and hence commutative), see

  • H.G. Moore, S.J. Pierce, and A. Yaqub, Commutativity in rings of zero divisors, Amer. Math. Monthly 75 (1968), 392

Thence, the question is:

Does there exist any non-commutative $\mathcal O$-ring? If so, can you provide a reference where this is discussed?

I've tried to track the citations of Cohn's paper, but couldn't find an answer to my question. (See also here and there.)

  • 1
    $\begingroup$ By "zero divisor" you mean "both left and right zero divisor"? $\endgroup$ – YCor Jun 20 at 10:44
  • 1
    $\begingroup$ Yes, let me make it clear in the OP. $\endgroup$ – Salvo Tringali Jun 20 at 10:44
  • 1
    $\begingroup$ @BenjaminSteinberg Sorry, I'm not sure to understand your comment. The OP cites a paper in the AMM where it's shown that any right artinian $\mathcal O$-ring is commutative. And yes, the Jacobson radical of any $\mathcal O$-ring is trivial. So what? I'm missing the point, I think. Clearly, you don't mean that an $\mathcal O$-ring is necessarily non-artinian. Do you mean that a non-commutative $\mathcal O$-ring is necessarily non-artinian and this can proved in a more direct way than done in the aforementioned AMM paper? $\endgroup$ – Salvo Tringali Jun 20 at 12:50
  • 1
    $\begingroup$ Sorry I missed that part of the OP. $\endgroup$ – Benjamin Steinberg Jun 20 at 13:24
  • 2
    $\begingroup$ The problem seems to reduce to does there exist a primitive unital ring where every non-identity element is a two-sided zero divisor. First note that if every nonidentity element of $R$ is a zero divisor, then the same is true for $R/I$ for any ideal $I$. Since $J(R)=0$ (the radical), $R$ is a subdirect product of primitive rings and hence is commutative iff each of these primitive quotients are. I don’t see how to use Jacobson’s density theorem to get a contradiction to noncommutativity if the primitive ring is not artinian. $\endgroup$ – Benjamin Steinberg Jun 20 at 20:34

[Sorry for answering my own question, and the more so because this is happening for the second time in 24 hours.]

The question might be open. In fact, a positive answer would imply an equally positive answer to a question stated in the introduction of Melvin Henriksen's paper

  • "Rings with a unique regular element", pp. 78-87 in B.J. Gardner (ed.), Rings, modules and radicals (Proc. Conf., Hobart/Aust. 1987), Pitman Res. Notes Math. Ser. 204, Longman Sci. Tech., Harlow, 1989,

where Henriksen writes:

We do not know if there is a ring with a unique regular ring [sic] that fails to be commutative.

In Henriksen's paper, a ring need not be unital; and a regular element is nothing else than a cancellative element of the multiplicative semigroup of the ring (loc. cit., Definition 2.1).

The question is marked as open by David Feldman in a 2012 post from the "Not especially famous, long-open problems which anyone can understand" big list (see also the comments under the same post), where Feldman writes:

Must a non-commutative ring (with identity) contain a non-zero-divisor aside from the identity?


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.