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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.)

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    $\begingroup$ By "zero divisor" you mean "both left and right zero divisor"? $\endgroup$
    – YCor
    Commented Jun 20, 2021 at 10:44
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    $\begingroup$ Yes, let me make it clear in the OP. $\endgroup$
    – Salvo Tringali
    Commented Jun 20, 2021 at 10:44
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    $\begingroup$ @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
    Commented Jun 20, 2021 at 12:50
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    $\begingroup$ Sorry I missed that part of the OP. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 20, 2021 at 13:24
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    $\begingroup$ 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
    Commented Jun 20, 2021 at 20:34

1 Answer 1

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[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?

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