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[I fear that I'm missing something obvious here, but I'll dare to ask anyway.]

As we all know, a division ring is a (unital, associative, non-zero) ring where every non-zero element is a unit. So, let an anti-division ring be a ring where any element other than the identity is a (two-sided) zero divisor.

There are many simple things one can actually say about these objects: Their class is closed under direct products; they all have characteristic $2$; every boolean ring is a member of the family; any anti-division ring is, in fact, a subdirect product of domains (see Jose Brox's answer to a related question on MSE); the only nilpotent element in such a ring is zero and, more generally, the Jacobson radical is trivial (as noted by Benjamin Steinberg in the comments below); if a ring in the family is semilocal, then it's a direct product of copies of the field with two elements (see znc's answer from the same thread on MSE); etc.

Question. Is it perhaps the case that a more precise characterization (say, than the one provided by Jose Brox for rings whose non-units are all zero divisors) is indeed possible?

If so, it's clear to me that this must be very well known (at least in some large circles) and I would much appreciate it if you could offer a reference.

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    $\begingroup$ The Jacobson radical must be zero as well since it x is in the radical 1-x is a unit $\endgroup$ Jun 18, 2021 at 20:02

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Sorry for answering my own question, but it turned out that what I'm calling "anti-division rings" in the OP were already studied by P.M. Cohn under the name of "$0$-rings" (though Cohn's work on this stuff is seemingly restricted to the commutative setting), see

  • P.M. Cohn, Rings of zero divisors, Proc. Amer. Math. Soc. 9 (1958), 914-919;
  • T. Porter, Cohn's rings of zero divisors, Arch. Math. 43 (1984), 340-343.

In particular, Cohn's paper (Sect. 3) contains a couple of structural results for commutative $0$-rings, and it seems unlikely that one can do any better (without throwing in further conditions on the ring).

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  • $\begingroup$ See also mathoverflow.net/a/395778/16537, where a reference is provided for the fact that every right (or left) artinian $0$-ring is boolean (and hence commutative). $\endgroup$ Jun 20, 2021 at 8:54

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