Is there any non-commutative ring such that every element other than the identity is a zero divisor?

Question

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.)

Answer 1

[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?

Answer 2

Time for a second attempt, here's hoping for the best. Algebras will be strictly non-unital, i.e. they have no global identity element.

Claim 1: Let $R$ be a $\mathbb{Z}/2$-algebra with the EV property: for every non-zero element $a \in R$ there is a non-zero $v \in R$ such that $av = v$. Then the equation $xy - x + y = 0$ has no solution in $R$ besides $x = y = 0$.

Proof. If $v \neq 0$ is such that $yv = v$, then $(xy - x + y)v = xyv - xv + yv = xv - xv + v = v \neq 0$ for any $x$, so $y$ must be $0$ and hence so must $x$. $\square$

Claim 2: Let $R$ be a $\mathbb{Z}/2$-algebra with the EV property and let $I$ be the two-sided ideal of $R[x]$ generated by $rx - r$, for all $r \in R$, and $x^2 - x$. If $S = R[x]/I$ then:

• $R$ embeds in $S$, and in particular any non-trivial zero-divisor in $R$ remains a non-trivial zero-divisor in $S$;
• $x \neq 0$ in $S$ and $sx = s = xs$ for all $s \in S$;
• if $0 \neq a \in R$, then $a + x$ is a zero divisor and not a unit in $S$.

In particular, $S$ is a ring with unit $x$.

Proof. The first two observations follow from the definition of $I$: every $p \in I$ is either $0$ or has degree at least $1$, so $I \cap R = 0$, and we don't have $x \in S$. Notice first that every element of $S$ is represented by something of the form $a$ or $a + x$ for $a \in R$. If $0 \neq v \in R$ is such that $av = v$, then $(a + x)v = av + xv = v + v = 0$. If an inverse to $a + x$ exists it must be of the form $b + x$ by degree reasons, since $(a + x)b = ab + b \neq x$. But if $(a + x)(b + x) = x$ then we must have $ab + a + b = 0$ in $R$, which forces $a = b = 0$ by Claim 1. $\square$

Now we just need to find a $\mathbb{Z}/2$-algebra that is non-commutative, only has zero divisors, and satisfies the EV property. Let's ignore the zero-divisors for now. Take $A_0$ to be a non-commutative $\mathbb{Z}/2$-algebra, for example the finite rank endomorphism algebra $\mathrm{End}^{\mathrm{f.r.}}((\mathbb{Z}/2)^{\mathbb{N}})$. Let $V_0 = \{\theta_a \mid a \in A_0 \setminus \{0\}\}$ be a new set of variables, $K_0 = \langle a \theta_a - \theta_a \mid a \in A_0 \rangle$ a two-sided ideal of $A_0[V_0]$, and $A_1 = A_0[V_0]/K_0$. Then the subring $A_0 \subset A_1$ has the EV property in $A_1$ since $a \theta_a = \theta_a$ for all $a \in A_0 \setminus \{0\}$ and $\theta_a \neq 0$ since $A_0 \cap K_0 = 0$. Proceed by induction to obtain $A_n \subset A_{n+1}$ with $A_n$ having the EV property within $A_{n+1}$. The union $A = \bigcup_{n = 0}^\infty A_n$ is a non-commutative $\mathbb{Z}/2$-algebra with the EV property. To introduce zero divisors, let $R = \bigoplus_{i = 1}^\infty A$ be the set of sequences of $A$ with only finitely many non-zero entries with the usual pointwise operations. We immediately get zero-divisors for every element (given $r \in R$, take $s$ with disjoint support from $r$; then $rs = 0$), and $R$ still has the EV property: given $r = (a_1, a_2, \dotsc, a_k, 0, 0, \dotsc)$, by the EV property in $A$ can choose $v_i \in A$ such that $a_i v_i = v_i$ (if any of the $a_i$ are zero, chose $v_i = 0$) and so for $s = (v_1, v_2, \dotsc, v_k, 0, 0, \dotsc)$ we get $rs = s$; $s$ is forced to be $0$ only when $r$ is.

And there we go! Claim 2 tells us that the ring $S$ constructed from $R$ has all the desired properties: it is unital, all of its non-unit elements are zero-divisors (the ones of the form $a$ are already zero-divisors since they are in $R$), and it's non-commutative by construction. Moreover, a formal consequence of there not being non-zero nilpotents is that every zero-divisor is a two-sided zero-divisor: if $ab = 0$ then $(ba)^2 = baba = 0$ and so $ba = 0$ as well.

Answer 3

Edit: This construction gives the zero ring, as Pace Nielsen pointed out, so it's not a valid answer.

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Given any ring $S$, let $A(S) = S \setminus \{1\}$. Let $R_0 = \mathbb{Z}/2[x]$, let $V_0 = \{\theta_{a, 0} \mid a \in A(R_0)\}$ be a set of new variables, and let $K_0 = (a\theta_{a, 0}, \theta_{a, 0} a \mid a \in A(R_0))$ be the ideal generated by $a \theta_{a, 0}$ and $\theta_{a, 0} a$ (for all $a \in A(R_0)$) in the free (non-commutative!) algebra $R_0\langle A(V_0) \rangle$. The ring $R_1 = R_0\langle V_0 \rangle/K_0$ has three key properties:

• the inclusion $R_0 \hookrightarrow R_0\langle V_0 \rangle$ extends to an inclusion $R_0 \hookrightarrow R_1$, since $a - b \notin K_0$ for any $a, b \in R_0$, so we can consider $R_0$ a subset of $R_1$;
• every element of $R_0$ that is not $1$ becomes a zero-divisor in $R_1$, since $a \theta_{a, 0} = 0$ in $R_1$;
• $R_1$ is strictly non-commutative, since for example $\theta_{x, 0} \theta_{x^2, 0} \neq \theta_{x^2, 0}\theta_{x, 0}$ for the two elements $x, x^2 \in R_0 = \mathbb{Z}/2[x]$.

Now repeat this construction: given $R_n$, set $R_{n+1} = R_n\langle V_n \rangle/K_n$ where $V_n = \{\theta_{a, n} \mid a \in A(R_n)\}$ is a set of new variables and $K_n = (a\theta_{a, n}, \theta_{a, n} a \mid a \in A(R_n))$ is the ideal that allows us to turn every $a \in A(R_n)$ into a zero-divisor. The three properties above hold with $0$ and $1$ replaced by $n$ and $n+1$ (for strict non-commutativity, it is enough to take two different elements of $R_n$ and their corresponding variables will not commute).

Finally, set $R = \bigcup_{n = 0}^\infty R_n$. Addition and multiplication are given by passing to the larger $R_n$ and $1 \in R_0$ is the unit. By construction each $x \in R$, except $1$, is a zero-divisor since $x \in R_n$ for a minimal $n$ is killed by $\theta_{x, n} \in R_{n+1}$ and so $x \theta_{x, n} = 0$ in $R$. Clearly $R$ is also non-commutative.

There is nothing special about the choice of $R_0$, I just needed a ring which had only one unit and at least two distinct non-unit elements to ensure that $R_1$ was non-commutative.