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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
    Jun 20, 2021 at 10:44
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    $\begingroup$ Yes, let me make it clear in the OP. $\endgroup$ 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? $\endgroup$ Jun 20, 2021 at 12:50
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    $\begingroup$ Sorry I missed that part of the OP. $\endgroup$ 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. $\endgroup$ Jun 20, 2021 at 20:34

3 Answers 3

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


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.

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  • $\begingroup$ Lorenzo, you might consider the following issue. Let $a=\theta_{x,0}$ and $b=\theta_{x^2,0}$. Notice that $abx\neq 0$, but $(abx)^2=0$. Thus, $1+abx$ is a different unit than $1$ in $R_1$. $\endgroup$ Jul 8 at 20:14
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    $\begingroup$ Pace, you are right, so we immediately get $R_2 = 0$. I'll try to see if this can be fixed, though I doubt it. $\endgroup$ Jul 8 at 20:33
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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.

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    $\begingroup$ Doesn't the finite rank endomorphism algebra contain nilpotent elements? $\endgroup$ Jul 14 at 4:35
  • $\begingroup$ Right, same mistake as the last time... I don't think fixing the choice of $A_0$ will work since it seems hard to prove that any $\theta_a \notin K_0$. Even if I only add one eigenvector at a time it remains hard. Argh! On the other hand, if I give up non-commutativity of $A$ then I don't know how to reinsert it later. I naively thought that the EV property for non-unital rings was easier to satisfy than having only zero divisors in a unital ring, but I really cannot come up with a single EV ring without identity. Any thoughts? $\endgroup$ Jul 14 at 8:42
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    $\begingroup$ If you are able to get such an example you would essentially be answering an old open question by Henriksen. Unfortunately, it is very difficult to avoid nilpotent elements in such constructions. $\endgroup$ Jul 14 at 19:36

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