Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Everyone knows that if in a ring A a unit a $\in$ A can´t be a zerodivisor. But could also be possible that "a" not be a zero divisor ( i.e does not exist a nonzero x $\in$ A , such that $ax=0$) but neither a unit ( in A ) in this case

My question is so simple in this case, considering the ring of fractions, we know that there exist an extension such that "a" is a unit . My question is, in this case, there exist other extension $J$ of $A$ , such that $a$ is a zero divisor in J ( i.e there exist a nonzero x $\in$ $J$ , such that $ax=0$) . So the question is obviously false if we consider "a" as a unit, but I think that here could be true.

Remark : I think that we may assume that A is commutative, but I´m not completely sure ( because we use the ring of fractions the commutative property is needed otherwise the sum in the fraction won´t be commutative). If you note that other properties are also needed please let me know ( like identity , etc).

share|cite|improve this question
What does "in this case" mean? What case? Did you mean to ask: "Suppose $A$ is a ring, and $a$ is a non-unit. Does there exist an extension $B$ of $A$ in which $a$ is a zero divisor?" – Arturo Magidin Dec 30 '11 at 18:43
What precisely do you mean by "extension"? Is $A$ supposed to be a subring of $J$ with the same identity element? Otherwise you could take $J=A\times A$, and $A\to A\times 0$, in which case every element of $A$ becomes a zero divisor in $J$. – Jonas Meyer Dec 30 '11 at 18:45

I believe you are asking if a is not a unit in the (unital, commutative, associative) ring A, then is there a ring B with AB a (unital) subring of B such that a is a zero divisor in B, that is, such that there is some nonzero b in B with ab = 0.

The answer is yes, there is always such a ring B when a is not a unit of A (and of course no when a is a unit of A).

Consider the ring $R=A[x]/(ax)$ and define a homomorphism $f:R\to B$ that is the identity on A and takes x to b. This is well-defined since ab = 0. In R we also have that ax = 0, and since $f(x) = b \neq 0$, we must also have $x \neq 0$. Hence we might as well assume $B=R$ in the first place, but then we'll still need to prove $x \neq 0$ in R.

Saying that $x = 0 $ in $R$ is the same as saying $x+(ax) = 0+(ax)$, which simply means $x \in (ax)$, where these ideals are all in $A[x]$. If $x \in (ax)$, then $x = rax$ for some $r \in A[x]$, but obviously by degree arguments, $r \in A$. Hence $ra=1$ and $a$ is a unit.

The background material is just ideals, quotient rings, and polynomial rings. Switching from B to R is called considering a "universal example".

As far as the non-commutative case goes, most everything already goes wrong and you can no longer invert some non-zero divisor elements.

share|cite|improve this answer

I'm assuming you are trying to ask:

Let $A$ be a commutative ring, and let $a\in A$ be an element that is neither a unit, nor a zero divisor. Does there exist a ring $C$ such that $A\subseteq C$ where $a$ is a zero divisor?

You can always construct a "universal" example: let $A$ be a ring, let $B=A[x]$ be the ring of polynomials with coefficients in $A$; then let $I=(ax)$ be the ideal generated by $ax$. Trivially, in $B/I$, we have $ax=0$. So the only question is whether $x\notin I$ and $A\cap I = \{0\}$.

If $x\in I$, then there exists $p(x)\in A[x]$ such that $x = axp(x)$. Writing $p(x) = b_0+b_1x+\cdots + b_nx^n$, we have that $ab_i=0$ if and only if $b_i=0$ (since $a$ is not a zero divisor in $A$), so $$x = ab_0x + ab_1x^2 + \cdots + ab_nx^{n+1}.$$ This requires $ab_0=1$, which cannot hold because we are assuming $a$ is not a unit. So $x\notin I$, hence the image of $x$ in $B/I$ is not zero.

Similarly, if $r\in A$ and $r\in I$, then $r=axp(x)$ for some $p(x)\in A[x]$. Proceeding as above, we see that we must have $r=0$.

Thus, $A$ embeds into $B/I$, and the images of $a$ and of $x$ are both nonzero, yet their product is equal to $0$. Thus, $C=B/I$ works as an extension of $A$ in which $a$ is a zero divisor.

As for an extension in which $A$ is a unit, let $S = \{a^n\mid n\in\mathbb{N}\}$. Then $S$ is a multiplicative subset of $A$ and we can consider the localization $S^{-1}A$. This localization contains $A$ because $S$ does not contain any zero divisors, and $a$ is a unit in $S^{-1}A$.

The noncommutative case is a bit more difficult, but the same idea can be used: construct the ring $A[x]$ and the ideal $I=(ax)$. The difficulty lies in the fact that the elements of $I$ are now of the form $$\sum_{i=1}^n q_i(x)axp_i(x)$$ i.e., polynomials of the form $b_0 + b_1x + \cdots + b_nx^n$, where $b_i\in (a)$, the ideal of $A$ generated by $a$. But this difficulty can be overcome.

Edited. You can also try to construct an extension in which $a$ has an inverse by considering $A\langle x\rangle$ (polynomials in a noncentral $x$) and moding out by the ideal generated by $ax-1$ and $xa-1$. But this need not work in general.

The problem of inverting a multiplicative set in a noncommutative ring is non trivial. You can find some results in Lam's Lectures on Modules and Rings. For example, given a ring $R$ and a multiplicative set $S$ in $R$ ($SS\subseteq S$, $1\in S$, $0\notin S$), then we say $R'$ is a right ring of fractions with respect to $S\subseteq R$ if there is a given ring homomorphism $\varphi\colon R\to R'$ such that $\phi(s)$ is a unit for each $s\in S$, every element of $R'$ has the form $\varphi(a)\varphi(s)^{-1}$ for some $a\in R$ and some $s\in S$, and $\mathrm{ker}\varphi = \{r\in R\mid rs = 0 \text{for some }s\in S\}$.

A multiplicative set $S$ is called right permutable or a right Ore set if for any $a\in R$ and $s\in S$, $aS\cap sR \neq\emptyset$. We say $S$ is right reversible if for any $a\in R$, if there exists $s'\in S$ such that $s'a = 0$, then there exists $s\in S$ such that $as=0$.

We say $S$ is a right denominator set if and only if it is both right permutable and right reversible.

Theorem. The ring $R$ has a right ring of fractions with respect to $S$ if and only if $S$ is a right denominator set.

Assume that $a\in R$ is not a zero divisor (on either side), not a unit, and not invertible (on either side). let $S=\{1,a,a^2,a^3,\ldots\}$. This is a multiplicative set. Since we are assuming that $a$ is not a zero divisor on either side, $S$ is right reversible by vacuity. But I believe that it may fail to be right permutable. For example, let $R=A\langle x\rangle$ be the ring of polynomials in a noncommuting variable $x$ with coefficients in a noncommutative ring $A$, and let $a=x$. If $b\in A$ is not central, then $bS\cap xR = \emptyset$, so $S$ is not right permutable.

But if $S$ is right permutable, then you can embed $R$ into a ring in which $a$ is a unit.

share|cite|improve this answer
I believe in the last paragraph that A need not embed in A⟨x⟩/I. Take A to be a 2x2 matrix ring (any connected ring with nontrivial idempotents), a to be E11 (any nontrivial idempotent). Note that (1-E11)*E11 = 0, so E22=(1-E11) is in I, but then E11 = E12*E22*E21 is also in I, so 1 = E11+E22 is in I, and I=A⟨x⟩. – Jack Schmidt Dec 30 '11 at 19:28
@Jack: \If anything works, then the above construction will, of course. But you may need some sort of Ore-like condition. Note, though, that if $a$ is a nontrivial idempotent, then $a$ is a zero divisor, since $a(1-a)=0$, and we are supposed to be excluding zero divisors. I'm looking it up in Lam's Modules and Rings and will post a correction shortly. Thanks. – Arturo Magidin Dec 30 '11 at 19:33

It seems to me that you are asking about a zero-divisor dual of the fraction-field or localization constructions. Regarding such you may find interesting the following paper of Cohn, which is excerpted below for convenience.

Note that rings where every element is a unit or a zero-divisor are sometimes called quasi-regular, e.g. finite products of local rings (Artinian), zero-dimensional rings, commutative regular (von Neumann) rings, etc. Searching on said terms should yield a springboard into the literature.

P. M. Cohn. Rings of Zero-Divisors.
Proc. AMS Vol. 9, No. 6, (Dec., 1958), pp. 909-914
enter image description here

share|cite|improve this answer
The AMS journals are available for free on the AMS homepage while they still are behind a paywall on JSTOR. Cohn's paper is available here. – t.b. Dec 30 '11 at 19:20
@t.b. Thanks, I've updated it (it was from an old post). – Bill Dubuque Dec 30 '11 at 19:25

What about any integral domain that's not a field? By definition then there exists an element that is non-zero element which is non-invertible, but since the ring is an integral domain this element isn't a zero divisor. For example, consider $\mathbb{Z}$.

share|cite|improve this answer
While the question is unclear, I'm pretty sure it isn't whether there can be a nonzero divisor that is not a unit... – Arturo Magidin Dec 30 '11 at 18:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.