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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).

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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

4 Answers 4

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.

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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.

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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
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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 $\mathbb{Z}$.

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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

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